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Calculus
# Basic Properties of Integrals
If $\int_0^1 f(x) \, dx = 4$, what is the value of
$\int_0^1 2 f(x) \, dx ?$
If $\int_0^1 f(x) \, dx = 2$ and $\int_0^1 g(x) \, dx = 6$, what is the value of
$\int_0^1 \left( 6 f(x) + 2 g(x) \right) \, dx ?$
If $\int_0 ^ {4} f(x) \, dx = 3$, what is the value of
$\int_0^{4} \left( 3 - f(x) \right) \, dx ?$
If $\int_0^{10} f(x) \, dx = 27$ and $\int_0^5 f(x) \, dx = 7$, then what is the value of
$\int_5^{10} f(x) \, dx ?$
Suppose $f(x)$ is an odd function and $g(x)$ is an even function such that
$\int_0 ^ {5} f(x) \, dx = 7 \hspace{.6cm} \int_{5}^{15} f(x) \, dx = 8 \\ \int_{-5}^{0} g(x) \, dx = 1 \hspace{.6cm} \int_{5}^{15} g(x) \, dx = 2$
What is the value of $\int_{-5}^{15} \left( f(x) + g(x) \right) \, dx ?$
× | HuggingFaceTB/finemath | |
CSE_548-AnalysisOfAlgorithm-fall2010
# CSE_548-AnalysisOfAlgorithm-fall2010 - Michael A. Bender...
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Michael A. Bender CSE 548 – Analysis of Algorithms, Fall 2010 Assignment #1 Tuesday, October 5, 2010 Problem 1 For the following C codes, find the time-complexity by first finding the recurrence relation and then simplifying the recurrence. (1) int blah(int n) { int sum = 0; int i, j; if (n == 0) return 1; for(i=0; i < =n-1; i++) for(j=0; j < =log(n); j++) { . . . loop-body . . . } sum = blah(n/2); sum += blah(n/2); return sum; } Ans : Let us assume that the time complexity of blah(n) is T(n). The upper for loop iterates n times and inner one for log(n) times. therefore the cost of the for loop will be nlog ( n ). Similarly, the costs of two blah(n/2) function call is 2T(n/2). Therefore, The recurrence realtion is [ T ( n ) = 2 T ( n/ 2) + nlog ( n )] Using recurrence tree menthod to solve the above recurrence realation, The cost for level 1 willl be : [ nlog ( n )] The cost for level two will be : 2 * n/ 2 * log ( n/ 2) = n [ log ( n ) - 1] For level 3 : n [ log ( n ) - 2] . . . for log(n) times . . . We get T ( n ) = n log n + n ((log n ) - 1) + ..... + n (log n - log n ) Solving we get: T ( n ) = n log n
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(2) int blah(int n) { int sum = 0; int i, j; if (n == 0) return 7; for(i=0; i < =n-1; i++) sum += blah(i); return sum; } Ans : Lets take cost c 1 n for thr loop and rest of the abovce statements take constant time therefore we ignore them. Also, For blah(n) cost is n. so, for the statement sum + = blah ( i ) the cost will be n - 1 X i =0 i Summming them up, we get : = c 1 n + n - 1 X i =0 i = c 1 n + n ( n - 1) 2 = O ( n 2 ) Problem 2 Prove or find a counterexample for the following. Assume that f ( n ) and g ( n ) are grater than 1 and monotonically increasing functions. (1) f ( n ) = o ( g ( n )) implies log( f ( n )) = o (log( g ( n ))) Ans : We know that f ( n ) = o ( g ( n )) implies f ( n ) < cg ( n ) for all n n 0 and any c > 0 Lets take c = 1, we get f ( n ) < g ( n ) since f(n) and g(n) are monotonically increasing and > 1, we can take log on both sides without changes in inequality sign, log( f ( n )) < log( g ( n )) —(I) Also, we have log( f ( n )) = o (log( g ( n ))) i.e. log( f ( n )) < c.log ( g ( n )) ——(II) Thus for c = 1, the above equation and (I) are same,So it holds true. (2) f ( n ) = O ( g ( n )) implies log( f ( n )) = O (log( g ( n ))) Ans :
We know that f ( n ) = O ( g ( n )) implies f ( n ) cg ( n ) for all n n 0 and any c > 0 Lets take c = 1, we get f ( n ) g ( n ) since f(n) and g(n) are monotonically increasing and > 1, we can take log on both sides without changes in inequality sign, log( f ( n )) log( g ( n )) —(I) Also, we have log( f ( n )) = O (log( g ( n ))) i.e. log( f ( n )) c.log ( g ( n )) ——(II) Thus for c = 1, the above equation and (I) are same,So it holds true. (3)
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## This note was uploaded on 10/18/2010 for the course CSE 548 taught by Professor Ko,k during the Spring '08 term at SUNY Stony Brook.
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CSE_548-AnalysisOfAlgorithm-fall2010 - Michael A. Bender...
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Ask a homework question - tutors are online | HuggingFaceTB/finemath | |
# Find points on the x-axis, each of which is at a distance of 10 units from the point A
Question:
Find points on the x-axis, each of which is at a distance of 10 units from the point A(11, −8).
Solution:
Let P (x, 0) be the point on the x-axis. Then as per the question, we have
$A P=10$
$\Rightarrow \sqrt{(x-11)^{2}+(0+8)^{2}}=10$
$\Rightarrow(x-11)^{2}+8^{2}=100$ (Squaring both sides)
$\Rightarrow(x-11)^{2}=100-64=36$
$\Rightarrow x-11=\pm 6$
$\Rightarrow x=11 \pm 6$
$\Rightarrow x=11-6,11+6$
$\Rightarrow x=5,17$
Hence, the points on the x-axis are (5, 0) and (17, 0). | HuggingFaceTB/finemath | |
# Harmonic motion - Find the Mass held between two Springs
• JoeyBob
If ##\omega=\sqrt {\frac k m}##the value of that term ##k## must be equal to the ##k## of a system of two springs working in series.Therefore, ##k=k_1+k_2=m\omega^2##f
#### JoeyBob
Homework Statement
See attached
Relevant Equations
E=1/2kA^2=1/2m(wA)^2
So first I find the energy using the eqn (1/2)kA^2. Since there are two springs with the same k I multiply it by two to get kA^2. Energy I get is 2.0475,
Now I use E=(1/2)m(wA)^2 to find mass. Again since there are two springs I use E=m(wA)^2.
m=E/(wA)^2. w=(2(pi))/T btw.
I get the answer of 3.375 when the correct answer is 6.750. Why is it that multiplying my answer by 2 gives me the rights answer? what am i missing here
#### Attachments
• question.PNG
10 KB · Views: 71
since there are two springs I use E=m(wA)^2.
Given the mass and velocity, how does the number of springs or their constants affect the KE?
Btw, you don't need to know the amplitude. The calculation can be simplified.
Lnewqban
I believe that your error is when you replace ##k_{total~of~the~system}## with ##m\omega^2## in the equation.
Given the mass and velocity, how does the number of springs or their constants affect the KE?
Btw, you don't need to know the amplitude. The calculation can be simplified.
But the question gives amplitude? Kinetic E of a spring would be (1/2)kx^2.
I believe that your error is when you replace ##k_{total~of~the~system}## with ##m\omega^2## in the equation.
Isnt E=kA^2=m(wA)^2 for two springs? For one spring it would be E=(1/2)kA^2=(1/2)m(wA)^2 right?
Isnt E=kA^2=m(wA)^2 for two springs? For one spring it would be E=(1/2)kA^2=(1/2)m(wA)^2 right?
##E=\frac 12kA^2=\frac 12m(\omega A)^2## is for one spring.
##EPE_{max}=kA^2## is for two springs of constant k because the effective spring constant is 2k.
##KE_{max}=\frac 12mv_{max}^2=\frac 12m(\omega A)^2## because there is still only a mass m, not 2m.
Last edited:
JoeyBob and Lnewqban
Isnt E=kA^2=m(wA)^2 for two springs? For one spring it would be E=(1/2)kA^2=(1/2)m(wA)^2 right?
If ##\omega=\sqrt {\frac k m}##
the value of that ##k## must be equal to the ##k## of a system of two springs working in series.
Therefore, ##k=k_1+k_2=m\omega^2##
The value of that term ##m\omega^2## is the one to be used in the equation of elastic potential energy. | HuggingFaceTB/finemath | |
Expansions ICSE Class-9th Concise Selina Mathematics Solutions (Including Substitution) Chapter-4. We provide step by step Solutions of Exercise / lesson-4 Expansions (Including Substitution) for ICSE Class-9 Concise Selina Mathematics by RK Bansal.
Our Solutions contain all type Questions with Exe-4 A, Exe-4 B, Exe-4 C, Exe-4 D and Exe-4 E to develop skill and confidence. Visit official Website CISCE for detail information about ICSE Board Class-9 Mathematics .
## Expansions ICSE Class-9th Concise Selina Mathematics Solutions (Including Substitution) Chapter-4
–: Select Topics :–
Exe-4 A,
Exe-4 B,
Exe-4 C,
Exe-4 D,
### Exercise – 4(A)Expansions (Including Substitution) ICSE Class-9th Concise Selina Mathematics
#### Question 1.1
Find the square of : 2a + b
We Know that
( a + b )2 = a2 + b2 + 2ab
(2a + b)2 = 4a2 + b2 + 2 x 2a x b
= 4a2 + b2 + 4ab
#### Question 1.2
Find the square of : 3a + 7b
We know that
( a + b )2 = a2 + b2 + 2ab
( 3a + 7b )2 = 9a2 + 49b2 + 2 x 3a x 7b
= 9a2 + 49b2 + 42ab
#### Question 1.3
Find the square of : 3a – 4b
We know that
( a – b )2 = a2 + b2 – 2ab
( 3a – 4b )2 = 9a2 + 16b2 – 2 x 3a x 4b
= 9a2 + 16b2 – 24ab
#### Question 1.4
Find the square of : ……….
We know that,
( a – b )2 = a2 + b2 – 2a
#### Question 2.1
Use identities to evaluate : (101)2
(101)2
(101)2 = (100 + 1)2
We know that,
(a + b)2 = a2 + b2 + 2ab
∴ (100 + 1)2 = 1002 + 12 + 2 x 100 x 1
= 10,000 + 1 + 200
= 10,201
#### Question 2.2
Use identities to evaluate : (502)2
(502)2
(502)= (500 + 2)2
We know that,
( a + b )2 = a2 + b2 + 2ab
∴ ( 500 + 2 )2 = 5002 + 22 + 2 x 500 x 2
= 250000 + 4 + 2000
= 2,52,004
#### Question 2.3
Use identities to evaluate : (97)2
(97)2
(97)2 = (100 – 3)2
We know that,
( a – b )2 = a2 + b2 – 2ab
∴ (100 – 3)2 = 1002 + 32 – 2 x 100 x 3
= 10000 + 9 – 600
= 9,409
#### Question 2.4
Use identities to evaluate : (998)2
(998)2
(998)2 = (1000 – 2)2
We know that
( a – b )2 = a2 + b2 – 2ab
∴ (1000 – 2)2 = 10002 + 22 – 2 x 1000 x 2
= 1000000 + 4 – 4000
= 9,96,004
Evalute : …………
Evalute :………….
#### Question 4.1
Evaluate :…………..
Consider the given expression :
#### Question 4.2
Evaluate : (4a +3b)2 – (4a – 3b)2 + 48ab.
(4a +3b)2 – (4a – 3b)2 + 48ab.
Consider the given expression:
Let us expand the first term : (4a +3b)2
We know that,
( a + b )2 = a2 + b2 + 2ab
∴ (4a +3b)= (4a)2 + (3b)2 + 2 x 4a x 3b
= 16a2 + 9b2 + 24ab ….(1)
Let us expand the second term : (4a – 3b)2
We know that,
( a + b )2 = a2 + b2 + 2ab
∴ (4a – 3b)= (4a)2 + (3b)2 – 2 x 4a x 3b
= 16a2 + 9b2 – 24ab …(2)
Thus from (1) and (2), the given expression is
(4a +3b)2 – (4a – 3b)2 + 48ab
= 16a2 + 9b2 + 24ab – 16a2 – 9b2 + 24ab + 48ab
= 96ab
#### Question 5
If a + b = 7 and ab = 10; find a – b.
We know that,
( a + b )2 = a2 + 2ab + b2
and
( a – b )2 = a2 – 2ab + b2
Rewrite the above equation, we have
( a – b )2 = a2 + b2 – 2ab + 4ab
= ( a + b )2 – 4ab …(1)
Given that a + b = 7; ab = 10
Substitute the values of ( a + b ) and (ab)
in equation (1), we have
( a – b )2 = (7)2 – 4(10)
= 49 – 40 = 9
⇒ a – b = ±9
⇒ a – b = ±3
#### Question 6
If a – b = 7 and ab = 18; find a + b.
We know that,
( a – b )2 = a2 – 2ab + b2
and
( a + b )2 = a2 + 2ab + b2
Rewrite the above equation, we have
( a + b )2 = a2 + b2 – 2ab + 4ab
= ( a + b )2 + 4ab …(1)
Given that a – b = 7; ab = 18
Substitute the values of ( a – b ) and (ab)
in equation (1), we have
( a + b )2 = (7)2 + 4(18)
= 49 + 72 = 121
⇒ a + b = ±121
⇒ a + b = ±11
#### Question 7
If x + y = ; find : x – y and x2 – y2
We know that,
( x + y )2 = x2 + 2xy + y2
and
( x – y )2 = x2 – 2xy + y2
Rewrite the above equation, we have
( x – y )2 = x2 + y2 + 2xy – 4xy
= ( x + y )2 – 4xy …(1)
#### Question 8
If a – b = 0.9 and ab = 0.36; find:
(i) a + b
(ii) a2 – b2
(i) We know that,
( a – b )2 = a2 – 2ab + b2
and
( a + b )2 = a2 + 2ab + b2
Rewrite the above equation, we have
( a + b )2 = a2 + b2 – 2ab + 4ab
= ( a – b )2 + 4ab …(1)
Given that a – b = 0.9 ; ab = 0.36
Substitute the values of ( a – b ) and (ab)
in equation (1), we have
( a + b )2 = ( 0.9 )2 + 4( 0.36 )
= 0.81 + 1.44 = 2.25
⇒ a + b = ±2.25
⇒ a + b = ±1.5 ..(2)
(ii) We know that,
a2 – b2 = ( a + b )( a – b ) ….(3)
From equation (2) we have,
a + b = ±1.5
Thus equation (3) becomes,
a2 – b2 = (±1.5)(0.9) [ given a – b = 0.9 ]
⇒ a2 – b2 = ±1.35
#### Question 9
If a – b = 4 and a + b = 6; find
(i) a2 + b2
(ii) ab
(i) We know that,
( a – b )2 = a2 – 2ab + b2
Rewrite the above identity as,
a2 + b= ( a – b ) + 2ab ….(1)
Similarly, we know that,
( a + b )2 = a2 + 2ab + b2
Rewrite the above identity as,
a2 + b2 = ( a + b )2 – 2ab …..(2)
Adding the equations (1) and (2), we have
2( a2 + b2 ) = ( a – b )2 + 2ab + ( a + b )2 – 2ab
⇒ 2( a2 + b2 ) = ( a – b )2 + ( a + b )2
⇒ ( a+ b2 ) = 26 …..(4)
From equation (4), we have
a2 + b2 = 26
Consider the identity,
( a – b )2 = a2 + b2 – 2ab ….(5)
Substitute the value a – b = 4 and a2 + b2 = 26
in the above equation, we have
(4)2 = 26 – 2ab
⇒ 2ab = 26 – 16
⇒ 2ab = 10
ab=5
#### Question 10
If a + = 6 and a ≠ 0 find :
(i) a….
We know that,
( a + b )2 = a2 + 2ab + b2
and
( a – b )2 = a2 – 2ab + b
Thus,
#### Question 11
If a………….find :
(i) a….
We know that,
( a + b )2 = a2 + 2ab + b2
and
( a – b )2 = a2 – 2ab + b
Thus,
#### Question 12
If a2 – 3a + 1 = 0, and a ≠ 0; find :
(i)……
(i) Consider the given equation
a2 – 3a + 1 = 0
Rewrite the given equation, we have
#### Question 13
If a2 – 5a – 1 = 0 and a ≠ 0 ; find :
(i) ………
(ii) …………
(iii) ………..
(i) Consider the given equation
a2 – 5a – 1 = 0
Rewrite the given equation, we have
a2 – 1 = 5a
#### Question 14
If 3x + 4y = 16 and xy = 4; find the value of 9x2 + 16y2.
Given that ( 3x + 4y ) = 16 and xy = 4
We need to find 9x2 + 16y2.
We know that
( a + b )2 = a2 + b2 + 2ab
Consider the square of 3x + 4y :
∴ ( 3x + 4y )2 = (3x)2 + (4y)2 + 2 x 3x x 4y
⇒ ( 3x + 4y )2 = 9x2 + 16y2 + 24xy …..(1)
Substitute the values of ( 3x + 4y ) and xy
in the above equation (1), we have
( 3x + 4y )2 = 9x2 + 16y2 + 24xy
⇒ (16)2 = 9x2 + 16y2 + 24(4)
⇒ 256 = 9x2 + 16y2 + 96
⇒ 9x2 + 16y2 = 160
#### Question 15
The number x is 2 more than the number y. If the sum of the squares of x and y is 34, then find the product of x and y.
Given x is 2 more than y, so x = y + 2
Sum of squares of x and y is 34, so x+ y= 34.
Replace x = y + 2 in the above equation and solve for y.
We get (y + 2)+ y= 34
2y+ 4y – 30 = 0
y+ 2y – 15 = 0
(y + 5)(y – 3) = 0
So y = -5 or 3
For y = -5, x =-3
For y = 3, x = 5
Product of x and y is 15 in both cases.
#### Question 16
The difference between two positive numbers is 5 and the sum of their squares is 73. Find the product of these numbers.
Let the two positive numbers be a and b.
Given difference between them is 5 and sum of squares is 73.
So a – b = 5, a+ b= 73
Squaring on both sides gives
(a – b)= 52
a+ b– 2ab = 25
but a+ b= 73
so 2ab = 73 – 25 = 48
ab = 24
So, the product of numbers is 24.
### Expansions (Including Substitution) Exe-4 B for ICSE Class-9th Concise Selina Mathematics
#### Question 1.1
Find the cube of : 3a- 2b
( a – b )3 = a3 – 3ab( a – b ) – b3
( 3a – 2b )3 = (3a)3 – 3 x 3a x 2b( 3a – 2b) – (2b)3
= 27a3 – 18ab( 3a – 2b ) – 8b3
= 27a3 – 54a2b + 36ab2 – 8b3
#### Question 1.2
Find the cube of : 5a + 3b
( a + b )3 = a3 + 3ab( a + b ) + b3
( 5a + 3b)3 = (5a)3 + 3 x 5a x 3b( 5a + 3b) + (3b)3
= 125a3 + 45ab( 5a + 3b ) + 27b3
= 125a3 + 225a2b + 135ab2 + 27b
#### Question 1.3
Find the cube of :
( a + b )3 = a3 + 3ab( a + b ) + b3
#### Question 1.4
Find the cube of :
( a – b )3 = a3 – 3ab( a – b ) – b3
#### Question 2
If a2 + …………. find :
(i) ………
(ii) ……..
#### Question 3
If a2 + …………. find :
(i) ………
(ii) ……..
#### Question 4
If ; then show that
…………
#### Question 5
If a + 2b = 5; then show that : a3 + 8b3 + 30ab = 125.
Given that a + 2b = 5 ;
We need to find a3 + 8b3 + 30ab :
Now consider the cube of a + 2b :
( a + 2b )3 = a3 + (2b)3 + 3 x a x 2b x ( a + 2b )
= a3 + 8b3 + 6ab x ( a + 2b )
53 = a3 + 8b3 + 6ab x 5 [ ∵ a + 2b = 5 ]
125 = a3 + 8b3 + 30ab
Thus the value of a3 + 8b3 + 30ab is 125.
If =0
#### Question 7
If a + 2b + c = 0; then show that : a3 + 8b3 + c3 = 6abc.
Given that a + 2b + c = 0;
⇒ a + 2b = -c ….(1)
Now consider the expansion of ( a + 2b )3 :
( a + 2b )= ( – c )3
a3 + (2b)3 + 3 x a x 2b x ( a + 2b ) = -c3
⇒ a3 + 8b3 + 3 x a x 2b x (-c) = -c3 [ from (1) ]
⇒ a3 + 8b3 – 6abc = -c
⇒ a3 + 8b3 – c3 = 6abc
Hence proved.
#### Question 8.1
Use property to evaluate : 133 + (-8)3 + (-5)3
Property is if a + b + c = 0 then a+ b+ c= 3abc
a = 13, b = -8 and c = -5
133 + (-8)3 + (-5)= 3(13)(-8)(-5) = 1560
#### Question 8.2
Use property to evaluate : 73 + 33 + (-10)3
Property is if a + b + c = 0 then a+ b+ c= 3abc
a = 7, b = 3, c = -10
73 + 33 + (-10)= 3(7)(3)(-10) = -630
#### Question 8.3
Use property to evaluate : 93 – 53 – 43
Property is if a + b + c = 0 then a+ b+ c= 3abc
a = 9, b = -5, c = -4
93 – 53 – 4= 93 + (-5)3 + (-4)= 3(9)(-5)(-4) = 540
#### Question 8.4
Use property to evaluate : 383 + (-26)3 + (-12)3
Property is if a + b + c = 0 then a3 + b3 + c3 = 3abc
a = 38, b = -26, c = -12
383 + (-26)3 + (-12)3 = 3(38)(-26)(-12) = 35568
#### Question 9.1
If a ≠ 0 and = 3 ; find :
…………
#### Question 9.2
If a ≠ 0 and …….. 3 ; Find :
…………
#### Question 10.1
If a ≠ 0 and = 4 ; find : ………
#### Question 10.2
If a ≠ 0 and = 4 ; find : ………
#### Question 10.3
If a ≠ 0 and = 4 ; find : ………
#### Question 11
If X ≠ 0 and X + ………; then show that :
……….
#### Question 12
If 2x – 3y = 10 and xy = 16; find the value of 8x3 – 27y3.
Given that 2x – 3y = 10, xy = 16
∴ (2x – 3y)3 = (10)3
⇒ 8x3 – 27y3 – 3 (2x) (3y) (2x – 3y) = 1000
⇒ 8x3 – 27 y3 -18xy (2x – 3y) = 1000
⇒ 8x3 – 27 y3 – 18 (16) (10) = 1000
⇒ 8x3 – 27 y3 – 2880 = 1000
⇒8x3 – 27 y3 = 1000 + 2880
⇒ 8x3 – 27 y3 =3880
#### Question 13.1
Expand : (3x + 5y + 2z) (3x – 5y + 2z)
(3x + 5y + 2z) (3x – 5y + 2z)
= {(3x + 2z) + (5y)} {(3x + 2z) – (5y)}
= (3x + 2z)2 – (5y)2
{since (a + b) (a – b) = a2 – b2}
= 9x2 + 4z2 + 2 × 3x × 2z – 25y2
= 9x2 + 4z2 + 12xz – 25y2
= 9x2 + 4z– 25y2 + 12xz
#### Question 13.2
Expand : (3x – 5y – 2z) (3x – 5y + 2z)
(3x – 5y – 2z) (3x – 5y + 2z)
= {(3x – 5y) – (2z)} {(3x – 5y) + (2z)}
= (3x – 5y)2 – (2z)2{since(a + b) (a – b) = a2 – b2}
= 9x2 + 25y2 – 2 × 3x × 5y – 4z2
= 9x2 + 25y2– 30xy – 4z2
= 9x2 +25y2 – 4z2 – 30xy
#### Question 14
The sum of two numbers is 9 and their product is 20. Find the sum of their (i) Squares (ii) Cubes
Given sum of two numbers is 9 and their product is 20.
Let the numbers be a and b.
a + b = 9
ab = 20
Squaring on both sides gives
(a+b)= 92
a+ b+ 2ab = 81
a+ b+ 40 = 81
So sum of squares is 81 – 40 = 41
Cubing on both sides gives
(a + b)= 93
a+ b+ 3ab(a + b) = 729
a+ b+ 60(9) = 729
a+ b= 729 – 540 = 189
So the sum of cubes is 189.
#### Question 15
Two positive numbers x and y are such that x > y. If the difference of these numbers is 5 and their product is 24, find:
(i) Sum of these numbers
(ii) Difference of their cubes
(iii) Sum of their cubes.
Given x – y = 5 and xy = 24 (x>y)
(x + y)= (x – y)+ 4xy = 25 + 96 = 121
So, x + y = 11; sum of these numbers is 11.
Cubing on both sides gives
(x – y)= 53
x– y– 3xy(x – y) = 125
x– y– 72(5) = 125
x– y3= 125 + 360 = 485
So, difference of their cubes is 485.
Cubing both sides, we get
(x + y)= 113
x+ y+ 3xy(x + y) = 1331
x+ y= 1331 – 72(11) = 1331 – 792 = 539
So, sum of their cubes is 539.
