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https://www.physicsforums.com/threads/proof-on-det-ab-det-a-det-b.718247/
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# Proof on det(AB) = det(A)*det(B)
unscientific
## Homework Statement
Let A and B be nxn matrices, show that det(A)=det(A)det(B). ## The Attempt at a Solution I'm not sure how to express each column as a linear combination of vectors purely from A, or even from B. It's more of the dot product form: aT°b
## Answers and Replies
Homework Helper
I think you should tell us what your "general definition of the determinant" is. It's probably easiest to go directly from there.
unscientific
I think you should tell us what your "general definition of the determinant" is. It's probably easiest to go directly from there.
But i think we should first express the columns of matrix AB in terms of the columns of matrix A? (I'm not sure where B comes in) then somehow show that it is a linear combination, then make use of result 13. Then I think result 14 comes in when you get something that is just jumbled up, which simply gives det (AB) = sgn(P)*det(A)*det(B)
det(A) = Ʃ sgn(P) AP(1)1AP(2)2...
unscientific
bumpp
Staff Emeritus
Homework Helper
As you noted, the first column of AB is
$$\begin{pmatrix} \sum_{k=1}^n a_{1k}b_{k1} \\ \sum_{k=1}^n a_{2k}b_{k1} \\ \vdots \\ \sum_{k=1}^n a_{nk}b_{k1} \end{pmatrix} = \begin{pmatrix} a_{11}b_{11} + a_{12}b_{21} + a_{13}b_{31} + \cdots + a_{1n}b_{n1} \\ a_{21}b_{11} + a_{22}b_{21} + a_{23}b_{31} + \cdots + a_{2n}b_{n1} \\ \vdots \\ a_{n1}b_{11} + a_{n2}b_{21} + a_{n3}b_{31} + \cdots + a_{nn}b_{n1} \end{pmatrix}.$$ Express that in the form ##c_1\vec{a}_1 + c_2\vec{a}_2 + \cdots + c_n\vec{a}_n## where ##\vec{a}_k## is the kth column of A. How are the ##c_i##'s related to the elements of B?
unscientific
As you noted, the first column of AB is
$$\begin{pmatrix} \sum_{k=1}^n a_{1k}b_{k1} \\ \sum_{k=1}^n a_{2k}b_{k1} \\ \vdots \\ \sum_{k=1}^n a_{nk}b_{k1} \end{pmatrix} = \begin{pmatrix} a_{11}b_{11} + a_{12}b_{21} + a_{13}b_{31} + \cdots + a_{1n}b_{n1} \\ a_{21}b_{11} + a_{22}b_{21} + a_{23}b_{31} + \cdots + a_{2n}b_{n1} \\ \vdots \\ a_{n1}b_{11} + a_{n2}b_{21} + a_{n3}b_{31} + \cdots + a_{nn}b_{n1} \end{pmatrix}.$$ Express that in the form ##c_1\vec{a}_1 + c_2\vec{a}_2 + \cdots + c_n\vec{a}_n## where ##\vec{a}_k## is the kth column of A. How are the ##c_i##'s related to the elements of B?
So, the first column of AB = b11##\vec{a}_1## + b21##\vec{a}_2## + .... + bn1##\vec{a}_n##
kth-column = b1k##\vec{a}_1## + b2k##\vec{a}_2## + .... + bnk##\vec{a}_n##
I'm just not used to the idea of having "columns within a column", it sort of makes it more than nxn-dimensional.
Also, the earlier problem of
det(A) = λdet(B) + μdet(C)
only had one "special column" in matrix B and C that had a coefficient, while the rest were simply ##\vec{a}_1, \vec{a}_2, .... \vec{a}_n##
Last edited:
Staff Emeritus
Homework Helper
Good! So you've done what the first part of the hint suggested, expressing the columns of AB as linear combinations of the columns of A. Now just follow through with the rest of the hint.
unscientific
Good! So you've done what the first part of the hint suggested, expressing the columns of AB as linear combinations of the columns of A. Now just follow through with the rest of the hint.
Also, the earlier problem of
det(A) = λdet(B) + μdet(C)
only had one "special column" in matrix B and C that had a coefficient, while the rest were simply ##\vec{a}_1, \vec{a}_2, .... \vec{a}_n##
while in the new form, it's literally in every column.
unscientific
I think I got it:
We can simply rearrange the columns of matrix AB so as to separate them into the vectors ##\vec{a}_1, \vec{a}_2, \vec{a}_3 ....## and by result of problem 14, that's not going to change the absolute value of the determinant.
That would solve the dimensional crisis. Explanation why the product of b's = det(B) Consider the kth bracket, any permutation within that bracket is equivalent to a permutation of Q(k).
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https://socratic.org/questions/how-much-work-would-it-take-to-push-a-6-kg-weight-up-a-12-m-plane-that-is-at-an-
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# How much work would it take to push a 6 kg weight up a 12 m plane that is at an incline of pi / 4 ?
Apr 2, 2016
51 Joules
#### Explanation:
Here,
Hypotenuse (h) = 12m
Angle of inclination ($\theta$)= π/4
Perpendicular (p) = ?
We have, $S \in \theta = \frac{p}{h}$
or, Sin (π/4) = p/12
or, $p = \frac{12}{\sqrt{2}}$
Hence, p = 8.5 m, which is the distance travelled.
Since work(W) is given by
W = F * S,
Where F = Force ( here weight) = 6 kg wt = 6 N
(Note kg wt, unlike kg, is same as Newton and is a unit of weight, not mass).
S is the distance traveled given by the perpendicular 8.5m
Therefore,
$W = 6 \cdot 8.5 = 51 \text{Joules}$
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https://metanumbers.com/50849
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# 50849 (number)
50,849 (fifty thousand eight hundred forty-nine) is an odd five-digits prime number following 50848 and preceding 50850. In scientific notation, it is written as 5.0849 × 104. The sum of its digits is 26. It has a total of 1 prime factor and 2 positive divisors. There are 50,848 positive integers (up to 50849) that are relatively prime to 50849.
## Basic properties
• Is Prime? Yes
• Number parity Odd
• Number length 5
• Sum of Digits 26
• Digital Root 8
## Name
Short name 50 thousand 849 fifty thousand eight hundred forty-nine
## Notation
Scientific notation 5.0849 × 104 50.849 × 103
## Prime Factorization of 50849
Prime Factorization 50849
Prime number
Distinct Factors Total Factors Radical ω(n) 1 Total number of distinct prime factors Ω(n) 1 Total number of prime factors rad(n) 50849 Product of the distinct prime numbers λ(n) -1 Returns the parity of Ω(n), such that λ(n) = (-1)Ω(n) μ(n) -1 Returns: 1, if n has an even number of prime factors (and is square free) −1, if n has an odd number of prime factors (and is square free) 0, if n has a squared prime factor Λ(n) 10.8366 Returns log(p) if n is a power pk of any prime p (for any k >= 1), else returns 0
The prime factorization of 50,849 is 50849. Since it has a total of 1 prime factor, 50,849 is a prime number.
## Divisors of 50849
2 divisors
Even divisors 0 2 2 0
Total Divisors Sum of Divisors Aliquot Sum τ(n) 2 Total number of the positive divisors of n σ(n) 50850 Sum of all the positive divisors of n s(n) 1 Sum of the proper positive divisors of n A(n) 25425 Returns the sum of divisors (σ(n)) divided by the total number of divisors (τ(n)) G(n) 225.497 Returns the nth root of the product of n divisors H(n) 1.99996 Returns the total number of divisors (τ(n)) divided by the sum of the reciprocal of each divisors
The number 50,849 can be divided by 2 positive divisors (out of which 0 are even, and 2 are odd). The sum of these divisors (counting 50,849) is 50,850, the average is 25,425.
## Other Arithmetic Functions (n = 50849)
1 φ(n) n
Euler Totient Carmichael Lambda Prime Pi φ(n) 50848 Total number of positive integers not greater than n that are coprime to n λ(n) 50848 Smallest positive number such that aλ(n) ≡ 1 (mod n) for all a coprime to n π(n) ≈ 5208 Total number of primes less than or equal to n r2(n) 8 The number of ways n can be represented as the sum of 2 squares
There are 50,848 positive integers (less than 50,849) that are coprime with 50,849. And there are approximately 5,208 prime numbers less than or equal to 50,849.
## Divisibility of 50849
m n mod m 2 3 4 5 6 7 8 9 1 2 1 4 5 1 1 8
50,849 is not divisible by any number less than or equal to 9.
• Arithmetic
• Prime
• Deficient
• Polite
• Prime Power
• Square Free
## Base conversion (50849)
Base System Value
2 Binary 1100011010100001
3 Ternary 2120202022
4 Quaternary 30122201
5 Quinary 3111344
6 Senary 1031225
8 Octal 143241
10 Decimal 50849
12 Duodecimal 25515
20 Vigesimal 6729
36 Base36 138h
## Basic calculations (n = 50849)
### Multiplication
n×y
n×2 101698 152547 203396 254245
### Division
n÷y
n÷2 25424.5 16949.7 12712.2 10169.8
### Exponentiation
ny
n2 2585620801 131476232110049 6685434926563881601 339947680580846815529249
### Nth Root
y√n
2√n 225.497 37.0477 15.0166 8.73487
## 50849 as geometric shapes
### Circle
Diameter 101698 319494 8.12297e+09
### Sphere
Volume 5.50726e+14 3.24919e+10 319494
### Square
Length = n
Perimeter 203396 2.58562e+09 71911.3
### Cube
Length = n
Surface area 1.55137e+10 1.31476e+14 88073.1
### Equilateral Triangle
Length = n
Perimeter 152547 1.11961e+09 44036.5
### Triangular Pyramid
Length = n
Surface area 4.47843e+09 1.54946e+13 41518
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https://brilliant.org/problems/dissimilar-terms/
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# Dissimilar Terms
Probability Level 4
We can rewrite the statement "$a$ and $b$ are distinct numbers" into a statement made of just variables and not-equal signs: "$a \ne b$".
Similarly, we can rewrite the statement "$a,b,$ and $c$ are distinct numbers" as "$a \ne b \ne c \ne a$".
Note that "$a \neq b, a \neq c, b \neq c$" is not valid because it is three statements, not one, and that "$(a-b)(a-c)(b-c) \neq 0$" is not valid either because it has something other than variables and not-equal signs, which in this case are the parentheses.
In the above two statements, the minimum numbers of not-equal signs $(\ne)$ used are 1 and 3, respectively.
What is the minimum number of not-equal signs used to rewrite the statement "$a,b,c,d$ are all distinct numbers" in such a way?
Bonus: Generalize this.
×
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https://mathhelpboards.com/threads/show-that-g-is-isomorphic-to-g-x-g.5123/#post-23278
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# Show that G is isomorphic to G x G
#### oblixps
##### Member
Let G be the group of positive rational numbers under multiplication. Show G is isomorphic to G x G.
i'm not sure how to start this. i am trying to come up with an explicit isomorphim but i can't seem to find one.
can someone give me some hints on how to approach this?
#### Opalg
##### MHB Oldtimer
Staff member
Let G be the group of positive rational numbers under multiplication. Show G is isomorphic to G x G.
i'm not sure how to start this. i am trying to come up with an explicit isomorphim but i can't seem to find one.
can someone give me some hints on how to approach this?
I think the easiest method would be to show that each of these groups is isomorphic to the direct sum of infinitely many copies of the integers.
To see that $G$ is isomorphic to $$\displaystyle \bigoplus_{n\in\mathbb{N}}\mathbb{Z}$$, notice that each element $g\in G$ can be uniquely expressed in the form $$\displaystyle g = \prod_{p\in\mathcal{P}}p^{g_p}$$, where $\mathcal{P}$ denotes the set of primes. Each $p\in\mathcal{P}$ is raised to an integer power, and all but finitely many of these powers are $0$. The map $g\mapsto (g_p)_{p\in\mathcal{P}}$ gives the required isomorphism.
If that seems a bit abstract, here's a concrete example. Suppose $g = \frac{63}{50}$. By factorising the numerator and denominator as products of primes, you see that $g = 2^{-1}3^25^{-2}7^111^013^0\ldots$. The map taking $g$ to $(-1,2,-2,1,0,0,\ldots)$ associates $g$ with an element of $$\displaystyle \bigoplus_{n\in\mathbb{N}}\mathbb{Z}$$.
It should be fairly obvious that if $$\displaystyle H = \bigoplus_{n\in\mathbb{N}}\mathbb{Z}$$ then $H$ is isomorphic to $H\times H$ (because "twice infinity is infinity"), and it follows that $G$ is isomorphic to $G\times G.$
#### oblixps
##### Member
i wouldn't have thought of taking the direct sum of infinitely many copies of Z. seems like an interesting example.
would the result also be true if we took an infinite direct product instead of direct sum?
#### Opalg
##### MHB Oldtimer
Staff member
would the result also be true if we took an infinite direct product instead of direct sum?
Direct product would not work here. The direct product of infinitely many copies of the integers is uncountable. So it could not be isomorphic to the group of positive rationals, which is countable.
#### oblixps
##### Member
ah i see. now forgetting about the group of rationals, if we just let G be the direct product of countably many copies of Z, we would also have G isomorphic to G x G right? it seems like it would be the same argument as in the case for a direct sum.
#### Opalg
##### MHB Oldtimer
Staff member
ah i see. now forgetting about the group of rationals, if we just let G be the direct product of countably many copies of Z, we would also have G isomorphic to G x G right? it seems like it would be the same argument as in the case for a direct sum.
Yes, that's correct. In both cases, if you take countably many copies and then countably many more copies, you still have countably many copies.
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https://www.bartleby.com/solution-answer/chapter-63-problem-36e-elementary-geometry-for-college-students-6th-edition/9781285195698/given-q-is-inscribed-in-isosceles-right-rst-the-perimeter-of-rst-is-842-find-tm/562e6af1-757c-11e9-8385-02ee952b546e
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Chapter 6.3, Problem 36E ### Elementary Geometry for College St...
6th Edition
Daniel C. Alexander + 1 other
ISBN: 9781285195698
#### Solutions
Chapter
Section ### Elementary Geometry for College St...
6th Edition
Daniel C. Alexander + 1 other
ISBN: 9781285195698
Textbook Problem
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# Given: ⊙ Q is inscribed in isosceles right Δ R S T . The perimeter of Δ R S T is 8 + 4 2 Find: TM To determine
To find:
To find TM.
Explanation
Given that, Q is inscribed in isosceles right ΔRST. The perimeter of ΔRST is 8+42.
The tangents segments from an external point are congruent. Therefore,
RP=RM=xSP=SN=yTN=TM=z
The diagrammatic representation is given below,
Perimeter of the ΔRST is 8+42.
That is RT+TS+RS=8+42
Given that ΔRST is an isosceles triangle. Therefore, RT=TS.
Then the above equation become the following,
RT+TS+RS=8+42RT+RT+RS=8+422RT+RS=8+42.........(1)
By using the Pythagorean theorem to get the following,
RT2+TS2=RS2RT2+RT2=RS22RT2=RS2. Since RT=TS
Taking square root on both sides to get the following,
RS=RT2...........(2)
Substitute equation (2) in (1) to get the following,
2RT+RT2=8+42RT(2+2)=8+42RT=4(2+2)(2+2)RT=4
Also, know that RT=TS=4
By using the Pythagorean theorem to get the following,
RT2+TS2=RS242+42=RS216+16=RS2RS=32RS=42.