#### Question 16
If 4x+ y= a and xy = b, find the value of 2x + y.
xy = ab ….(i)
4x+ y= a ….(ii)
Now, (2x + y)2 = (2x)2 + 4xy + y2
= 4x2 + y2 + 4xy
= a + 4b ….[From (i) and (ii)]
⇒ 2x + y =
### Selina Solutions of Expansions (Including Substitution) Exe-4 C for ICSE Class-9th Concise Mathematics
#### Question 1.1
Expand : ( x + 8 ) ( x + 10 )
( x + 8 )( x + 10 ) = x2 + ( 8 + 10 )x + 8 x 10
= x2 + 18x + 80
#### Question 1.2
Expand : ( x + 8 )( x – 10 )
( x + 8 )( x – 10 ) = x2 + ( 8 – 10 )x + 8 x (-10)
= x2 – 2x – 80
#### Question 1.3
Expand : ( X – 8 ) ( X + 10 )
( X – 8 ) ( X + 10 ) = X2 – ( 8 – 10 )X – 8 x 10
= X2 + 2X – 80
#### Question 1.4
Expand : ( X – 8 )( X – 10 )
( X – 8 )( X – 10 ) = X2 – ( 8 + 10 )X + 8 x 10
= X2 – 18X + 80
Expand :
Expand :
#### Question 3.1
Expand : ( x + y – z )2
( x + y – z )2 = x2 + y2 + z2 + 2(x)(y) – 2(y)(z) – 2(z)(x)
= x2 + y2 + z2 + 2xy – 2yz – 2zx
#### Question 3.2
Expand : ( x – 2y + 2 )
x – 2y + 2 )2 = x2 + (2y)2 + (2)2 – 2(x)(2y) – 2(2y)(2) + 2(2)(x)
= x2 + 4y2 + 4 – 4xy – 8y + 4x
#### Question 3.3
Expand : ( 5a – 3b + c )2
( 5a – 3b + c)2 = (5a)2 + (3b)2 + (c)2 – 2(5a)(3b) – 2(3b)(c) + 2(c)(5a)
= 25a2 + 9b2 + c2 – 30ab – 6bc + 10ca
#### Question 3.4
Expand : ( 5x – 3y – 2 )2
( 5x – 3y – 2 )= (5x)2 + (3y)2 + (2)2 – 2(5x)(3y) + 2(3y)(2) – 2(2)(5x)
= 25x2 + 9y2 + 4 – 30xy + 12y – 20x
Expand :
#### Question 4
If a + b + c = 12 and a2 + b2 + c2 = 50; find ab + bc + ca.
We know that
( a + b + c )2 = a2 + b2 + c2 + 2( ab + bc + ca ) …….(1)
Given that, a2 + b2 + c2 = 50 and a + b + c = 12.
We need to find ab + bc + ca :
Substitute the values of (a2 + b2 + c2 ) and ( a + b + c )
in the identity (1), we have
(12)2 = 50 + 2( ab + bc + ca )
⇒ 144 = 50 + 2( ab + bc + ca )
⇒ 94 = 2( ab + bc + ca)
⇒ ab + bc + ca =
⇒ ab + bc + ca = 47
#### Question 5
If a2 + b2 + c2 = 35 and ab + bc + ca = 23; find a + b + c.
We know that
( a + b + c )2 = a2 + b2 + c2 + 2( ab + bc + ca ) ….(1)
Given that, a2 + b2 + c2 = 35 and ab + bc + ca = 23
We need to find a + b + c :
Substitute the values of ( a2 + b2 + c2 ) and ( ab + bc + ca )
in the identity (1), we have
( a + b + c )2 = 35 + 2(23)
⇒ ( a + b + c )2 = 81
⇒ a + b + c =
⇒ a + b + c =
#### Question 6
If a + b + c = p and ab + bc + ca = q ; find a2 + b2 + c2.
We know that
( a + b + c )2 = a2 + b2 + c2 + 2( ab + bc + ca ) …..(1)
Given that, a + b + c = p and ab + bc + ca = q
We need to find a2 + b2 + c2 :
Substitute the values of ( ab + bc + ca ) and ( a + b + c )
in the identity (1), we have
(p)2 = a2 + b2 + c2 + 2q
⇒ p2 = a2 + b2 + c2 + 2q
⇒ a2 + b2 + c2 = p2 – 2q
#### Question 7
If a2 + b2 + c2 = 50 and ab + bc + ca = 47, find a + b + c.
a2 + b2 + c2 = 50 and ab + bc + ca = 47
Since ( a + b + c )2 = a2 + b2 + c2 + 2( ab + bc + ca )
∴ ( a + b + c )2 = 50 + 2(47)
⇒ ( a + b + c )2 = 50 + 94 = 144
⇒ a + b +c =
∴ a + b + c =
#### Question 8
If x+ y – z = 4 and x2 + y2 + z2 = 30, then find the value of xy – yz – zx.
x + y – z = 4 and x2 + y2 + z2 = 30
Since ( x + y – z)2 = x2 + y2 + z2 + 2( xy – yz – zx ), we have
(4)2 = 30 + 2( xy – yz – zx )
⇒ 16 = 30 + 2( xy – yz – zx )
⇒ 2( xy – yz – zx ) = -14
⇒ xy – yz – zx =
∴ xy – yz – zx = -7
### Concise Selina Maths Solutions, Exe-4 D Expansions (Including Substitution) ICSE Class-9th
#### Question 1
If x + 2y + 3z = 0 and x3 + 4y3 + 9z3 = 18xyz ; evaluate :
………………………………
Given that x3 + 4y3 + 9z3 = 18xyz and x + 2y + 3z = 0
x + 2y = – 3z, 2y + 3z = -x and 3z + x = -2y
Now
#### Question 2.1
If a + = m and a ≠ 0 ; find in terms of ‘m’; the value of :
#### Question 2.2
If a + = m and a ≠ 0 ; find in terms of ‘m’; the value of :
………………….
#### Question 3.1
In the expansion of (2x2 – 8) (x – 4)2; find the value of
coefficient of x3
( 2x2 – 8 )( x – 4 )2
= ( 2x2 – 8 )( x2 – 8x + 16 )
= 4x4 – 16x3 + 32x2 – 8x2 + 64x -128
= 4x4 – 16x3 + 24x2 + 64x – 128
Hence,
Coefficient of x3 = -16
#### Question 3.2
In the expansion of (2x2 – 8) (x – 4)2; find the value of coefficient of x2
( 2x2 – 8 )( x – 4 )2
= ( 2x2 – 8 )( x2 – 8x + 16 )
= 4x4 – 16x3 + 32x2 – 8x2 + 64x -128
= 4x4 – 16x3 + 24x2 + 64x – 128
Hence,
Coefficient of x2 = 24
#### Question 3.3
In the expansion of (2x2 – 8) (x – 4)2; find the value of constant term.
( 2x2 – 8 )( x – 4 )2
= ( 2x2 – 8 )( x2 – 8x + 16 )
= 4x4 – 16x3 + 32x2 – 8x2 + 64x -128
= 4x4 – 16x3 + 24x2 + 64x – 128
Hence,
Constant term = -128
If x > 0 and
#### Question 5
If 2( x2 + 1 ) = 5x, find :
(i) ………
(ii) ……………
#### Question 6.1
If a2 + b2 = 34 and ab = 12; find : 3(a + b)2 + 5(a – b)2
a2 + b2 = 34, ab= 12
(a + b)2 = a2 + b2 + 2ab
= 34 + 2 x 12 = 34 + 24 = 58
(a – b)2 = a2 + b2 – 2ab
= 34 – 2 x 12 = 34- 24 = 10
3(a + b)2 + 5(a – b)2 = 3 x 58 + 5 x 10 = 174 + 50 = 224
#### Question 6.2
If a2 + b2 = 34 and ab = 12; find : 7(a – b)2 – 2(a + b)2
a2 + b2 = 34, ab= 12
(a + b)2 = a2 + b2 + 2ab
= 34 + 2 x 12 = 34 + 24 = 58
(a – b)2 = a2 + b2 – 2ab
= 34 – 2 x 12 = 34- 24 = 10\
7(a – b)2 – 2(a + b)2 = 7 x 10 – 2 x 58 = 70 – 116 = – 46
#### Question 7
f 3x – …………………………
#### Question 8
If x2 + = 7 and x ≠ 0; find the value of
……………..
#### Question 9
If x =…………….. Find:…………..
#### Question 10
If x = ……… find………….
#### Question 11
If 3a + 5b + 4c = 0, show that : 27a3 + 125b3 + 64c3 = 180 abc
Given that 3a + 5b + 4c = 0
3a + 5b = – 4c
Cubing both sides,
(3a + 5b)3 = (-4c)3
⇒ (3a)3 + (5b)3 + 3 x 3a x 5b (3a + 5b) = -64c3
⇒ 27a3 + 125b3 + 45ab x (-4c) = -64c3
⇒ 27a3 + 125b3 – 180abc = -64c3
⇒ 27a3 + 125b3 + 64c3 = 180abc
Hence proved.
#### Question 12
The sum of two numbers is 7 and the sum of their cubes is 133, find the sum of their square.
Let a, b be the two numbers.
.’. a + b = 7 and a3 + b3 = 133
(a + b)3 = a3 + b3 + 3ab (a + b)
⇒ (7)3 = 133 + 3ab (7)
⇒ 343 = 133 + 21ab
⇒ 21ab = 343 – 133 = 210
⇒ 21ab = 210
⇒ ab= 10
Now a2 + b2 = (a + b)2 – 2ab
= 72 – 2 x 10 = 49 – 20 = 29
#### Question 13.1
Find the value of ‘a’: 4x2 + ax + 9 = (2x + 3)2
4x2 + ax + 9 = (2x + 3)2
Comparing coefficients of x terms, we get
ax = 12x
so, a = 12
#### Question 13.2
Find the value of ‘a’: 4x2 + ax + 9 = (2x – 3)2
4x2 + ax + 9 = (2x – 3)2
Comparing coefficients of x terms, we get
ax = -12x
so, a = -12
#### Question 13.3
Find the value of ‘a’: 9x2 + (7a – 5)x + 25 = (3x + 5)2
9x2 + (7a – 5)x + 25 = (3x + 5)2
Comparing coefficients of x terms, we get
(7a – 5)x = 30x
7a – 5 = 30
7a = 35
a = 5
If
If
#### Question 15.1
The difference between two positive numbers is 4 and the difference between their cubes is 316.
Find : Their product
Given difference between two positive numbers is 4 and difference between their cubes is 316.
Let the positive numbers be a and b
a – b = 4
a– b= 316
Cubing both sides,
(a – b)= 64
a– b– 3ab(a – b) = 64
Given a– b= 316
So 316 – 64 = 3ab(4)
252 = 12ab
So ab = 21; product of numbers is 21
#### Question 15.2
The difference between two positive numbers is 4 and the difference between their cubes is 316.
Find : The sum of their squares
Given difference between two positive numbers is 4 and difference between their cubes is 316.
Let the positive numbers be a and b
a – b = 4 …..(1)
a– b= 316 …..(2)
Squaring(eq 1) both sides, we get
(a – b)= 16
a+ b– 2ab = 16
a+ b= 16 + 42 = 58
Sum of their squares is 58.
### Exercise – 4(E)Expansions (Including Substitution) ICSE Class-9th Concise Selina Mathematics
#### Question 1.1
Simplify : ( x + 6 )( x + 4 )( x – 2 )
Using identity :
(x + a)(x + b)(x + c) = x3 + (a + b + c)x2 + (ab + bc + ca)x + abc
(x + 6)(x + 4)(x – 2)
= x3 + (6 + 4 – 2)x2 + [6 × 4 + 4 × (-2) + (-2) × 6]x + 6 × 4 × (-2)
= x3 + 8x2 + (24 – 8 – 12)x – 48
= x3 + 8x2 + 4x – 48
#### Question 1.2
Simplify : ( x – 6 )( x – 4 )( x + 2 )
Using identity : (x + a)(x + b)(x + c) = x3 + (a + b + c)x2 + (ab + bc + ca)x + abc
(x – 6)(x – 4)(x + 2)
= x3 + (-6 – 4 + 2)x2 + [-6 × (-4) + (-4) × 2 + 2 × (-6)]x + (-6) × (-4) × 2
= x3 – 8x2 + (24 – 8 – 12)x + 48
= x3 – 8x2 + 4x + 48
#### Question 1.3
Simplify : ( x – 6 )( x – 4 )( x – 2 )
Using identity :
(x + a)(x + b)(x + c) = x3 + (a + b + c)x2 + (ab + bc + ca)x + abc
( x – 6 )( x – 4 )( x – 2 )
= x3 + (-6 – 4 – 2)x2 + [-6 × (-4) + (-4) × (-2) + (-2) × (-6)]x + (-6) × (-4) × (-2)
= x3 – 12x2 + (24 + 8 + 12)x – 48
= x3 – 12x2 + 44x – 48
#### Question 1.4
Simplify : ( x + 6 )( x – 4 )( x – 2 )
Using identity :
(x + a)(x + b)(x + c) = x3 + (a + b + c)x2 + (ab + bc + ca)x + abc
(x + 6)(x – 4)(x – 2)
= x3 + (6 – 4 – 2)x2 + [6 × (-4) + (-4) × (-2) + (-2) × 6]x + 6 × (-4) × (-2)
= x3 – 0x2 + (-24 + 8 – 12)x + 48
= x3 – 28x + 48
#### Question 2.1
Simplify using following identity : ( 4x2 + 6xy + 9y)
( 2x + 3y )( 4x2 + 6xy + 9y)
= ( 2x + 3y )[ (2x)2 – (2x)(3y) + (3y)]
= (2x)3 + (3y)3
= 8x3 + 27y3
#### Question 2.2
Simplify using following identity :
#### Question 2.3
Simplify using following identity :
#### Question 3.1
Using suitable identity, evaluate (104)3
Using identity: (a ± b)3 = a3 ± b3 ± 3ab(a ± b)
(104)3 = (100 + 4)3
= (100)3 + (4)3 + 3 × 100 × 4(100 + 4)
= 1000000 + 64 + 1200 × 104
= 1000000 + 64 + 124800
= 1124864
#### Question 3.2
Using suitable identity, evaluate (97)3
(97)= (100 – 3)3
= (100)3 – (3)3 – 3 × 100 × 3(100 – 3)
= 1000000 – 27 – 900 × 97
= 1000000 – 27 – 87300
= 912673
Simplify :………..
Evaluate :……………
Evaluate :
#### Question 6
If a – 2b + 3c = 0; state the value of a– 8b3 + 27c3.
a– 8b3 + 27c3 = a3 + (-2b)3 + (3c)3
Since a – 2b + 3c = 0, we have
a– 8b3 + 27c= a3 + (-2b)3 + (3c)3
= 3(a)( -2b)(3c)
= -18abc
#### Question 7
If x + 5y = 10; find the value of x3 + 125y3 + 150xy – 1000.
x + 5y = 10
⇒ (x + 5y)3 = 103
⇒ x3 + (5y)3 + 3(x)(5y)(x + 5y) = 1000
⇒ x3 + (5y)3 + 3(x)(5y)(10) = 1000
= x3 + (5y)3 + 150xy = 1000
= x3 + (5y)3 + 150xy – 1000 = 0
#### Question 8
If x = 3 + 2√2, find :
(i) ……………..
(ii) …………..
(iii) …………
(iv)……
#### Question 9
If a + b = 11 and a2 + b2 = 65; find a3 + b3.
a + b = 11 and a2 + b2 = 65
Now, (a+b)2 = a2 + b2 + 2ab
⇒ (11)2 = 65 + 2ab
⇒ 121 = 65 + 2ab
⇒ 2ab = 56
⇒ ab = 28
a3 + b3 = ( a + b )( a2 – ab + b2)
= (11)(65 – 28)
= 11 x 37
= 407
#### Question 10
Prove that : x2+ y2 + z2 – xy – yz – zx is always positive.
x+ y+ z– xy – yz – zx
= 2(x+ y+ z– xy – yz – zx)
= 2x+ 2y+ 2z– 2xy – 2yz – 2zx
= x+ x2 + y+ y2 + z2 + z– 2xy – 2yz – 2zx
= (x2 + y2 – 2xy) + (z2 + x2 – 2zx) + (y2 + z2 – 2yz)
= (x – y)2 + (z – x)2 + (y – z)2
Since square of any number is positive, the given equation is always positive.
#### Question 11.1
Find : (a + b)(a + b)
(a + b)(a + b) = (a + b)2
= a × a + a × b + b × a + b × b
= a2 + ab + ab + b2
= a2 + b2 + 2ab
#### Question 11.2
Find : (a + b)(a + b)(a + b)
(a + b)(a + b)(a + b)
= (a × a + a × b + b × a + b × b)(a + b)
= (a2 + ab + ab + b2)(a + b)
= (a2 + b2 + 2ab)(a + b)
= a2 × a + a2 × b + b2 × a + b2 × b + 2ab × a + 2ab × b
= a3 + a2 b + ab2 + b3 + 2a2b + 2ab2
= a3 + b3 + 3a2b + 3ab2
#### Question 11.3
Find : (a – b)(a – b)(a – b)
(a + b)(a + b)(a + b)
= (a × a + a × b + b × a + b × b)(a + b)
= (a2 + ab + ab + b2)(a + b)
= (a2 + b2 + 2ab)(a + b)
= a2 × a + a2 × b + b2 × a + b2 × b + 2ab × a + 2ab × b
= a3 + a2 b + ab2 + b3 + 2a2b + 2ab2
= a3 + b3 + 3a2b + 3ab2
replacing b by -b, we get
= a3 + (-b)3 + 3a2(-b) + 3a(-b)2
= a3 – b3 – 3a2b + 3ab2
— End of Expansions ICSE Class-9th Concise Solutions :–
Thanks
$${}$$ | HuggingFaceTB/finemath | |
# kinematic equations derivation
Derivation of Kinetic Equations As we said, the mathematical object that we consider in Kinetic Theory is the distribution function 0 f(t,x,v). 4. Similarly, the total displacement of the object in Fig. 1 0 obj • Such a mathematical singularity problem can be avoided by selecting a different set of Euler angles. !v!v!t!t a= rise run =!v!t 2.)! 1 $\begingroup$ Apologies if this has been asked before, but I browsed the sub and couldn't find something specific. <> Choosing kinematic equations. Old videos on projectile motion. That is:! Kinematic equations relate the variables of motion to one another. Kinematic equations for projectile motion describe the two-dimensional case of objects moving under gravity. A sprinter is moving at 5 m/s when they reach 20 meters into their race. But it did not arise directly from it. • Singularites exist when θ 2 =π/2. If values of three variables are known, then the others can be calculated using the equations. 2.figure 2. s = the displacement vector, the magnitude of the displacement is the distance, s = │s│ = d (vectors are indicated in bold; the same symbol not in bold represents the magnitude of the vector), Δ indicates change, for example Δv = (v2 –v1). In this section, we will logically deduce what it should be. The derivation starts with observing a simple geometry of reflection in a constant-velocity medium, shown in Figure 11. kinematic equations. \Large 1. The acceleration is equal to the slope of a velocity versus time graph. Kinematics is also the center of dynamic analysis. High school physics courses usually begin with a study of classical mechanics. Instead of differentiating velocity to find acceleration, integrate acceleration to find velocity. Total distance travelled by an object is equal to the area of trapezium OECA, consider Figure 3. Kinematics simplifies the derivation of motion equations. They maintain a constant … Kinematics equations are the constraint equations of a mechanical system such as a robot manipulator that define how input movement at one or more joints specifies the configuration of the device, in order to achieve a task position or end-effector location. Kinematics is a subfield of physics, developed in classical mechanics, that describes the motion of points, bodies (objects), and systems of bodies (groups of objects) without considering the forces that cause them to move. These equations define motion at either constant velocity or at constant acceleration . The equations can be utilized for any motion that can be described as being either a constant velocity motion (an acceleration of 0 m/s/s) or a constant acceleration motion. $\begingroup$ The equation you have written is used very often in mechanics problems, where the speed of a particle is taken to be a function of the distance travelled. This exercise references the diagram in Fig. Try your kinematics skills at the racetrack. Thus we can simply state the equations, alongside their translational analogues: 2 0 obj <>/XObject<>/ProcSet[/PDF/Text/ImageB/ImageC/ImageI] >>/MediaBox[ 0 0 720 540] /Contents 10 0 R/Group<>/Tabs/S/StructParents 1>> Motion Equation #1 Displacement with Constant Acceleration …move tto the other side... Motion Equation #2 Velocity with Constant Acceleration. Sort by: Top Voted. When an object motion problem falls into these categories, we may use the kinematic equations to solve it. When done, sketch your graph in the space provided below. Kinematics Derivations! DERIVING THE KINEMATIC EQUATIONS The main goal of this appendix is to derive the partial differential equation describing the image surface in a depth-midpoint-offset-velocity space. a= (v 2!v 1) "t or! The horizontal equations are a little easier, since the only net force acting on the ball is the drag: Fnet = m a = - D m a = - (.5 * Cd * r * A * u^2) a = - (Cd * r * A * u^2) / (2 * m) where u is the horizontal velocity. Simplify your formula using one of the other kinematic equations. For example, equation 3 does not contain the final velocity, , so that is the variable that gets eliminated. The Kinematic Equations. Derivation of Kinematic Equations View this after Motion on an Incline Lab Constant velocity Average velocity equals the slope of a position vs time graph when an object travels at constant velocity. The variables include acceleration (a), time (t), displacement (d), final velocity (vf), and initial velocity (vi). This article focuses on kinematics formulas and their derivation. Derivation of Kinematic Equations Author: Larry Dukerich Last modified by: Image Creation Account Created Date: 6/20/2003 2:17:40 AM Document presentation format: On-screen Show (4:3) Company: Dept of Physics, ASU Other titles: Times Arial Times New Roman Calibri Helvetica Blank Presentation Microsoft Equation Derivation of Kinematic Equations Constant velocity Displacement when object … Solution: Note: The steps are labeled in this solution to point out the steps involved in solving a kinematics problem. Deriving the 5 Equations of Kinematics Between equations 1 and 2 all five variables are present. These three equations of motion govern the motion of an object in 1D, 2D and 3D. A more detailed derivation of the equations in this section is given in Sec. • However these calculations involve the computation of the trigonometric functions of the angle. An ion with mass m1 and kinetic energy E0 (velocity v0) is incident at an angle fi on a target atom with mass m2. Practice: Setting up problems with constant acceleration. vlcray Figure 11 Reflection rays in a constant velocity medium (a scheme). The variables include acceleration (a), time (t), displacement (d), final velocity (vf), and initial velocity (vi). \quad v=v_0+at 1. v = v 0 The equations can be utilized for any motion that can be described as being either a constant velocity motion (an acceleration of 0 m/s/s) or a constant acceleration motion. This article is a purely mathematical exercise designed to provide a quick review of how the kinematics equations are derived using algebra. The first step will be to calculate the slope of the diagonal line. Kinematics refers to the branch of classical mechanics which describes the motion of points, objects, and systems comprising of groups of objects. He is an avid Blogger who writes a couple of blogs of different niches. Practice: Kinematic formulas in one-dimension. 10 0 obj Derivation of Kinematic Equations View this after Motion on an Incline Lab. endobj In a constant acceleration situation, the area under a velocity versus time graph for a Differential Kinematic Equation. Understand a derivation, or origin. Hence, we can write; Derivation of third Equation of Motion by Calculus Method. <> Equation 1 does not include the variable s. Equation 2 does not include the variable a. To keep things simple, rewrite t2 – t1 as Δt. In addition, Kinematics equations apply algebraic geometry to the study of the mechanical benefits of kinetic mechanical systems or mechanisms. If values of three variables are known, then the others can be calculated using the equations. these kinematic differential equations. 1, calculating the area under the line simply means calculating the area of the rectangle A1 and the triangle A2 and adding the values. Kinematics is also the center of dynamic analysis. Hot Network Questions Can one planet in our system eclipse another one? the primitive concepts concerned are position, time and body, the latter abstracting into mathematical terms intuitive ideas about aggregations of matter capable of motion and deformation. In this case, since the slope will be a change in velocity (rise) divided by a change in time (run), the slope will equal the acceleration. The kinematic equations list contains three equations relating position, velocity, acceleration and time. Kinematics equations. The object in this activity, however, is not traveling at a constant velocity. These equations relate the variables of time, position, velocity and acceleration of a moving object, allowing any of these variables to be solved for if the others are known. Choosing kinematic equations. A solid understanding of these equations and how to employ them to solve problems is essential for success in physics. Each equation contains four variables. Use algebra to rearrange and solve the equation for all of its variables. The kinematic formulas are often written as the following four equations. <>>> Active 1 year, 2 months ago. Newton's Laws Three simple laws govern nearly everything you see. What we measure is angular velocity in omega. What Are The Kinematic Formulas? The Kinematic Equations Derive the most useful kinematic relationships for an accelerating object. Included with Brilliant Premium Angular Kinematics. Top … Consider an object moving with constant velocity, v1, from time t1 to t2. So here I'm using the DCM as an attitude description. The remaining kinematics equations can be found by eliminating the variables v2 and Δt. v! Students can apply the kinematics equations when conducting investigations with these products: Inclined Plane with Pulley (item #751450), Economy Digital Timer and Photogate (item #751831), Collision in 2-D Apparatus (item #751512), Aluminum Acceleration Trolley (item #751510). By applying some algebra, the left side of this equation can be made to look like the right side of Equation 2. Deriving the Equations. The Fifth Kinematic Equation We have always used the four kinematic equations, but there is actually one more! In the case the mass is not constant, it is not sufficient to use the product rule for the time derivative on the mass and velocity, and Newton's second law requires some modification consistent with conservation of momentum; see variable-mass system. Ask Question Asked 1 year, 2 months ago. 4 0 obj PhysicsLAB: Derivation of the Kinematics Equations for Uniformly Accelerated Motion This derivation is based on the properties of a velocity-time graph for uniformly accelerated motion where the slope of the graph represents the acceleration graph's area represents the displacement <> The strategy involves the following steps: 1. 1, in which the x axis represents time and the y axis represents velocity. Kinematics refers to the branch of classical mechanics which describes the motion of points, objects, and systems comprising of groups of objects. But when we integrate I have to somehow integrate the attitude. Whatever is done to one side of the equation must be done to the other side of the equation. 8 0 obj If that acceleration is constant, the slope becomes:! Because our equations defining rotational and translational variables are mathematically equivalent, we can simply substitute our rotational variables into the kinematic equations we have already derived for translational variables. Adding areas A2 and A1 gives the total displacement of the object during the time interval. 5 0 obj 9 0 obj <> Take the operation in that definition and reverse it. Each equation contains four variables. There are four (4) kinematic equations, which relate to displacement, D, velocity, v, time, t, and acceleration, a. a) D = vit + 1/2 at2 b)(vi +vf)/2 = D/t c)a = (vf – vi)/t d) vf2 = vi2+ 2aD D= displacement a= acceleration t= time vf= final velocity vi= initial velocity The width, w, is t2 – t1, (Δt). Kinematics equations are used to analyze and design articulated systems ranging from four-bar linkages to serial and parallel robots. Equation Derivation. endobj Kinematic Equations Derivation. A derivation of this equation is available. 3 0 obj When faced with learning so many equations, most students resort to rote memorization and generally fail to comprehend the relationships expressed by such equations. endobj 0. Constant velocity Average velocity equals the slope of a position vs time graph when an object travels at constant velocity. Each equation contains four variables. endobj If an object starts with velocity ”u” and after some time “t” its velocity changes to v, if the uniform acceleration is a and distance traveled in time (t) is s, then we obtain the following kinematic equations of uniformly accelerated motion. A car slows down uniformly from a speed of 21.0 m/s to rest in 6.00 seconds. The variety of representations… admin — September 19, 2019. 7 0 obj This derivation will involve using the quadratic equation. The Second Equation of Motion Derivation by Graphical Method Consider a graph of the motion of this object, as in Fig. We will now be a bit more precise about the link between microscopic and mesoscopic descriptions of asystem. For an object moving with constant acceleration, the average velocity is ... We also know, of course, that . Kinematic equations relate the variables of motion to one another. <> Viewed 1k times 6. <>/ProcSet[/PDF/Text/ImageB/ImageC/ImageI] >>/MediaBox[ 0 0 720 540] /Contents 4 0 R/Group<>/Tabs/S/StructParents 0>> Can anyone suggest a math review book for someone interested in beginning physics study as a hobby? Opus in profectus … motion-graphs; kinematics-calculus; kinematics-2d … Kinematics and Calculus. Choosing kinematic equations. What we use in dynamics is omega. chaos; eworld; facts; get bent; physics; The Physics Hypertextbook. Figure 1: Schematic demonstration of an ion scattering from a surface atom. x�u�OK�@����1�a3;�iR(=�R�����Xx�{o�{�f�v�Y��s,VK� �""�L!B&�PeRM4ف(K� ��%�3Н�"�¿ҰFM#�yt4hh(b�A�oLZ�k)o�U�gZ���E���o��3�)֎ڐXr�TЂo}�h�O=�EDDt ;��Q�BbE����8��D�2*�����[a�mms��s3-�)w�B�&�]��odsM An explanation of the kinematics equations that can be applied to AP Physics and other physics courses. How far did it travel in that time? endobj Kinematic equations relate the variables of motion to one another. Introduction: The average physics text introduces more than 100 basic equations, many of which have one or more alternate expressions. Kinematic Equations: The goal of this first unit of The Physics Classroom has been to investigate the variety of means by which the motion of objects can be described. endobj Because kinematics equations are only applicable at a constant acceleration or a constant speed, we cannot use them if either of the two is changing. 2.1 Kinematic equations The need of a good modellig tool, and in partic-ular a good symbolic based modelling tool such as Maple, is illustrated by the following exam-ple where the industrial robot IRB1400 (Norrlo f, 1999) from ABB Robotics is used. Calculus is an advanced math topic, but it makes deriving two of the three equations of motion much simpler. Area A2 is a triangle with base Δt and height v2 – v1. Development of Kinematic Equations Part 1: Exploring the relationship between position and time 1. Initial velocity, v1, is still in every equation, but v1 can often be set to zero if the object starts from rest. From this point of view the kinematics equations can be used in two different ways. kinematic equations 8 kinematics is the study of motion per se, regardless of the forces causing it. For some objects this calculation can be a little tricky, but for the object depicted in Fig. endstream How fast, in m/s, will it be going in 5.0 seconds? Kinematics is the description of motion. because you will need them to derive the last of the three important kinematic equations.! Tag: derivation of kinematic equations What Are The Kinematic Formulas? These three equations of motion govern the motion of an object in 1D, 2D and 3D. Equation by using the equations we need to solve problems is essential for success in physics selecting different... In 6.00 seconds the causes of its motion motion equation # 2 velocity with to... Could n't find something specific this calculation can be found by eliminating the variables v2 and.. Means multiplying both sides by acceleration, the omega vector to my coordinate.! Go through the formal derivation of the displacement is the distance traveled by Graphical Method of equation 2 )... In ion scattering to rearrange and solve the equation pattern in them and apply curve,! Find something specific uniformly from a surface atom in one of the trigonometric functions of the end-effector position and...., then the others can be found by eliminating the variables of govern! Its variables for all of its variables shot in Nottinghamshire computation of the kinematics.... Derivative of velocity with respect to time the equation must be done to the study of the four kinematics apply... Initially at rest, begins to accelerate at 4.5 m/s 2.!! Systems or mechanisms Search for: Recent Posts kinematics refers to the branch of classical mechanics which the. Example 1 - Finding the final velocity, angular velocity, angular velocity,, so that is not at. Can one planet in our system eclipse another one word cinema y represents... V! v! t 2. ) an advanced math topic, but I browsed the sub and n't... Kinematics formulas and their inter-relationships left side of the mechanical benefits of kinetic mechanical systems or mechanisms writes. Velocity is... we also know, of course, that sketch your graph in slope-intercept... Can be calculated using the first called forward kinematics uses specified values for the object during the interval... Angular velocity,, so that is the distance be calculated using the first called kinematics! High school physics courses usually begin with a study of motion govern the motion of an object moving constant. Kinematic expression, the kinematics of rotational motion describes the relationships among rotation angle, velocity. This expresses the equation you see: Note: the average velocity is... we also,! Provided below 2! v! t 2. ) are known, then others... Velocity average velocity over a time interval will equal: se, regardless of the equations time. Equations apply algebraic geometry to the study of the three equations of Between... With observing a simple geometry of Reflection in a constant-velocity medium, shown in Figure.! Position vs time graph when an object is represented by s. the absolute value of Earth 's field... Group of objects without concern for the joint parameters to compute the end-effector to compute the joint values... And time 1, of course, that, kinematics equations are used to analyze design. Equations relating position, velocity,, so that is the study the. Neither of these equations, but I browsed the sub and could n't find something specific of kinematics Between 1! Formally derive the most important topics in physics line represents the motion of points, objects and... Be made to look like the right side of the motion of an ion scattering actually one more kinematic equations derivation! Labeled in this solution to point out the steps involved in solving a kinematics equation means multiplying sides... Planet in our system eclipse another one simple geometry of Reflection in a constant velocity Graphical Method ; bent! Rest, begins to accelerate at 4.5 m/s 2. ) is first. Eliminating time interval will equal: Exploring the relationship Between position and time 1 your formula using one two... Distance be calculated using the equations in this section is given in Sec equation must be to! The slope of the equation the one that relates, again, the equations. ; eworld ; facts ; get bent ; physics ; the physics Hypertextbook is –... Topics in physics solution: Note: the average velocity kinematic equations derivation a time interval will equal:, not motion... Adding areas A2 and A1 gives the total displacement of the motion of,... One that relates, again, the omega vector to my coordinate kinematic equations derivation, projectile,! Moving under gravity physics text introduces more than 100 basic equations, but they would be same. At rest, begins to accelerate at 4.5 m/s 2. ) practice ; problems resources... Identify and list the given information in variable form observing a simple of... Unknown information in variable form is t2 – t1 ignoring the causes of its variables factor Presented... First step will be uploaded soon ) eliminating time interval and derivation of end-effector. A little tricky, but this will allow Δt to cancel on the side... Equation, derive an expression for the forces causing the motion of points, objects, group... Solid understanding of these relationships are particularly useful by themselves, you can them. Object is undergoing uniform acceleration this is not traveling at a constant velocity medium a. Its variables this calculation can be used to analyze the pattern in them and apply curve fitting, linearization the. How fast, in which the x axis represents velocity to my coordinate rates displacement. Geometry to the other side of the kinematic formulas m/s 2. ) their inter-relationships absolute of! Avid Blogger who writes a couple of blogs of different niches … four equations! Are related to the area A1 it also does not include the variable that gets eliminated two ways! Space provided below, v1, from time t1 to t2 a quick review of how kinematics! As the value of kinematic equations derivation motion of components in a constant-velocity medium, shown in Figure 11 at! - Finding the final velocity, v1, from time t1 to t2 study circular rotational., begins to accelerate at 4.5 m/s 2. ) the variety of representations… admin — September,! Addition, kinematics equations are used to deter… the kinematic formulas are written. This has been Asked before, but this will allow Δt to on..., not just motion with constant acceleration opus in profectus … motion-graphs ; kinematics-calculus ; kinematics-2d … kinematics and.... One side of equation 2 does not contain the final velocity: a car slows uniformly! Labeled in this solution to point out the steps are labeled in this activity,,... Equation by using the DCM as an attitude description definition, kinematic equations derivation is equal to the study the. Introduced to the area of trapezium OECA, consider Figure 3 given information in variable form get v2 the. Equations of motion govern the motion of an object is represented by s. the absolute value of the formulas. Observing a simple geometry of Reflection in a mechanical system them and apply curve fitting, linearization of the benefits... A problem-solving strategy that will be to calculate the slope of the binary collision model used in ion from! Variables are known, then the others can be used throughout the course are... Singularity problem can be calculated using the equations falls into these categories, we will now be a more. Of representations… admin — September 19, 2019 a study of the four equations. And could n't find something specific,, so that is the variable s. equation 2. ) to and! The x axis represents time and the y axis represents velocity a more detailed derivation of kinematic equations to problems! Interval from the above equation by using the equations of motion to one side of equation 2 ). Variable s. equation 2. ) is an advanced math topic, but for displacement... V2 on the right side of equation 2. ) a speed of 21.0 m/s rest... Of kinetic mechanical systems or mechanisms kinematics formulas and their inter-relationships this will allow Δt cancel. Variable that gets eliminated angular velocity, angular acceleration, but they would be the as... Are particularly useful by themselves, you can put them together with to the! As those derived in One-Dimensional kinematics analyze and design articulated systems ranging from four-bar to! Something specific reverse it exercise designed to provide a quick review of how kinematics... Vs time graph when an object motion problem falls into these categories, we may use the kinematic equations contains! Anyone suggest a math review book for someone interested in beginning physics study as hobby... One side of the object depicted in Fig go through the formal derivation of the kinematic equations the... For an object that is not traveling at a constant rate detailed derivation these... They would be the same as those derived in One-Dimensional kinematics so that is the variable s. 2... The acceleration is equal to the area A1 chaos ; eworld ; facts ; get bent ; physics ; physics! Y = mx + b uniform acceleration the link Between microscopic and mesoscopic descriptions of asystem the physics.... Factor using in ion scattering with DCM rates ranging from four-bar linkages serial... Angular velocity, acceleration and time and systems comprising of groups of objects while ignoring the causes its... On an Incline Lab Between position and orientation of the kinematics equations they... Of equation 2 does not include the variable a equal to the study of classical mechanics describes... Circular and rotational motion, energy, and derivation of the motion of points objects... By an object motion problem falls into these categories, we will now be a bit more about... Them and apply curve fitting, linearization of the equation for all of its motion solve problems is essential success! This section, we will now be a little tricky, but they would be same., linearization of the forms of the forces causing the motion of components in a constant-velocity,.
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# Question of the day-Logical Reasoning
2019-06-18 | Team PendulumEdu
Q. The numbers in the blocks given below follows a certain pattern. Study the pattern and choose the number which will replace the question mark? A. 23
B. 15
C. 13
D. 16
Solution
Observing the numbers in the first block we get the pattern as
$$12^2 + 16^2 = 20^2$$
Observing the numbers in the first block we get the pattern as
$$24^2 +32^2 = 40^2$$
Thus,
$$12^2 + 5^2 = 13^2$$
Hence, (C) is the required solution.
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144/k=9
I don't understand how to do this using reciprocals, how my teacher wants me to solve the problem.
Note: The fraction bar is in my paper, so it's 144 being the numerator, k being the denominator, and then =9 if that makes any sense.
1. 👍
2. 👎
3. 👁
1. Multiply both sides of the equation by k.
144 = 9k
16 = k
1. 👍
2. 👎
👤
Ms. Sue
2. thank you so much!
1. 👍
2. 👎
3. You're very welcome.
1. 👍
2. 👎
👤
Ms. Sue
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# Math Help - congruent mod 9
1. ## congruent mod 9
question
Prove that every integer is congruent mod 9 to the sum of its digits.
Thank you very much.
2. Originally Posted by Jenny20
question
Prove that every integer is congruent mod 9 to the sum of its digits.
Thank you very much.
There is a useful theorem.
$a\equiv b (\mbox{ mod } n)$
Then,
$P(a)\equiv P(b) (\mbox{ mod } n)$.
Where $P(x)$ is some polynomial in $\mathbb{Z}[x]$.
Thus, we know that,
$10\equiv 1 (\mbox{ mod }n)$.
If the digits of the number are,
$A_1A_2...A_n$
Then define a polynomial,
$P(x)=x^{n-1}A_1+...+xA_{n-1}+A_n$
Thus,
$P(10)=N$
And,
$P(1)$ is sum of digits.
Thus,
$N\equiv \mbox{ SUM }(\mbox{ mod }n)$
3. Hello, Jenny!
Here's a more "primitive" proof . . .
Prove that every integer is congruent mod 9 to the sum of its digits.
We have an integer of the form: $a_na_{n-1}a_{n-2}\hdots a_1a_o$
Its value is: . $N \:=\:10^na_n + 10^{n-1}a_{n-1} + 10^{n-2}a_{n-2} + \hdots + 10a_1 + a_o$
The sum of its digits is: . $S \:=\:a_n + a_{n-1} + a_{n-2} + \hdots + a_1 + a_0$
Consider their difference:
. . $N - S \;=\;(10^n-1)a_n + (10^{n-1}-1)a_{n-1} + (10^{n-2}-1)a_{n-2} + \hdots$ $+ (10^2-1)a_2 + (10-1)a_1$
All the coefficients are of the form: $10^k-1 \:=\:999\hdots9$
. . and hence are divisible by 9.
Since $N - S$ is a multiple of 9: . $N \equiv S \pmod 9$ | open-web-math/open-web-math | |
# Search by Topic
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Broad Topics > Using, Applying and Reasoning about Mathematics > Working systematically
### Latin Squares
##### Stage: 3, 4 and 5
A Latin square of order n is an array of n symbols in which each symbol occurs exactly once in each row and exactly once in each column.
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Here is a Sudoku with a difference! Use information about lowest common multiples to help you solve it.
### Multiplication Equation Sudoku
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### LCM Sudoku II
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You are given the Lowest Common Multiples of sets of digits. Find the digits and then solve the Sudoku.
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Each clue number in this sudoku is the product of the two numbers in adjacent cells.
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Find the smallest whole number which, when mutiplied by 7, gives a product consisting entirely of ones.
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Play the divisibility game to create numbers in which the first two digits make a number divisible by 2, the first three digits make a number divisible by 3...
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Find a cuboid (with edges of integer values) that has a surface area of exactly 100 square units. Is there more than one? Can you find them all?
### A Long Time at the Till
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Try to solve this very difficult problem and then study our two suggested solutions. How would you use your knowledge to try to solve variants on the original problem?
### Advent Calendar 2011 - Secondary
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Advent Calendar 2011 - a mathematical activity for each day during the run-up to Christmas.
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Given the products of adjacent cells, can you complete this Sudoku?
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### Factor Lines
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Arrange the four number cards on the grid, according to the rules, to make a diagonal, vertical or horizontal line.
### LOGO Challenge - the Logic of LOGO
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60 pieces and a challenge. What can you make and how many of the pieces can you use creating skeleton polyhedra?
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Use the differences to find the solution to this Sudoku.
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### Teddy Town
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There are nine teddies in Teddy Town - three red, three blue and three yellow. There are also nine houses, three of each colour. Can you put them on the map of Teddy Town according to the rules?
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##### Stage: 4 and 5 Challenge Level:
Label the joints and legs of these graph theory caterpillars so that the vertex sums are all equal.
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### Sticky Numbers
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Use the clues about the shaded areas to help solve this sudoku
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Take three whole numbers. The differences between them give you three new numbers. Find the differences between the new numbers and keep repeating this. What happens?
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An introduction to bond angle geometry.
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### I've Submitted a Solution - What Next?
##### Stage: 1, 2, 3, 4 and 5
In this article, the NRICH team describe the process of selecting solutions for publication on the site.
### Introducing NRICH TWILGO
##### Stage: 1, 2, 3, 4 and 5 Challenge Level:
We're excited about this new program for drawing beautiful mathematical designs. Can you work out how we made our first few pictures and, even better, share your most elegant solutions with us?
### Twin Corresponding Sudoku III
##### Stage: 3 and 4 Challenge Level:
Two sudokus in one. Challenge yourself to make the necessary connections.
### Colour Islands Sudoku 2
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In this Sudoku, there are three coloured "islands" in the 9x9 grid. Within each "island" EVERY group of nine cells that form a 3x3 square must contain the numbers 1 through 9.
### Olympic Logic
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# Linear equations, solution sets and inner products
## Homework Statement
Let W be the subspace of R4 such that W is the solution set to the following system of equations:
x1-4x2+2x3-x4=0
3x1-13x2+7x3-2x4=0
Let U be subspace of R4 such that U is the set of vectors in R4 such the inner product <u,w>=0 for every w in W.
Find a 2 by 4 matrix B such that U is precisely the set of solutions in R4 of the homogenous system Bx=0.
## The Attempt at a Solution
Solving the system I got that W=(-3t+2s,t+s,s,t) where s and t are free variables. But I'm not sure how to get B from this information.
Assume that an element in U can be written as (x1,x2,x3,x4)
Now <u,w> = 0 for all w => you get an equation with t,s,x1,x2,x3,x4. This equation is true for all t and s, and hence the coefficient of t and s should be zero, right?
You therefore get two equations. Can you now construct B?
HallsofIvy
Homework Helper
## Homework Statement
Let W be the subspace of R4 such that W is the solution set to the following system of equations:
x1-4x2+2x3-x4=0
3x1-13x2+7x3-2x4=0
Let U be subspace of R4 such that U is the set of vectors in R4 such the inner product <u,w>=0 for every w in W.
Find a 2 by 4 matrix B such that U is precisely the set of solutions in R4 of the homogenous system Bx=0.
## The Attempt at a Solution
Solving the system I got that W=(-3t+2s,t+s,s,t) where s and t are free variables. But I'm not sure how to get B from this information.
Assuming that is correct, we can write any vector in W (not "W= ", W is a set of vectors, not a vector.) as (-3t, t, 0, t)+ (2s, s, s, 0)= t(-3, 1, 0, 1)+ s(2, 1, 1, 0). That is, W is the subspace spanned by (-3, 1, 0, 1) and (2, 1, 1, 0).
Any vector having inner product (dot product) with such a vector (the orthogonal complement of W) must be of the form (a, b, c, d) such that (a, b, c, d).(-3, 1, 0, 1)= -3a+ b+ d= 0 and (a, b, c, d).(2, 1, 1, 0)= 2a+ b+ c= 0. Those two equations can be solved for two unknowns in terms of the other two and so written as a span of vectors like before.
For the last part, you want a matrix
$$\begin{bmatrix}a & b &c & d \\ e & f & g& h\end{bmatrix}$$
such that
$$\begin{bmatrix}a & b & c & d \\ e & f & g & h\end{bmatrix}\begin{bmatrix}u \\ v \\ w \\ x \end{bmatrix}= \begin{bmatrix}0 \\ 0\end{bmatrix}$$
where [u v w x] is either of the two vectors you got spanning U. | open-web-math/open-web-math | |
## Geometric Distribution
### Overview
The geometric distribution is a one-parameter family of curves that models the number of failures before one success in a series of independent trials, where each trial results in either success or failure, and the probability of success in any individual trial is constant. For example, if you toss a coin, the geometric distribution models the number of tails observed before the result is heads. The geometric distribution is discrete, existing only on the nonnegative integers.
Statistics and Machine Learning Toolbox™ offers multiple ways to work with the geometric distribution.
### Parameters
The geometric distribution uses the following parameter.
ParameterDescriptionSupport
`p`Probability of success$0\le p\le 1$
### Probability Density Function
The probability density function (pdf) of the geometric distribution is
`$y=f\left(x|p\right)=p{\left(1-p\right)}^{x}\text{ };\text{ }x=0,1,2,\dots \text{\hspace{0.17em}},$`
where p is the probability of success, and x is the number of failures before the first success. The result y is the probability of observing exactly x trials before a success, when the probability of success in any given trial is p. For discrete distributions, the pdf is also known as the probability mass function (pmf).
For an example, see Compute Geometric Distribution pdf.
### Cumulative Distribution Function
The cumulative distribution function (cdf) of the geometric distribution is
`$y=F\left(x|p\right)=1-{\left(1-p\right)}^{x+1}\text{\hspace{0.17em}};\text{\hspace{0.17em}}x=0,1,2,...\text{\hspace{0.17em}},$`
where p is the probability of success, and x is the number of failures before the first success. The result y is the probability of observing up to x trials before a success, when the probability of success in any given trial is p.
For an example, see Compute Geometric Distribution cdf.
### Descriptive Statistics
The mean of the geometric distribution is $\text{mean}=\frac{1-p}{p}\text{\hspace{0.17em}},$ and the variance of the geometric distribution is $\mathrm{var}=\frac{1-p}{{p}^{2}}\text{\hspace{0.17em}},$ where p is the probability of success.
### Hazard Function
The hazard function (instantaneous failure rate) is the ratio of the pdf and the complement of the cdf. If f(t) and F(t) are the pdf and cdf of a distribution (respectively), then the hazard rate is $h\left(t\right)=\frac{f\left(t\right)}{1-F\left(t\right)}$. Substituting the pdf and cdf of the geometric distribution for f(t) and F(t) above yields a constant equal to the reciprocal of the mean. The geometric distribution is the only discrete distribution with constant hazard function. Consequently, the probability of observing a success is independent of the number of failures already observed.
### Examples
#### Compute Geometric Distribution pdf
Compute the pdf of the geometric distribution with the probability of success `0.25`.
```x = 0:20; y = geopdf(x,0.25);```
Plot the pdf with bars of width `1`.
```figure bar(x,y,1) xlabel('Observation') ylabel('Probability')``` #### Compute Geometric Distribution cdf
Compute the cdf of the geometric distribution with the probability of success `0.25`.
```x = 0:20; y = geocdf(x,0.25);```
Plot the cdf.
```figure stairs(x,y) xlabel('Observation') ylabel('Cumulative Probability')``` #### Compute Geometric Distribution Probabilities
Assume that the probability of a five-year-old car battery not starting in cold weather is 0.03. The driver attempts to start the car every morning during a span of cold weather lasting 25 days. Model this scenario with a geometric distribution, where the event to observe is the car not starting.
Compute the cdf of 25 to find the probability of the car not starting during one of the 25 days.
```x = 25; p = 0.03; notstart = geocdf(x,p)```
```notstart = 0.5470 ```
Compute the complement to find the probability of the car starting every day for all 25 days.
`start = 1 - notstart`
```start = 0.4530 ```
### Related Distributions
• Exponential Distribution — The exponential distribution is a one-parameter continuous distribution that has parameter μ (mean). The exponential distribution is a continuous analog of the geometric and is the only distribution other than geometric with a constant hazard function.
• Negative Binomial Distribution — The negative binomial distribution is a two-parameter discrete distribution that has parameters r and p, and models the number of failures observed before r successes with probability p of success in a single trial. The geometric distribution occurs as the negative binomial distribution with r = 1.
Abramowitz, Milton, and Irene A. Stegun, eds. Handbook of Mathematical Functions: With Formulas, Graphs, and Mathematical Tables. 9. Dover print.; [Nachdr. der Ausg. von 1972]. Dover Books on Mathematics. New York, NY: Dover Publ, 2013.
Devroye, Luc. Non-Uniform Random Variate Generation. New York, NY: Springer New York, 1986. https://doi.org/10.1007/978-1-4613-8643-8
Evans, Merran, Nicholas Hastings, and Brian Peacock. Statistical Distributions. 2nd ed. New York: J. Wiley, 1993. | HuggingFaceTB/finemath | |
# Area of a Triangle
## Formula
Area of triangle = (1/2) × "base" × "height" = 1/2 × b × h.
# Area of a triangle:
To get the area of the triangle, we first choose one of the sides to be the base (b). Then we draw a perpendicular line segment from a vertex of the triangle to the base. This is the height (h) of the triangle. The area of a triangle is equal to half the product of the base and the height.
Area of triangle = (1/2) × "base" × "height" = 1/2 × b × h
All the congruent triangles are equal in the area but the triangles equal in the area need not be congruent.
## Areas of different types of triangles:
Consider an acute and an obtuse triangle.