The value of RT=4,TS=4,andRS=42
We know that,
RS=RP+PS=x+yTS=TN+NS=z+yRT=RM+MT=x+z
Substitute the value of RT=4,TS=4,andRS=42 in the above equation to get the following,
x+y=42
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https://www.univerkov.com/the-30-cm-conductor-is-placed-horizontally-what-value-should-the-induction-of-the-magnetic-field-have-so-that-the-gravity/
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| 4.1875
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| 418
| 1,585
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# The 30 cm conductor is placed horizontally. What value should the induction of the magnetic field have so that the gravity
The 30 cm conductor is placed horizontally. What value should the induction of the magnetic field have so that the gravity force of a conductor with a mass of 6 g is balanced by the Ampere force? A current of 5 A flows through the conductor.
L = 30cm = 0.3m.
m = 6 g = 0.006 kg.
g = 10 m / s2.
I = 5 A.
B -?
Ft = Famp – 1 Newton’s law.
The force of gravity Ft, which acts on the conductor, is expressed by the formula: Ft = m * g, where m is the mass of the conductor, g is the acceleration of gravity.
The Ampere force Famp is expressed by the formula: Famp = I * L * B * sinα, where ∠α is the angle between the magnetic induction and the direction of the current in the conductor, I is the current in the conductor, L is the length of the conductor, V is the magnetic induction of the field.
We will assume that the conductor is parallel to the vector of magnetic induction B, therefore ∠α = 00, sin00 = 1.
Famp = I * L * V.
m * g = I * L * B.
We express the magnetic induction of the field B by the ratio: B = m * g / I * L.
B = 0.006 kg * 10 m / s2 / 5 A * 0.3 m = 0.04 T.
Answer: the induction of the magnetic field is B = 0.04 T. One of the components of a person's success in our time is receiving modern high-quality education, mastering the knowledge, skills and abilities necessary for life in society. A person today needs to study almost all his life, mastering everything new and new, acquiring the necessary professional qualities.
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https://cracku.in/blog/ibps-po-and-ibps-rrb-po-percentage-questions-pdf/
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| 4.46875
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0
192
# Percentage Questions for IBPS PO and IBPS RRB Prelims
Question 1: Suresh gave 17.39% of his monthly income to his wife and spent 31.57% of the remaining amount for household expenses. Out of the remaining amount he gave 15.38% to the charity. Again from the remaining amount he gave 18.18% to his friend. Finally if Suresh is left with ___ then monthly income of Suresh is ___ .
Which of the following values can fill in the blanks appropriately?
i) 10035, 25645
ii) 8991, 22977
iii) 16083, 41101
iv) 14085, 35995
a) Only (ii) and (iv)
b) Only (i) and (iv)
c) All (i), (ii), (iii) and (iv)
d) Only (i), (ii) and (iv)
e) Only (ii), (iii) and (iv)
Solution:
Suresh gave 17.39% of his monthly income to his wife.
Let the monthly income of Suresh = 23p
Amount given to his wife = $\frac{4}{23}\times$23p = 4p
Amount remaining = 23p – 4p = 19p
Suresh spent 31.57% of the remaining amount for household expenses.
Amount spent for household expenses = $\frac{6}{19}\times$19p = 6p
Amount remaining = 19p – 6p = 13p
Out of the remaining amount he gave 15.38% to charity.
Amount given charity = $\frac{2}{13}\times$13p = 2p
Amount remaining = 13p – 3p = 11p
Again from the remaining amount he gave 18.18% to his friend.
Amount given to friend = $\frac{2}{11}\times$11p = 2p
Final amount remaining = 11p – 2p = 9p
Ratio of the final amount remaining with Suresh and his monthly income respectively = 9p : 23p
= 9 : 23
i) 10035, 25645
Ratio of the final amount remaining with Suresh and his monthly income respectively = 10035 : 25645
= 9 : 23
Given values satisfy the condition of the question.
ii) 8991, 22977
Ratio of the final amount remaining with Suresh and his monthly income respectively = 8991 : 22977
= 9 : 23
Given values satisfy the condition of the question.
iii) 16083, 41101
Ratio of the final amount remaining with Suresh and his monthly income respectively = 16083 : 41101
= 9 : 23
Given values satisfy the condition of the question.
iii) 14085, 35995
Ratio of the final amount remaining with Suresh and his monthly income respectively = 14085 : 35995
= 9 : 23
Given values satisfy the condition of the question.
Hence, the correct answer is Option C
Question 2: If $(x+5) \text{ of } 11.11\% = y \text{ of } 12.5\% -10$ and $(x-25) = 75\% \text{ of } (y-40)$, then find out the value of $40\% of (y-x)$.
a) 39
b) 51
c) 28
d) 13
e) None of the above
Solution:
$(x+5) \text{ of } 11.11\% = y \text{ of } 12.5\% – 10$
$(x+5)\times\frac{1}{9} = y\times\frac{1}{8} – 10$
$(x+5) = y\times\frac{9}{8} – 90$
$x = y\times\frac{9}{8} – 90 -5$
$x = y\times\frac{9}{8} – 95$ Eq.(i)
$(x-25) = 75\% \text{ of } (y-40)$
$(x-25) = \frac{3}{4} \times (y-40)$
Put the value of ‘x’ in the above equation from Eq.(i).
$(y\times \frac{9}{8}-95-25)=\frac{3}{4}\times (y-40)$
After solving the above equation, y = 240.
Put the value of ‘y’ in Eq.(i).
$x = 240\times\frac{9}{8} – 95$
x = 270-95
= 175
value of $40\% of (y-x)$ = $40\% of (240-175)$
= 40% of 65
= 26
Hence, option e is the correct answer.
Question 3: Ketan spent 41.67% of his monthly salary on house rent. Out of the remaining, he spent 20% on food. Out of the remaining, he spent 10% on education. After that the remaining amount was invested in FD and MF in the ratio of 7:5 respectively. If the amount spent on education by him is Rs. 4480, then find out the difference between the amount spent on house rent and the amount invested on FD from his monthly salary.
a) Rs. 15240
b) Rs. 18640
c) Rs. 16480
d) Rs. 12680
e) None of the above
Solution:
Let’s assume the monthly salary of Ketan is 12z.
Ketan spent 41.67% of his monthly salary on house rent.
Amount spent on house rent = 12z of (5/12) = 5z
Out of the remaining, he spent 20% on food.
Amount spent on food = (12z-5z) of 20%
= 7z of 20%
= 1.4z
Out of the remaining, he spent 10% on education.
Amount spent on education = (7z-1.4z) of 10%
= 5.6z of 10%
= 0.56z
If the amount spent on education by him is Rs. 4480.
0.56z = 4480
z = 8000
After that the remaining amount was invested in FD and MF in the ratio of 7:5 respectively.
Amount invested on FD = 5.04z of (7/12)
= 2.94z
Difference between the amount spent on house rent and the amount invested on FD from his monthly salary = 5z – 2.94z
= 2.06z
Put the value of ‘z’ in the equation.
= $2.06\times8000$
= Rs. 16480
Hence, option c is the correct answer.
Question 4: The monthly savings of A is 58.33% less than the monthly expenditure of B. The monthly income of A is 25% more than the monthly income of B. If the monthly savings of B is Rs. 760 and the sum of the monthly expenditure of A and B together is Rs. 2490, then find out the difference between the monthly income of A and B.
a) Rs. 500
b) Rs. 200
c) Rs. 300
d) Rs. 400
e) None of the above
Solution:
The monthly savings of A is 58.33% less than the monthly expenditure of B.
Let’s assume the monthly expenditure of B is 12y.
monthly savings of A = (5/12) of 12y = 5y
The monthly income of A is 25% more than the monthly income of B.
Let’s assume the monthly income of B is 4z.
monthly income of A = 5z
monthly expenditure of A = (5z-5y)
monthly savings of B = (4z-12y)
The monthly savings of B is Rs. 760.
(4z-12y) = 760
z-3y = 190
z = (190+3y) Eq.(i)
The sum of the monthly expenditure of A and B together is Rs. 2490.
(5z-5y)+12y = 2490
(5z+7y) = 2490
Put Eq.(i) in the above equation.
$5\times(190+3y)+7y = 2490$
950+15y+7y = 2490
15y+7y = 2490-950 = 1540
22y = 1540
y = 70
Put the value of ‘y’ in Eq.(i).
z = (190+210) = 400
Difference between the monthly income of A and B = (5z-4z) = 400
Hence, option d is the correct answer.
Question 5: The ratio of income of Rahul and Amit is 9:13 respectively. The ratio of savings and expenditure of Amit is 6:7 respectively. If the savings of Amit is 20% more than the savings of Rahul, then the expenditure of Amit is how much percent more or less than the expenditure of Rahul?
a) 75% less
b) 25% more
c) 25% less
d) 75% more
e) None of the above
Solution:
The ratio of income of Rahul and Amit is 9:13 respectively.
Let the income of Rahul and Amit are 9p and 13p respectively.
The ratio of savings and expenditure of Amit is 6:7.
Let the savings and expenditure of Amit are 6q and 7q respectively.
Income of Amit = 13p
$\Rightarrow$ 6q + 7q = 13p
$\Rightarrow$ 13q = 13p
$\Rightarrow$ q = p……..(1)
The savings of Amit is 20% more than the savings of Rahul.
Let the savings of Rahul = t
Savings of Amit = $\frac{120}{100}$t
$\Rightarrow$ 6q = $\frac{6}{5}$t
$\Rightarrow$ t = 5q
$\Rightarrow$ t = 5p
Savings of Rahul = t = 5p
Expenditure of Rahul = Income of Rahul – Savings of Rahul
= 9p – 5p
= 4p
Expenditure of Amit = 7q = 7p
Required percentage = $\frac{7p-4p}{4p}\times$100
= $\frac{3p}{4p}\times$100
= 75% more
Hence, the correct answer is Option D
Question 6: A man spent 20% of his monthly income on house rent. Out of the remaining amount he spent 22% for shopping. Out of the remaining amount one fourth was spent on food and the remaining amount was saved. If the amount spent on shopping was ₹9504, then what was the amount saved by the man?
a) ₹26732
b) ₹27892
c) ₹24382
d) ₹25272
e) None of the above
Solution:
Let the monthly income of man = p
Man spent 20% of his monthly income on house rent.
Remaining amount after spending on house rent = 80% of p
= $\frac{80}{100}$p
Amount spent on shopping was ₹9504.
22% of $\frac{80}{100}$p = 9504
$\frac{22}{100}\times\frac{80}{100}$p = 9504
p = ₹54000
Amount saved by the man = 54000 of (100-20)% of (100-22)% of (3/4)
= 54000 of 80% of 78% of (3/4)
= 54000$\times\frac{80}{100}\times\frac{78}{100}\times\frac{3}{4}$
= ₹25272
Hence, the correct answer is Option D
Question 7: A is 25% less than B 14.28% more than C. If the sum of A and C is 195, then find out the value of B.
a) 180
b) 120
c) 150
d) 90
e) Cannot be determined
Solution:
Let’s assume the value of C is 7y.
A is 25% less than B 14.28% more than C.
Value of B = (8/7) of 7y = 8y
Value of A = 8y of 75% = 6y
If the sum of A and C is 195.
6y+7y = 195
13y = 195
y = 15
value of B = 8y = 120
Hence, option b is the correct answer.
Question 8: In 2010, the number of bears in the zoo was increased by 35% and next year these were decreased by 16.67%. Now the number of bears in the zoo is 405, then find out the 20% of the number of bears in the zoo two years ago.
a) 66
b) 54
c) 60
d) 78
e) None of the above
Solution:
Let’s assume the number of bears in the zoo two years ago is ‘y’.
y of (100+35)% of (100-16.67)% = 405
$y\times1.35\times\frac{5}{6} = 405$
6.75y = 2430
y = 360
20% of the number of bears in the zoo two years ago = 20% of 360
= 72
Hence, option e is the correct answer.
Question 9: The price of icecream has increased by 40%. Lekha has decided to spend only 4% more than what he initially did on buying icecream. What is the percentage decrease in Lekha’s rice consumption?
a) 19%
b) 60%
c) 15%
d) 30%
e) 25%
Solution:
Let the initial price of icecream be Rs. 100 per unit and Lekha’s consumption be 10 units.
∴ Initial amount spent = 100 × 10 = Rs. 1,000
New price of icecream = 140% of 100 = Rs. 140 and new total amount spent = 104% of 1000 = Rs. 1,040
therefore , the new consumption will be $\ \frac{\ 1040}{150}$ = 6.93 (let us take approximate value as 7)
Decrease in consumption = 3 units
% decrease =$\ \frac{\ 3}{10}\times\ 100$ = 30%
Question 10: Three students has birthday on same day they received 126, 786 and 986 gifts respectively from their friends What percentage of the total gifts did the person with more gifts get?
a) 52%
b) 55%
c) 75%
d) 90%
e) 80%
Solution:
Total number of gifts = (126+786+986) = 1898
Required percentage = $\ \frac{\ 986}{1898}\times\ 100$
=> 51.95% which is approximately 52%
Question 11: If 35% of a number P is 82 less than the 45% of a number Q and the sum of numbers P & Q is 1320, then find out the sum of the 55% of P and 25% of Q.
a) 546
b) 568
c) 584
d) 522
e) None of the above
Solution:
35% of P = 45% of Q – 82
0.35P = 0.45Q – 82 Eq.(i)
Sum of numbers P & Q is 1320.
P+Q = 1320
Q = 1320-P Eq.(ii)
Put Eq.(ii) in Eq.(i).
0.35P = 0.45(1320-P) – 82
0.35P = 594-0.45P – 82
0.35P+0.45P = 594 – 82
0.8P = 512
P = 640
Put the value of P in Eq.(ii).
Q = 1320-640 = 680
Sum of the 55% of P and 25% of Q = 55% of 640 + 25% of 680
= 352+170
= 522
Hence, option d is the correct answer.
Question 12: The present population of a village P is 19683. The population of the village P increases by 11.11% every year. After 4 years, out of the total population of village P, 67.3% are adults and the rest are children. Out of the total number of adults, 30% are female while out of the total number of children, 60% are female. Find the total number of male(children+adults) after 4 years.
a) 16486
b) 17156
c) 18057
d) 16874
e) None of these
Solution:
Present population of the village P = 19683.
11.11% = $\dfrac{1}{9}$
Population of P after 4 years = $19683\times\dfrac{10}{9}\times\dfrac{10}{9}\times\dfrac{10}{9}\times\dfrac{10}{9}$
$=30000$
Total number of adults = 67.3% of 30000 = 20190.
Total number of children = 30000-20190 = 9810.
Number of male adults = (100-30)% of 20190 = 70% of 20190 = 14133.
Number of male children = (100-60)% of 9810 = 40% of 9810 = 3924.
Total number of male children and adults together = 14133+3924 = 18057.
Question 13: The population of a village M increases by 12.5% every year. After 4 years, the population of village M will be 465831. Out of the present population of village M, 25% are females and the remaining are males. Out of the female population, 43.75% are children while out of the male population, 58.33% are children. Find the total number of adults in the village M at present.
a) 124786
b) 131776
c) 134786
d) 142784
e) None of these
Solution:
After 4 years, the population of M = 465831.
Present population of M = $465831\times\dfrac{9}{8}\times\dfrac{9}{8}\times\dfrac{9}{8}\times\dfrac{9}{8}$
$=290816$
Number of females = 25% of 290816.