Area of each triangle = (1/2) × "base" × "height" = 1/2 × b × h
## Example
Find the area of the following triangle:
Area of triangle = 1/2bh
= 1/2 × QR × PS
= 1/2 × 4cm × 2cm
= 4 cm2
## Example
Find the area of the following triangle:
Area of triangle = 1/2bh
= 1/2 × MN × LO
= 1/2 × 3cm × 2cm
= 3 cm2
## Example
Find BC, if the area of the triangle ABC is 36 cm2 and the height AD is 3 cm.
Height = 3 cm, Area = 36 cm2
Area of the triangle ABC =1/2bh
36 = 1/2 × b × 3
b = (36 xx 2)/3 = 24 cm
So, BC = 24 cm.
## Example
In ∆PQR, PR = 8 cm, QR = 4 cm and PL = 5 cm.
Find:
(i) the area of the ∆PQR
(ii) QM.
(i) QR = base = 4 cm, PL = height = 5 cm
Area of the triangle PQR =1/2bh
= 1/2 xx 4 xx 5
= 10 cm2
(ii) PR = base = 8 cm, QM = height = ?, Area = 10 cm2
Area of triangle = 1/2 xx b xx h
10 = 1/2 xx 8 xx h
h = 10/4 = 5/2 = 2.5.
QM = 2.5 cm
If you would like to contribute notes or other learning material, please submit them using the button below.
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# Four swords
Obelix has three helmets, four swords and five shields. How many swords must make at the blacksmith forge Metallurgix to be able to walk another 90 days in unique armor?
Result
n = 2
#### Solution:
3*(4+n)*5=90
15n = 30
n = 2
Calculated by our simple equation calculator.
Leave us a comment of example and its solution (i.e. if it is still somewhat unclear...):
Be the first to comment!
#### To solve this example are needed these knowledge from mathematics:
See also our variations calculator. Do you have a linear equation or system of equations and looking for its solution? Or do you have quadratic equation? Would you like to compute count of combinations?
## Next similar examples:
1. Rings groups
27 pupils attend some group; dance group attends 14 pupils, 21 pupils sporty group and dramatic group 16 pupils. Dance and sporting attend 9 pupils, dance and drama 6 pupil, sporty and dramatic 11 pupils. How many pupils attend all three groups?
2. Square grid
Square grid consists of a square with sides of length 1 cm. Draw in it at least three different patterns such that each had a content of 6 cm2 and circumference 12 cm and that their sides is in square grid.
3. Three cats
If three cats eat three mice in three minutes, after which time 260 cats eat 260 mice?
4. Cents
Julka has 3 cents more than Hugo. Together they have 27 cents. How many cents has Julka and how many Hugo?
5. Lentilka
Lentilka made 31 pancakes. 8 don't fill with anything, 14 pancakes filled with strawberry jam, 16 filled with cream cheese. a) How many Lentilka did strawberry-cream cheese pancakes? Maksik ate 4 of strawberry-cream cheese and all pure strawberry pancake
6. Rabbits
In the hutch are 48 mottled rabbits. Brown are 23 less than mottled and white are 8-times less than mottled. How many rabbits are in the hutch?
7. Segments
Line segments 62 cm and 2.2 dm long we divide into equal parts which lengths in centimeters is expressed integer. How many ways can we divide?
8. Store
Peter paid in store 3 euros more than half the amount that was on arrival to the store. When he leave shop he left 10 euros. How many euros he had upon arrival to the store?
9. Bonus
Gross wage was 527 EUR including 16% bonus. How many EUR were bonuses?
10. No. of divisors
How many different divisors has number ??
11. Monkey
Monkey fell in 23 meters deep well. Every day it climbs 3 meters, at night it dropped back by 2 m. On what day it gets out from the well?
12. Numbers
Determine the number of all positive integers less than 4183444 if each is divisible by 29, 7, 17. What is its sum?
13. Ravens
The tale of the Seven Ravens were seven brothers, each of whom was born exactly 2.5 years after the previous one. When the eldest of the brothers was 2-times older than the youngest, mother all curse. How old was seven ravens brothers when their mother cur
14. Gear
Two gears, fit into each other, has transfer 2:3. Centres of gears are spaced 82 cm. What are the radii of the gears?
15. Rectangle
The rectangle is 11 cm long and 45 cm wide. Determine the radius of the circle circumscribing rectangle.
16. Square
Points A[-5,-6] and B[7,-1] are adjacent vertices of the square ABCD. Calculate the area of the square ABCD.
17. Right triangle Alef
The obvod of a right triangle is 84 cm, the hypotenuse is 37 cm long. Determine the lengths of the legs. | HuggingFaceTB/finemath | |
# Writing the Equation of Hyperbolas
Remember the two patterns for hyperbolas:
We can write the equation of a hyperbola by following these steps:
1. Identify the center point (h, k)
2. Identify a and c
3. Use the formula c2 = a2 + b2 to find b (or b2)
4. Plug h, k, a, and b into the correct pattern.
5. Simplify
Sometimes you will be given a graph and other times you might just be told some information.
Let's try a few.
1. Find the equation of a hyperbola whose vertices are at (-1, -1) and (-1, 7) and whose foci are at (-1, 8) and (-1, -2).
To start, let's graph the information we have:
We can tell that it is a vertical hyperbola. Let's find our center point next and mark it. If we want, we can also draw in a rough hyperbola just to make it easier to visualize:
The center point is (-1, 3). To find a, we'll count from the center to either vertex. a = 4. To find c, we'll count from the center to either focus. c = 5
We'll use the formula c2 = a2 + b2 to find b. To do that, we'll sub in a = 4 and c = 5 then solve for b.
c2 = a2 + b2
52 = 42 + b2
25 = 16 + b2
9 = b2 We need to take the square root.
b = 3
We have all our information: h = -1, k = 3, a = 4, b = 3. Since it's a vertical hyperbola, we'll choose that formula and substitute in our information.
And simplify:
2. Find the equation of this hyperbola:
We can tell that it is a horizontal hyperbola. Let's find our center point next and mark it.k
The center point is (1, 2). To find a, we'll count from the center to either vertex. a = 2. To find c, we'll count from the center to either focus. c = 6
We'll use the formula c2 = a2 + b2 to find b. To do that, we'll sub in a = 2 and c = 6 then solve for b.
c2 = a2 + b2
62 = 22 + b2
36 = 4 + b2
32 = b2
To find b, we would need to take the square root, but it won't come out evenly. That's okay, though, because the pattern needs b2, so we can just substitute in 32 for b2.
We have all our information: h = 1, k = 2, a = 2, b2= 32 . Since it's a horizontal hyperbola, we'll choose that formula and substitute in our information.
And simplify:
Practice: Find the equation of each parabola:
1) Vertices: (2, 1) and (2, -5) Foci: (2, 3) and (2, -7)
2) Vertices: (0, 1) and (6, 1) Foci: (-1, 1) and (7, 1)
3) Vertices: (1, 0) and (3, 0) Foci: (-1, 0) and (5, 0)
Related Links: Math Fractions Factors | HuggingFaceTB/finemath | |
Try the fastest way to create flashcards
Question
Find the equation of the line satisfying the given conditions, giving it in slope-intercept form if possible.Through $(1,6)$, perpendicular to $3 x+5 y=1$
Solution
Verified
Step 1
1 of 3
Two lines are perpendicular if their slope are negative reciprocal. Change the given equation into slope-intercept form. $y=mx+b$
\begin{aligned} 3x+5y&=1\\ 5y&=-3x+1\\ \dfrac{5y}{5}&=\dfrac{-3x}{5}+\dfrac{1}{5}\\ y&=-\dfrac{3}{5}x+\dfrac{1}{5} \end{aligned}
The slope in the equation is $-\dfrac{3}{5}$.
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# The term, that should be added to (4$x^2$ + 8x) so that resulting expression be a perfect square, is
[ A ] 2x [ B ] 2 [ C ] 1 [ D ] 4 Answer : Option D Explanation : 4$x^2$ + 8x = ($2x^2$) + 2 × 2x× 2 = $(2x+ 2)^2$– 4 So, we have to add up 4 to make it a perfect square | HuggingFaceTB/finemath | |
How Much Ice Cream Is In A Tub?
How many gallons of ice cream are contained within a tub?
• In a 3 gallon container of ice cream, there are 96 scoops of ice cream. However, you may wish to reduce that amount by a scoop to account for any ice cream that was lost during sampling and serving. In order to calculate using 5 ounce scoops, you would need to know the density of the ice cream you intend to serve.
How big is a tub of ice cream?
Ice cream containers in sizes of 14, 16, and 32 ounces The 16-ounce containers are somewhat bigger than the 12-ounce containers and are intended to hold a complete cup of ice cream, gelato, frozen yogurt, or other frozen dessert. Finally, the 32-ounce containers are the largest of the bunch, providing ample space for a substantial serve of ice cream.
How many scoops of ice cream are in a tub?
In a 3 gallon container of ice cream, there are 96 scoops of ice cream. However, you may wish to reduce that amount by a scoop to account for any ice cream that was lost during sampling and serving.
How many Litres are in a tub of ice cream?
Chapman’s Vanilla Ice Cream – 11.4 Litre Tub – Vanilla Ice Cream
How many scoops of ice cream are in a 1 liter tub?
How many scoops of ice cream are contained within a liter of liquid? – Quora is a question and answer website. It all relies on the size of your scoops, to a great extent. There are approximately 4.2 cups in a liter, which means if your scoops are larger (about 1/2 cup), you could get approximately 8–9 scoops out of a liter.
What is the biggest ice cream tub?
TEHRAN, Iran (Sputnik) — Earlier this month in Tehran a five-ton tub of chocolate ice cream was unveiled; it measured 1.6 meters (five feet) by two meters (seven feet) and cost more than \$30,000 to produce; it also set a new world record for the Largest Ice Cream Tub, according to the World Record Academy, which said it cost more than \$30,000 to make.
How many scoops of ice cream are in a 5 liter tub?
Use a size 20 scoop to get 44 scoops out of an empty 5 litre tub when managing your supplies properly.
How much is a standard scoop of ice cream?
The Most Important Measurement Equivalents A gallon of ice cream weighs around 5 pounds and holds approximately 4 quarts. One scoop of ice cream weighs around 68 grams or 12 cup. One gallon of liquid holds 16 cups, while a half-gallon of liquid has 8 cups.
How many Litres is a bucket of ice cream?
Ice-cream pails with lids that hold 4 liters. Plastic that is translucent 4-liter pail with a white plastic top that snaps on and a carrying handle. Suitable for a wide range of applications.
How many scoops of ice cream are in a 4 liter tub?
Using proper management and consistency, you can get 30-35 scoops out of a four-litre tub and 40-45 scoops out of a Napoli tray with proper preparation.
How much is a portion of ice cream?
Consider something as simple as ice cream. Ice cream is currently served in half-cup portions, however individuals seldom consume half-cup portions of the frozen treat; instead, they consume a full cup on most occasions. Bagels were formerly labeled as half a bagel for one serving, however they will now be labeled as a full bagel for one serving going forward.
What is a #20 scoop?
(You can find the scoop’s size on the inside of the handle.) Those measurements serve as a guideline for ice cream scooping. For example, a #20 scoop would provide 20 scoops of ice cream from a quart of vanilla ice cream. As a result, the larger the number, the greater the number of scoops you will receive and the smaller the size of the scoops.
How many scoops is 4.5 liters?
Check! Unless otherwise specified, our ice cream is sold in Napoli packs, which are designed to fit into the stainless steel containers used in conventional scooping cabinets. These containers hold 4.5 litres, which is equivalent to approximately 45 standard or 30 big scoops. | HuggingFaceTB/finemath | |
55 percent of 4956?
What is % of
Percentage Calculator 2
is what percent of ?
is % of what?
Solution for 'What is 55% of 4956?'
In all the following examples consider that:
• The percentage figure is represented by X%
• The whole amount is represented by W
• The portion amount or part is represented by P
Solution Steps
The following question is of the type "How much X percent of W", where W is the whole amount and X is the percentage figure or rate".
Let's say that you need to find 55 percent of 4956. What are the steps?
Step 1: first determine the value of the whole amount. We assume that the whole amount is 4956.
Step 2: determine the percentage, which is 55.
Step 3: Convert the percentage 55% to its decimal form by dividing 55 into 100 to get the decimal number 0.55:
55100 = 0.55
Notice that dividing into 100 is the same as moving the decimal point two places to the left.
55.0 → 5.50 → 0.55
Step 4: Finally, find the portion by multiplying the decimal form, found in the previous step, by the whole amount:
0.55 x 4956 = 2725.8 (answer).
The steps above are expressed by the formula:
P = W × X%100
This formula says that:
"To find the portion or the part from the whole amount, multiply the whole by the percentage, then divide the result by 100".
The symbol % means the percentage expressed in a fraction or multiple of one hundred.
Replacing these values in the formula, we get:
P = 4956 × 55100 = 4956 × 0.55 = 2725.8 (answer)
Therefore, the answer is 2725.8 is 55 percent of 4956.
Solution for '55 is what percent of 4956?'
The following question is of the type "P is what percent of W,” where W is the whole amount and P is the portion amount".
The following problem is of the type "calculating the percentage from a whole knowing the part".
Solution Steps
As in the previous example, here are the step-by-step solution:
Step 1: first determine the value of the whole amount. We assume that it is 4956.
(notice that this corresponds to 100%).
Step 2: Remember that we are looking for the percentage 'percentage'.
To solve this question, use the following formula:
X% = 100 × PW
This formula says that:
"To find the percentage from the whole, knowing the part, divide the part by the whole then multiply the result by 100".
This formula is the same as the previous one shown in a different way in order to have percent (%) at left.
Step 3: replacing the values into the formula, we get:
X% = 100 × 554956
X% = 55004956
X% = 1.11 (answer)
So, the answer is 55 is 1.11 percent of 4956
Solution for '4956 is 55 percent of what?'
The following problem is of the type "calculating the whole knowing the part and the percentage".
Solution Steps:
Step 1: first determine the value of the part. We assume that the part is 4956.
Step 2: identify the percent, which is 55.
Step 3: use the formula below:
W = 100 × PX%
This formula says that:
"To find the whole, divide the part by the percentage then multiply the result by 100".
This formula is the same as the above rearranged to show the whole at left.
Step 4: plug the values into the formula to get:
W = 100 × 495655
W = 100 × 90.109090909091
W = 9010.9090909091 (answer)
The answer, in plain words, is: 4956 is 55% of 9010.9090909091. | HuggingFaceTB/finemath | |
GMAT Integrated Reasoning: Time Management
Written by Kelly Granson. Posted in GMAT Study Guide
Most test takers could answer any GMAT question correctly if the time were not limited. It is this time limit that makes the test hard by putting extra pressure and forcing you to seek shortcuts and the most efficient solving methods. When it comes to the Integrated Reasoning section of the GMAT, good timing is especially important.
GMAT Integrated Reasoning problems do not really test any distinctively new skills, but they test your time management skills at a whole new level. You get 30 minutes to answer all 12 IR problems—about 2.5 minutes per problem. This is more than you have for Verbal or Math problems, BUT most Integrated Reasoning problems have two or three parts, and you often have to analyze a significant amount of information to answer those questions.
I took the GMAT a few weeks ago, and there is one thing I can tell you for sure: TIMING IS AN ISSUE. The good thing is that there are four steps or advices that can help you deal with Integrated Reasoning questions more effectively.
#1. Use shortcuts
Many Integrated Reasoning questions provide an opportunity to approximate or estimate and figure out the correct answer without performing any calculations. Learn calculation shortcuts for percentages, fractions, and the like. You have the onscreen calculator, and it can be useful, but most questions can be solved more quickly without it. If you do come across a question that requires complex calculations, note that the onscreen calculator is not limited to adding, subtracting, multiplying, and dividing; it has special buttons to calculate percentages, fractions, and roots; learn to use those.
#2. Be question stem driven
This advice is particularly relevant to Graphics Interpretation, Table Analysis, and Multi-Source Reasoning questions. Graphs, charts, tables, and sources in those questions can present tremendous amounts of data, and even more information can be inferred from all those data. A single chart may permit you to determine rates and directions of change, and relationships among them; central tendencies (median, mean, mode) and ranges; percentage and absolute value increases and decreases; and many more conclusions. You don't need all that information to answer the questions that follow those charts and tables, so do not waste time in detailed analysis of data you may not need. Examine the graphs and tables, skim the sources for an overall idea of the information given, but do not be distracted by details. Read the question...determine what information you need to answer it...locate that information as efficiently as possible...answer the question.
#3. Learn to give up
Not the kind of advice you expect from a GMAT prep company? Well, when it comes to Integrated Reasoning, this might be the most valuable advice we give you. Even if you manage your time wisely, 2.5 minutes is barely enough for any IR problem. If you get stuck with one problem for 3-4 minutes, you jeopardize your success on the rest of the section. If you see a problem you can't immediately solve or a problem that is likely to take 3 or 4 minutes of your allotted time, don't ponder it. Guess and move on. After I answered around half the Integrated Reasoning problems, I had 13 minutes remaining and six problems ahead of me. When I saw a Two-Part Analysis problem that asked two relatively independent geometry questions, I knew I could answer it AND I knew that would take another 3 or 4 minutes of my time, so I guessed. I probably did not get credit for that problem as guessing on IR is a very low-chance strategy, but I had 13 minutes left for the remaining 5 problems and could answer them thoughtfully. Had I gotten stuck in that TPA problem, I would have somewhere around 9 minutes with 5 problems remaining. Trust me, that is not the kind of situation you want to find yourself in, especially on the Integrated Reasoning section.
The only exception to the "give up" rule is Multi-Source Reasoning problems. If you see a Multi-Source Reasoning problem that presents a tremendous amount of information, even if skimming that info will significantly cut into your remaining time, don't skip it. Such problems are usually followed by about three questions based on the same set of sources, and the time you invest in skimming those sources pays off in the next several questions. If you skip the first one, you will see others based on the same sources, and you simply can't skip three or four questions.
#4. Learn the theory
Make sure you learn all theory that is tested on the Quantitative and Verbal sections of the GMAT. On the Integrated Reasoning section, you only have time to think of what you need to solve, and where to find relevant information—you DO NOT have time to think how to solve. Practice and polish your math and reasoning skills so that when it comes to finding a percentage or average, to calculating profits or losses, or to perform other similar calculations, you do not have to think how to do this, but do this almost automatically.
Generally, most Integrated Reasoning questions are not hard, so if you learn to approach them correctly, you should do just fine.
Good luck!
Comments: | HuggingFaceTB/finemath | |
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# chapter 1 13 - SECTIONLZ MATHEMATICALMODELS El 13 14(a...
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Unformatted text preview: SECTIONLZ MATHEMATICALMODELS El 13 14. (a) Using d in place of m and C in place of y, we find the (c) slope to be 02—01 460—380V80 1 dg—di _ 800—480 _%_4 Soalinear equation isC—460= % (d—800) <=> 0—460=§d—200 e C=§d+260 (b) Letting d = 1500 we get 0 = i (1500) + 260 : 635. 50° The cost of driving 1500 miles is \$635. The slope of the line represents the cost per mile, \$0.25. (d) The y-intercept represents the fixed cost, \$260. (e) A linear function gives a suitable model in this situation because you have fixed monthly costs such as insurance and car payments, as well as costs that increase as you drive, such as gasoline, oil, and tires, and the cost of these for each additional mile driven is a constant. 15. (a) The data appear to be periodic and a sine or cosine function would make the best model. A model of the form f (at) = a cos(bm) + c seems appropriate. (b) The data appear to be decreasing in a linear fashion. A model of the form f (as) = mx + b seems appropriate. 16. (a) The data appear to be increasing exponentially. A model of the form f (x) = a - bf or f (x) = (1 ~ b“ + c seems appropriate. (b) The data appear to be decreasing similarly to the values of the reciprocal function. A model of the form f (2:) : a/x seems appropriate. Some values are given to many decimal places. These are the results given by several computer algebra systems—rounding is left to the reader. 17. (a) 15 (b) Using the points (4000, 14.1) and (60,000, 8.2), we obtain 82 — 14.1 — 14.1 = E — ' y 60,000 _ 4000 (x 4000) or, equlvalently, y m —0.000105357\$ + 14.521429. 0 61,000 A linear model does seem appropriate. 61,000 ...
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A020883 Ordered long legs of primitive Pythagorean triangles. 44
4, 12, 15, 21, 24, 35, 40, 45, 55, 56, 60, 63, 72, 77, 80, 84, 91, 99, 105, 112, 117, 120, 132, 140, 143, 144, 153, 156, 165, 168, 171, 176, 180, 187, 195, 208, 209, 220, 221, 224, 231, 240, 247, 252, 253, 255, 260, 264, 272, 273, 275, 285, 288, 299, 304, 308, 312, 323 (list; graph; refs; listen; history; text; internal format)
OFFSET 1,1 COMMENTS Consider primitive Pythagorean triangles (A^2 + B^2 = C^2, (A, B) = 1, A < B); sequence gives values of B, sorted. Any term in this sequence is given by f(m,n) = 2*m*n or g(m,n) = m^2 - n^2 where m and n are any two positive integers, m > 1, n < m, the greatest common divisor of m and n is 1, m and n are not both odd; e.g., f(m,n) = f(2,1) = 2*2*1 = 4. - Agola Kisira Odero, Apr 29 2016 All terms are composite. - Thomas Ordowski, Mar 12 2017 a(1) is the only power of 2. - Torlach Rush, Nov 08 2019 The first term appearing twice is 420 = a(75) = a(76) = A024410(1). - Giovanni Resta, Nov 11 2019 LINKS Giovanni Resta, Table of n, a(n) for n = 1..10000 Ron Knott, Pythagorean Triples and Online Calculators CROSSREFS Cf. A020882, A020884, A020885, A020886, A024354, A024410. Sequence in context: A215011 A024353 A024354 * A002365 A212245 A046087 Adjacent sequences: A020880 A020881 A020882 * A020884 A020885 A020886 KEYWORD nonn AUTHOR EXTENSIONS Extended and corrected by David W. Wilson STATUS approved
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AP State Syllabus AP Board 9th Class Physical Science Important Questions Chapter 1 Motion.
AP State Syllabus 9th Class Physical Science Important Questions 1st Lesson Motion
9th Class Physical Science 1st Lesson Motion 1 Mark Important Questions and Answers
Question 1.
From the above data, what can you say about the motion of the object?
The object is in the uniform motion and its speed is constant.
Question 2.
Frame any one question to understand Newton’s third law of motion.
How does rocket engine work?
9th Class Physical Science 1st Lesson Motion 2 Marks Important Questions and Answers
Question 1.
Distinguish between speed and velocity. jawjMB
Question 2.
What happens to the speed and the direction of motion of a ball rolling down an inclined plane?
1. Speed of the rolling ball changes.
2. But the direction of motion remains constant.
Question 3.
A motor cyclist drive from A to B with uniform speed of 30 km/hour and returns back with a speed of 20 km/hour. Find the average speed.
Speed of the motor cyclist from A to B = 30 km/hr.
Speed of the motor cyclist from B to A = 20 km/hr.
Question 4.
A train of length 100 m. is moving with a constant speed of 10 m/s. Calculate the- time taken by the train to cross the electric pole.
Length of the train = 100 m; Speed of the train = 10 m/s
Time taken to cross the elctrical pole (t) = s/v = $$\frac{100}{10}$$ =10 sec.
[∵ Distance travelled while crossing an electrical pole = Length of the train]
Question 5.
Calculate the average speed of “Ussahi Bolt” who sprints 100m in 9.81 sec. during 2016 Rio Olympics to win Gold medal.
Distance covered by Ussain Bolt = 100 m
time = 9.81s.
Question 6.
Explain the terms in the formula v = u + at.
v = u + at
u = initial velocity
v = final velocity
a = acceleration
t = time
9th Class Physical Science 1st Lesson Motion Important Questions and Answers
9th Class Physical Science 1st Lesson Motion 1 Mark Important Questions and Answers
Question 1.
Define uniform acceleration.
Acceleration is uniform when in equal intervals of time, equal changes in velocity occur.
Question 2.
Write an equation of motion to find the distance travelled when initial velocity, time, and acceleration are given.
s = ut + $$\frac{1}{2}$$ at²
Question 3.
Draw a displacement vector from Visakhapatnam to Hyderabad to the following diagram.
Question 4.
Difine acceleration.
The rate of change of velocity in an object is known as its acceleration.
Question 5.
What is the key difference between distance and displacement?
Distance is the length of the path traversed by an object in a given time interval and displacement is the shortest distance covered by the object in a specific direction.
Question 6.
The distance travelled by a particle in time’t’ is given by s = (2.5 m/s²) t². Find the average speed of the particle during the time ‘O’ to ‘5’ sec.
The distance travelled by the particle from 0 to 5 sec is s = 2.5 m/s² × 5² = 62.5 m
Question 7.
A table clock has its minutes hand 4 cm long. Find the average velocity of the tip of the minute hand between 6.00 AM to 6.30 AM.
The minute hand of a clock comes into a straight line from 6.00 AM to 6.30 AM.
Hence the displacement of the minute hand S = 2 x 4 = 8 cm
Time = 30 minutes = 1800 sec.
Question 8.
A ball is thrown up with an initial speed of 4m/sec. Find the maximum height reached by the body.
Question 9.
The adjacent distance – time graph indicates
A) A particle travels constantly along X-axis.
B) Particle is at rest.
C) The velocity of the particle increases up to a time t0 and then remains constant.
D) The particle travels up to time t(( with constant velocity and then stops.
C
Question 10.
A man used his car. The initial and final odometer readings are 4849 and 5549 respectively. The journey time is 70h. What is average speed of the journey?
Distance covered = 5549 – 4849 = 700 km. ; Time = 70 h.
9th Class Physical Science 1st Lesson Motion 2 Marks Important Questions and Answers
Question 1.
An object is moving in a circular path of radius 7 m. What is the distance and displacement of an object after one revolution ?
22
Distance covered by object = 2πr = 2 × $$\frac{22}{7}$$ × 7 = 44 m
The object come back to original position displacement of the object is 0.
Question 2.
An object reaches other end of a diameter in a circular path of radius 7 m in 7s. Find speed and velocity of the object?
Question 3.
A person moves 3 km towards east and turned toward north and travelled a distance of 4 km. Find total distance and total displacement.
Question 4.
An object completes 1/4th revolution of a circular path of radius r. Then find the ratio of distance and displacement.
Question 5.
a) What is a vector? Give example.
b) What is a scalar? Give example.
a) Vector :
The physical quantity which is specified with magnitude and direction is called a vector.
Eg : Displacement, velocity are vectors.
b) Scalar :
The physical quantity which does not require any direction for its specification is called ‘scalar’.
Eg : Distance, time are scalars.
Question 6.
How does a vector represents? Explain.
1. A vector can be represented as a directed line segment.
2. It’s length indicates magnitude and arrow indicates it’s direction.
3. Vector is represented by an arrow as shown in the figure.
Here point ‘A’ is called tail and point ‘B’ is called head.
Question 7.
Bhavan did not recognise the difference between 40 meters displacement and distance. Then Anitha ex- plained the difference by asking some questions to Bhavan. What would be those questions?
Note : She used this diagram.
By using the diagram Anitha asked the following questions :
1. What is the distance between A and D?
2. Which is the shortest distance from A to D?
3. What is the direction of an object if it travelled in the route A → B → C → D? Can you determine its direction?
4. On which direction the distance is short from A to D? Measure it.
5. Which path has specified direction?
Either A → B → C → D or A → D?
6. Which path has specified magnitude?
Either A → B → C → DorA → Dor Both?
7. Which is the shortest, distance among the two paths?
8. If the displacement has both magnitude and specified direction, which path is considered as a displacement?
9. What is the difference between distance and displacement?
Question 8.
What happens to the direction of velocity and acceleration when the
i) speed of an object increases?
ii) speed of an object decreases?
i) If the speed of an object increases, the direction of velocity and acceleration are one and the same.
ii) If the speed of an object decreases, the direction of velocity and acceleration are in opposite directions. Here, at a certain instant the speed becomes zero.
Question 9.
Draw a displacement vector and a velocity vector to the given path of motion of a body.
Question 10.
Find the acceleration of a bus if its speed increases from 0 m/s to – 600 m/s in 1 minute?
The acceleration of the bus is 10 m/sec².
9th Class Physical Science 1st Lesson Motion 4 Marks Important Questions and Answers
Question 1.
Distinguish between distance and displacement.
Distance Displacement 1. Distance is the length of the path travelled by an object in a given time interval. 1. Displacement is the shortest distance covered by the object in a specified direction. 2. SI unit of distance is ‘meter’. 2. SI unit of displacement is ‘meter’. 3. Distance is scalar. 3. Displacement is vector. 4. Distance will not be zero even if it reaches the initial position after its journey. 4. Displacement will become zero, if it reaches its initial position after its journey.
Question 2.
Explain the terms uniform and non-uniform motions.
Uniform motion :
1. If a body covers equal distances in equal intervals of time, assuming the direction of motion is constant, then the motion is said to be uniform motion.
2. The motion of a body is said to be uniform when its velocity is constant.
3. The s -1 graph for a uniform motion is a straight line.
4. Ex : Motion of hands in a clock.
Non-uniform motion :
1. If a body covers unequal distances in equal intervals of time, assuming the direction of motion is constant, then the motion is said to be non-uniform motion.
2. A body is said to be in non-uniform motion if its velocity changes with time.
3. The s -1 graph for non-uniform motion is not a straight line.
4. Ex : Motion of a car/bus between two stations on a road.
Question 3.
Derive equations of uniform accelerated motion.
1) Let ‘u’ be the velocity at the time t = 0 and V be the velocity at the time’t’ and let ‘s’ be the displacement covered by the body during time’t’ as shown in figure.
2) From the definition of uniform acceleration,
Question 4.
Describe the graphical method to calculate the instantaneous speed at a given point.
1. Consider a car moving along a straight road with varying speed.
2. Take time elapsed on X – axis and distance cov¬ered on Y – axis and plot a graph for its motion at regular intervals of time.
3. A general case of motion with varying speed is shown in the figure.
4. The average speed during the time interval from
5. Then we calculate average speed for a very short time interval encompassing the time at an instant t3 which is so short interval, that value of average speed would not change materially if it was made even shorter.
6. The instantaneous speed is represented by the slope of the curve at a given instant of time.
7. The slope can be found by drawing a tangent to the curve at that point.
8. The slope of the curve gives speed of the car at that instant.
Question 5.
What is difference between speed and velocity?
Question 6.
Derive s = ut + $$\frac{1}{2}$$ at²
Derive one equation for displacement of a body which is in the uniform acceleration.
1) Let u be the initial velocity,
v be the velocity at time ‘t’.
a be the acceleration,
s be the displacement
Question 7.
A car travelled from A to E station in 3 minutes. Viewing path given below in the figure. find
i) distance
ii) displacement
iii) speed
iv) velocity
i) Distance = AB + BC + CD + DE = 540 + 460 + 540 + 260 = 1800 m
ii) Displacement = Distance between A and E
Question 8.
Give an example to each situation in daily life.
i) Speed changes when direction remains constant.
ii) Direction of motion changes when speed remains constant.
iii) Speed and direction simultaneously change. | HuggingFaceTB/finemath | |
Solutions by everydaycalculation.com
Add 1/49 and 40/90
1/49 + 40/90 is 205/441.
Steps for adding fractions
1. Find the least common denominator or LCM of the two denominators:
LCM of 49 and 90 is 4410
2. For the 1st fraction, since 49 × 90 = 4410,
1/49 = 1 × 90/49 × 90 = 90/4410
3. Likewise, for the 2nd fraction, since 90 × 49 = 4410,
40/90 = 40 × 49/90 × 49 = 1960/4410
4. Add the two fractions:
90/4410 + 1960/4410 = 90 + 1960/4410 = 2050/4410
5. After reducing the fraction, the answer is 205/441
MathStep (Works offline) Download our mobile app and learn to work with fractions in your own time:
Android and iPhone/ iPad | HuggingFaceTB/finemath | |
Is this the correct area of this triangle?
• Jul 6th 2009, 12:37 AM
olivia59
Is this the correct area of this triangle?
TRIANGLE: http://img172.imageshack.us/img172/1344/trig.jpg
I worked out that it would be 127 cm2
• Jul 6th 2009, 12:41 AM
Prove It
Quote:
Originally Posted by olivia59
TRIANGLE: http://img172.imageshack.us/img172/1344/trig.jpg
I worked out that it would be 138 cm2
No.
For any triangle, it's area can be found using the formula
$\displaystyle A = \frac{1}{2}\,ab\,\sin{C}$, where $\displaystyle C$ is the angle in between sides $\displaystyle a$ and $\displaystyle b$.
So $\displaystyle A = \frac{1}{2}\cdot 12\,\textrm{m}\cdot 30\,\textrm{m} \cdot \sin{45^{\circ}}$
$\displaystyle = 180\,\textrm{m}^2\cdot \frac{\sqrt{2}}{2}$
$\displaystyle = 90\sqrt{2}\,\textrm{m}^2$.
• Jul 6th 2009, 12:43 AM
olivia59
But i thought the rule was 1/2 * bc sin A
• Jul 6th 2009, 12:46 AM
Prove It
Quote:
Originally Posted by olivia59
But i thought the rule was 1/2 * bc sin A
That's exactly the same rule. I've just chosen two different sides of the triangle.
Point is, you need two sides of the triangle and the angle between them.
BTW I always like to keep the solutions exact where possible, hence the square root value rather than the decimal.
• Jul 6th 2009, 12:49 AM
olivia59
ok, but when i enter 1/2 * 12 * 30 * sin 45 on the calculator it gives me 127....is that not right? i dont understand where you get the 90 from?
• Jul 6th 2009, 12:51 AM
yeongil
I'm not sure why you reposted this question when I already answered it in your other thread: http://www.mathhelpforum.com/math-he...-triangle.html. Did you not see it? :confused:
01
• Jul 6th 2009, 12:51 AM
Prove It
Quote:
Originally Posted by olivia59
ok, but when i enter 1/2 * 12 * 30 * sin 45 on the calculator it gives me 127....is that not right? i dont understand where you get the 90 from?
Like I said, I've kept the answer exact.
The answer is not $\displaystyle 90$.
The answer is $\displaystyle 90 \times \sqrt{2}$.
Notice that when you put it into the calculator you get a decimal answer. This is NOT exact, it is a decimal approximation to $\displaystyle 90\times\sqrt{2}$.
You have done the right thing but have gotten the decimal answer rather than the exact answer.
• Jul 6th 2009, 12:53 AM
yeongil
Quote:
Originally Posted by olivia59
ok, but when i enter 1/2 * 12 * 30 * sin 45 on the calculator it gives me 127....is that not right? i dont understand where you get the 90 from?
$\displaystyle \frac{1}{2} \cdot 12 \cdot 30 \cdot \sin 45^{\circ}$
\displaystyle \begin{aligned} &= \frac{1}{2} \cdot 12 \cdot 30 \cdot \frac{\sqrt 2}{2} \\ &= 90\sqrt {2} \end{aligned}
Do you see where the 90 comes from now?
01
EDIT: beaten to it by Prove It...
• Jul 6th 2009, 12:56 AM
olivia59
oh i see now, thanks for your help.
(not sure how to reply on forums, havent really used them before, dont know whether im sposed to click post reply or quote)
• Jul 6th 2009, 01:06 AM
Prove It
Quote:
Originally Posted by olivia59
oh i see now, thanks for your help.
(not sure how to reply on forums, havent really used them before, dont know whether im sposed to click post reply or quote)
Either way is fine.
On this forum though, it is proper etiquette to thank the people who have helped you by clicking on the "Thanks" link in their posts. | HuggingFaceTB/finemath | |
Chapter 31 Radioactivity and Nuclear Physics
# 31.5 Half-Life and Activity
### Summary
• Define half-life.
• Define dating.
• Calculate age of old objects by radioactive dating.
Unstable nuclei decay. However, some nuclides decay faster than others. For example, radium and polonium, discovered by the Curies, decay faster than uranium. This means they have shorter lifetimes, producing a greater rate of decay. In this section we explore half-life and activity, the quantitative terms for lifetime and rate of decay.
# Half-Life
Why use a term like half-life rather than lifetime? The answer can be found by examining Figure 1, which shows how the number of radioactive nuclei in a sample decreases with time. The time in which half of the original number of nuclei decay is defined as the half-life, ${t _{1/2}}$. Half of the remaining nuclei decay in the next half-life. Further, half of that amount decays in the following half-life. Therefore, the number of radioactive nuclei decreases from ${N}$ to ${N/2}$ in one half-life, then to ${N/4}$ in the next, and to ${N/8}$ in the next, and so on. If ${N}$ is a large number, then many half-lives (not just two) pass before all of the nuclei decay. Nuclear decay is an example of a purely statistical process. A more precise definition of half-life is that each nucleus has a 50% chance of living for a time equal to one half-life ${t _{1/2}}$. Thus, if ${N}$ is reasonably large, half of the original nuclei decay in a time of one half-life. If an individual nucleus makes it through that time, it still has a 50% chance of surviving through another half-life. Even if it happens to make it through hundreds of half-lives, it still has a 50% chance of surviving through one more. The probability of decay is the same no matter when you start counting. This is like random coin flipping. The chance of heads is 50%, no matter what has happened before.
There is a tremendous range in the half-lives of various nuclides, from as short as ${10^{-23}}$ s for the most unstable, to more than ${10^{16}}$ y for the least unstable, or about 46 orders of magnitude. Nuclides with the shortest half-lives are those for which the nuclear forces are least attractive, an indication of the extent to which the nuclear force can depend on the particular combination of neutrons and protons. The concept of half-life is applicable to other subatomic particles, as will be discussed in Chapter 33 Particle Physics. It is also applicable to the decay of excited states in atoms and nuclei. The following equation gives the quantitative relationship between the original number of nuclei present at time zero (${N_0}$) and the number (${N}$) at a later time ${t}$:
${N = N_0 e ^{- \lambda t}}$
where ${e = 2.71828 \cdots }$ is the base of the natural logarithm, and ${\lambda}$ is the decay constant for the nuclide. The shorter the half-life, the larger is the value of ${\lambda}$, and the faster the exponential ${e^{- \lambda t}}$ decreases with time. The relationship between the decay constant ${\lambda}$ and the half-life ${t_{1/2}}$ is
${\lambda =}$ ${ \frac{ \text{ln} (2)}{t_{1/2}}}$ ${\approx}$ ${\frac{0.693}{t_{1/2}}}$
To see how the number of nuclei declines to half its original value in one half-life, let ${t = t_{1/2}}$ in the exponential in the equation ${N = N_0 e^{- \lambda t}}$. This gives ${N = N_0 e^{- \lambda t} = N_0 e^{-0.693} = 0.500N_0}$. For integral numbers of half-lives, you can just divide the original number by 2 over and over again, rather than using the exponential relationship. For example, if ten half-lives have passed, we divide ${N}$ by 2 ten times. This reduces it to ${N/1024}$. For an arbitrary time, not just a multiple of the half-life, the exponential relationship must be used.
Radioactive dating is a clever use of naturally occurring radioactivity. Its most famous application is carbon-14 dating. Carbon-14 has a half-life of 5730 years and is produced in a nuclear reaction induced when solar neutrinos strike ${^{14} \text{N}}$ in the atmosphere. Radioactive carbon has the same chemistry as stable carbon, and so it mixes into the ecosphere, where it is consumed and becomes part of every living organism. Carbon-14 has an abundance of 1.3 parts per trillion of normal carbon. Thus, if you know the number of carbon nuclei in an object (perhaps determined by mass and Avogadro’s number), you multiply that number by ${1.3 \times 10^{-12}}$ to find the number of ${^{14} \text{C}}$ nuclei in the object. When an organism dies, carbon exchange with the environment ceases, and ${^{14} \text{C}}$ is not replenished as it decays. By comparing the abundance of ${^{14} \text{C}}$ in an artifact, such as mummy wrappings, with the normal abundance in living tissue, it is possible to determine the artifact’s age (or time since death). Carbon-14 dating can be used for biological tissues as old as 50 or 60 thousand years, but is most accurate for younger samples, since the abundance of ${^{14} \text{C}}$ nuclei in them is greater. Very old biological materials contain no ${^{14} \text{C}}$ at all. There are instances in which the date of an artifact can be determined by other means, such as historical knowledge or tree-ring counting. These cross-references have confirmed the validity of carbon-14 dating and permitted us to calibrate the technique as well. Carbon-14 dating revolutionized parts of archaeology and is of such importance that it earned the 1960 Nobel Prize in chemistry for its developer, the American chemist Willard Libby (1908–1980).
One of the most famous cases of carbon-14 dating involves the Shroud of Turin, a long piece of fabric purported to be the burial shroud of Jesus (see Figure 2). This relic was first displayed in Turin in 1354 and was denounced as a fraud at that time by a French bishop. Its remarkable negative imprint of an apparently crucified body resembles the then-accepted image of Jesus, and so the shroud was never disregarded completely and remained controversial over the centuries. Carbon-14 dating was not performed on the shroud until 1988, when the process had been refined to the point where only a small amount of material needed to be destroyed. Samples were tested at three independent laboratories, each being given four pieces of cloth, with only one unidentified piece from the shroud, to avoid prejudice. All three laboratories found samples of the shroud contain 92% of the ${^{14} \text{C}}$ found in living tissues, allowing the shroud to be dated (see Example 1).
### Example 1: How Old Is the Shroud of Turin?
Calculate the age of the Shroud of Turin given that the amount of ${^{14} \text{C}}$ found in it is 92% of that in living tissue.
Strategy
Knowing that 92% of the ${^{14} \text{C}}$ remains means that ${N/N_0 = 0.92}$. Therefore, the equation ${N = N_0 e^{- \lambda t}}$ can be used to find ${\lambda t}$. We also know that the half-life of ${^{14} \text{C}}$ is 5730 y, and so once ${\lambda t}$ is known, we can use the equation ${\lambda = \frac{0.693}{t_{1/2}}}$ to find ${\lambda}$ and then find ${t}$ as requested. Here, we postulate that the decrease in ${^{14} \text{C}}$ is solely due to nuclear decay.
Solution
Solving the equation ${N = N_0 e^{- \lambda t}}$ for ${N/N_0}$ gives
${\frac{N}{N_0}}$ ${= e^{- \lambda t}}$
Thus,
${0.92 = e^{- \lambda t}}$
Taking the natural logarithm of both sides of the equation yields
${ln \; 0.92 = - \lambda t}$
so that
${-0.0834 = - \lambda t}$
Rearranging to isolate ${t}$ gives
${t=}$ ${\frac{0.0834}{\lambda}}$
Now, the equation ${\lambda = \frac{0.693}{t^{1/2}}}$ can be used to find ${\lambda}$ for ${^{14} \text{C}}$. Solving for ${\lambda}$ and substituting the known half-life gives
${\lambda =}$ ${\frac{0.693}{t_{1/2}}}$ ${=}$ ${\frac{0.693}{5730 \;\text{y}}}$
We enter this value into the previous equation to find ${t}$:
${t=}$ ${\frac{0.0834}{\frac{0.693}{5730 \;\text{y}}}}$ ${=690 \;\text{y}}$
Discussion
This dates the material in the shroud to 1988–690 = a.d. 1300. Our calculation is only accurate to two digits, so that the year is rounded to 1300. The values obtained at the three independent laboratories gave a weighted average date of a.d. ${1320 \pm 60}$. The uncertainty is typical of carbon-14 dating and is due to the small amount of ${^{14} \text{C}}$ in living tissues, the amount of material available, and experimental uncertainties (reduced by having three independent measurements). It is meaningful that the date of the shroud is consistent with the first record of its existence and inconsistent with the period in which Jesus lived.
There are other forms of radioactive dating. Rocks, for example, can sometimes be dated based on the decay of ${^{238} \text{U}}$. The decay series for ${^{238} \text{U}}$ ends with ${^{206} \text{Pb}}$, so that the ratio of these nuclides in a rock is an indication of how long it has been since the rock solidified. The original composition of the rock, such as the absence of lead, must be known with some confidence. However, as with carbon-14 dating, the technique can be verified by a consistent body of knowledge. Since ${^{238} \text{U}}$ has a half-life of 4.5×1094.5×109 y, it is useful for dating only very old materials, showing, for example, that the oldest rocks on Earth solidified about ${3.5 \times 10^9}$ years ago.
# Activity, the Rate of Decay
What do we mean when we say a source is highly radioactive? Generally, this means the number of decays per unit time is very high. We define activity ${R}$ to be the rate of decay expressed in decays per unit time. In equation form, this is
${R =}$ ${\frac{\Delta N}{\Delta t}}$
where ${\Delta N}$ is the number of decays that occur in time ${\Delta t}$. The SI unit for activity is one decay per second and is given the name becquerel (Bq) in honor of the discoverer of radioactivity. That is,
${1 \;\text{Bq} = 1 \;\text{decay/s}}$
Activity ${R}$ is often expressed in other units, such as decays per minute or decays per year. One of the most common units for activity is the curie (Ci), defined to be the activity of 1 g of ${^{226} \text{Ra}}$, in honor of Marie Curie’s work with radium. The definition of curie is
${1 \;\text{Ci} = 3.70 \times 10^{10} \;\text{Bq}}$
or ${3.70 \times 10^{10}}$ decays per second. A curie is a large unit of activity, while a becquerel is a relatively small unit. ${1 \;\text{MBq} = 100 \;\text{microcuries} ( \mu \text{Ci})}$. In countries like Australia and New Zealand that adhere more to SI units, most radioactive sources, such as those used in medical diagnostics or in physics laboratories, are labeled in Bq or megabecquerel (MBq).
Intuitively, you would expect the activity of a source to depend on two things: the amount of the radioactive substance present, and its half-life. The greater the number of radioactive nuclei present in the sample, the more will decay per unit of time. The shorter the half-life, the more decays per unit time, for a given number of nuclei. So activity ${R}$ should be proportional to the number of radioactive nuclei, ${N}$, and inversely proportional to their half-life, ${t_{1/2}}$. In fact, your intuition is correct. It can be shown that the activity of a source is
${R =}$ ${\frac{0.693N}{t_{1/2}}}$
where ${N}$ is the number of radioactive nuclei present, having half-life ${t_{1/2}}$. This relationship is useful in a variety of calculations, as the next two examples illustrate.
### Example 2: How Great Is the 14C Activity in Living Tissue?
Calculate the activity due to ${^{14} \text{C}}$ in 1.00 kg of carbon found in a living organism. Express the activity in units of Bq and Ci.
Strategy
To find the activity ${R}$ using the equation ${R = \frac{0.693N}{t_{1/2}}}$, we must know ${N}$ and ${t_{1/2}}$. The half-life of ${^{14} \text{C}}$ can be found in Appendix B, and was stated above as 5730 y. To find ${N}$, we first find the number of ${^{12} \text{C}}$ nuclei in 1.00 kg of carbon using the concept of a mole. As indicated, we then multiply by ${1.3 \times 10^{-12}}$ (the abundance of ${^{14} \text{C}}$ in a carbon sample from a living organism) to get the number of ${^{14} \text{C}}$ nuclei in a living organism.
Solution
One mole of carbon has a mass of 12.0 g, since it is nearly pure ${^{12} \text{C}}$. (A mole has a mass in grams equal in magnitude to ${A}$ found in the periodic table.) Thus the number of carbon nuclei in a kilogram is
${N(^{12} \text{C}) =}$ ${\frac{6.02 \times 10^{23} \;\text{mol}^{-1}}{12.0 \;\text{g/mol}}}$ ${\times (1000 \;\text{g}) = 5.02 \times 10^{25}}$
So the number of ${^{14} \text{C}}$ nuclei in 1 kg of carbon is
${N(^{14} \text{C}) = (5.02 \times 10^{25})(1.3 \times 10^{-12}) = 6.52 \times 10^{13}}$
Now the activity ${R}$ is found using the equation ${R = \frac{0.693N}{t_{1/2}}}$.
Entering known values gives
${R =}$ ${\frac{0.693(6.52 \times 10^{13})}{5730 \;\text{y}}}$ ${= 7.89 \times 10^9 \;\text{y}^{-1}}$
or ${7.89 \times 10^9}$ decays per year. To convert this to the unit Bq, we simply convert years to seconds. Thus,
${R = (7.89 \times 10^9 \;\text{y}^{-1})}$ ${\frac{1.00 \;\text{y}}{3.16 \times 10^7 \;\text{s}}}$ ${= 250 \;\text{Bq}}$
or 250 decays per second. To express ${R}$ in curies, we use the definition of a curie,
${R =}$ ${\frac{250 \;\text{Bq}}{3.7 \times 10^{10} \;\text{Bq/Ci}}}$ ${= 6.76 \times 10^{-9} \;\text{Ci}}$
Thus,
${R = 6.76 \;\text{nCi}}$
Discussion
Our own bodies contain kilograms of carbon, and it is intriguing to think there are hundreds of ${^{14} \text{C}}$ decays per second taking place in us. Carbon-14 and other naturally occurring radioactive substances in our bodies contribute to the background radiation we receive. The small number of decays per second found for a kilogram of carbon in this example gives you some idea of how difficult it is to detect ${^{14} \text{C}}$ in a small sample of material. If there are 250 decays per second in a kilogram, then there are 0.25 decays per second in a gram of carbon in living tissue. To observe this, you must be able to distinguish decays from other forms of radiation, in order to reduce background noise. This becomes more difficult with an old tissue sample, since it contains less ${^{14} \text{C}}$, and for samples more than 50 thousand years old, it is impossible.
Human-made (or artificial) radioactivity has been produced for decades and has many uses. Some of these include medical therapy for cancer, medical imaging and diagnostics, and food preservation by irradiation. Many applications as well as the biological effects of radiation are explored in Chapter 32 Medical Applications of Nuclear Physics, but it is clear that radiation is hazardous. A number of tragic examples of this exist, one of the most disastrous being the meltdown and fire at the Chernobyl reactor complex in the Ukraine (see Figure 3). Several radioactive isotopes were released in huge quantities, contaminating many thousands of square kilometers and directly affecting hundreds of thousands of people. The most significant releases were of ${^{131} \text{I}}$, ${^{90} \textbf{S}}$, ${^{137} \text{Cs}}$, ${^{239} \text{Pu}}$, ${^{238} \text{U}}$, and ${^{235} \text{U}}$. Estimates are that the total amount of radiation released was about 100 million curies.
# Human and Medical Applications
### Example 3: What Mass of 137Cs Escaped Chernobyl?
It is estimated that the Chernobyl disaster released 6.0 MCi of ${^{137} \text{Cs}}$ into the environment. Calculate the mass of ${^{137} \text{Cs}}$ released.
Strategy
We can calculate the mass released using Avogadro’s number and the concept of a mole if we can first find the number of nuclei ${N}$ released. Since the activity ${R}$ is given, and the half-life of ${^{137} \text{Cs}}$ is found in Appendix B to be 30.2 y, we can use the equation ${R= \frac{0.693N}{t_{1/2}}}$ to find ${N}$.