Number of adult females = (100-43.75)% of 25% of 290816 = $\dfrac{9}{16}\times\dfrac{1}{4}\times290816 = 40896$.
Number of males = 75% of 290816.
Number of adult males = (100-58.33)% of 75% of 290816 = $\dfrac{5}{12}\times\dfrac{3}{4}\times290816 = 90880$.
Total number of adults = 40896+90880 = 131776.
Question 14: Four natural numbers P, Q, R, and S are given such that their total sum is equal to 7700. If Q is increased by 36.36% then it will become 30/11 times of the sum of P and S. If Q is decreased by 8.33% it will become 11 times the difference between P and S (Considering P>S). Find the value of S if S is 100% more of R?
a) 990
b) 1000
c) 880
d) 2000
e) Can’t be determined
Solution:
ATQ,
P+Q+R+S = 7700 —————(i)
If Q is increased by 36.36% then it will become 30/11 times of the sum of P and S i.e 36.36% = 4/11, so, if it is increased by 36.36% i.e. 1+ 4/11 = 15/11.
15/11Q= 30/11(P+S)
Q=2(P+S) ———(ii)
If Q is decreased by 8.33% it will become 11 times the difference of P, and S i.e. 8.33% = 1/12
11/12 Q = 11 (P-S)
Q = 12(P-S) ————(iii)
From eq (i) and (ii), we get
2(P+S) = 12(P-S) ——–(iiib)
Now using eq (iii) in the above eq.,we get,
5P= 7S —(iv)
Again using eq (ii) in the above eqn, we get,
Q = 2(7S/5+S)
Q = 24S/5
Q = 4.8S ——(v)
S is 100% more of R
S= 2R —-(vi)
So, from eq (i), we get,
P+Q+R+S = 2000
7S/5+24S/5+S/2+S = 7700
From eq (iiib), we get
S = 1000.
Question 15: In an essay writing class a boy writes a 6 page long essay with 25 lines in each page and 21 characters in each line. If he writes the same content in a different notebook with 20 lines in each page and 15 characters in each line, then the required numbers of pages to write the same essay will be how much percent more or less than the previous notebook?
a) 11%
b) 111%
c) 83.33%
d) 66.67%
e) None of the above
Solution:
According to the question,
6 x 25 x 21 = 20 x 15 x number of pages, solving this we get,
Number of pages = 10.5
So number of pages required = 10.5 pages
Difference = 10.5 – 6 = 4.5
Percentage increase = $\frac{4.5}{6} \times 100$ = 75% more.
Hence (E)
Question 16: A person spends 12.5% of his monthly income on a house loan. Out of his remaining monthly salary, he spends 16.67% on clothes. He distributes the remaining salary to his wife and his son in the ratio 5:2 respectively. His wife spends 20% of the amount obtained by her and saves the rest. His son invests 10% of the amount obtained by him in mutual funds and saves the rest. If the difference between the amount saved by his wife and his son is Rs.4543, then find the amount spent by him on clothes.
a) Rs.2171
b) Rs.2469
c) Rs.2631
d) Rs.2891
e) None of these
Solution:
Let the monthly salary of the person be Rs.48x.
Amount spent on house loan = 12.5% of 48x = Rs.6x.
Remaining salary = 48x-6x = Rs.42x.
Amount spent on clothes = 16.67% of 42x = $\dfrac{1}{6}\times42x = Rs.7x$.
Remaining salary = Rs.42x-7x = Rs.35x.
Amount obtained by his wife = $\dfrac{5}{7}\times35x = Rs.25x$.
Amount spent by his wife = 20% of Rs.25x = Rs.5x.
Amount saved by his wife = Rs.25x – Rs.5x = Rs.20x.
Amount obtained by his son = $\dfrac{2}{7}\times35x = Rs.10x$
Amount invested in mutual funds = 10% of Rs.10x = Rs.x.
Amount saved = Rs.10x-x = Rs.9x.
Given, 20x – 9x = 4543
11x = 4543
x = 413.
Hence, The amount spent on clothes = Rs.7x = Rs.2891.
Question 17: If number A is 37.5% of B and B is 50% of C.Then A will be how much percent of C?
a) 0.625%
b) 6.25%
c) 62.5%
d) 18.75%
e) None of the above
Solution:
ATQ,
A = 37.5/100 B
A = 3/8 B —-(a)
B = 50/100 C
Or
B = ½ C —–(b)
Putting the value of B in eq (a)
A = 3/8 x (½ C)
A = 3/16 C
So, in percentage
$A = (\dfrac{3}{16} \times 100) of C$
A = 18.75% of C.
Question 18: The population of village A in 2010 is 16384. In 2011, the population of the village A decreased by 12.5% and in 2012, it increased by 12.5% as compared to 2011. In 2013, it again decreased by 12.5% and in 2014, it again increased by 12.5%. The population of another village B in 2010 is 43923. In 2011, the population of the village B increased by 9.09% and in 2012, it decreased by 9.09%. In 2013, it again increased by 9.09% and in 2014, it again decreased by 9.09%. Find the difference between the population of villages A and B in 2014.
a) 24384
b) 17364
c) 27324
d) 21464
e) None of these
Solution:
Population of village A in 2010 = 16384.
Population of village A in 2014 = (100-12.5)% of (100+12.5)% of (100-12.5)% of (100+12.5)% of 16384.
$=\dfrac{7}{8}\times\dfrac{9}{8}\times\dfrac{7}{8}\times\dfrac{9}{8}\times16384 = 15876$.
Population of village B in 2010 = 43923.
Population of village B in 2014 = (100+9.09)% of (100-9.09)% of (100+9.09)% of (100-9.09)% of 43923.
$=\dfrac{12}{11}\times\dfrac{10}{11}\times\dfrac{12}{11}\times\dfrac{10}{11}\times43923 = 43200$.
Hence, Difference = 43200 – 15876 = 27324.
Question 19: The ratio between the first and third numbers is 3:1 respectively and the second number is 44.44% more than the first number. The sum of the second and third numbers is 1216. Find out the fourth number which is 83.33% of the first number.
a) 660
b) 540
c) 630
d) 570
e) None of the above
Solution:
The ratio between the first and third numbers is 3:1 respectively.
Let’s assume the first and third numbers are 9y and 3y respectively.
The second number is 44.44% more than the first number.
Second number = (13/9) of 9y = 13y
The sum of the second and third numbers is 1216.
13y+3y = 1216
16y = 1216
y = 76
First number = 9y = $9\times76$ = 684
Fourth number = 83.33% of 684
= ⅚ of 684
= 570
Hence, option d is the correct answer.
Question 20: A person spends 20% of his monthly salary on house rent. Out of the remaining, he spends 15% on food. Out of the remaining salary, he spends 25% on travel. He spends 33.33% of the remaining salary on clothes and distributes the remaining salary to his two sons equally. If each son got Rs.1581, then find the amount spent on house rent.
a) Rs.1860
b) Rs.1720
c) Rs.1240
d) Rs.1450
e) None of these
Solution:
Let his monthly salary be Rs.100x.
Amount spent on house rent = 20% of 100x = Rs.20x.
Remaining salary = 100x-20x = Rs.80x.
Amount spent on food = 15% of 80x = Rs.12x.
Remaining salary = 80x-12x = Rs.68x.
Amount spent on travel = 25% of 68x = Rs.17x.
Remaining salary = 68x-17x = Rs.51x.
Amount spent on clothes = 33.33% of 51x = Rs.17x.
Remaining salary = 51x-17x = Rs.34x.
Amount obtained by each son = Rs.17x.
Given, 17x = 1581 ⇒ x = 93.
Amount spent on house rent = $20\% of 100\times93 = 20\% of 9300 = Rs.1860$.
|
https://math.stackexchange.com/questions/1611212/if-g-a-b-a-b-in-a-a-group-then-is-a-a-group/1611239
|
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# If $G = \{a + b: a, b \in A\}$ a group, then is $A$ a group?
Let $A$ be a subset of an abelian group $H$. Then if $G = \{ a + b : a, b \in A\}$ is a group and $A$ is closed under taking negatives, then is $A$ also a group?
• Have you tried an examples? For instance, what about if $H$ is the group of order $2$, where there's only four subsets to test? Jan 13, 2016 at 20:00
• A group $H\subset G$ is a subgroup of $G$ if and only if $a,b\in H\rightarrow ab^{-1}\in H$ Jan 13, 2016 at 20:00
Probably the simplest counterexample is given by $G=(\mathbb{Z},+)$, with $A$ the set of odd numbers. More generally, if $G$ has a subgroup $H$ of index $2$, then $A=G\setminus H$ will do.
• So essentially you notice that [with $H=(\mathbb{Z},+)$] $G=2\mathbb{Z}$ when $A$ is the set of odd numbers. While $2\mathbb{Z}$ is a subgroup of $H$, $A$ however isn't. This seems to work as an example when ever you take $A=H\setminus K$ where $K$ is a normal subgroup with index 2. Jan 14, 2016 at 18:00
Since A is closed, we have
$a+0=a$, Since $a\in A$ due to closure property, $0\in A$,hence existence of identity.
$a+a^{-1}=0$. $0\in$A, due to closure property, $a^{-1}\in A$. Hence existence of inverse.
Also check for associative property. Let $c\in A$
$c+(a+b)=c+a+b$
$(c+a)+b=c+a+b$
Since $c+(a+b)=(c+a)+b$, Associative property holds.
Hence A must be a group.
|
https://www.gcflcm.com/gcf-of-14-and-26
|
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| 4.40625
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# What is the Greatest Common Factor of 14 and 26?
Greatest common factor (GCF) of 14 and 26 is 2.
GCF(14,26) = 2
We will now calculate the prime factors of 14 and 26, than find the greatest common factor (greatest common divisor (gcd)) of the numbers by matching the biggest common factor of 14 and 26.
GCF Calculator and
and
## How to find the GCF of 14 and 26?
We will first find the prime factorization of 14 and 26. After we will calculate the factors of 14 and 26 and find the biggest common factor number .
### Step-1: Prime Factorization of 14
Prime factors of 14 are 2, 7. Prime factorization of 14 in exponential form is:
14 = 21 × 71
### Step-2: Prime Factorization of 26
Prime factors of 26 are 2, 13. Prime factorization of 26 in exponential form is:
26 = 21 × 131
### Step-3: Factors of 14
List of positive integer factors of 14 that divides 14 without a remainder.
1, 2, 7
### Step-4: Factors of 26
List of positive integer factors of 26 that divides 14 without a remainder.
1, 2, 13
#### Final Step: Biggest Common Factor Number
We found the factors and prime factorization of 14 and 26. The biggest common factor number is the GCF number.
So the greatest common factor 14 and 26 is 2.
Also check out the Least Common Multiple of 14 and 26
|
https://web2.0calc.com/questions/please-help_5737
|
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+0
+1
168
3
Suppose p and q are inversely proportional. If p=25 when q= 6, find the value of p when q=15.
#1
0
We compute $25 \cdot \frac{6}{15}=\boxed{10}$.
Aug 5, 2020
#2
0
Sorry, I forgot the LaTeX. We multiplied $$p$$ by $$\frac{6}{15}$$, so we multiply $$q$$ by $$\frac{15}{6}$$, to get $$\boxed{10}$$
Aug 5, 2020
#3
0
If p and q are inversely proportional, then there is a constant k such that: p = k / q
If p = 25 then q = 6 ---> 25 = k / 6 ---> 25 · 6 = k ---> k = 150
For this problem, the formula is: p = 150 / q
Snce q = 15 ---> p = 150 / 15 = 10
Aug 5, 2020
|
https://tools.carboncollective.co/compound-interest/55365-at-35-percent-in-27-years/
|
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# What is the compound interest on $55365 at 35% over 27 years? If you want to invest$55,365 over 27 years, and you expect it will earn 35.00% in annual interest, your investment will have grown to become $182,914,044.87. If you're on this page, you probably already know what compound interest is and how a sum of money can grow at a faster rate each year, as the interest is added to the original principal amount and recalculated for each period. The actual rate that$55,365 compounds at is dependent on the frequency of the compounding periods. In this article, to keep things simple, we are using an annual compounding period of 27 years, but it could be monthly, weekly, daily, or even continuously compounding.
The formula for calculating compound interest is:
$$A = P(1 + \dfrac{r}{n})^{nt}$$
• A is the amount of money after the compounding periods
• P is the principal amount
• r is the annual interest rate
• n is the number of compounding periods per year
• t is the number of years
We can now input the variables for the formula to confirm that it does work as expected and calculates the correct amount of compound interest.
For this formula, we need to convert the rate, 35.00% into a decimal, which would be 0.35.
$$A = 55365(1 + \dfrac{ 0.35 }{1})^{ 27}$$
As you can see, we are ignoring the n when calculating this to the power of 27 because our example is for annual compounding, or one period per year, so 27 × 1 = 27.
## How the compound interest on $55,365 grows over time The interest from previous periods is added to the principal amount, and this grows the sum a rate that always accelerating. The table below shows how the amount increases over the 27 years it is compounding: Start Balance Interest End Balance 1$55,365.00 $19,377.75$74,742.75
2 $74,742.75$26,159.96 $100,902.71 3$100,902.71 $35,315.95$136,218.66
4 $136,218.66$47,676.53 $183,895.19 5$183,895.19 $64,363.32$248,258.51
6 $248,258.51$86,890.48 $335,148.99 7$335,148.99 $117,302.15$452,451.14
8 $452,451.14$158,357.90 $610,809.03 9$610,809.03 $213,783.16$824,592.20
10 $824,592.20$288,607.27 $1,113,199.47 11$1,113,199.47 $389,619.81$1,502,819.28
12 $1,502,819.28$525,986.75 $2,028,806.03 13$2,028,806.03 $710,082.11$2,738,888.14
14 $2,738,888.14$958,610.85 $3,697,498.98 15$3,697,498.98 $1,294,124.64$4,991,623.63
16 $4,991,623.63$1,747,068.27 $6,738,691.90 17$6,738,691.90 $2,358,542.16$9,097,234.06
18 $9,097,234.06$3,184,031.92 $12,281,265.98 19$12,281,265.98 $4,298,443.09$16,579,709.07
20 $16,579,709.07$5,802,898.18 $22,382,607.25 21$22,382,607.25 $7,833,912.54$30,216,519.79
22 $30,216,519.79$10,575,781.93 $40,792,301.71 23$40,792,301.71 $14,277,305.60$55,069,607.31
24 $55,069,607.31$19,274,362.56 $74,343,969.87 25$74,343,969.87 $26,020,389.45$100,364,359.32
26 $100,364,359.32$35,127,525.76 $135,491,885.09 27$135,491,885.09 $47,422,159.78$182,914,044.87
We can also display this data on a chart to show you how the compounding increases with each compounding period.
As you can see if you view the compounding chart for $55,365 at 35.00% over a long enough period of time, the rate at which it grows increases over time as the interest is added to the balance and new interest calculated from that figure. ## How long would it take to double$55,365 at 35% interest?
Another commonly asked question about compounding interest would be to calculate how long it would take to double your investment of $55,365 assuming an interest rate of 35.00%. We can calculate this very approximately using the Rule of 72. The formula for this is very simple: $$Years = \dfrac{72}{Interest\: Rate}$$ By dividing 72 by the interest rate given, we can calculate the rough number of years it would take to double the money. Let's add our rate to the formula and calculate this: $$Years = \dfrac{72}{ 35 } = 2.06$$ Using this, we know that any amount we invest at 35.00% would double itself in approximately 2.06 years. So$55,365 would be worth $110,730 in ~2.06 years. We can also calculate the exact length of time it will take to double an amount at 35.00% using a slightly more complex formula: $$Years = \dfrac{log(2)}{log(1 + 0.35)} = 2.31\; years$$ Here, we use the decimal format of the interest rate, and use the logarithm math function to calculate the exact value. As you can see, the exact calculation is very close to the Rule of 72 calculation, which is much easier to remember. Hopefully, this article has helped you to understand the compound interest you might achieve from investing$55,365 at 35.00% over a 27 year investment period.
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https://tutorialspoint.dev/language/cpp/set-variable-without-using-arithmetic-relational-conditional-operator
|
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# Set a variable without using Arithmetic, Relational or Conditional Operator
Given three integers a, b and c where c can be either 0 or 1. Without using any arithmetic, relational and conditional operators set the value of a variable x based on below rules –
```If c = 0
x = a
Else // Note c is binary
x = b.
```
Examples:
```Input: a = 5, b = 10, c = 0;
Output: x = 5
Input: a = 5, b = 10, c = 1;
Output: x = 10
```
## Recommended: Please try your approach on {IDE} first, before moving on to the solution.
Solution 1: Using arithmetic operators
If we are allowed to use arithmetic operators, we can easily calculate x by using any one of below expressions –
```x = ((1 - c) * a) + (c * b)
OR
x = (a + b) - (!c * b) - (c * a);
OR
x = (a * !c) | (b * c);
```
`#include ` `using` `namespace` `std; ` ` ` `int` `calculate(``int` `a, ``int` `b, ``int` `c) ` `{ ` ` ``return` `((1 - c) * a) + (c * b); ` `} ` ` ` `int` `main() ` `{ ` ` ``int` `a = 5, b = 10, c = 0; ` ` ` ` ``int` `x = calculate(a, b, c); ` ` ``cout << x << endl; ` ` ` ` ``return` `0; ` `} `
Output:
```5
```
Solution 2: Without using arithmetic operators
The idea is to construct an array of size 2 such that index 0 of the array stores value of variable ‘a’ and index 1 value of variable b. Now we return value at index 0 or at index 1 of the array based on value of variable c.
`#include ` `using` `namespace` `std; ` ` ` `int` `calculate(``int` `a, ``int` `b, ``int` `c) ` `{ ` ` ``int` `arr[] = {a, b}; ` ` ``return` `*(arr + c); ` `} ` ` ` `int` `main() ` `{ ` ` ``int` `a = 5, b = 10, c = 1; ` ` ` ` ``int` `x = calculate(a, b, c); ` ` ``cout << x << endl; ` ` ` ` ``return` `0; ` `} `
Output:
```10
```
## tags:
C C++ cpp-operator cpp-operator C CPP
|
https://en.m.wikibooks.org/wiki/FHSST_Biology/How_to_Change_Units
|
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FHSST Biology/How to Change Units
How to Change Units
Firstly you obviously need some relationship between the two units that you wish to convert between. Let us demonstrate with a simple example. We will consider the case of converting millimeters (mm) to meters (m)—the SI unit of length. We know that there are 1000 mm in 1 m which we can write as
${\begin{matrix}1000{\mbox{ mm}}=1{\mbox{ m}}\end{matrix}}$ .
Now multiplying both sides by
${\frac {1}{1000{\mbox{ mm}}}}$
we get
${\begin{matrix}{\frac {1}{1000{\mbox{ mm}}}}1000{\mbox{ mm}}={\frac {1}{1000{\mbox{ mm}}}}1{\mbox{ m}}\end{matrix}}$ ,
which simply gives us
${\begin{matrix}1={\frac {1{\mbox{ m}}}{1000{\mbox{ mm}}}}\end{matrix}}$ .
This is the conversion ratio from millimeters to meters. You can derive any conversion ratio in this way from a known relationship between two units. Let's use the conversion ratio we have just derived in an example:
Question: Express 3800 mm in meters.
${\begin{matrix}3800{\mbox{ mm}}&=&3800{\mbox{ mm}}\times 1\\&=&3800{\mbox{ mm}}\times {\frac {1{\mbox{ m}}}{1000{\mbox{ mm}}}}\\&=&3.8{\mbox{ m}}\\\end{matrix}}$ .
Note that we wrote every unit in each step of the calculation. By writing them in and canceling them properly, we can check that we have the right units when we are finished. We started with mm and multiplied by ${\frac {\mbox{m}}{\mbox{mm}}}$
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https://metanumbers.com/50647
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## 50647
50,647 (fifty thousand six hundred forty-seven) is an odd five-digits prime number following 50646 and preceding 50648. In scientific notation, it is written as 5.0647 × 104. The sum of its digits is 22. It has a total of 1 prime factor and 2 positive divisors. There are 50,646 positive integers (up to 50647) that are relatively prime to 50647.
## Basic properties
• Is Prime? Yes
• Number parity Odd
• Number length 5
• Sum of Digits 22
• Digital Root 4
## Name
Short name 50 thousand 647 fifty thousand six hundred forty-seven
## Notation
Scientific notation 5.0647 × 104 50.647 × 103
## Prime Factorization of 50647
Prime Factorization 50647
Prime number
Distinct Factors Total Factors Radical ω(n) 1 Total number of distinct prime factors Ω(n) 1 Total number of prime factors rad(n) 50647 Product of the distinct prime numbers λ(n) -1 Returns the parity of Ω(n), such that λ(n) = (-1)Ω(n) μ(n) -1 Returns: 1, if n has an even number of prime factors (and is square free) −1, if n has an odd number of prime factors (and is square free) 0, if n has a squared prime factor Λ(n) 10.8326 Returns log(p) if n is a power pk of any prime p (for any k >= 1), else returns 0
The prime factorization of 50,647 is 50647. Since it has a total of 1 prime factor, 50,647 is a prime number.
## Divisors of 50647
2 divisors
Even divisors 0 2 1 1
Total Divisors Sum of Divisors Aliquot Sum τ(n) 2 Total number of the positive divisors of n σ(n) 50648 Sum of all the positive divisors of n s(n) 1 Sum of the proper positive divisors of n A(n) 25324 Returns the sum of divisors (σ(n)) divided by the total number of divisors (τ(n)) G(n) 225.049 Returns the nth root of the product of n divisors H(n) 1.99996 Returns the total number of divisors (τ(n)) divided by the sum of the reciprocal of each divisors
The number 50,647 can be divided by 2 positive divisors (out of which 0 are even, and 2 are odd). The sum of these divisors (counting 50,647) is 50,648, the average is 25,324.
## Other Arithmetic Functions (n = 50647)
1 φ(n) n
Euler Totient Carmichael Lambda Prime Pi φ(n) 50646 Total number of positive integers not greater than n that are coprime to n λ(n) 50646 Smallest positive number such that aλ(n) ≡ 1 (mod n) for all a coprime to n π(n) ≈ 5190 Total number of primes less than or equal to n r2(n) 0 The number of ways n can be represented as the sum of 2 squares
There are 50,646 positive integers (less than 50,647) that are coprime with 50,647. And there are approximately 5,190 prime numbers less than or equal to 50,647.
## Divisibility of 50647
m n mod m 2 3 4 5 6 7 8 9 1 1 3 2 1 2 7 4
50,647 is not divisible by any number less than or equal to 9.
## Classification of 50647
• Arithmetic
• Prime
• Deficient
### Expressible via specific sums
• Polite
• Non-hypotenuse
• Prime Power
• Square Free
## Base conversion (50647)
Base System Value
2 Binary 1100010111010111
3 Ternary 2120110211
4 Quaternary 30113113
5 Quinary 3110042
6 Senary 1030251
8 Octal 142727
10 Decimal 50647
12 Duodecimal 25387
20 Vigesimal 66c7
36 Base36 132v
## Basic calculations (n = 50647)
### Multiplication
n×i
n×2 101294 151941 202588 253235
### Division
ni
n⁄2 25323.5 16882.3 12661.8 10129.4
### Exponentiation
ni
n2 2565118609 129915562190023 6579833478238094881 333248826172324791438007
### Nth Root
i√n
2√n 225.049 36.9985 15.0016 8.72792
## 50647 as geometric shapes
### Circle
Diameter 101294 318224 8.05856e+09
### Sphere
Volume 5.44189e+14 3.22342e+10 318224
### Square
Length = n
Perimeter 202588 2.56512e+09 71625.7
### Cube
Length = n
Surface area 1.53907e+10 1.29916e+14 87723.2
### Equilateral Triangle
Length = n
Perimeter 151941 1.11073e+09 43861.6
### Triangular Pyramid
Length = n
Surface area 4.44292e+09 1.53107e+13 41353.1
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https://www.physicsforums.com/threads/reaction-rate-question.155713/
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# Reaction Rate Question
Question
The rate of appearance of I2 in aqueous solution, in the reaction of the I- ion with hydrogen peroxide, was found to change from 3.7*10^-5 mol/dm^3 to 7.9*10^-5 mol/dm^3 over a time interval of 1.0 to 3.0s.
2H+(aq) + 2I-(aq) + H2O2(aq) --> I2(aq) + 2H2O(l)
a)What is the rate of appearance of H2 during this period?
b) At what rate should the concentration of the H+ ion be changing during this same time period?
c) Is this mechanism likely to occur in one step?
Attempt
a) rate(H2) = rate(I2)
rate(H2)=delta[H2]/delta t
rate(H2)=(7.9*10^-5 mol/dm^3-3.7*10^-5 mol/dm^3)/(3.0s-1.0s)
rate(H2)=2.1*10^-5 mol/(dm^3*s)
b)mol(H+)/mol(H2)=2/1 = 2
Therefore rate(H+)=2*(-2.1*10^-5 mol/(dm^3*s))
=-4.2*10^-5 mol/(dm^3*s)
c) No, it is not likely to occur in one step, as the overall reaction is not elementary: it has a molecularity of 5, and a molecularity of 3 or less must exist to ensure a one-step (elementary) mechanism.
I am unsure about the whole process. Please tell me if I did this correctly.
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https://365datascience.com/question/is-deltaodds-the-difference-between-odds-2-and-odds-1-or-the-quotient-between-odds-2-and-odds-1/
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Super learner
26 Sept 2022
Posted on:
25 Sept 2022
0
# Resolved:Is DeltaOdds the difference between Odds_2 and Odds_1 or the quotient between Odds_2 and Odds_1?
Usually with "Delta" we mean the difference between two objects. Taking the example of the video, Odds_2 = 1.042 * Odds_1 = e^(0.042) * Odds_1 when the difference between the corresponding SATs is 1. If we take the difference, we obtain DeltaOdds = Odds_2 - Odds_1 = 0.042 Odds_1, so it isn't e^b_k = e^0.042 = 1.042, but it also depends on Odds_1.
But if with "Delta" here you mean the ratio between Odds_2 and Odds_1, it is correct, just a bit confusing to call it "Delta". Am I right?
Instructor
Posted on:
26 Sept 2022
0
Hey Alessandro,
Thank you for this comment!
I agree that deltas are typically reserved for differences between two quantities, rather than their fraction. In this example, however, it is used to represent the fraction between the odds. It's a matter of choice :)
Kind regards,
365 Hristina
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https://openhs.ca/learn.php?p=3
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# Getting the Ball Rolling: Domain, Range, Functions, and Relations
🧑Zayn Alazawi 📅June 18, 2020 📊Grade 11 Math
Introduction:
In the realm of mathematics, functions and relations serve as fundamental concepts that form the backbone of various mathematical disciplines. Understanding these concepts is crucial for tackling complex problems in algebra, calculus, and other branches of mathematics. In this article, we will delve into the definitions and properties of functions, relations, domain, and range.
Functions:
A function is a mathematical object that assigns to every element in a set $$X$$ exactly one element in a set $$Y$$. In simpler terms, a function relates each input to exactly one output. The input values, often denoted by $$x$$, belong to the domain of the function, and the corresponding output values, denoted by $$f(x)$$, belong to the codomain. A function is typically represented as $$f:X\rightarrow Y$$:, where $$f$$ is the function, $$X$$ is the domain, and $$Y$$ is the codomain.
Functions are represented using function notation. It's a shorthand method that makes it more convenient to represent and work with functions. The most common form of function notation is $$f(x)$$, where:
1. $$f$$: This is the name of the function. It can be any letter or symbol and is often chosen to represent the nature of the function (e.g., $$g(x)$$, $$h(x)$$, etc.).
2. $$x$$: This represents the input variable. It's the value you plug into the function to get the corresponding output.
3. $$f(x)$$: This is the output of the function given the input $$x$$. It's the result of applying the function to the specific value of $$x$$.
For example, if you have a function $$f$$ defined as $$f(x) = 2x+3$$, you could use function notation to express that if $$x$$ is 4, then $$f(4)$$ would be $$2(4)+3=11$$. So, $$f(4)=11$$.
Function notation is particularly useful when dealing with more complex functions or when expressing relationships in mathematical formulas. It helps in understanding and communicating how a function transforms its input into output, and it's a standard convention used in various branches of mathematics.
Relations:
A relation, on the other hand, is a set of ordered pairs $$(x,y)$$, where $$x$$ is in the domain and $$y$$ is in the codomain. In contrast to functions, a single input value in the domain can be associated with more than one output value in the codomain in a relation. Relations can be represented graphically as points on a coordinate plane, showcasing the pairing of input and output values.
Domain:
The domain of a function or relation is the set of all possible input values for which the function or relation is defined. It is essentially the x-values that make the function or relation meaningful. The domain is a critical aspect as it determines the scope of the function or relation. For instance, a square root function is defined only for non-negative real numbers, and thus its domain is restricted to non-negative values.
Range:
The range of a function or relation is the set of all possible output values that the function or relation can produce. It represents the y-values that the function or relation can attain based on its inputs. Determining the range is crucial for understanding the behavior of a function or relation and its output variability. In some cases, the range may be a subset of the codomain, while in others, it may coincide with the entire codomain.
Conclusion:
In conclusion, functions and relations play pivotal roles in mathematics, offering a structured way to model and understand the relationships between sets of values. The concepts of domain and range further refine our understanding by specifying the input and output values, respectively. A strong grasp of these foundational concepts is essential for students and professionals alike, as they serve as the building blocks for more advanced mathematical studies and applications in diverse fields.
## ✍️Practice
1. Consider the function $$f(x)=3x−2$$.
a) Find $$f(4)$$.
b) Determine the input value ($$x$$) when $$f(x)=7$$.
2. a) $$f(4)=3×4−2=10$$
b) $$3x−2=7$$, solve for $$x$$$$9 = 3x$$ so $$x=3$$
3. Let $$R$$ be the relation defined by $$R=\{(1,2),(3,4),(5,6),(1,4)\}$$
a) Is $$R$$ a function? Why or why not?
b) Find the range of $$R$$.
4. a) No, $$R$$ is not a function because it has the same input (1) paired with different outputs (2 and 4).
b) Range of $$R$$ is $$\{2,4,6\}$$.
5. Given the function $$g(x)=x^2−1$$
a) Determine the domain of $$g(x)$$.
b) Find the range of $$g(x)$$.