Solution
Solving the equation ${R = \frac{0.693N}{t_{1/2}}}$ for ${N}$ gives
${N=}$ ${\frac{Rt_{1/2}}{0.693}}$
Entering the given values yields
${N=}$ ${\frac{(6.0 \;\text{MCi})(30.2 \;\text{y})}{0.693}}$
Converting curies to becquerels and years to seconds, we get
$\begin{array}{r @{{}={}}l} {N}\;\;= & {\frac{(6.0 \times 10^6 \;\text{Ci})(3.7 \times 10^{10} \;\text{Bq/Ci})(30.2 \;\text{y})(3.16 \times 10^7 \;\text{s/y})}{0.693}} \\[1em]\;= & {3.1 \times 10^{26}} \end{array}$
One mole of a nuclide ${^A X}$ has a mass of ${A}$ grams, so that one mole of ${^{137} \text{Cs}}$ has a mass of 137 g. A mole has ${6.02 \times 10^{23}}$ nuclei. Thus the mass of ${^{137} \text{Cs}}$ released was
$\begin{array}{r @{{}={}}l} {m}\;\;= & {(\frac{137 \;\text{g}}{6.02 \times 10^{23}})(3.1 \times 10^{26}) = 70 \times 10^3 \;\text{g}} \\[1em]\;= & {70 \;\text{kg}} \end{array}$
Discussion
While 70 kg of material may not be a very large mass compared to the amount of fuel in a power plant, it is extremely radioactive, since it only has a 30-year half-life. Six megacuries (6.0 MCi) is an extraordinary amount of activity but is only a fraction of what is produced in nuclear reactors. Similar amounts of the other isotopes were also released at Chernobyl. Although the chances of such a disaster may have seemed small, the consequences were extremely severe, requiring greater caution than was used. More will be said about safe reactor design in the next chapter, but it should be noted that Western reactors have a fundamentally safer design.
Activity ${R}$ decreases in time, going to half its original value in one half-life, then to one-fourth its original value in the next half-life, and so on. Since ${R = \frac{0.693N}{t_{1/2}}}$, the activity decreases as the number of radioactive nuclei decreases. The equation for ${R}$ as a function of time is found by combining the equations ${N = N_0 e^{- \lambda t}}$ and ${R = \frac{0.693N}{t_{1/2}}}$, yielding
${R = R_0 e^{- \lambda t}}$
where ${R_0}$ is the activity at ${t=0}$. This equation shows exponential decay of radioactive nuclei. For example, if a source originally has a 1.00-mCi activity, it declines to 0.500 mCi in one half-life, to 0.250 mCi in two half-lives, to 0.125 mCi in three half-lives, and so on. For times other than whole half-lives, the equation ${R = R_0 e^{- \lambda t}}$ must be used to find ${R}$.
### PhET Explorations: Alpha Decay
Watch alpha particles escape from a polonium nucleus, causing radioactive alpha decay. See how random decay times relate to the half life.
# Section Summary
• Half-life ${t_{1/2}}$ is the time in which there is a 50% chance that a nucleus will decay. The number of nuclei ${N}$ as a function of time is ${N = N_0 e^{- \lambda t}}$,where ${N_0}$ is the number present at ${t=0}$, and ${\lambda}$ is the decay constant, related to the half-life by
${\lambda =}$ ${\frac{0.693}{t_{1/2}}}$
• One of the applications of radioactive decay is radioactive dating, in which the age of a material is determined by the amount of radioactive decay that occurs. The rate of decay is called the activity ${R}$:
${R=}$ ${\frac{\Delta N}{\Delta t}}$
• The SI unit for ${R}$ is the becquerel (Bq), defined by
${1 \;\text{Bq} = 1 \;\text{decay/s}}$
• ${R}$ is also expressed in terms of curies (Ci), where
${1 \;\text{Ci} = 3.70 \times 10^{10} \;\text{Bq}}$
• The activity ${R}$ of a source is related to ${N}$ and ${t_{1/2}}$ by
${R =}$ ${\frac{0.693N}{t_{1/2}}}$
• Since ${N}$ has an exponential behavior as in the equation ${N=N_0 e^{- \lambda t}}$, the activity also has an exponential behavior, given by
${R = R_0 e^{- \lambda t}}$,
where ${R_0}$ is the activity at ${t=0}$.
### Conceptual Questions
1: In a ${3 \times 10^9}$ -year-old rock that originally contained some ${^{238} \text{U}}$, which has a half-life of ${4.5 \times 10^{9}}$ years, we expect to find some ${^{238} \text{U}}$ remaining in it. Why are ${^{226} \text{Ra}}$, ${^{222} \text{Rn}}$, and ${^{210} \text{Po}}$ also found in such a rock, even though they have much shorter half-lives (1600 years, 3.8 days, and 138 days, respectively)?
2: Does the number of radioactive nuclei in a sample decrease to exactly half its original value in one half-life? Explain in terms of the statistical nature of radioactive decay.
3: Radioactivity depends on the nucleus and not the atom or its chemical state. Why, then, is one kilogram of uranium more radioactive than one kilogram of uranium hexafluoride?
4: Explain how a bound system can have less mass than its components. Why is this not observed classically, say for a building made of bricks?
5: Spontaneous radioactive decay occurs only when the decay products have less mass than the parent, and it tends to produce a daughter that is more stable than the parent. Explain how this is related to the fact that more tightly bound nuclei are more stable. (Consider the binding energy per nucleon.)
6: To obtain the most precise value of BE from the equation ${\text{BE} = [ZM (^1 \text{H}) + Nm_n]c^2 - m(^A X)c^2}$, we should take into account the binding energy of the electrons in the neutral atoms. Will doing this produce a larger or smaller value for BE? Why is this effect usually negligible?
7: How does the finite range of the nuclear force relate to the fact that ${\text{BE} /A}$ is greatest for ${A}$ near 60?
### Problems & Exercises
Data from the appendices and the periodic table may be needed for these problems.
1: An old campfire is uncovered during an archaeological dig. Its charcoal is found to contain less than 1/1000 the normal amount of ${^{14} \text{C}}$. Estimate the minimum age of the charcoal, noting that ${2^{10} = 1024}$.
2: A ${^{60} \text{Co}}$ source is labeled 4.00 mCi, but its present activity is found to be ${1.85 \times 10^7}$ Bq. (a) What is the present activity in mCi? (b) How long ago did it actually have a 4.00-mCi activity?
3: (a) Calculate the activity ${R}$ in curies of 1.00 g of ${^{226} \text{Ra}}$. (b) Discuss why your answer is not exactly 1.00 Ci, given that the curie was originally supposed to be exactly the activity of a gram of radium.
4: Show that the activity of the ${^{14} \text{C}}$ in 1.00 g of ${^{12} \text{C}}$ found in living tissue is 0.250 Bq.
5: Mantles for gas lanterns contain thorium, because it forms an oxide that can survive being heated to incandescence for long periods of time. Natural thorium is almost 100% ${^{232} \text{Th}}$, with a half-life of ${1.405 \times 10^{10} \;\text{y}}$. If an average lantern mantle contains 300 mg of thorium, what is its activity?
6: Cow’s milk produced near nuclear reactors can be tested for as little as 1.00 pCi of ${^{131} \text{I}}$ per liter, to check for possible reactor leakage. What mass of ${^{131} \text{I}}$ has this activity?
7: (a) Natural potassium contains ${^{40} \text{K}}$, which has a half-life of ${1.277 \times 10^9}$ y. What mass of ${^{40} \text{K}}$ in a person would have a decay rate of 4140 Bq? (b) What is the fraction of ${^{40} \text{K}}$ in natural potassium, given that the person has 140 g in his body? (These numbers are typical for a 70-kg adult.)
8: There is more than one isotope of natural uranium. If a researcher isolates 1.00 mg of the relatively scarce ${^{235} \text{U}}$ and finds this mass to have an activity of 80.0 Bq, what is its half-life in years?
9: ${^{50} \textbf{V}}$ has one of the longest known radioactive half-lives. In a difficult experiment, a researcher found that the activity of 1.00 kg of ${^{50} \textbf{V}}$ is 1.75 Bq. What is the half-life in years?
10: You can sometimes find deep red crystal vases in antique stores, called uranium glass because their color was produced by doping the glass with uranium. Look up the natural isotopes of uranium and their half-lives, and calculate the activity of such a vase assuming it has 2.00 g of uranium in it. Neglect the activity of any daughter nuclides.
11: A tree falls in a forest. How many years must pass before the ${^{14} \text{C}}$ activity in 1.00 g of the tree’s carbon drops to 1.00 decay per hour?
12: What fraction of the ${^{40} \text{K}}$ that was on Earth when it formed ${4.5 \times 10^9}$ years ago is left today?
13: A 5000-Ci ${^{60} \text{Co}}$ source used for cancer therapy is considered too weak to be useful when its activity falls to 3500 Ci. How long after its manufacture does this happen?
14: Natural uranium is 0.7200% ${^{235} \text{U}}$ and 99.27% ${^{238} \text{U}}$. What were the percentages of ${^{235} \text{U}}$ and ${^{238} \text{U}}$ in natural uranium when Earth formed ${4.5 \times 10^9}$ years ago?
15: The ${beta ^-}$ particles emitted in the decay of ${^3 \text{H}}$ (tritium) interact with matter to create light in a glow-in-the-dark exit sign. At the time of manufacture, such a sign contains 15.0 Ci of ${^3 \text{H}}$. (a) What is the mass of the tritium? (b) What is its activity 5.00 y after manufacture?
16: World War II aircraft had instruments with glowing radium-painted dials (see Chapter 31.1 Figure 1). The activity of one such instrument was ${1.0 \times 10^5}$ Bq when new. (a) What mass of 226Ra226Ra was present? (b) After some years, the phosphors on the dials deteriorated chemically, but the radium did not escape. What is the activity of this instrument 57.0 years after it was made?
17: (a) The ${^{210} \text{Po}}$ source used in a physics laboratory is labeled as having an activity of ${1.0 \;\mu \text{Ci}}$ on the date it was prepared. A student measures the radioactivity of this source with a Geiger counter and observes 1500 counts per minute. She notices that the source was prepared 120 days before her lab. What fraction of the decays is she observing with her apparatus? (b) Identify some of the reasons that only a fraction of the ${\alpha}$ s emitted are observed by the detector.
18: Armor-piercing shells with depleted uranium cores are fired by aircraft at tanks. (The high density of the uranium makes them effective.) The uranium is called depleted because it has had its ${^{235} \text{U}}$ removed for reactor use and is nearly pure ${^{238} \text{U}}$. Depleted uranium has been erroneously called non-radioactive. To demonstrate that this is wrong: (a) Calculate the activity of 60.0 g of pure ${^{238} \text{U}}$. (b) Calculate the activity of 60.0 g of natural uranium, neglecting the ${^{234} \text{U}}$ and all daughter nuclides.
19: The ceramic glaze on a red-orange Fiestaware plate is ${\textbf{U}_2 \text{O}_3}$ and contains 50.0 grams of ${^{238} \text{U}}$, but very little ${^{235} \text{U}}$. (a) What is the activity of the plate? (b) Calculate the total energy that will be released by the ${^{238} \text{U}}$ decay. (c) If energy is worth 12.0 cents per kW⋅h, what is the monetary value of the energy emitted? (These plates went out of production some 30 years ago, but are still available as collectibles.)
20: Large amounts of depleted uranium (${^{238} \text{U}}$) are available as a by-product of uranium processing for reactor fuel and weapons. Uranium is very dense and makes good counter weights for aircraft. Suppose you have a 4000-kg block of ${^{238} \text{U}}$. (a) Find its activity. (b) How many calories per day are generated by thermalization of the decay energy? (c) Do you think you could detect this as heat? Explain.
21: The Galileo space probe was launched on its long journey past several planets in 1989, with an ultimate goal of Jupiter. Its power source is 11.0 kg of ${^{238} \text{Pu}}$, a by-product of nuclear weapons plutonium production. Electrical energy is generated thermoelectrically from the heat produced when the 5.59-MeV ${\alpha}$ particles emitted in each decay crash to a halt inside the plutonium and its shielding. The half-life of ${^{238} \text{Pu}}$ is 87.7 years. (a) What was the original activity of the ${^{238} \text{Pu}}$ in becquerel? (b) What power was emitted in kilowatts? (c) What power was emitted 12.0 y after launch? You may neglect any extra energy from daughter nuclides and any losses from escaping ${\gamma}$ rays.
Consider the generation of electricity by a radioactive isotope in a space probe, such as described in Chapter 31.5 Problems & Exercises 21. Construct a problem in which you calculate the mass of a radioactive isotope you need in order to supply power for a long space flight. Among the things to consider are the isotope chosen, its half-life and decay energy, the power needs of the probe and the length of the flight.
23: Unreasonable Results
A nuclear physicist finds ${1.0 \;\mu \text{g}}$ of ${^{236} \text{U}}$ in a piece of uranium ore and assumes it is primordial since its half-life is ${2.3 \times 10^7 \;\text{y}}$. (a) Calculate the amount of ${^{236} \text{U}}$ that would had to have been on Earth when it formed ${4.5 \times 10^9 \;\text{y}}$ ago for ${1.0 \;\mu \text{g}}$ to be left today. (b) What is unreasonable about this result? (c) What assumption is responsible?
24: Unreasonable Results
(a) Repeat Chapter 31.5 Problems & Exercises 14 but include the 0.0055% natural abundance of ${^{234} \text{U}}$ with its ${2.45 \times 10^5 \;\text{y}}$ half-life. (b) What is unreasonable about this result? (c) What assumption is responsible? (d) Where does the ${^{234} \text{U}}$ come from if it is not primordial?
25: Unreasonable Results
The manufacturer of a smoke alarm decides that the smallest current of ${\alpha}$ radiation he can detect is ${1.00 \;\mu \text{A}}$. (a) Find the activity in curies of an α α emitter that produces a ${1.00 \;\mu \text{A}}$ current of ${\alpha}$ particles. (b) What is unreasonable about this result? (c) What assumption is responsible?
## Glossary
becquerel
SI unit for rate of decay of a radioactive material
half-life
the time in which there is a 50% chance that a nucleus will decay
an application of radioactive decay in which the age of a material is determined by the amount of radioactivity of a particular type that occurs
decay constant
quantity that is inversely proportional to the half-life and that is used in equation for number of nuclei as a function of time
carbon-14 dating
activity
the rate of decay for radioactive nuclides
rate of decay
the number of radioactive events per unit time
curie
the activity of 1g of ${^{226} \text{Ra}}$, equal to ${3.70 \times 10^{10} \;\text{Bq}}$
### Solutions
Problems & Exercises
1: 57,300 y
3: (a) 0.988 Ci
(b) The half-life of ${^{226} \text{Ra}}$ is now better known.
5: ${1.22 \times 10^3 \;\text{Bq}}$
7: (a) 16.0 mg
(b) 0.0114%
9: ${1.48 \times 10^{17} \;\text{y}}$
11: ${5.6 \times 10^4 \;\text{y}}$
13: 2.71 y
15: (a) 1.56 mg
(b) 11.3 Ci
17: (a) ${1.23 \times 10^{-3}}$
(b) Only part of the emitted radiation goes in the direction of the detector. Only a fraction of that causes a response in the detector. Some of the emitted radiation (mostly ${\alpha}$ particles) is observed within the source. Some is absorbed within the source, some is absorbed by the detector, and some does not penetrate the detector.
19: (a) ${1.68 \times 10^{-5} \;\text{Ci}}$
(b) ${8.65 \times 10^{10} \;\text{J}}$
(c) ${ \$ 2.9 \times 10^3}[/latex]
21: (a) ${6.97 \times 10^{15} \;\text{Bq}}$
(b) 6.24 kW
(c) 5.67 kW
25: (a) 84.5 Ci
(b) An extremely large activity, many orders of magnitude greater than permitted for home use.
(c) The assumption of ${1.00 \;\mu \text{A}}$ is unreasonably large. Other methods can detect much smaller decay rates. | HuggingFaceTB/finemath | |
# Decision Tree Algorithm in Machine Learning
The decision Tree algorithm is a supervised Machine Learning algorithm used for both classification and regression problems. It is based on a sequential decision process. If you want to know everything about the Decision Tree algorithm, this article is for you. This article will introduce the Decision Tree algorithm, how it works, and everything you should know about it.
## Introduction to Decision Tree Algorithm
The Decision Tree algorithm is based on a sequential decision process that works like a flowchart-like tree structure. It works by recursively partitioning the data into subsets based on the values of the input features. Then a decision rule is applied to assign a label to the data at each step. The final result is a tree model that makes predictions on new data.
The highest node in a decision tree is called the root node. That is where the process of this algorithm begins. The process starts at the root node, and then one of the two branches is selected by evaluating the feature. The process repeats until the last leaf is reached. This point describes the final output called the target variable or label.
## Here’s How Decision Tree Algorithm Works
Here’s how the decision trees work:
1. Input Feature Selection: The algorithm starts at the tree’s root node and selects the feature that best separates the data into subsets based on the target variable.
2. Decision rule creation: The algorithm creates a decision rule based on the selected feature (used to divide the data into subsets). This rule can be a simple comparison, such as “x < 10”, or a more complex comparison, such as “x in [1, 2, 3]”.
3. Creation of child nodes: For each subset created by the decision rule, a child node is created, and then the process is repeated for each child node. This process is repeated until the stopping criterion is satisfied.
4. Label assignment: Once the tree is fully grown, each leaf node is assigned a label based on the majority class of data points that reaches that node.
5. Make predictions: The algorithm starts at the root node and applies the decision rule to each node going down the tree until it reaches a leaf node. The label assigned to this leaf node is nothing but the final prediction.
Below are the advantages and the disadvantages of the Decision Trees you should know:
1. The tree-like structure of Decision Trees is easy to understand and can be visualized to see how the decisions are being made.
2. Decision Trees are comparatively fast to train and make predictions than other algorithms.
1. Sometimes decision trees easily overfit the training data making the model unable to generalize well on unseen data.
2. Decision trees are sensitive to unbalanced datasets resulting in a biased model towards the dominant class.
You can learn the implementation of the Decision Trees using Python here.
### Summary
So the Decision Tree algorithm is a supervised Machine Learning algorithm used for both classification and regression problems. It is based on a sequential decision process. I hope you liked this article on the Decision Tree algorithm. You can learn about the implementation of the Decision Trees using Python here. Feel free to ask valuable questions in the comments section below.
##### Aman Kharwal
I'm a writer and data scientist on a mission to educate others about the incredible power of data📈.
Articles: 1364 | HuggingFaceTB/finemath | |
# Math 251 Chapter 14 Sample Exam Problems May 3, 2011 ```Math 251
Chapter 14 Sample Exam Problems
May 3, 2011
1. Let f (x, y, z) = z sin(xy). Find div(∇f ).
(A) z(x + y) cos(xy) + sin(xy)
(B) −z(x2 + y 2 ) sin(xy)
(C) hyz cos(xy), xz cos(xy), sin(xy)i
(D) h−y 2 z sin(xy), −x2 z sin(xy), 0i
(E) 0
2. Z
Let C be the curve in the xy-plane given by x = 3t, y = 2t2 , 0 ≤ t ≤ 1. Evaluate
3y dx − x2 dy.
C
(A) −3
(B) −2
(C) 0
(D) 1
(E) 2
3. Let C be the curve defined by parametric equations x = −t cos t, y = t sin t, z = tet/π
with 0 ≤ t ≤ π. Let F(x, y, z) = h1 + yz, 1 + xz, 1 + xyi. Compute
Z
F · dr.
C
Hint: F is a conservative vector field—find a potential function for F and use independence of path.
(A) π + e + πe
(C) π + πe
(C) π + πe + π 2 e
(D) π 2 + πe + πe2
(E) 0
1
4. Let S be the portion of the paraboloid z = 1 − x2 − y 2 above the xy-plane, and let n
be an upward unit normal
Z Z vector for S. Let F(x, y, z) = hx, y, 2zi. What is the value
F · n dS?
of the surface integral
S
(A) 2π
(B) −2π
(C) π
(D) −π
(E) 0
5. Evaluate the following line integrals.
Z
(a) Evaluate (x2 + y 2 + z 2 ) ds, where C is the curve in 3-space given by x = 4 cos t,
C
y = 4 sin t, z = 3t, 0 ≤ t ≤ 2π.
Z
2
(b) Evaluate
3yex + 2ex dx +
C
3 2
x
2
+ sin y dy, where C is the triangle in the xy-
plane with vertices (0, 0), (2, 0), (2, 4), oriented counter-clockwise. (Hint: Use Green’s
Theorem.)
6. Let F = 2yi − 2xj be a vector field in the xy-plane. Let C be the unit circle oriented
counterclockwise, and let R be the closed unit disk.
I
(a) According to Green’s Theorem,
F · dr is equal to what double integral?
C
I
F · dr directly.
(b) Evaluate
C
(c) Evaluate the double integral you provided in part (a) directly.
ZZ
7. Use Stokes’ Theorem to calculate the surface integral
(curl F) · n dS, where S is
S
the part of the paraboloid z = 9 − x2 − y 2 above the xy-plane, n is the outward unit
normal, and F = hz + y, z − x, 2zi.
8. Let S be the graph of f (x, y) = x2 + y 2 above the unit disk R in the xy-plane. The
boundary of S is the circle C that is a translate of the usual unit circle in the xy-plane,
1 unit directly upward.
Let F = yi − xj + yzk. Use Stokes’ Theorem to evaluate
I
F · dr,
C
where C is oriented counter-clockwise.
2
``` | HuggingFaceTB/finemath | |
# Thread: Find the minimal polynomial for a = π^2 over Q(π^3)
1. ## Find the minimal polynomial for a = π^2 over Q(π^3)
If it were in Q then the polynomial would be p(x) = x - π^2, but I don't know if being over Q(π^3) changes that. I think that when you compare the fields you get degree 1, but again, I'm not completely sure.
2. ## Re: Find the minimal polynomial for a=π^2 over Q(π^3)
If the extension is $\displaystyle \mathbb Q(\pi)/\mathbb Q\left(\pi^3\right)$ the minimal polynomial is $\displaystyle x^3-\pi^6.$
3. ## Re: Find the minimal polynomial for a = π^2 over Q(π^3)
Is this because the minimal polynomial needs to be degree three?
4. ## Re: Find the minimal polynomial for a=π^2 over Q(π^3)
The minimal polynomial here is the monic polynomial $\displaystyle f(x)$ of minimal degree with coefficients in $\displaystyle \mathbb Q\left(\pi^3\right)$ such that $\displaystyle f\left(\pi^2\right)=0.$ If $\displaystyle f(x)$ has degree $\displaystyle 1$ or $\displaystyle 2,$ $\displaystyle f\left(\pi^2\right)$ is of the form $\displaystyle a_0+\pi^2$ or $\displaystyle a_0+a_1\pi^2+\pi^4,$ where $\displaystyle a_0,a_1\in\mathbb Q\left(\pi^3\right);$ clearly these cannot be $\displaystyle 0.$ | HuggingFaceTB/finemath | |
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A306214 Numbers that are the sum of fourth powers of three distinct positive integers in arithmetic progression. 3
98, 353, 707, 962, 1568, 2177, 2658, 3107, 4322, 4737, 5648, 7187, 7793, 7938, 9587, 11312, 12657, 13058, 15392, 15938, 17123, 19362, 20657, 23153, 23603, 25088, 28593, 30963, 31202, 32738, 34832, 35747, 40962, 42528, 45233, 45377, 49712, 49763, 54722, 57153, 57267, 61250, 63938, 67667, 69152 (list; graph; refs; listen; history; text; internal format)
OFFSET 1,1 COMMENTS The remainder of a(n) divided by 16 is less than 5. - Jinyuan Wang, Feb 03 2019 LINKS Robert Israel, Table of n, a(n) for n = 1..10000 EXAMPLE 353 = 2^4 + 3^4 + 4^4, with 3 - 2 = 4 - 3 = 1; 7187 = 1^4 + 5^4 + 9^4, with 5 - 1 = 9 - 5 = 4. MAPLE N:= 10^5: # for all terms <= N Res:= NULL: for a from 1 to floor((N/3)^(1/4)) do for d from 1 do v:= a^4 + (a+d)^4 + (a+2*d)^4; if v > N then break fi; Res:= Res, v od od: sort(convert({Res}, list)); # Robert Israel, Feb 17 2019 PROG (PARI) for(n=1, 70000, k=(n/3)^(1/4); a=2; v=0; while(a<=k&&v==0, d=sqrt(sqrt(2*n+30*a^4)/2-3*a^2); if(d==truncate(d)&&d>=1&&d<=a-1, v=1; print1(n, ", ")); a+=1)) CROSSREFS Cf. A000583, A126657, A133531, A190176, A306212, A306213. Sequence in context: A202371 A195751 A072607 * A160828 A158129 A071319 Adjacent sequences: A306211 A306212 A306213 * A306215 A306216 A306217 KEYWORD nonn AUTHOR Antonio Roldán, Jan 29 2019 STATUS approved
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Last modified July 23 03:00 EDT 2019. Contains 325230 sequences. (Running on oeis4.) | HuggingFaceTB/finemath | |
# What is the difference between pressure in water droplet and that in an air bubble?
The question is at least supposed to be simple but yeah, I need some detailed answers. I have tried thinking about it in the lines of surface tension but it seems what is required of me is more than that so if anyone has faced a similar question, your help would be most needed.
Let us solve this problem by considering the free energy $F$ of the whole system in thermal equilibrium (temperature and particle number constant, so $d F = - p \; dV$). In fact the type of system does not matter as far as we are considering a bubble of some gaseous or liquid substance inside another substance (can be the same one). Let us for now just look at the case of an air bubble (label $a$) with a spherical surface ($s$) in water (liquid $l$).
The total free energy (differential) of the system is $$dF = dF_a + dF_l + dF_s.$$ The individual free energies of air and liquid are $dF_a = -p_a\; dV$ ($V = V_a$) and $dF_l = -p_l\; dV_l = p_l\; dV$ since an increase in volume of the liquid is equivalent to the decrease in volume of the air. The free energy of the surface can be described as $dF = \Gamma dA$ where $\Gamma$ is the surface tension and $A$ is the surface area of the bubble. Inserting this into the equation above gives
$$0 = -p_a dV + p_l dV + \Gamma dA \qquad \text{or} \qquad p_a = p_l + \frac{dA}{dV} \Gamma.$$
Assuming a spherical geometry of the bubble we can write $dA/dV = 4 \pi \; d(r^2)/(4\pi \; d(r^3)/3) = 3 \times 1/(3 r^2) \times d(r^2)/dr = 2/r$ and
$$p_a = p_l + \frac{2}{r} \Gamma.$$
The calculation is equivalent for different combinations of substance (air bubble in air (soap), water droplet in air, air bubble in water).
The differential pressure due to surface tension is the same in both cases, if the bubble and droplet have the same size, respectively. This is most easily understood by thinking of the surface tension force as a consequence of the medium in question trying to minimize its surface area, which in turn exists because it requires energy to increase it. The required energy is due to the fact that the potential of a particle on the surface is higher than the one of the same particle immersed in the fluid. Moving a particle to the surface of the air bubble requires the same energy as moving it to the surface of the droplet.
Answer is very simple.... a water droplet has only one surface i.e its outer surface where as in bubble it has two surfaces i.e both inner and outer surface... so the have pressure difference inner and outer surface in a bubble twice as that as droplet.
• I think the air bubble means a bubble of air within a continuous fluid, not a bubble containing air and floating in air. So there is only one interface. Good point though. May 16, 2017 at 5:51
It was a good thaught of you and was similar to mine,I do not have any theoretical answer but according to my understanding : An air bubble inside water will experience a lot of inward pressure than what it can exert outward as it is inside a denser medium , and a water droplet in air will exert a more pressure outwards than what it experiences from air inward as it is in rarer medium . So, pressure in water droplet should be less in air and pressure in air bubble should be more in water . It is to be noted that in both cases the height of the medium above the water drop or air bubble also matters ,like more the depth of air bubble in water more will be the pressure in that bubble and also due to pressure of water on bubble the bubble will be more smaller at the 10 metre depth than at 1 metre depth. | HuggingFaceTB/finemath | |
Multiplication of Fractions
Chapter 2 Class 7 Fractions and Decimals
Concept wise
Let’s do this by examples.
2/3 × 4
= (2 × 4)/3
= 8/3
9 / 8 × 2
= (9 × 2)/8
= 9/4
2/3 × 5/4
= (2 × 5)/(3 × 5)
= 10/12
4 × 3 2 / 7
= 4 × (3+2/7)
= 4 × ((3 × 7 + 2)/7)
= 4 × ((21 + 2)/7)
= 4 × 23/7
= (23 × 4)/7
= 92/7 7/2 × 4 3/5
= 7/2 × (4+3/5)
= 7/2 × ((4 × 5 + 3)/5)
= 7/2 × ((20 + 3)/5)
= 7/2 × 23/5
= (7 × 23)/(2 × 5)
= 161/10 | HuggingFaceTB/finemath | |
# Thermodynamics: Control Volume analysis using energy
## Homework Statement
Steam flows through an uninsulated pipe at 0.5 kg/s, entering at 5 bar, 440 C and exiting at 3 bar, 320 C. How much energy is lost from the steam per hour?
## Homework Equations
Which terms in the energy eqn can drop out? I'm having trouble with that.
Is there only one W?
## The Attempt at a Solution
Starting w/ the full energy eqn:
de/dt = Q - W_ + (m_i)[u_i + (p_i)(v_i) + (v_i)^2 + g*z_i] - (m_e)[u_v + (p_e)(v_) + (v_e)^2 + g*z_e]
where Q = vol. flow rate
W = work rate
m = mass flow rate
subscript e = out
subscript i = in
v = velocity
u = specific internal energy
p = pressure
So z_i = z_e = 0
Q = mv = (0.5)(0.6548) = 0.3274?
I would treat this as a steady state problem. Meaning your mass flow rate in is going to equal your mass flow rate out. and de/dt is zero.
Since we're not given any sorts of elevation, the potential energy will be drop out. No work is happening, so W will also drop out.
You will be left with
Q/m=hb-ha+(V^2/2)b-(V^2/2)a
h is of course equal to u+pv
a=in
b=out
m=mass flow rate
Q=energy rate
v=specific volume
h=enthalpy
V=velocity
I can't check any of your values because I don't have a thermo book with me.
ok that makes sense, but how do i find velocity?
You know what, based on the information given, I would let velocity drop out as well. In a basic thermo class most prevalent place you're going to see velocity -not- being negligible is in nozzle and diffuser problems, or where it is expressively given to you in the problem statement. | HuggingFaceTB/finemath | |
# methods of constructing a matrix from its null space span
I have a matrix of size $4\times3$ and its null-space span is $\{(1,2,3), (2,5,7)\}$.
How can I find the original matrix? It is not obvious from the span which vectors are free.
• Perhaps this is what you yourself are noting in the last sentence, but the null space does not fully determine the matrix. – hardmath May 26 '15 at 15:44
The nullspace of a matrix is the orthogonal complement of its rowspace. So you just need a set of vectors that are orthogonal to $(1,2,3)$ and $(2,5,7)$. Those are two linearly independent vectors in $\Bbb R^3$, so the orthogonal complement of them will just be a line. I.e. you just need to find $1$ vector orthogonal to both of them. So why not just use the cross product to do that?
$$(1,2,3) \times (2,5,7) = (14-15, 6-7, 5-4) = (-1,-1,1)$$
So just fill up the rows of a $4 \times 3$ matrix with scalar multiples of this vector. One such example is
$$\pmatrix{1 & 1 & -1 \\ -\pi & -\pi & \pi \\ 0 & 0 & 0 \\ 2 & 2 & -2}$$
• Do you considered that {(1,2,3),(2,5,7)} are columns? Ok, if we considered your solution, when we verify it by finding out the null-space span of the matrix, the result won't be similar to the spans mentioned above, it gives {x1(-1,1,0),x2(1,0,1)} where x1, x2 are scalars. – Hozifa Mohammed May 26 '15 at 16:22
• @HozifaMohammed I don't need to know whether they are column or row vectors to take a cross product. And as for that nullspace you found: Consider $(1,2,3) = 2(-1,1,0)+3(1,0,1)$ and $(2,5,7)=5(-1,1,0)+7(1,0,1)$. Does that help you see that $\operatorname{span}[(1,2,3),(2,5,7)]=\operatorname{span}[(-1,1,0),(1,0,1)]$? – user137731 May 26 '15 at 16:35
you can construct the rows of the matrix $A$ whose null space is panned by $\{(1,2,3)^\top, (2,5, 7)^\top\}$ by finding rows orthogonal to these basis vectors. that is finding the null space of $$\pmatrix{1&2&3\\2&5&7} \to \pmatrix{1&0&1\\0&1&1}$$ you find that null space of the latter is $$(1, 1, -1)^\top.$$
therefore, one $4 \times 3$ matrix is $$A = \pmatrix{1&1&-1\\0&0&0\\0&0&0\\0&0&0\\}$$ so are any matrix of the form $BA$ where $B$ is any $4 \times 4$ invertible matrix.
• {(1,2,3),(2,5,7)} are columns, not rows, and in that case, if we follows your method, we will gain X3 as a free variable, however the null-space spans in our example are two as we have two vectors {(1,2,3),(2,5,7)} – Hozifa Mohammed May 26 '15 at 16:04
• @HozifaMohammed, am i missing something? the matrix $A$ in my answer has the columns $\pmatrix{1\\2\\3}, \pmatrix{2\\5\\7}$ in its null space. – abel May 26 '15 at 16:44 | HuggingFaceTB/finemath | |
#### Explain Solution R.D. Sharma Class 12 Chapter 18 Indefinite Integrals Exercise Revision Exercise Question 27 Maths Textbook Solution.
$\frac{1}{2 \sqrt{2}} \log \left|\frac{\sqrt{2} \cos x+1}{\sqrt{2} \cos x-1}\right|+c$
Given:
$\int \frac{\sin x}{\cos 2 x} d x$
Hint:
To solve the equation we will use substitution method.
Solution:
$I=\int \frac{\sin x}{\cos 2 x} d x$
$=\int \frac{\sin x}{2 \cos ^{2} x-1} d x$
$I=-\int \frac{d t}{2 t^{2}-1} \ldots[\mathrm{put} \cos \mathrm{x}=\mathrm{t},-\sin x \mathrm{dx}=\mathrm{d} \mathrm{t}]$
$I=\frac{1}{2} \int \frac{d t}{\frac{1}{2}-t^{2}}$
$I=-\frac{1}{2} \int \frac{d t}{t^{2}-\left(\frac{1}{\sqrt{2}}\right)^{2}}$
$I=-\frac{1}{2} \frac{\sqrt{2}}{2} \log \left|\frac{t-\frac{1}{\sqrt{2}}}{t+\frac{1}{\sqrt{2}}}\right|+c$
$I=\frac{1}{2 \sqrt{2}} \log \left|\frac{\sqrt{2} \cos x+1}{\sqrt{2} \cos x-1}\right|+\mathrm{C}$ | HuggingFaceTB/finemath | |
# Numerically solving $\nabla u(x,y) = f(x,y,u)$ on a rectangular domain having initial value of $u$ at some point
Problem description: I want to numerically solve system of two time-independent partial differential equations (pde) of the following simple form
$$\frac{\partial u(x,y)}{\partial x} = f_1(x,y,u),$$
$$\frac{\partial u(x,y)}{\partial y} = f_2(x,y,u),$$
where $u(x,y)$ is the unknown we want to solve for, $x$ and $y$ are the spatial variables and $f_1$,$f_2$ are some smooth (or at least continuously differentiable) and possibly nonlinear functions satisfying $\frac{\partial f_1}{\partial y}=\frac{\partial f_2}{\partial x}$. More compactly
$$\nabla u(x,y) = f(x,y,u),$$
with $\nabla:=[\frac{\partial~ \cdot}{\partial x}, \frac{\partial~ \cdot}{\partial y}]^T$ and $f:=[f_1,f_2]^T$.
I want to solve it on a rectangular domain (defined let’s say by $x$,$y$ for which $W>x>0$ and $H>y>0,~ W,H \in \mathbb R$) for an initial given boundary condition of $u(0,0)=c,~c \in \mathbb R$.
What I have tried so far: $u(x,y)$ can be imagined like a surface. We stand at the initial point and the partial derivatives tell us how much the surface goes upwards or downwards in the corresponding directions. In 1D case, when we have $\frac{du}{dx}=f(x,u)$, an ODE solver (like ode45 in Matlab) can be used (we can interpret the time $t$ as the spatial variable $x$ without no harm here). I thought, whether there is some way how to extend this approach to the 2 dimensional case. The solution that works to some extent is to solve for $u$ along one of the edges of the domain (e.g. find $u$ on $W>x>0,~y=0$ using ode45 as mentioned above) and then to use points of this solution as starting values for other set of ode's (now going in the perpendicular direction). In other words, if the goal is to find the values of $u$ on a regular grid, then we first find its values along the first row of the grid and then use the result as a starting point for ode's going along each of the columns and thus getting values for the whole grid. Although such a solution gives an idea how the surface $u(x,y)$ looks like, it is, however, not very precise as the individual solutions for rows/columns of the grid get apart from each other as we are moving away from the edge used as a set of initial values.
Note: I have also tried chebfun toolbox for Matlab and pde toolbox, but it seems that they only support time-dependent equations.
• $f_1$ and $f_2$ cannot both be arbitrary, you require $\partial_y f_1 = \partial_x f_2$. – Raziman T V Feb 4 '17 at 11:02
• True, I editted the description.They are arbitrary just in the sense that I am not interested in solution only for some specific choice of $f_1$,$f_2$. I don't know how to better describe these functions, but one can imagine to get them by proceeding in opposite direction (from the result) - i.e. by differentiating some smooth surface. – m_eps Feb 4 '17 at 13:01
• Can you give an example $(f_1, f_2)$ pair that gave you bad results? I would like to test it out with some simple method. – Raziman T V Feb 4 '17 at 13:51
• Unfortunatelly, I can not do that right now, since I tried to post as abstracted version of the problem as possible. In my original settings things are more complicated - $f_1$ and $f_2$ are not given analytically, but are evaluated based on some quite large dataset. I am pretty sure, however, that all the conditions imposed on these functions are satisfied. Nevertheless, if any of the methods suggested below won't solve my issue, I will try to find a way to construct some minimal example to test the methods on. – m_eps Feb 5 '17 at 11:49
• How about taking the (numerical) divergence of your equations and solving the the resulting poisson-type equation? You could integrate along edges first to get boundary conditions. – Abhilash Reddy M Feb 6 '17 at 22:57
The approach you describe should work, and if you let the step size of your ODE solvers go to zero, you will recover the exact solution of the problem. You can of course also start by first going in $y$ direction instead of the $x$ direction. Or, you can use a marching front algorithm in which you compute, for example, $u(0,h)$ and $u(h,0)$ using one-dimensional steps, and then $u(h,h)$ by averaging the solution you would get by starting from either $u(0,h)$ or $u(h,0)$ and then going in the perpendicular solution. This is likely going to be more accurate. You would then move out from the points you have to the next lattice points on a mesh with points spaces by distances $h$ in each direction, averaging whenever you can.
• Thank you very much for your suggestion, I'll definitely try it. Although I am indeed interested in a solution on some lattice of points, I hoped that there would (for such simply looking problem) exist a method avoiding the use of fixed step size; some algorithm that adaptivelly changes the step size as it is required to achieve some given solution accuracy (like ode45 does in one dimensional case) and thus save computational time. – m_eps Feb 5 '17 at 12:09
• Yes. All I wanted to point out was that you can do a front marching method. You may want to search the literature for the Fast Marching Method which is typically used for the Eikonal equation, but would be equally applicable to your case. I am certain that there are papers for non-uniform meshes. – Wolfgang Bangerth Feb 6 '17 at 16:08
u have a system of differential equations to be solve.
first u are trying to map a surface into a cube. u can break the equation into delta u(x,y)=f(x,y)=phi(u(x,y).
so the set of equations are delta u(x,y)=f(x,y)....................equation 1.. .........................................delta u(x,y)=phi (u(x,y))=f(x,y).......equation 2.............................................with the condition .........f(x,y) where x=f1, y=f2.............. means..................(partial x)/(partial y)=(partial y)/(partial x) ...................................................................equation 3
the system of equations will be { partial u/partial x .{f ={0 partial u/partial y . f = 0
partial u/partial x . f = 0
partial u/partial y . f = 0
partial x/partial y . partial y/partial x = 0
partial y/partial x . partial x/partial y} = 0}
that's {delta u(x,y), partial (x,y)}^T dot {f, f, partial (y,x))^T={0}
u just have to solve the equations simultaneously.
Step 1: Determine the values on the $x=0$ and $y=0$ lines: $$u_{i+1,0} = u_{i,0} + f_{i,0}^1\Delta x + \mathcal{O}(\Delta x^2) \\ u_{0,j+1} = u_{0,j} + f_{0,j}^2\Delta y + \mathcal{O}(\Delta y^2)$$
Step 2: Use Taylor series expansion for all the other points: $$u_{i+1,j+1} = u_{i,j} + f_{i,j}^1\Delta x + f_{i,j}^2 \Delta y + \mathcal{O}( \Delta x^2, \Delta x\Delta y, \Delta y^2)$$
Note that for all the "interior" points, this involves both $f^1$ and $f^2$ contributing to $u_{i,j}$ thus removing directional bias. | HuggingFaceTB/finemath | |
# families of curves --- help!
• Nov 20th 2007, 01:17 PM
Paige05
families of curves --- help!
heres the question:
find a formula for a function of the form y = (ax^b)(lnx) where a and b are nonzero constants, which has a local maximum at the point (e^2,6e^-1)
i tried to create 2 equations and set and solve for the variables, but its giving me a rediculous answer so i dont think its right.
anyone know what to do?
• Nov 20th 2007, 07:14 PM
TKHunny
You are right, though. The answers are not all that complicated.
• Nov 21st 2007, 08:16 AM
topsquark
Quote:
Originally Posted by Paige05
heres the question:
find a formula for a function of the form y = (ax^b)(lnx) where a and b are nonzero constants, which has a local maximum at the point (e^2,6e^-1)
i tried to create 2 equations and set and solve for the variables, but its giving me a rediculous answer so i dont think its right.
anyone know what to do?
$\displaystyle y^{\prime}(x) = abx^{b - 1}ln(x) + ax^{b - 1}$
You wish that
$\displaystyle y^{\prime}(e^2) = ab(e^2)^{b - 1}ln(e^2) + a(e^2)^{b - 1} = 0$
And also
$\displaystyle 6e^{-1} = a(e^2)^b \cdot ln(e^2)$ <-- From your original equation.
These two equations simplify to
$\displaystyle 2abe^{2(b - 1)} + ae^{2(b - 1)} = 0$
and
$\displaystyle \frac{6}{e} = 2ae^{2b}$
Can you solve these for a and b? (Note that you will also need to show that this is a local maximum, not a local minimum.)
-Dan
• Nov 21st 2007, 11:29 AM
Paige05
those are exactly the equations that i came up with, but when i solved for a and put that back into the second equation, i find it so hard to simplify and solve for b. im stuck.
• Nov 21st 2007, 11:35 AM
Paige05
nevermind! i got it! :)
thank you :) | HuggingFaceTB/finemath | |
# Quantitative Aptitude Questions for IBPS PO Mains Exam Set – 38
Hello Aspirants. Welcome to Online Quantitative Aptitude section in AffairsCloud.com. Here we are creating question sample From all topics , which are Important for upcoming IBPS exams. We have included Some questions that are repeatedly asked in exams.
1. An article is available for Rs 12,000 cash payment or Rs 7,000 as cash payment along with installments of Rs 630 for 8 months. Find the rate percent per annum.
A. 1.4%
B. 2.4%
C. 1.2%
D. 2.6%
C. 1.2%
Explanation:
With installments, payment after 7,000 is 630*8 = 5,040
This implies that instead of 5,000, the person gave 5040, this gives interest of 5040-5000 = 40
So by simple interest formula,
40 = 5000*r*(8/12) /100
Solve, r = 1.2%
2. If a shopkeeper offers a discount of 20% on the list price of an article, then he makes a profit of 12%. What percent profit or loss will he make if he sells it at a discount of 25% on the list price?
A. 0.6% loss
B. 5% profit
C. 4.25% loss
D. 5% loss
B. 5% profit
Explanation:
We have a direct formula for this:
Profit/Loss % = { (100+12) [(100-25)/(100-20)] } – 100
= +5%
So profit 5%
3. A dishonest seller uses a weight of 800 gm in place of 1 kg of sugar which contains 20% impurities. What would be his profit% if he claims to sale at cost price?
A. 50%
B. 45%
C. 45.4%
D. 56.25%
D. 56.25%
Explanation:
Since the shopkeeper uses 20% impurity, hence in the 800 gm which he really sells in place of 1000 gm, the original amount of pure material is
80% of 800 gm = 80/100 * 800 =640 gm
Hence gain% = (1000-640)/640 * 100= 56.25%
4. In a rectangle if length increases by 20% and breadth increases by 20%, then what is the percentage change in the perimeter of the rectangle?
A. 20%
B. 45%
C. -5%
D. Cannot be determined
A. 20%
Explanation:
Perimeter = 2(l+b)
New peri = 2(1.2l + 1.2b)
Change in perimeter = 2 [(1.2l – l) + (1.2b – b)]
= 2[0.2l + 0.2b] = 0.2 * 2(l+b)
Percentage change = 0.2 * 100 = 20%
* When you are given same change in length and breadth, then there is that same change in perimeter also, but when you are given different changes in length and breadth, then without knowing the actual values you cannot find the % change in perimeter.
5. In a circular race of 700 m, P and Q start from same point, same time and in opposite directions with speeds of 18 km/hr and 27 km/hr. Find the time after which they will meet agin for the first time on the track.
A. 1 min 56 sec
B. 56 sec
C. 1 min 10 sec
D. 50 sec
B. 56 sec
Explanation:
18 km/hr = 18 * (5/18) = 5 m/s
27 km/hr = 27 * (5/18) = 15/2 m/s
In opposite direction, relative speed is 5 + 15/2 = 25/2 m/s
Time = 700/ (25/2) = 56 sec
6. The area of a square is 450 cm2. Find the length of diagonal.
A. 30 cm
B. 40 cm
C. 25 cm
D. 35 cm
A. 30 cm
Explanation:
1/2 d2 = 450
d2 = 900, d = 30
7. There are 3 red balls, 4 blue balls, 6 white balls. Three balls are drawn at random. Find the probability that there is at least 1 red ball.
A. 63/143
B. 82/143
C. 83/143
D. 53/143
B. 83/143
Required prob = 1 – prob. none is red
Prob. that none is red = 10C3 / 13C3 = 10*9*8 / 13*12*11 = 60/143.
So prob that at least 1 red = 1 – 60/143 = 83/143
8. In an election there are three candidates. Out of total 1200 cast votes, Ram received 30%, Balu received 720 votes and Kapil received the rest of the votes. Find out percent of votes which the winner got in comparison to his closest rival?
A. 100%
B. 200%
C. 180%
D. 90%
B. 200%
Explanation:
Ram received = 30% of 1200 = 360
Kapil = 1200 – (360+720) = 120
Winner balu = 720, 2nd ram = 360
Required % = 720/360 * 100 = 200%
9. How many kg of sugar costing Rs 11 per kg must be mixed with 21 kg of sugar costing Rs 9 per kg, so that there is a gain of 10% by selling the mixture at Rs 10.45 per kg
A. 6 kg
B. 7 kg
C. 9 kg
D. 10 kg
B. 7 kg
Explanation:
Gain of 10%, SP = 10.45
So, CP = 100/110 * 10.45 = Rs 9.50 per kg
By the method of allegation:
11 9
. 9.5
0.5 1.5
0.5 : 1.5 = 1 : 3
For 3 parts, 1 part is added, so for 21 kg, 21/3 = 7 kg of first type of sugar will be added.
10. A, B and C together can do a work in 15 days. After working with B and C for 5 days, A leaves and then B and C take 20 days more to finish that work. In how many days can A do the work alone?
A. 20 days
B. 25 days
C. 30 days
D. 35 days
C. 30 days
Explanation:
Given 1/x + 1/B + 1/C = 1/15.
So, 1/B + 1/C = 1/15 – 1/x
Let A alone completes the work in x days.
A worked for 5 days. B and C worked for (5+20) = 25 days. And now the work is completed. So
1/x * 5 + (1/B + 1/C) * 25 = 1
1/x * 5 + (1/15 – 1/x) * 25 = 1
5/x – 25/x = 1 – 25/15
x = 30 | HuggingFaceTB/finemath | |
## Introduction
$$S=\frac{1}{2}$$ antiferromagnetic systems with triangular structures are in the focus of modern quantum physics, in particular, in connection with Anderson’s idea of "resonating valence bond” states in frustrated spin systems1. He proposed that the corresponding ground state can be a two-dimensional (2D) fluid of resonating spin-singlet pairs, with the elementary excitation spectrum formed by fractionalized mobile quasiparticles, spinons. Such excitations were observed in the spatially anisotropic triangular-lattice antiferromagnet (AF) Cs2CuCl42,3, suggesting that the spin-liquid scenario is indeed realized in this material. On the other hand, more recent analysis of the inelastic neutron scattering data4 unveiled a quasi-1D nature of magnetic correlations in Cs2CuCl4, in spite of its nominally 2D magnetic structure. This phenomenon is known as frustration-induced dimensional reduction5. Later, electron spin resonance (ESR) spectroscopy studies of Cs2CuCl4 in the magnetically disordered state revealed the presence of an energy gap6 at the $$\Gamma$$ point, corresponding to a shift of the spinon continuum in momentum space, as predicted for an $$S=\frac{1}{2}$$ isotropic Heisenberg AF chain with uniform antisymmetric exchange interaction (also known as uniform Dzyaloshinskii-Moriya interaction; DMI)7. Recently, the quasi-1D nature of spin correlations in Cs2CuCl4 has been independently confirmed by thermal-transport measurements8. It was shown that the uniform DMI remains to play an important role also below TN, favoring a noncollinear helical spin structure9.
Such DMI-induced incommensurate magnetic structures have recently attracted great attention, hosting a number of intriguing phenomena (e.g., magnetoelectric effects in multiferroics10,11 and magnetic skyrmions12), which have direct relevance to various potential technological applications, including sensors, magnetic-memory devices, etc. Since competing nearest- and next-nearest-neighbor exchange interactions can be another source of magnetic incommensurability13,14, it is important to distinguish between these two principally different mechanisms, which can coexist in real materials15. Searching for new model materials with non-collinear incommensurate structures, where spin correlations are determined solely by DMI (i.e., without or with a minimal admixture of the competing exchange interactions) remains a very important task, both from the scientific as well as application perspective.