6. a) Domain of $$g(x)=x^2−1$$ is all real numbers, $$X\epsilon\mathbb{R}$$.
b) Range of $$g(x)$$ is all real numbers greater than or equal to -1, $$Y\epsilon\mathbb{R}|Y\geq-1$$.
7. Consider the function $$h(x)=\sqrt{x+1}$$.
a) Determine the domain of $$h(x)$$.
b) Find the range of $$h(x)$$.
8. a) Domain of $$h(x)=\sqrt{x+1}$$ is $$x≥−1$$ because the expression inside the square root cannot be negative.
b) Range of $$h(x)$$ is $$y≥0$$ because the square root of any real number is non-negative.
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https://yutsumura.com/the-union-of-two-subspaces-is-not-a-subspace-in-a-vector-space/
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# The Union of Two Subspaces is Not a Subspace in a Vector Space ## Problem 274
Let $U$ and $V$ be subspaces of the vector space $\R^n$.
If neither $U$ nor $V$ is a subset of the other, then prove that the union $U \cup V$ is not a subspace of $\R^n$. Add to solve later
## Proof.
Since $U$ is not contained in $V$, there exists a vector $\mathbf{u}\in U$ but $\mathbf{u} \not \in V$.
Similarly, since $V$ is not contained in $U$, there exists a vector $\mathbf{v} \in V$ but $\mathbf{v} \not \in U$.
Seeking a contradiction, let us assume that the union is $U \cup V$ is a subspace of $\R^n$.
The vectors $\mathbf{u}, \mathbf{v}$ lie in the vector space $U \cup V$.
Thus their sum $\mathbf{u}+\mathbf{v}$ is also in $U\cup V$.
This implies that we have either
$\mathbf{u}+\mathbf{v} \in U \text{ or } \mathbf{u}+\mathbf{v}\in V.$
If $\mathbf{u}+\mathbf{v} \in U$, then there exists $\mathbf{u}’\in U$ such that
$\mathbf{u}+\mathbf{v}=\mathbf{u}’.$ Since the vectors $\mathbf{u}$ and $\mathbf{u}’$ are both in the subspace $U$, their difference $\mathbf{u}’-\mathbf{u}$ is also in $U$. Hence we have
$\mathbf{v}=\mathbf{u}’-\mathbf{u} \in U.$
However, this contradicts the choice of the vector $\mathbf{v} \not \in U$.
Thus, we must have $\mathbf{u}+\mathbf{v}\in V$.
In this case, there exists $\mathbf{v}’ \in V$ such that
$\mathbf{u}+\mathbf{v}=\mathbf{v}’.$
Since both $\mathbf{v}, \mathbf{v}’$ are vectors of $V$, it follows that
$\mathbf{u}=\mathbf{v}’-\mathbf{v}\in V,$ which contradicts the choice of $\mathbf{u} \not\in V$.
Therefore, we have reached a contradiction. Thus, the union $U \cup V$ cannot be a subspace of $\R^n$.
## Related Question.
In fact, the converse of this problem is true.
Problem. Let $W_1, W_2$ be subspaces of a vector space $V$. Then prove that $W_1 \cup W_2$ is a subspace of $V$ if and only if $W_1 \subset W_2$ or $W_2 \subset W_1$.
For a proof, see the post “Union of Subspaces is a Subspace if and only if One is Included in Another“. Add to solve later
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###### More in Linear Algebra ##### Quiz 2. The Vector Form For the General Solution / Transpose Matrices. Math 2568 Spring 2017.
(a) The given matrix is the augmented matrix for a system of linear equations. Give the vector form for the...
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https://socratic.org/questions/what-is-the-total-number-of-grams-of-nai-needed-to-make-1-0-liter-of-a-0-010-m-s
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| 4.34375
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| 304
| 1,008
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# What is the total number of grams of NaI needed to make 1.0 liter of a 0.010 M solution?
Jun 7, 2016
1.50g of NaI
#### Explanation:
Molarity is represented by the following equation: In our case, we already have the molarity and the volume of solution, both of which have the appropriate units.
Now we can rearrange the equation to solve for the number of moles, which will allow us to determine the mass. We can do this by multiplying by liters of solution on both sides of the equation. The liters of solution will cancel out on the right side, leaving the number of moles being equal to the molarity times volume like so:
Moles of solute = $l i t e r s o f s o l u t i o n \times M o l a r i t y$
Moles of solute = (1.0 L)$\times$(0.010 M) = 0.010 moles of NaI
Now we have to convert the 0.010 moles of NaI into grams of NaI. This can be done by multiplying 0.010 moles by the molecular weight of NaI, which is $149.89 \frac{g}{m o l}$.
$0.010$moles $\times$$149.89 \frac{g}{m o l}$ = 1.50g NaI
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https://socratic.org/questions/562ce35811ef6b338b114f66
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| 4.21875
| 4
| 728
| 2,309
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# Question 14f66
Jan 13, 2016
Here's what I got.
#### Explanation:
Your starting point here will be the rate law for a second-order and a third-order reaction, respectively.
As you know, the rate law for a given reaction establishes a relationship between the rate of the reaction, the rate constant, and the concentrations of the reactants.
• Second-order reaction
For a second-order reaction, the rate law takes the form
color(blue)("rate" = k * ["A"] * ["B"])" ", where
$k$ - the rate constant
SIDE NOTE Since your goal here is to determine the units of the rate constant, you don't have to worry about having
"rate" = k * ["A"]^2
since the units will come out the same regardless if you have one reactant or two reactants.
Now, the rate of a reaction is a measure of the change in concentration of partial pressure of the reactants (or products) per unit of time.
In your case, the units for the rate of a reaction will be
$\textcolor{b l u e}{{\text{M"/"s" = "mol"/("L" * "s") = "mol L"^(-1)"s}}^{- 1}} \to$ for concentrations
$\textcolor{b l u e}{{\text{kPa"/"s" = "kPa s}}^{- 1}} \to$ for partial pressures
So, rearrange the rate law equation to find the units of $k$
k = "rate"/(["A"] * ["B"]) = (color(red)(cancel(color(black)("mol"))) color(red)(cancel(color(black)("L"^(-1)))) "s"^(-1))/(color(red)(cancel(color(black)("mol"))) color(red)(cancel(color(black)("L"^(-1)))) * "mol L"^(-1)) = color(green)("L mol"^(-1)"s"^(-1))
and
k = "rate"(["A"] * ["B"]) = (color(red)(cancel(color(black)("kPa"))) "s"^(-1))/(color(red)(cancel(color(black)("kPa"))) * "kPa") = color(green)("kPa"^(-1)"s"^(-1))
• Third-order reaction
For a third-order reaction, the rate law could look like this
color(blue)("rate" = k * ["A"] * ["B"] * ["C"])
This time, the units of the rate constant will be
k = "rate"/(["A"] * ["B"] * ["C"]) = (color(red)(cancel(color(black)("mol"))) color(red)(cancel(color(black)("L"^(-1)))) "s"^(-1))/(color(red)(cancel(color(black)("mol"))) color(red)(cancel(color(black)("L"^(-1)))) * "mol L"^(-1) * "mol L"^(-1))
$k = \textcolor{g r e e n}{{\text{L"^(2) "mol"^(-2)"s}}^{- 1}}$
and
k = "rate"/(["A"] * ["B"] * ["C"]) = (color(red)(cancel(color(black)("kPa"))) "s"^(-1))/(color(red)(cancel(color(black)("kPa"))) * "kPa" * "kPa") = color(green)("kPa"^(-2)"s"^(-1))#
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https://answers.everydaycalculation.com/simplify-fraction/1607-2518
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| 3.59375
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Solutions by everydaycalculation.com
## Reduce 1607/2518 to lowest terms
1607/2518 is already in the simplest form. It can be written as 0.638205 in decimal form (rounded to 6 decimal places).
#### Steps to simplifying fractions
1. Find the GCD (or HCF) of numerator and denominator
GCD of 1607 and 2518 is 1
2. Divide both the numerator and denominator by the GCD
1607 ÷ 1/2518 ÷ 1
3. Reduced fraction: 1607/2518
Therefore, 1607/2518 simplified to lowest terms is 1607/2518.
MathStep (Works offline) Download our mobile app and learn to work with fractions in your own time:
|
http://staging.physicsclassroom.com/class/estatics/Lesson-3/Newton-s-Laws-and-the-Electrical-Force
|
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| 4.46875
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Static Electricity - Lesson 3 - Electric Force
# Newton's Laws and the Electrical Force The attractive or repulsive interaction between any two charged objects is an electric force. Like any force, its effect upon objects is described by Newton's laws of motion. The electric force - Felect - joins the long list of other forces that can act upon objects. Newton's laws are applied to analyze the motion (or lack of motion) of objects under the influence of such a force or combination of forces. The analysis usually begins with the construction of a free-body diagram in which the type and direction of the individual forces are represented by vector arrows and labeled according to type. The magnitudes of the forces are then added as vectors in order to determine the resultant sum, also known as the net force. The net force can then be used to determine the acceleration of the object.
In some instances, the goal of the analysis is not to determine the acceleration of the object. Instead, the free-body diagram is used to determine the spatial separation or charge of two objects that are at static equilibrium. In this case, the free-body diagram is combined with an understanding of vector principles in order to determine some unknown quantity in the midst of a puzzle involving geometry, trigonometry and Coulomb's law. In this last section of Lesson 3, we will explore both types of applications of Newton's laws to static electricity phenomenon.
### Electric Force and Acceleration
Suppose that a rubber balloon and a plastic golf tube are both charged negatively by rubbing them with animal fur. Suppose that the balloon is tossed up into the air and the golf tube is held beneath it in an effort to levitate the balloon in midair. This goal would be accomplished when the spatial separation between charged objects is adjusted such that the downward gravity force (Fgrav) and the upward electric force (Felect) are balanced. This would present a difficult task of manipulation as the balloon would constantly move from side to side and up and down under the influences of both the gravity force and the electric force. When the golf tube is held too far below the balloon, the balloon would fall and accelerate downward. This would in turn decrease the separation distance and lead to an increase in the electric force. As the Felect increases, it would likely exceed the Fgrav and the balloon would suddenly accelerate upward. And finally, if the point of charge on the golf tube is not directly under the point of charge of the balloon (a likely scenario), the electric force would be exerted at an angle to the vertical and the balloon would have a sideways acceleration. The likely result of such an effort to levitate the balloon would be a variety of instantaneous accelerations in a variety of directions. Suppose that at some instant in the process of trying to levitate the balloon, the following conditions existed:
A 0.90-gram balloon with a charge of -75 nC is located a distance of 12 cm above a plastic golf tube that has a charge of -83 nC.
How could one apply Newton's laws to determine the acceleration of the balloon at this instant?
Like any problem involving force and acceleration, the problem would begin with the construction of a free-body diagram. There are two forces acting upon the balloon. The force of gravity on the balloon is directed downward. The electric force on the balloon is exerted upward since the balloon and golf tube are like-charged and the golf tube is held below the balloon. These two forces are shown in the free-body diagram at the right. The second step involves determining the magnitude of these two forces. The force of gravity is determined by multiplying the mass (in kilograms) by the acceleration of gravity.
Fgrav = m • g = (0.00090 kg) • (9.8 m/s/s)
Fgrav = 8.82 x 10-3 N, down
The electric force is determined using Coulomb's law. As shown below, the appropriate unit on charge is the Coulomb (C) and the appropriate unit on distance is meters (m). Use of these units will result in a force unit of the Newton. The demand for these units emerges from the units on Coulomb's constant.
Felect = k • Q1 • Q2 /d2
Felect = (9 x 109 N•m2/C2) • (-75 x 10-9 C) • (-83 x 10-9 C) / (0.12)2
Felect = 3.89 x 10-3 N, up
The net force is the vector sum of these two forces. The upward and downward forces are added together as vectors.
Fnet = ·F = Fgrav (down) + Felect (up)
Fnet = 8.82 x 10-3 N, down + 3.89 x 10-3 N, up
Fnet = 4.93 x 10-3 N, down
The final step of this problem involves the use of Newton's second law to determine the acceleration of the object. The acceleration is the net force divided by the mass (in kilograms).
a = Fnet / m = (4.93 x 10-3 N, down) / (0.00090 kg)
a = 5.5 m/s/s, down
The above analysis illustrates how Newton's law and Coulomb's law can be applied to determine an instantaneous acceleration. The next analysis involves a case in which two objects are in a state of static equilibrium.
### Electric Force and Static Equilibrium
Suppose that two rubber balloons are hung from the ceiling by two long strings such that they hang vertically. Then suppose that each balloon is given 10 average-strength rubs with animal fur. The balloons, having a greater attraction for electrons than animal fur, would acquire a negative charge. The balloons would have the same type of charge and they would subsequently repel each other. The result of their repulsion is that the strings and suspended balloons would now make an angle with the vertical. The angle of the string with the vertical would be mathematically related to the quantity of charge on the balloons. As the balloons acquire a greater quantity of charge, the force of repulsion between them would increase and the angle that the string makes with the vertical would also increase. Like any situation involving electrostatic force, this situation can be analyzed using vector principles and Newton's laws.
Suppose that the following conditions existed.
Two 1.1-gram balloons are suspended from 2.0-meter long strings and hung from the ceiling. They are then rubbed ten times with animal fur to impart an identical charge Q to each balloon. The balloons repel each other and each string is observed to make an angle of 15 degrees with the vertical. Determine the electric force of repulsion, the charge on each balloon (assumed to be identical), and the quantity of electrons transferred to each balloon as a result of 10 rubs with animal fur.
Because of the complexity of the physical situation, it would be wise to represent it using a diagram. The diagram will serve as a means of identifying the known information for this situation. The diagram below depicts the two balloons with the string of length L and the angle "theta". The mass (m) of the balloons is known; it is expressed here in kilogram (the standard unit of mass). The distance between the balloons (a variable in Coulomb's law) is marked on the diagram and represented by the variable d. The vertical line extending from the pivot point on the ceiling is drawn; this vertical line is one side of a right triangle formed by the horizontal line connecting the balloons and the string extending from balloon to ceiling. This right triangle will be useful as we analyze the situation using vector principles. Note that the vertical line bisects the line segment connecting the balloons; thus, one side of the right triangle has a distance of d/2. The application of Newton's laws to this situation begins with the construction of a free-body diagram for one of the balloons. There are three forces acting upon the balloons: the tension force, the force of gravity and the electrostatic force of repulsion. These three forces are represented for the balloon on the right. (See diagram below.) Note that the tension force is directed at an angle to the vertical. In physics, such situations are treated by resolving the force vector into horizontal and vertical components. This is shown below; the components are labeled as Fx and Fy. These components are related to the angle that the string makes with the vertical by trigonometric functions. Since the balloon is at equilibrium, the forces that act upon the balloon must balance each other. This would mean that the vertical component of the tension force (Fy) must balance the downward force of gravity (Fgrav). And the horizontal component of the tension force (Fx) must balance the rightward electrostatic force (Felect). Since the mass of the balloon is known, the force of gravity acting upon it can be determined.
Fgrav = m •g = (0.0011 kg) • (9.8 m/s/s)
Fgrav = 0.01078 N
The force of gravity is equal to the vertical component of the tension force (Fy = 0.0108 N ). The Fy component is related to the Fx component and the angle theta by the tangent function. This relationship can be used to determine the horizontal component of the tension force. The work is shown below.
Tangent(theta) = Fx / Fy
Tangent(15 degrees) = Fx / (0.01078 N)
Fx = (0.01078 N) • Tangent(15 degrees)
Fx = 0.00289 N
The horizontal component of the tension force is equal to the electrostatic force. Thus,
Felect = 0.00289 N
Now that the electrostatic force has been determined using Newton's laws and vector principles, Coulomb's law can now be applied to determine the charge on the balloon. It is assumed that the balloons have the same quantity of charge since they are charged in the same manner with 10 average-strength rubs. Since Q1 is equal to Q2, the equation can be rewritten as This equation can be algebraically rearranged in order to solve for Q. The steps are shown below.
F • d2 = k • Q2
Q2 = F • d2 / k
Q = SQRT(F • d2 / k)
The value of d must be known to complete the solution. This demands that the right triangle be analyzed in order to determine the length of the side opposite the 15-degree angle. This length is one-half the distance d. Since the length of the hypotenuse is known, the sine function is used.
Sine(Theta) = opposite side / hypotenuse side
Sine(15 degrees) = opposite side / (2.0 m)
opposite side = (2.0 m) • Sine(15 degrees)
opposite side = d/2 = 0.518 m
Doubling this distance yields a value of d of 1.035 m. Now substitutions can be made in order to determine the value of Q.
Q = SQRT(F • d2 / k)
Q = SQRT [(0.00289 N) • (1.035 m)2 / (9 x 109 N•m2/C2)]
Q = 5.87 x 10-7 C (negative)
The charge on an object is related to the number of excess (or deficient) electrons in the object. Using the charge of a single electron (-1.6 x 10-19 C), the number of electrons on this object can be determined:
# excess electrons = (-5.87 x 10-7 C) / (-1.6 x 10-19 C/electron)
# excess electrons = 3.67 x 1012 electrons
During the charging process, more than three trillion electrons were transferred from the animal fur to each of the balloons. Wow!
### Configurations of Three or More Charges
In each of the examples above, we have explored the interaction of two charged objects. Newton's laws and Coulomb's law were combined to analyze the situations. But what if there are three or more charges present? Coulomb's law can only consider the interaction between Q1 and Q2. Does the law for electric force have to be rewritten to account for a Q3? No!
Electrical forces result from mutual interactions between two charges. In situations involving three or more charges, the electric force on a single charge is merely the result of the combined effects of each individual charge interaction of that charge with all other charges. If a particular charge encounters two or more interactions, then the net electric force is the vector sum of those individual forces. As an example of this approach, suppose that four charges (A, B, C, and D) are present and that they are spatially arranged to form a square. Charges A and D are both negatively charged and occupy opposite corners of the square and Charges B and C are both positively charged and occupy the remaining two corners as shown. If one is concerned with the net electric force acting upon charge A, then the electric forces between A and each of the other three charges must be calculated. That is, FBA, FCA and FDA must first be determined by the application of Coulomb's law to each of these pairs of charges. The notation FBA is used to denote the force of B on A.
FBA = k • QA • QB / dBA2
FCA = k • QA • QC / dCA2
FDA = k • QA • QD / dDA2
The direction of each of these three forces can be determined by applying the basic rules of charge interaction: oppositely charged objects attract and like-charged objects repel. Applied to this scenario, one would reason that the forces FBA, FCA and FDA are directed as shown in the diagram below. Charge B attracts A and Charge C attracts A since these are pairs of oppositely charged objects. But Charge D repels A since they are a pair of like-charged objects. So the magnitudes of the individual forces are determined through Coulomb's law calculations. The direction of the individual forces are determined by applying the rules of charge interactions. And once the magnitude and direction of the three force vectors are known, the three vectors can be added using rules of vector addition in order to determine the net electric force. This is illustrated in the diagram above.
Use your understanding of charge to answer the following questions. When finished, click the button to view the answers.
1. A positively charged object with a charge of +85 nC is being used to balance the downward force of gravity on a 1.8-gram balloon that has a charge of -63 nC. How high above the balloon must the object be held in order to balance the balloon? (NOTE: 1 nC = 1 x 10-9 C)
2. Balloon A and Balloon B are charged in a like manner by rubbing with animal fur. Each acquires an excess of 25 trillion electrons. If the mass of the balloons is 1 gram, then how far below Balloon B must Balloon A be held in order to levitate Balloon B? Assume the balloons act as point charges.
3. Two 1.2-gram balloons are suspended from light strings attached to the ceiling at the same point. The net charge on the balloons is -540 nC. The balloons are distanced 68.2 cm apart when at equilibrium. Determine the length of the string.
4. ZINGER: Three charges are placed along the X-axis. Charge A is a +18 nC charge placed at the origin. Charge B is a -27 nC charge placed at the 60 cm location. Where along the axis (at what x-coordinate?) must positively charged C be placed in order to be at equilibrium?
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https://www.examrace.com/Sample-Objective-Questions/Quantitative-Analysis-Questions/Mathematics-Basic-Algebra.html
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| 4.25
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# Basic Algebra: Practice Questions Competitive Exams
Doorsteptutor material for competitive exams is prepared by world's top subject experts: get questions, notes, tests, video lectures and more- for all subjects of your exam.
1. The sales price of a car is \$ 12,590, which is 20 % off the original price. What is the original price?
1. \$ 14,310.40
2. \$ 14,990.90
3. \$ 15,290.70
4. \$ 15,737.50
5. \$ 16,935.80
2. Solve the following equation for A: 2A/3 = 8 + 4A
1. -2.4
2. 2.4
3. 1.3
4. -1.3
5. 0
3. If Lee is 6 years older than her sister, Sue, and John is 5 years older than Lee, and the total of their ages is 41. Then how old is Sue?
1. 8
2. 10
3. 14
4. 19
5. 21
4. Alita wants to invest \$ 4,000 at 6 % simple interest rate for 5 years. How much interest will he receive?
1. \$ 240
2. \$ 480
3. \$ 720
4. \$ 960
5. \$ 1,200
5. Jim is able to sell a hand-carved statue for \$ 670 which was a 35 % profit over his cost. How much did the statue originally cost him?
1. \$ 496.30
2. \$ 512.40
3. \$ 555.40
4. \$ 574.90
5. \$ 588.20
6. The city council has decided to add a 0.3 % tax on motel and hotel rooms. If a traveler spends the night in a motel room that costs \$ 55 before taxes, how much will the city receive in taxes from him?
1. 10 cents
2. 11 cents
3. 15 cents
4. 17 cents
5. 21 cents
7. If Lily can type a page in p minutes, what piece of the page can she do in 5 minutes?
1. 5/p
2. p − 5
3. p + 5
4. p/5
5. 1 − p + 5
8. If Larry can paint a house in 4 hours, and John can paint the same house in 6 hour, how long will it take for both of them to paint the house together?
1. 2 hours and 24 minutes
2. 3 hours and 12 minutes
3. 3 hours and 44 minutes
4. 4 hours and 10 minutes
5. 4 hours and 33 minutes
9. Employees of a discount appliance store receive an additional 20 % off of the lowest price on an item. If an employee purchases a dishwasher during a 15 % off sale, how much will he pay if the dishwasher originally cost \$ 450?
1. \$ 280.90
2. \$ 287
3. \$ 292.50
4. \$ 306
5. \$ 333.89
10. A student receives his grade report from a local community college, but the GPA is smudged. He took the following classes: a 2 hour credit art, a 3 hour credit history, a 4 hour credit science course, a 3 hour credit mathematics course, and a 1 hour science lab. He received a B in the art class, an A in the history class, a C in the science class, a B in the mathematics class, and an A in the science lab. What was his GPA if the letter grades are based on a 4 point scale (A = 4, B = 3, C = 2, D = 1, F = 0) ?
1. 2.7
2. 2.8
3. 3.0
4. 3.1
5. 3.2
11. Ben arrived at work at 8: 15 A. M. And left work at 10: 30 P. M. If Ben gets paid by the hour at a rate of \$ 10 and time and for any hours worked over 8 in a day. How much did Ben get paid?
1. \$ 120.25
2. \$ 160.75
3. \$ 173.75
4. \$ 180
5. \$ 182.50
12. Lily worked 22 hours this week and made \$ 132. If she works 15 hours next week at the same pay rate, how much will she make?
1. \$ 57
2. \$ 90
3. \$ 104
4. \$ 112
5. \$ 122
13. If 8x + 5x + 2x + 4x = 114, the 5x + 3 =
1. 12
2. 25
3. 33
4. 47
5. 86
14. You need to purchase a textbook for MBA school. The book cost \$ 80.00, and the sales tax where you are purchasing the book is 8.25 % . You have \$ 100. How much change will you receive back?
1. \$ 5.20
2. \$ 7.35
3. \$ 13.40
4. \$ 19.95
5. \$ 21.25
15. You purchase a car making a down payment of \$ 3,000 and 6 monthly payments of \$ 225. How much have you paid so far for the car?
1. \$ 3225
2. \$ 4350
3. \$ 5375
4. \$ 6550
5. \$ 6398
16. Grace has 16 apples in her pocket. She has 8 red ones, 4 green ones, and 4 blue ones. What is the minimum number of apples she must take out of her pocket to ensure that she has one of each color?
1. 4
2. 8
3. 12
4. 13
5. 16
17. If r = 5 z then 15 z = 3 y, then r =
1. y
2. 2 y
3. 5 y
4. 10 y
5. 15 y
18. If 300 apples cost you x dollars. How many apples can you purchase for 50 cents at the same rate?
1. N/A
2. 150x
3. 6x
4. N/A
5. 1500x
19. Your supervisor instructs you to purchase 240 pens and 6 staplers for the nurse՚s station. Pens are purchase in sets of 6 for \$ 2.35 per pack. Staplers are sold in sets of 2 for 12.95. How much will purchasing these products cost?
1. \$ 132.85
2. \$ 145.75
3. \$ 162.90
4. \$ 225.25
5. \$ 226.75
20. If y = 3, then y3 (y3 − y) =
1. 300
2. 459
3. 648
4. 999
5. 1099
|
https://www.esaral.com/q/construct-a-quadrilateral-abcd-in-which-ab-2-9-cm-66908/
|
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Construct a quadrilateral ABCD in which AB = 2.9 cm,
Question:
Construct a quadrilateral ABCD in which AB = 2.9 cm, BC = 3.2 cm, CD = 2.7 cm, DA = 3.4 cm and ∠A = 70°.
Solution:
Steps of construction:
Step 1: Draw $A B=2.9 \mathrm{~cm}$
Step 2: Make $\angle A=70^{\circ}$
Step 3: With $A$ as the centre, draw an arc of $3.4 \mathrm{~cm}$. Name that point as $D$.
Step 4: With $D$ as the centre, draw an arc of $2.7 \mathrm{~cm}$.
Step 5: With $B$ as the centre, draw an arc of $3.2 \mathrm{~cm}$, cutting the previous arc at $C$.
Step 6: Join $C D$ and $B C$.
Then, $A B C D$ is the required quadrilateral.
|
https://learnmathonline.org/calculus2/Work.html
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Calculus II Home
## Work
Suppose $$f$$ is a continuous function on $$[a,b]$$. Suppose a particle is moving on the $$x$$-axis under a force field that exerts force $$f(x)$$ when the particle is at $$x$$. Then the work done $$W$$ in moving the particle from $$x=a$$ to $$x=b$$ is $W=\int_a^b f(x)\;dx.$
Break $$[a,b]$$ into $$n$$ subintervals $$[x_0,x_1],[x_1,x_2],\ldots,[x_{n-1},x_n]$$ where $$x_i=a+i \Delta x$$ and $$\Delta x=(b-a)/n$$. Choose $$x_i^*$$ from $$[x_{i-1},x_i]$$, $$i=1,2,\ldots,n$$. When the particle is on $$[x_{i-1},x_i]$$, the force acting on the particle is approximately $$f(x_i^*)$$. Then the work done $$W_i$$ in moving the particle from $$x_{i-1}$$ to $$x_i$$ is approximately $$f(x_i^*) \Delta x$$. Therefore the total work done $$W$$ is $W\approx \sum_{i=1}^n f(x_i^*) \Delta x.$ This approximation of $$W$$ gets better as $$n\to \infty$$. Thus $W=\lim_{n\to \infty} \sum_{i=1}^n f(x_i^*) \Delta x =\int_a^b f(x)\;dx.$
Example. A force of $$\sin\left( \frac{\pi x}{4} \right)$$ Newtons is acting on a particle moving on a line when it is $$x$$ meters from the origin. How much work is required in moving the particle from $$x=1$$ to $$x=9$$?
Solution. Here the force is $$f(x)=\sin\left( \frac{\pi x}{4} \right)$$ N when the particle is $$x$$ m from the origin on the $$x$$-axis. Thus required work is $\int_1^9 \sin\left( \frac{\pi x}{4} \right)\;dx =\left.-\frac{4}{\pi}\cos\left( \frac{\pi x}{4} \right) \right\vert_1^9 =-\frac{4}{\pi}\cos\left( \frac{9\pi}{4} \right)+\frac{4}{\pi}\cos\left( \frac{\pi}{4} \right) =-\frac{4}{\pi\sqrt{2}}+\frac{4}{\pi\sqrt{2}}=0.$
Now we discuss the problems of finding the work required to stretch or compress a spring.
Hooke's Law: The force $$f$$ required to maintain a spring stretched or compressed $$x$$ units beyond its natural length is $$f(x)=kx$$ where $$k>0$$ is the spring constant.
Suppose the natural length of a spring is $$x_0$$ m and $$k$$ is the spring constant. Then the work required to stretch it from $$x_1$$ m to $$x_2$$ m is $\int_{x_1-x_0}^{x_2-x_0} f(x)\;dx =\int_{x_1-x_0}^{x_2-x_0} kx\;dx \;J,$ where $$J=Nm$$.
Example. A spring has a natural length of $$20$$ cm and $$15$$ N force is required to keep it stretched to a length of $$25$$ cm. How much work is required to stretch it from $$20$$ cm to $$30$$ cm?
Solution. Since $$15$$ N force is required to stretch $$0.25-0.2=0.05$$ m, by Hooke's Law, $15=f(0.05)=k\cdot 0.05 \implies k=\frac{15}{0.05}=300 \;\text{N/m}.$ Then $$f(x)=300x$$. The work required to stretch the spring from $$0.2$$ m to $$0.3$$ m is $\int_{0.2-0.2}^{0.3-0.2} 300x\;dx=\int_0^{0.1} 300x\;dx=\left. 150x^2 \right\vert_0^{0.1}=1.5 \;J.$
Now we discuss the problems of finding the work required to pump liquid out of a tank.
Example. Consider a spherical tank of radius $$6$$ m. Suppose the tank is filled with a liquid with density $$1200$$ kg/$$m^3$$ to a height of $$4$$ m. Find the work required to empty the tank through a whole at the top of the tank. Assume $$g=9.8$$ m/s$$^2$$.
Solution. First we approximate the required work $$W$$ by approximating the work required to pump a thin layer of liquid to the top of the tank. Draw the origin at the bottom of the sphere and the $$y$$-axis vertically up through the center of the sphere.
Since the tank is filled to a height of $$4$$ m, we partition the liquid into $$n$$ layers by breaking $$[0,4]$$ into $$n$$ subintervals $$[y_0,y_1],[y_1,y_2],\ldots,[y_{n-1},y_n]$$ where $$y_i=i \Delta y$$ and $$\Delta y=4/n$$. The $$i$$th layer of the liquid is approximately a circular cylinder of radius $$\sqrt{6^2-(6-y_i)^2}=\sqrt{12y_i-y_i^2}$$ and height $$\Delta y$$. So the volume $$V_i$$ of the $$i$$th layer is $$V_i\approx \pi (12y_i-y_i^2) \Delta y$$ and its mass $$m_i$$ is $m_i =1200V_i \approx 1200 \pi (12y_i-y_i^2) \Delta y.$ The force $$f_i$$ on the $$i$$th layer to overcome the gravitational force on it is $f_i= m_i g \approx 9.8\cdot 1200 \pi (12y_i-y_i^2) \Delta y=11760 \pi (12y_i-y_i^2) \Delta y.$ Then the work $$W_i$$ required in pumping the $$i$$th layer to the top of the tank is $W_i\approx f_i(12-y_i) \approx 11760 \pi (12y_i-y_i^2)(12-y_i) \Delta y=11760 \pi (144y_i-24y_i^2+y_i^3) \Delta y.$ Therefore the total work $$W$$ required is $W= \sum_{i=1}^n W_i \approx \sum_{i=1}^n 11760 \pi (144y_i-24y_i^2+y_i^3) \Delta y.$ This approximation of $$W$$ gets better as $$n\to \infty$$. Thus \begin{align*} W=\lim_{n\to \infty} \sum_{i=1}^n 11760 \pi (144y_i-24y_i^2+y_i^3) \Delta y &=\int_0^4 11760 \pi (144y-24y^2+y^3)\;dy\\ &=11760 \pi\int_0^4 (144y-24y^2+y^3)\;dy\\ &=11760 \pi \left.\left(72y^2-8y^3+\frac{y^4}{4}\right) \right\vert_0^4\\ &=8279040 \pi \;J. \end{align*}
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https://www.statisticshowto.com/rank-histogram/
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# Rank Histogram / Talagrand Diagram
Statistics Definitions > Rank Histogram
## What is a Rank Histogram? Rank histograms (sometimes called verification rank histograms or Talagrand diagrams) are a way to show how reliable an ensemble forecast is compared to a set of newly observed data. In other words, they show the bias for the model. If an ensemble forecast is accurate, the rank histogram — a graph of observed data — will be flat (represented by the red line in the image to the left). Deviations from a uniform distribution(i.e. histogram blocks that are above or below the red line) mean that the model isn’t completely accurate. These types of diagrams are not commonly used outside of ensemble forecasting.
## How Talagrand Diagrams are Constructed
The Talagrand is basically a histogram. Usually, you would use observed data to make the histogram, using the actual data points to define the bins(class intervals). The difference with this type of histogram is that you use the forecast data to create the bins, and then you use observed (new) data to fill those bins. Meteorologist Peter Houtekamer suggests the following steps for creating a rank histogram:
Step 1: Place forecasts in order. For this example, let’s say the forecast values are: {0.25, 0.7, 1.49, 2.17, 4.2}.
Step 2: Define bins based on the list of forecast values from Step 1. There are 6 bins for this list of data:
1. Values below 0.25 (the lowest value),
2. Values between 0.25 and 0.7,
3. Values between 0.7 and 1.49,
4. Values between 1.49 and 2.17,
5. Values between 2.17 and 4.2,
6. Values above 4.2 (the highest value).
Step 3: Place observations in the appropriate bin. Ideally, a large number of observations should be taken over multiple days from different locations.
Step 4: Make a histogram using the new data. Houtekamer states that this histogram should be a uniform distribution, since each bin should represent an equally likely scenarios.
References:
Houtekamer,P. (n.d.). Ensemble forecasts. Retrieved 1/14/2017 from: http://collaboration.cmc.ec.gc.ca/cmc/cmoi/product_guide/docs/lib/ens_en.pdf
CITE THIS AS:
Stephanie Glen. "Rank Histogram / Talagrand Diagram" From StatisticsHowTo.com: Elementary Statistics for the rest of us! https://www.statisticshowto.com/rank-histogram/
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https://number.rocks/as-simplified/604/201
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# Simplify 604/201 to lowest terms
/
#### Solution for what is 604/201 in simplest fraction
604/201 =
Now we have: what is 604/201 in simplest fraction = 604/201
Question: How to reduce 604/201 to its lowest terms?
Step by step simplifying fractions:
Step 1: Find GCD(604,201) = 1.
Step 2: Can't simplify any further
Therefore, 604/201 is simplified fraction for 604/201
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https://www.assignmentexpert.com/blog/the-midpoint-theorem-and-formula/
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# The Midpoint Theorem and Formula
Once a while in mathematics, we really need to find the midpoint between two other points, that is, the point that is exactly in the middle of the two other points. A good example is if you have to find the point at which a line bisects or divides a given line segment into two equal parts. The midpoint formula is quite simple and you should endeavour to know how to derive it for future use.
The midpoint formula can be conceived in terms of finding the middle number that exists between two given numbers such as 10 and 15. By adding the two numbers and dividing by 2, we obtain the exact middle number as follows: Exactly the same way, the midpoint formula works. Let us consider the following question:
Find the midpoint between the points (-2, 4) and (6,-12).
Solution:
If we apply the midpoint formula, we have the following: Hence, the coordinates of the midpoint = (2,-4)
The Midpoint Formula:
The midpoint formula is, therefore, given as: This formula can be applied for solving different questions such as the followings:
1. If a line segment has its endpoints given as (-1.8, 3.9) and (8.2,-1.1), determine whether the equation y=2x-4.9 is a bisector of this line segment.
Solution:
The exact midpoint of the line segment can only be determined by applying the midpoint formula and not by graphical method. Hence, by applying the midpoint formula, we have: Now, check to see if this point is on the line y=2x-4.9.
y=2x-4.9
y=2(3.2)-4.9=6.4-4.9=1.5. But, y, ought to be equal to 1.4 as the y-coordinate of the midpoint indicates. Therefore, y=2x-4.9 is not a bisector of the line segment.
2. Find the bisector of the line segment given above.
To find the perpendicular bisector of this line segment, you will need to solve the problem in a multi-parts manner. Note that this is one of the typical questions you will surely come across as you proceed in your learning of maths particularly the straight lines. The procedure for solving this problem is simple and straight forward and it is given as follows:
First of all, determine the midpoint of the line segment by applying the midpoint formula given above. The slope of the line segment is needed here; therefore, we obtain the slope as follows:
Slope, this is a negative reciprocal which must be flipped.
So, the slope, m=2 (since negative reciprocal means that it should be changed).
The equation of the bisector of the line segment is therefore obtained as follows:
y – y1 = slope(x – x1)
y – 1.4 = 2(x – 3.2)
y = 2x – 5.
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Filed under Math.
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http://www.themathpage.com/Arith/percent-with-a-calculator.htm
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S k i l l
i n
A R I T H M E T I C
Lesson 14
# PERCENTWITH A CALCULATOR
For the very first Lesson on percent, see Lesson 4.
In this Lesson, we will answer the following:
1. Every statement of percent can be expressed verbally in what way? "One number is some percent of another number." ____ is ____ % of ____.
For example,
8 is 50% of 16.
Every statement of percent therefore involves three numbers. 8 is called the Amount. 50% is the Percent. 16 is called the Base. The Base always follows "of."
Example. "\$78 is 12% of how much?" Which number is unknown -- the Amount, the Percent, or the Base?
Answer. We do not know the Base, the number that follows "of."
2. What is a percent problem? Given two of those numbers, to find the third.
We have seen (Lesson 4) that to find
8% of \$600,
for example, we multiply. We can now recognize that \$600 is the Base -- it follows "of," and 8% is the Percent. We are looking for the Amount. We can state the rule as follows:
1. Amount = Base × Percent
This is Rule 1. To find the Amount, multiply. There is also a rule for finding the Base and finding the Percent.
2. Base = Amount ÷ Percent 3. Percent = Amount ÷ Base
Notice that we multiply only to find the Amount. In the other two cases, we divide.
(This follows from the relationship between multiplication and division, which we saw in Lesson 11.)
3. How do we use a calculator to solve a percent problem? Apply one of the three rules.
Example 1. How much is 37.5% of \$48.72?
Solution. We have the Percent, and we have the Base -- it follows "of." We are missing the Amount. Apply Rule 1: Multiply
Base × Percent.
Press
4 8 . 7 2 × 3 7 . 5 %
Press the percent key % last. And when you press the percent key, do not press = . (At any rate, that is true for simple calculators.)
18.27
If your calculator does not have a percent key, then express the percent as a decimal (Lesson 4), and press = . Press
4 8 . 7 2 × . 3 7 5 =
Example 2. \$250 is 62.5% of how much?
Solution. The Base -- the number that follows "of" is unknown. Apply Rule 2: Divide:
Amount ÷ Percent.
Press
2 5 0 ÷ 6 2 . 5 %
Do not press = . The answer is displayed:
400
Now, for calculators that instead of a division key ÷ have the division slash / , the percent key % will not be effective in finding the Base or the Percent. To find the Base, do not press the percent key. Press equals = . Then multiply by 100.
Again, \$250 is 62.5% of how much? Press
2 5 0 / 6 2 . 5 =
See:
4
On multiplying by 100, the answer is 400.
Or, change 62.5% to the decimal .625 (Lesson 4), and press
2 5 0 / . 6 2 5 =
See:
400
Example 3. \$51.03 is what percent of \$405?
Solution. The Percent is unknown. Apply Rule 3:
Amount ÷ Base.
Press
5 1 . 0 3 ÷ 4 0 5 %
See
12.6
\$51.03 is 12.6% of \$405.
For calculators without a % key, press = .
5 1 . 0 3 ÷ 4 0 5 =
Similarly, with only the division slash, press = .
5 1 . 0 3 / 4 0 5 =
In either case, see
0.126
Then multiply by 100 by moving the decimal point two places right.
Again, we multiply in only one of the three problems; namely, to find the Amount.
Now in Lesson 12, we saw how to round off a decimal. The following examples will require that.
Example 4. How much is 9.7% of \$84.60?
Solution. The Amount is missing. Multiply
Base × Percent.
Press
8 4 . 6 × 9 . 7 %
It is not necessary to press the 0 of 84.60.
On the screen, see this:
8.2062
Since this is money, we must round off to two decimal digits. In the third decimal place is a 6; therefore add 1 to the second place:
\$8.21
Example 5. \$84.60 is 9.7% of how much? (Compare this with Example 4.)
Solution. Here, the Base is missing. Divide:
8 4 . 6 ÷ 9 . 7 %
On the screen, see
872.165
Again, this is money, so we must approximate it to two decimal digits:
\$872.16
Example 6. \$48.60 is what percent of \$96.40?
Solution. The Percent is missing. (Compare Example 3.) Divide:
Percent = Amount ÷ Base.
4 8 . 6 ÷ 9 6 . 4 %
Again, it is not necessary to press the 0's on the end of decimals.
On the screen, see this decimal:
50.4149
Let us round this off to one decimal digit. Since the digit in the second place is 1 (less than 5), this is approximately
50.4%.
Example 7. Michelle paid \$82.68 for a pair of shoes -- but that included a tax of 6%. What was the actual price of the shoes before the tax?
Solution. The actual price, the base, was 100%. When the 6% tax was added, the price became 106% of that base. So the question is:
\$82.68 is 106% of how much?
To find the Base, press
8 2 . 6 8 ÷ 1 0 6 %
See
78
The actual price was \$78.
Equivalently, the calculation is
8 2 . 6 8 ÷ 1 . 0 6
*
For problems of percent increase or decrease, see Lesson 31.
Summary Amount = Base × Percent Base = Amount ÷ Percent Percent = Amount ÷ Base
At this point, please "turn" the page and do some Problems.
or
Continue on to the next Lesson.
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|
https://answers.everydaycalculation.com/simplify-fraction/640-84
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Solutions by everydaycalculation.com
## Reduce 640/84 to lowest terms
The simplest form of 640/84 is 160/21.
#### Steps to simplifying fractions
1. Find the GCD (or HCF) of numerator and denominator
GCD of 640 and 84 is 4
2. Divide both the numerator and denominator by the GCD
640 ÷ 4/84 ÷ 4
3. Reduced fraction: 160/21
Therefore, 640/84 simplified to lowest terms is 160/21.
MathStep (Works offline) Download our mobile app and learn to work with fractions in your own time:
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https://mathleaks.com/study/divide_decimals
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Dividing decimals is similar to dividing whole numbers. The difference is that a decimal point is in play. It is crucial to understand where to place it within the quotient. The methods for dividing decimals will be developed in this lesson.
### Catch-Up and Review
Here are a few recommended readings before getting started with this lesson.
## LaShay's Piggy Bank
LaShay is fascinated by numbers and how she can buy her favorite things. She is currently saving money to buy a used drone. Help LaShay answer the following questions to support her math skills. Doing so will help her improve her saving plan.
a How many one-hundred dollar bills does she need to make
b How many ten dollar bills does she need to make
c How many dimes does she need to make
d How many pennies does she need to make
## Compatible Numbers
Compatible numbers are numbers that help make doing mental math more manageable. They are typically used to estimate quotients. Consider finding the estimation of the following quotient.
It might seem difficult to divide these numbers using mental math. However, both numbers can be rounded to compatible numbers — to and to An estimate for is then found by mentally dividing by which is equal to That means the given quotient is estimated to be about The compatible numbers previously chosen are not the only options. Any numbers easily divisible and close to the original numbers are worth exploring. For example, rather than rounding to try Notice that is a multiple of This makes evenly divisible by The estimate for the quotient is then Compatible numbers are not used to get precise answers. They are used to get quick and rough approximations. The pairs of numbers in the multiplication table are also compatible numbers. This is why it is a good habit to choose a pair from the multiplication table when calculating a quotient. The following table shows some compatible numbers for different quotients.
Quotients Exact Value Use Compatible Numbers Estimate
Keep in mind that the estimates hold value as they are close to the exact values of the quotients.
## Estimating the Number of Dimes in Piggy Bank
LaShay decides to use one of the all-time classic methods of saving money — a piggy bank! She then weighs it. All of the coins in the bank add up to grams. There is something interesting about LaShay's piggy bank. LaShay is nit-picky about what coins she puts in the bank — only dimes! The weight of one dime is grams. Use compatible numbers to estimate the number of dimes in the piggy bank.
### Hint
Divide the total weight of all the money in the piggy bank by the weight of a dime. Then, round the weight of a dime to a whole number.
### Solution
The total weight of all the money in the piggy bank should be divided by the weight of one dime to find the exact number of the dimes.
This quotient can be estimated using compatible numbers. First, round the divisor to a whole number. The digit in the tenths place is which is less than This is why the divisor is rounded to
Recall that compatible numbers are numbers that make doing mental math more straightforward. One way to know that the numbers will be compatible is to round the dividend to a number that is a multiple of In this case, it can be because it is close to and it is a multiple of
The quotient of and is LaShay's now knows that her piggy bank contains about
## Dividing Decimals
The process of dividing decimals is similar to the process of dividing whole numbers using long division. What differentiates these processes is how the decimal point is placed when dividing decimals. Consider dividing the following decimal numbers.
The following steps can be followed to divide the decimals.
1
Multiply the Divisor and the Dividend by the Same Power of
expand_more
When dividing decimals, the divisor is changed to a whole number. This requires multiplying the divisor and the dividend by the same power of First, write the division using long division notation. The dividend is and the divisor is In this case, the divisor must be multiplied by so that the decimal point moves to the right and the divisor becomes a whole number. The dividend is also multiplied by the same power of 2
Divide in the Same Way as Dividing Whole Numbers
expand_more
Start by dividing the whole number part of the dividend by the divisor. 3
Place the Decimal Point and Continue the Division
expand_more
Now place the decimal point in the quotient above the decimal point in the dividend. Then bring down the digit in the tenths place and continue dividing as usual. The quotient of and is
## How Many Hours Should LaShay Work?
LaShay needs to buy the drone. She has already saved up to this point. She then decided to mow her neighbor's lawn to make the rest of money she needs. Her neighbor agrees to pay her per hour.
a Estimate the number of hours LaShay needs to work to earn enough to buy the drone.
b Find the exact number of hours. Round the answer to one decimal place.
### Hint
a Subtract the amount that LaShay saved from the price of the drone. Use compatible numbers to estimate the quotient of the difference and
b Start by converting the divisor into a whole number. Use the standard division algorithm to divide the decimals.
### Solution
a Start by finding the amount LaShay needs. It is the difference between the price of the drone, and the amount LaShay has saved,
The number of hours LaShay needs to work to earn this difference can be found by dividing it by the hourly wage,
Compatible numbers will be used to estimate the quotient. First, round the divisor to a whole number. For the digit in the tenths place is which is less than This is why is rounded to
Find a number that is compatible with This means that the number should be a multiple of and close to It can be either or Since is closer, the dividend can be replaced with
The quotient of and is Therefore, LaShay needs to work about hours to earn Note that the goal here is to find an answer quickly, which leads to finding imprecise answers. The exact value will probably be different from this estimation.
b Recall the quotient from Part A that represents the number of hours LaShay needs to work.
This is a division of two decimal numbers. The first step in dividing decimals is to convert the divisor into a whole number. Since the divisor has decimal places, it will be multiplied by Make sure the divided is multiplied by the same number as the divisor. Now it is a division of two whole numbers. Use long division to calculate the quotient. The quotient is with a remainder of This means that LaShay has to work about hours to save enough to buy the drone.
## How Far Does the Drone Go?
a LaShay managed to save enough money to buy the drone. External credits: macrovector
The drone can go meters in minutes. How far can the drone go in one minute?
b LaShay notices a bee nest on a tree branch while flying the drone in the backyard of the house.
Her grandfather, who used to be a beekeeper, tells LaShay the following.
So you want to approximate the number of bees in that hive huh? Well, count the number of bees that leave the hive in one minute. Then multiply it by and divide by
LaShay counts bees that leave the nest in one minute. Find the number of bees.
### Hint
a Decide if the divisor is a whole number. Then, use long division to divide the decimals.
b What can be done when the divisor is not a whole number? Multiply it by a power of
### Solution
a Divide by to find how far the drone can travel.
The steps for dividing a decimal by a whole number is the same as the steps for dividing whole numbers. However, keep in mind that the decimal point in the quotient is placed above the decimal point of the dividend. The result of the division is Therefore, the drone can go meters in one minute.
b Now, it is time to find the number of bees in the nest. The number of bees that leave the hive in one minute is multiplied by and then divided by
LaShay counts bees that leave the nest in one minute. Multiply this number by
Now, divide this number by using long division.
In this case, the divisor is a decimal. The first step in dividing decimals is to convert the divisor into a whole number. Since the divisor has decimal places, it will be multiplied by Make sure the divided is multiplied by the same number as the divisor. Now that this is a division of two whole numbers, they can be divided as usual. This means that there are about bees in the nest.
## Finding the Quotient of Two Decimal Numbers
Divide the decimal numbers. Make sure that the decimal point is placed correctly. Round the answer to two decimal places, if necessary. ## Money in the Piggy Bank
When dividing decimals, the divisor is converted into a whole number by multiplying it by a power of However, the dividend must also be multiplied by the same power of to keep the value of the quotient the same. Now take a look at LaShay's piggy bank problem. a How many one-hundred dollar bills does she need to make
b How many ten dollar bills does she need to make
c How many dimes does she need to make
d How many pennies does she need to make
### Hint
a Use long division to divide by
b Use long division to divide by
c A dime is worth Divide the decimal by Rewrite it so that the divisor is
d A penny is worth Divide the decimal by
### Solution
a The number of one-hundred dollar bills is found by dividing by
Long division can be used to find this quotient. The quotient is LaShay needs one-hundred dollar bills to make
b Use the same reasoning as in Part A. Divide by to find the number of ten dollar bills.
This is also equal to LaShay can make with ten dollar bills.
c Recall that a dime is worth To find the number of dimes needed to make divide by
In this case, the divisor is a decimal number. The quotient must be rewritten so that the divisor is a whole number. Multiplying the divisor by will give the desired divisor. Keep in mind that the dividend must also be multiplied by the same number.
When a number is divided by the result will be the number itself. Check using the division algorithm to see if it really is. The result is indeed Therefore, dimes are need to make
d A penny is worth and LaShay wants to make with pennies. Then, must be divided by to find the number of pennies she needs.
Like in the previous part, the divisor is a decimal number. Multiply the divisor and the dividend by to make the divisor a whole number.
After multiplication, the quotient becomes the same quotient as in the previous part. Since divided by is the quotient, is also
This means that LaShay can make using pennies. On a final note, examine the table created using the quotients mentioned in these examples.
Dividend Divisor Quotient
The numbers in the first columns get smaller downwards by a factor of The same is true for the numbers in the second column. However, the quotient always stays the same. Therefore, when the dividend and divisor both increase by the same factor of the quotient remains the same.
{{ subexercise.title }}
|
https://www.triangle-calculator.com/?a=17.09&b=19.1&c=8.54
|
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| 3.75
| 4
| 603
| 1,607
|
17.09 19.1 8.54 triangle
Acute scalene triangle.
Sides: a = 17.09 b = 19.1 c = 8.54
Area: T = 72.97442845896
Perimeter: p = 44.73
Semiperimeter: s = 22.365
Angle ∠ A = α = 63.47881592307° = 63°28'41″ = 1.10879028817 rad
Angle ∠ B = β = 89.96327642535° = 89°57'46″ = 1.57701464404 rad
Angle ∠ C = γ = 26.55990765158° = 26°33'33″ = 0.46435433315 rad
Height: ha = 8.54399981966
Height: hb = 7.64112863445
Height: hc = 17.0989996391
Median: ma = 12.07769936242
Median: mb = 9.55549646781
Median: mc = 17.61326701553
Inradius: r = 3.26328788102
Circumradius: R = 9.55500020167
Vertex coordinates: A[8.54; 0] B[0; 0] C[0.01111065574; 17.0989996391]
Centroid: CG[2.85503688525; 5.69766654637]
Coordinates of the circumscribed circle: U[4.27; 8.54222267893]
Coordinates of the inscribed circle: I[3.265; 3.26328788102]
Exterior(or external, outer) angles of the triangle:
∠ A' = α' = 116.5221840769° = 116°31'19″ = 1.10879028817 rad
∠ B' = β' = 90.03772357465° = 90°2'14″ = 1.57701464404 rad
∠ C' = γ' = 153.4410923484° = 153°26'27″ = 0.46435433315 rad
How did we calculate this triangle?
Now we know the lengths of all three sides of the triangle and the triangle is uniquely determined. Next we calculate another its characteristics - same procedure as calculation of the triangle from the known three sides SSS. 1. The triangle circumference is the sum of the lengths of its three sides 2. Semiperimeter of the triangle 3. The triangle area using Heron's formula 4. Calculate the heights of the triangle from its area. 5. Calculation of the inner angles of the triangle using a Law of Cosines
|
https://en.wikibooks.org/wiki/MATLAB_Programming/Vectoring_Mathematics
|
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| 4.25
| 4
| 664
| 2,340
|
# MATLAB Programming/Vectoring Mathematics
MATLAB is a vector programming language. The most efficient use of MATLAB will involve taking advantage of the built-in capabilities for manipulating data instead of using constructs such as loops.
## Basic Math
Most arithmetic operators will work as expected on vectors:
```>> a = [2 43 943 78];
>> 5 * a
ans =
10 215 4715 390
>> a / 2
ans =
1.0000 21.5000 471.5000 39.0000
>> 0.2 + a
ans =
2.2000 43.2000 943.2000 78.2000
```
Likewise, all of these operations can be done with matrices for the expected results.
Most MATLAB functions, such as `sin` or `log`, will return a vector or matrix of the same dimensions as its input. So to compute the sine of all the integers between 0 and 10, it suffices to run
```>> sin(0:10)
```
and the returned vector will contains ten values.
## Per Element Operations
Operators such as the carrot (^) or multiplication between vectors may not work as expected because MATLAB sees vectors the same as any other matrix, and so performs matrix power and matrix multiplication. All operators can be prefixed by a `.` to make it explicit that an operation should be performed on each element of the matrix. For example, to compute the differences of the sine and cosine function squared for all integers between 1 and 4, one can use:
```>> (sin(1:4) - cos(1:4)).^2
ans =
0.0907 1.7568 1.2794 0.0106
```
as opposed to
```>> (sin(1:4) - cos(1:4))^2
??? Error using ==> mpower
Matrix must be square.
```
which results from MATLAB's attempt to square a 1x4 vector using matrix multiplication.
Using `.*` or `./` allows one to divide each element of a matrix or vector by the elements of another matrix or vector. To do this, both vectors must be of the same size.
## Converting Loops to Vector-based mathematics
Since MATLAB is a vector language, an artificial algorithm such as
```x = [];
v = [5,2,4,6];
for i=1:4
x(i) = v(i) * ((i+32)/2 - sin(pi/2*i));
if(x(i) < 0)
x(i) = x(i) + 3;
end
end
```
can be done far more efficiently in MATLAB by working with vectors instead of loops:
```i = 1:4;
v = [5,2,4,6];
x = v .* ((i+32)/2 - sin(pi/2*i));
x(x<0) = x(x<0) + 3;
```
Internally, MATLAB is of course looping through the vectors, but it is at a lower level then possible in the MATLAB programming language.
|
http://ibpsexamguide.org/verbal-reasoning/daily-reasoning-quiz/reasoning-quiz-12.html
|
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| 3.5
| 4
| 1,073
| 3,529
|
# Reasoning Quiz – 12
(Q. 1–5)Read the following information and answer the questions given below:
There are groups of three couples comprising J, K, L, M, N, and P. Each female member of this group likes three different cars, viz Alto, Swift and Maruti 800, but not necessarily in the same order. All of them are seared at three different tables in a coffee shop for taking coffee with their partners. The tables are numbered, I, II and III, but not necessarily in the same order.Those who like the same car do not sit together at the same table and no two females like the same car. K and N like the same car but not Swift. M likes Maruti 800 and she does not sit at table no. II.N and J are sitting at table no. I and III respectively and both like the same car. L sits at table no. I and he like Swift.
1.Which of the following groups sits at table no. II?
a)J, K
b)K, P
c)P, L
d)Can’t be determined
e)None of these
2.Which of the following groups represents the group of males?
a) J, K, L
b) M, K, L
c) N, P, J
d) J, M, L
e) None of these
3.Who among the following like Swift?
a) N, J
b) M
c) L, M
d) P, L
e) None of these
4.P likes which of the following cars?
a) Alto
b) Swift
c) Maruti 800
d) Can’t be determined
e) None of these
5.Which of the following statements is/are not true?
a) J and M are sitting at table no. III
b) L and P both like Alto
c) L and P are sitting at table no. II
d) All are true
e) Both b and c
(Q. 6–10) Study the following information carefully and answer the questions given below.
489 – 541 – 654 – 953 – 983
6. If in each number, the first and the last digits are interchanged, which of the following will be the second highest number ?
a) 489
b) 541
c) 654
d) 953
e) 783
7. If in each number, all the three digits are arranged in ascending order, which of the following will be the lowest number ?
a) 489
b) 541
c) 654
d) 953
e) 783
8. Which of the following numbers will be obtained if the first digit of lowest number is subtracted with the second digit of highest number after adding one to each of the given numbers ?
a) 1
b) 2
c) 3
d) 4
e) 5
9. If five is subtracted from each of the numbers, which of the following numbers will be the difference between the second digit of second highest number and the second digit of the highest number ?
a) Zero
b) 3
d) 4
e) 2
10. If in each number the first and the second digits are interchanged, which will be the third highest number ?
a) 489
b) 541
c) 654
d) 953
e) 783
Q(11 to 15) Read the following information carefully and answer the questions given below:
Six friends, Rohan, Shyam, Ram,Saurabh, Mohan and Gaurav, are living on different floors in the same building. This building has six floors numbered 1 to 6. (The ground floor is number 1 and the floor above it number 2 and so on, and top most floor is numbered 6) . Ram and Shyam live on odd number’s floors. Two people live between Mohan and Gaurav. Gaurav does not live on even number floor and Gaurav lives on the floor, above Mohan. Sohan is exactly between the Gaurav and Ram. Rohan does not live on second floor.
Q-11. How many persons live between Rohan and Mohan?
a. Three
b. Two
c. Four
d. Five
e. None of these
Q-12. On which floor does Gaurav live?
a. Third
b. Second
c. Sixth
d. Fifth
e. First
Q-13. How many persons live between Shyam and Saurabh?
a. Two
b. Three
c. Four
d. Five
e. None of these
Q-14. Who is immediately above and below Ram’s floor?
a. Saurabh and Mohan
b. Shyam and Gaurav
c. Gaurav and Rohan
d. Saurabh and Rohan
e. None of these
|
https://www.proofwiki.org/wiki/Set_of_Successive_Numbers_contains_Unique_Multiple
|
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| 4.71875
| 5
| 523
| 1,559
|
# Set of Successive Numbers contains Unique Multiple
## Theorem
Let $m \in \Z_{\ge 1}$.
Then $\set {m, m + 1, \ldots, m + n - 1}$ contains a unique integer that is a multiple of $n$.
That is, in any set containing $n$ successive integers, $n$ divides exactly one of those integers.
## Proof
Let $S_m = \set {m, m + 1, \ldots, m + n - 1}$ be a set containing $n$ successive integers.
The proof proceeds by induction on $m$, the smallest number in $S_m$.
### Basis for the Induction
The case $m = 1$ is verified as follows:
$S_1 = \set {1, 2, \ldots, n}$
This contains a multiple of $n$, namely, $n$ itself.
If $S_1$ is a singleton then this element is trivially unique.
If there are any other integers in $S_1$, they are less than $n$ and more than $0$, and so cannot be a multiple of $n$.
This is the basis for the induction.
### Induction Hypothesis
Fix $m \in \N$ with $m \ge 1$.
Assume that:
$S_m = \set {m, m + 1, \ldots, m + n - 1}$
contains a unique multiple of $n$.
This is our induction hypothesis.
### Induction Step
This is our induction step:
Consider the set:
$S_{m + 1} = \set {m + 1, m + 2, \ldots, m + n - 1, m + n}$
As $n \equiv 0 \pmod n$, $m + n$ is a multiple of $n$ if and only if $m$ is a multiple of $n$.
That means that we can replace $m + n$ with $m$ and instead analyze:
${S_{m + 1} }' = \set {m + 1, m + 2, \ldots, m + n - 1, m}$
But ${S_{m + 1} }' = S_m$, which contains a unique multiple of $n$ by the induction hypothesis.
The result follows by the Principle of Mathematical Induction.
$\blacksquare$
|
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