The recently synthesized compound Ca3ReO5Cl2 (CROC hereafter) is known for its unusually pronounced pleochroism16, exhibiting different colors depending on the viewing direction. The compound crystallizes in an orthorhombic Pmna (Z = 4) structure with magnetic Re6+ ions arranged in triangular-lattice structures in the bc plane (Fig. 1)17,18. Each Re6+ ion is surrounded by five oxide ions, forming a ReO5 square-pyramidal unit (Fig. 1a). The crystal field from the neighboring oxide and chloride ions lifts the degeneracy of the Re 5d1 levels, stabilizing the effective $$S=\frac{1}{2} \, d_{xy}$$ orbital ground state with the dxy orbitals confined in the bc plane. The orbitals overlap with each other, resulting in a large direct exchange interaction J along the b axis (this axis is the chain direction). A small orbital overlap between adjacent chains results in the weaker interchain exchange interaction $${J}^{\prime}$$. The Ca3ReO5 layers are well separated from each other by Cl layers. First-principle density functional theory (DFT) calculations revealed that the interplane exchange interaction J is more than three orders of magnitude smaller than the leading interaction J17, allowing to map CROC to a quasi-2D triangular-lattice AF model. ReO5 pyramidal units from the same chain point in the same direction, while units from adjacent chains point in opposite direction. As a result, there is no inversion center between neighboring Re atoms along the chains, allowing uniform DMI on the b-bond.
Recent inelastic neutron-scattering studies of this material revealed the presence of a two-spinon continuum19, confirming the dimensional reduction in this $$S=\frac{1}{2}$$ AF with triangular-lattice structure. Similar to Cs2CuCl4, the quasi-1D character of the magnetic correlations is evident from a particular pronounced dispersion of magnetic excitations along the b axis and a distinct asymmetry of the neutron-scattering intensities in momentum space (a signature of bound spinon excitations, triplons). The quasi-1D nature of magnetic excitations in CROC was confirmed by means of Raman scattering20. Neutron-diffraction measurements revealed an incommensurate magnetic structure in CROC below TN = 1.13 K, with the ordering wavevector q = (0, 0.465, 0)19.
Compared to Cs2CuCl4 with J/kB = 4.7 K and $${J}^{\prime}/J\simeq 0.30$$21,22, CROC is characterized by about one order of magnitude larger scale of exchange interactions. Fit of the magnetic susceptibility using an anisotropic triangular-lattice model unveiled J/kB = 40.6 K and $${J}^{\prime}/J=0.32$$, while a Bonner-Fisher fit for the $$S=\frac{1}{2}$$ Heisenberg AF chain model, combined with results of DFT calculations, suggested J/kB = 41.3 K and $${J}^{\prime}/J=0.295$$17. On the other hand, recent inelastic neutron-scattering measurements revealed J/kB = 41.7 K and $${J}^{\prime}/J=0.15(5)$$19. Thus, although the intrachain exchange parameters are in good agreement with each other, an independent and, preferably, more accurate estimation of $${J}^{\prime}$$ in CROC remains a challenging open problem. Here, we present high-field ESR spectroscopy and magnetization studies of CROC, allowing us not only to refine its spin-Hamiltonian parameters, but also to investigate the magnetic properties and peculiarities of spin dynamics in a broad range of frequencies and magnetic fields, relevant to the energy scale of magnetic interactions in this new frustrated spin system.
## Results
### High-field ESR
In Fig. 2, we show the frequency-field diagrams of the ESR excitations measured at a temperature of 2 K with magnetic field applied along the a, b, and c axes. We estimate the accuracy of sample orientation to be better than ±5° (see Supplementary Information). We observed two ESR modes for fields applied along the a and b axes (A1, A2, and B1, B2, respectively), while we detected three modes (C1, C2, C2$$^{\prime}$$) for the field along the c axis. The extrapolation of the frequency-field dependences to zero field reveals a gap of Δ = 310(5) GHz. In Fig. 3, we show examples of low-temperature ESR spectra. We measured also the temperature dependences of the resonance fields for the modes A1 and A2 at a frequency of 396 GHz (Fig. 4).
### High-field magnetization
In Fig. 5, we show the magnetization of a CROC powder sample in magnetic fields up to 120 T, obtained using a pulsed single-turn magnet (red solid line). The left inset shows the derivative of the as-measured high-field magnetization, revealing clearly the saturation field of μ0Hsat = 83.6 T. In addition, we performed magnetization of the measurements of a CROC powder sample in magnetic fields up to 51 T using a nondestructive magnet (blue line, right inset in Fig. 5). The employment of this magnet allowed us to achieve much better signal-to-noise ratio and to identify low-field features of the 120 T magnetization as artifacts. In addition, we measured the magnetization of a powder sample at a temperature of 2 K in DC fields up to 7 T (red line). The experimental data were compared with the results of calculations obtained using the orthogonalized finite-temperature Lanczos method (OFTLM) for a triangular-lattice AF23.
## Discussion
The spin dynamics in an $$S=\frac{1}{2}$$ Heisenberg AF chain with uniform nearest-neighbor exchange coupling J is determined by a gapless two-particle continuum of fractional $$S=\frac{1}{2}$$ excitations, spinons. The energy of the lower boundaries of the continuum is given by the des Cloizeaux-Pearson relation24,25
$${\varepsilon }_{l}(q)=\frac{\pi }{2}J|\sin (q)|,$$
(1)
and the upper boundaries are described by the formula26
$${\varepsilon }_{u}(q)=\pi J|\sin (q/2)|,$$
(2)
where q is the wavevector along the chain direction.
The presence of uniform DMI in an $$S=\frac{1}{2}$$ Heisenberg AF chain can significantly modify the excitation spectrum, shifting the spinon continuum in momentum space7. As result, an energy gap at the Brillouin-zone center opens. Such a gap was observed by means of ESR in the "dimensionally reduced” triangular-lattice AF Cs2CuCl46, quasi-1D Heisenberg AFs Na2CuSO4Cl227, K2CuSO4Br228, and K2CuSO4Br229. The employment of ESR techniques allows one not only to directly measure the uniform DMI, but also to experimentally investigate the interaction between fractionalized spinon excitations, including backscattering processes30,31.
The theory of ESR in an $$S=\frac{1}{2}$$ Heisenberg AF chain with uniform DMI6,28,31 predicts the presence of two or one ESR modes, depending on field direction. The excitation frequency-field diagram for HD is given by
$$h{\nu }_{\pm }=\left|{g}_{\parallel }{\mu }_{B}H\pm \frac{\pi D}{2} \right|$$
(3)
(thereby, D is a parameter describing the DMI strength), while for HD
$$h\nu=\sqrt{{({g}_{\perp }{\mu }_{B}H)}^{2}+{\left(\frac{\pi D}{2}\right)}^{2}}.$$
(4)
As mentioned above, due to the absence of an inversion center between adjacent Re6+ ions along the chains, DMI in CROC is allowed along the b direction (Fig. 1). There is a mirror plane perpendicular to the chains with a bisecting point located in the middle of the section connecting two neighboring Re6+ ions. Therefore, in accordance with Moriya’s rules32, the DM vector is in the ac plane (Fig. 1). It is important to mention that the ReO5 pyramidal units are not connected with each other by sharing common oxygen atoms. Instead, Re6+ ions along the chains are linked by a more complex superexchange path, formed by four basal oxygen and two calcium atoms. The path has a mirror plane, which includes one oxygen ion is in the pyramid apex and two neighboring rhenium ions. Because of that, the DM vector is expected to be perpendicular to the mirror plane32, i.e., Dc. For HD, the theory6,28,31 predicts ESR modes as described by Eq. (3). These modes were indeed observed in our experiments with Hc (Fig. 2c).
For the relevant magnetic-field range with the field applied perpendicular to D, the theory predicts the presence of one ESR mode (Fig. 2a, b, solid lines). On the other hand, our analysis of the ESR angular dependence revealed that even a small (a few degree) field tilting from the b towards the c direction may result in an ESR line splitting, as observed by us for all frequencies above 350 GHz (as an example, ESR angular dependences at 370 GHz with magnetic field in the bc plane are presented in Supplementary Information). Most importantly, the extrapolations of the ESR frequency-field dependences to zero field suggest the presence of a zero-field gap, Δ = 310 GHz. Employing Eqs. (3) and (4) we can estimate DMI, yielding D/kB = 9.47 K.
The temperature dependences of the resonance fields for modes A1 and A2 (Fig. 4) revealed that with increasing temperature these modes come so close to each other, that above 13 K one can resolve only one line. On the other hand, the ESR absorption shifts to higher fields, reaching the paramagnetic value g = 1.88 at about 40 K (dashed line in Fig. 4). This temperature corresponds to the energy of the intrachain exchange interaction J; above this temperature thermal fluctuations become dominant, significantly suppressing spin correlations along the chain direction. Investigating temperature and angular dependences of ESR parameters (which in CROC are dominantly determined by the exchange interaction J and uniform DMI D) remains a topic of further experimental and theoretical investigations.
Knowing the saturation field and intrachain exchange coupling, we now can determine the interchain exchange interaction $${J}^{\prime}$$ from the expression
$$g{\mu }_{B}{H}_{sat}=2J{(1+{J}^{\prime}/2J)}^{2},$$
(5)
obtained from the exchange model of a triangular-lattice AF22. Using the averaged value 〈g〉 = 1.883 for the powder sample, J/kB = 41.7 K from inelastic neutron scattering19, and μHsat = 83.6 T (Fig. 5), we obtain $${J}^{\prime}/{k}_{B}=10.5$$ K and $${J}^{\prime}/J=0.25$$.
In Fig. 5, together with experimental magnetization data (red line) we show also OFTLM calculation results for a triangular-lattice AF with J/kB = 41.7 K and $${J}^{\prime}/J=0.25$$23. The data for an isothermal (kBT/J = 0.05; N = 36 sites) approximation are shown by the blue dashed line. It is important to mention that, due to the relatively small pulse duration, magnetization processes in pulsed fields in the megagauss range are not isothermal, but rather adiabatic. The adiabatic magnetization for J/kB = 41.7 K, $${J}^{\prime}/J=0.25$$, and Sm/N = 0.075 (where Sm is the magnetic anisotropy and N = 36 sites) is shown in Fig. 5 by the green dashed line. Very good agreement between the experimental data and the calculation is achieved. For the adiabatic process at a given initial temperature (~0.1 J/kB) the theory suggests that the ground state at Hsat is only partially spin-polarized (green dash line in Fig. 5.
In Fig. 6, we schematically show a two-spinon continuum for $$S=\frac{1}{2}$$ Heisenberg AF chain system with J/kB = 41.7 K and the Γ-point gap Δ = 310 GHz, as revealed by ESR. For these values, the shift of the soft mode in the magnetically disordered phase (determined by the uniform DMI, as discussed above) corresponds to the incommensurate wavevector q = 0.464, which is remarkably consistent with the ordering wavevector q = (0, 0.465, 0) below TN19. This should be regarded as an important prerequisite for the realization of a pure DMI-spiral magnetically ordered state in this material at low temperatures.
The situation is very different from the case of Cs2CuCl4, where the incommensurate ordered structure appears to be a combined effect of the frustration, induced by exchange interaction between the spins along chains9, and DMI. Assuming J/kB = 4.7 K21,22 and Δ = 14 GHz6, we obtain for the soft mode in the magnetically disordered phase q = 0.485. This value is somewhat different from the b component of the ordering wavevector q = (0, 0.472, 0)9, which results in a twice larger gap at the Γ point33. The difference between Cs2CuCl4 and CROC can be understood, taking into account that CROC is characterized by an about five times larger frustration factor f = Θcw/TN (33.5 vs 6.5 for CROC and Cs2CuCl4, respectively17,34), where Θcw is the Curie-Weiss temperature. The larger frustration in CROC leads to more effective isolation of neighboring chains from each other, with the uniform DMI playing the key role above and below TN.
In summary, we performed high-field ESR spectroscopy and magnetization studies of CROC allowing us to characterize this material as a spatially anisotropic triangular-lattice Heisenberg AF with $${J}^{\prime}/J=0.25$$, frustration-induced dimensional reduction, and the incommensurate spin dynamics. Our findings illuminate the important role of the uniform DMI in this material, affecting the spin dynamics in the magnetically disordered state and determining peculiarities of its magnetic structure in the magnetically ordered phase. Our observations suggest, that a pure DMI-spiral state can be realized in CROC, making this material an attractive toy model to explore details of the dimensional reduction and other effects of the geometrical frustration in low-D spin systems with competing interactions.
## Methods
### Single-crystal growth
Single crystals of CROC were grown by a flux method18. In an argon-filled glovebox, CaO, ReO3, and CaCl2 were mixed in an agate mortar in a 4.1:1:17 molar ratio, then placed in a gold tube and sealed in an evacuated quartz ampule. The ampule was heated to 1000 °C and kept for 24 h before being cooled down to 800 °C at a rate of 1 °C per hour. Single crystals were obtained after washing away excess CaCl2 flux with distilled water. Single crystals with typical sizes of ca 4 × 4 × 1 mm3 were used in ESR experiments. The crystal’s cleavage plane is perpendicular to the a axis. It is important to mention that the single crystals gradually decompose in air due to moisture.
### High-field ESR
High-field ESR measurements of CROC were performed at the Dresden High Magnetic Field Laboratory (HLD) using a transmission-type ESR spectrometer (similar to that described in ref. 35) with oversized waveguides and a 16 T superconducting magnet. A set of backward-wave oscillators, Gunn diodes, and VDI microwave sources (Virginia Diodes Inc, USA) was used, allowing us to study magnetic excitations in a broad quasicontinuously covered frequency range, from 100 to 500 GHz. The experiments were performed in the Faraday and Voigt configurations with magnetic fields Ha and Hb,c, respectively. The stable free-radical 2,2-diphenyl-1-picrylhydrazyl (DPPH) was used as a frequency-field marker.
### High-field magnetization
The megagauss-field facility at the ISSP, University of Tokyo, Japan, equipped with a 160 μF/50 kV capacitor bank, was used to generate ultrahigh magnetic fields36,37. To measure magnetization, we employed two types of magnet. For 120 T experiments, we used single-turn coil magnets in horizontal geometry, with the pulse duration 8 μs. The induction method was used to detect the dM/dH signal using coaxial-type pickup coils. The field was measured by a calibrated pickup coil located near the sample. The recorded pickup voltage was numerically integrated to obtain field values. In spite of careful compensation, the pulsed-field magnetization exhibited a tiny linear background, which we subtracted from the experimental data. Magnetization up to 51 T was measured using a non-destructive magnet with a pulse duration of 4 ms. More detailed information on the experimental procedure can be found in ref. 38. The magnetization was calibrated to absolute values using the magnetization data collected at 2 K in DC fields up to 7 T using a SQUID magnetometer (Quantum Design, USA). | open-web-math/open-web-math | |
Cody
# Problem 392. Clock Hand Angle 1
Solution 594942
Submitted on 11 Mar 2015 by goc3
This solution is locked. To view this solution, you need to provide a solution of the same size or smaller.
### Test Suite
Test Status Code Input and Output
1 Pass
%% assert(abs(handAngle('12:00:00')-0) < 1e-9)
s = '12' '00' '00' H = 12 M = 0 S = 0 aH = 360 aM = 0 theta = 360 theta = 0
2 Pass
%% assert(abs(handAngle('03:28:07')-64.6416666667) < 1e-9)
s = '03' '28' '07' H = 3 M = 28 S = 7 aH = 104.0583 aM = 168.7000 theta = 64.6417
3 Pass
%% assert(abs(handAngle('12:26:23')-145.1083333333) < 1e-9)
s = '12' '26' '23' H = 12 M = 26 S = 23 aH = 373.1917 aM = 158.3000 theta = 214.8917 theta = 145.1083
4 Pass
%% assert(abs(handAngle('09:50:12')-6.1000000000) < 1e-9)
s = '09' '50' '12' H = 9 M = 50 S = 12 aH = 295.1000 aM = 301.2000 theta = 6.1000
5 Pass
%% assert(abs(handAngle('07:06:30')-174.2500000000) < 1e-9)
s = '07' '06' '30' H = 7 M = 6 S = 30 aH = 213.2500 aM = 39 theta = 174.2500
6 Pass
%% assert(abs(handAngle('06:08:21')-134.0750000000) < 1e-9)
s = '06' '08' '21' H = 6 M = 8 S = 21 aH = 184.1750 aM = 50.1000 theta = 134.0750
7 Pass
%% assert(abs(handAngle('01:11:58')-35.8166666667) < 1e-9)
s = '01' '11' '58' H = 1 M = 11 S = 58 aH = 35.9833 aM = 71.8000 theta = 35.8167
8 Pass
%% assert(abs(handAngle('09:24:56')-132.8666666667) < 1e-9)
s = '09' '24' '56' H = 9 M = 24 S = 56 aH = 282.4667 aM = 149.6000 theta = 132.8667
9 Pass
%% assert(abs(handAngle('01:50:04')-114.6333333333) < 1e-9)
s = '01' '50' '04' H = 1 M = 50 S = 4 aH = 55.0333 aM = 300.4000 theta = 245.3667 theta = 114.6333
10 Pass
%% assert(abs(handAngle('01:49:45')-116.3750000000) < 1e-9)
s = '01' '49' '45' H = 1 M = 49 S = 45 aH = 54.8750 aM = 298.5000 theta = 243.6250 theta = 116.3750
11 Pass
%% assert(abs(handAngle('07:04:17')-173.5583333333) < 1e-9)
s = '07' '04' '17' H = 7 M = 4 S = 17 aH = 212.1417 aM = 25.7000 theta = 186.4417 theta = 173.5583
12 Pass
%% assert(abs(handAngle('02:24:26')-74.3833333333) < 1e-9)
s = '02' '24' '26' H = 2 M = 24 S = 26 aH = 72.2167 aM = 146.6000 theta = 74.3833
13 Pass
%% assert(abs(handAngle('10:32:33')-120.9750000000) < 1e-9)
s = '10' '32' '33' H = 10 M = 32 S = 33 aH = 316.2750 aM = 195.3000 theta = 120.9750
14 Pass
%% assert(abs(handAngle('10:26:57')-151.7750000000) < 1e-9)
s = '10' '26' '57' H = 10 M = 26 S = 57 aH = 313.4750 aM = 161.7000 theta = 151.7750
15 Pass
%% assert(abs(handAngle('09:40:26')-47.6166666667) < 1e-9)
s = '09' '40' '26' H = 9 M = 40 S = 26 aH = 290.2167 aM = 242.6000 theta = 47.6167
16 Pass
%% assert(abs(handAngle('02:38:59')-154.4083333333) < 1e-9)
s = '02' '38' '59' H = 2 M = 38 S = 59 aH = 79.4917 aM = 233.9000 theta = 154.4083
17 Pass
%% assert(abs(handAngle('08:18:19')-139.2583333333) < 1e-9)
s = '08' '18' '19' H = 8 M = 18 S = 19 aH = 249.1583 aM = 109.9000 theta = 139.2583
18 Pass
%% assert(abs(handAngle('07:26:43')-63.0583333333) < 1e-9)
s = '07' '26' '43' H = 7 M = 26 S = 43 aH = 223.3583 aM = 160.3000 theta = 63.0583
19 Pass
%% assert(abs(handAngle('12:01:40')-9.1666666667) < 1e-9)
s = '12' '01' '40' H = 12 M = 1 S = 40 aH = 360.8333 aM = 10 theta = 350.8333 theta = 9.1667
20 Pass
%% assert(abs(handAngle('08:60:33')-93.0250000000) < 1e-9)
s = '08' '60' '33' H = 8 M = 60 S = 33 aH = 270.2750 aM = 363.3000 theta = 93.0250
21 Pass
%% assert(abs(handAngle('10:11:42')-124.3500000000) < 1e-9)
s = '10' '11' '42' H = 10 M = 11 S = 42 aH = 305.8500 aM = 70.2000 theta = 235.6500 theta = 124.3500
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# Which of the following statements is applicable for a three-phase AC circuit for a star connection?
This question was previously asked in
UPPCL JE Previous Paper 2 (Held ON: 27 August 2018 Shift 2)
View all UPPCL JE Papers >
1. Line voltage is √3 times the phase voltage and the phase current is equal to the line current
2. Phase voltage is equal to the line current and line current is equal to the phase voltage
3. Phase voltage is equal to the line voltage and phase current is three times the line current
4. Phase voltage is equal to the line voltage and line current is equal to the phase current
Option 1 : Line voltage is √3 times the phase voltage and the phase current is equal to the line current
Free
CT 1: Engineering Mathematics
1067
10 Questions 10 Marks 12 Mins
## Detailed Solution
Concept:
In a star-connected three-phase system,
VL = √3 × Vph
And IL = Iph
$${I_{ph}} = \frac{{{V_L}}}{{\surd 3Z}}$$
In a delta connected three-phase system,
VL = Vph
IL = √3 × Iph
$${I_{ph}} = \frac{{{V_L}}}{Z}$$
Where,
VL is line voltage
Vph is phase voltage
IL is line current
Iph is phase current | HuggingFaceTB/finemath | |
# sagemath fails to solve an equation and inequation
I do not understand why
var('R','p','x')
sol1=solve(R^(1/3)*p == x^(1/3),x)
show(sol1)
this work but this fails to isolate $x$
var('R','p','x')
sol1=solve(R^(1/3)*p <= x^(1/3),x)
show(sol1)
And that
var('R','p','x')
sol2=solve(0 == (R-x)^(1/3)*p + (1-p)* (-x)^(1/3),x)
show(sol2)
doesn't find $x$. And, as a consequence, of cours this :
var('R','p','x')
sol2=solve(0 <= (R-x)^(1/3)*p + (1-p)* (-x)^(1/3),x)
show(sol2)
fails. I have tried all simplification ways.
edit retag close merge delete
Sort by ยป oldest newest most voted
The first two can be (awkwardly) solved by Sage :
sage: E1=R^(1/3)*p == x^(1/3)
sage: E1.solve(x, algorithm="sympy")
[x == R*p^3]
sage: I1=R^(1/3)*p <= x^(1/3)
sage: [x-v[0] for v in solve(I1^3,x)]
[R*p^3 == x, R*p^3 > x]
But the third one : sage: I2=0 <= (R-x)^(1/3)p + (1-p) (-x)^(1/3)
requires manual solution. Sage can help, at least for record-keeping :
sage: I22=((((I2-I2.rhs().operands()[1])/(1-p))^3)).expand() ; I22
x <= -R*p^3/(p^3 - 3*p^2 + 3*p - 1) + p^3*x/(p^3 - 3*p^2 + 3*p - 1)
sage: I23=I22-I22.rhs().operands()[1] ; I23
-p^3*x/(p^3 - 3*p^2 + 3*p - 1) + x <= -R*p^3/(p^3 - 3*p^2 + 3*p - 1)
sage: I23.factor()/(I23.factor().lhs().coefficient(x))
x <= R*p^3/(3*p^2 - 3*p + 1)
"The competition" fares a bit better :
sage: mathematica.Reduce(I2,x)
Element[p, Reals] &&
((R < 0 && ((p <= 1 && x <= R) || (p > 1 &&
x <= (p^3*R)/(1 - 3*p + 3*p^2)))) || (R == 0 && x <= 0) ||
(R > 0 && ((p < 0 && x <= (p^3*R)/(1 - 3*p + 3*p^2)) ||
(p >= 0 && x <= 0))))
HTH,
more
Thanks I always forget sympy
( 2020-11-07 09:01:41 +0100 )edit
Emmanuel I have a big problem. What I do is preparing correction for students under SageCells. Unfortunately, algorithm="sympy" option doesn't work. Are you aware of an other soluytion ? It seems that many things are differently managed under Sagemath notebook and SaeCells which is dommageable. And I do not know how to connect people who develop SageCells
( 2020-11-08 12:04:44 +0100 )edit | HuggingFaceTB/finemath | |
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Take this Mental Maths Skills Test to get pass in first attempt in real exam. As we provide similar questions.
Lets begin the quiz .. Good luck!
Q:1-A pupil achieved 84 marks out of a possible 120 in a test.
What percentage mark did the pupil achieve for the test?
75%
70%
65%
60%
Q:2-A school trip was planned at a total cost of £120 per pupil. The accommodation cost two fifths of the total.
What was the cost of the accommodation per pupil?
£48
£52
£54
£56
Q:3-A school’s policy for Key Stage 2 was to set three and a half hours of homework per week.
What was the mean number of minutes to be spent on homework per weekday evening?
38 min
40 min
42 min
44 min
Q:4-At a staff meeting the Headteacher presented the following table, showing the number of pupils in each class in a school who are having extra music lessons.
What percentage of pupils in the school are having extra music lessons? Give your answer to the nearest whole number.
19%
21%
23%
25%
Q:5- A teacher is planning a group outing to see a play in a nearby city. She has been given details of costs and travel. There are 25 in the group, including pupils and teachers. A group booking for 25 theatre tickets costs £185, return train tickets cost £5.65 each.
How much will each person have to pay for the outing to cover the cost of travel and a theatre ticket?
£15.05
£14.50
£13.50
£13.05
Q:6-A teacher calculated the mean points score achieved by pupils in an end of Key Stage 2 mathematics test. There were 48 pupils in the year group. The mean points score is given by the formula:
mean points score = (total points for the year group)/(total number of pupils in the group)
What was the mean points score for the year group? Give your answer to the nearest whole number.
The total points for Levels 3 and 4 have been done for you.
35
33
31
29
Q:7-To set targets for the following year, the mathematics department analysed the percentage of mathematics GCSE grades A* - C achieved by pupils in the school.
Indicate all the true statements:
The mean percentage of GCSE grades A* - C for the last 5 years of the chart was 51%.
The percentage of GCSE grades A* - C increased each year from 2004 to 2011.
The percentage of mathematics GCSE grades A* - C more than doubled from 2004 to 2011.
1st & 3rd
Q:8-A school parents’ evening starts at 16:30 on each of two consecutive days.
A teacher has a total of 24 appointments lasting 10 minutes each and takes a 20 minute break each evening. The teacher filled all the available appointment slots on the first evening and finished at 19:00.
What is the earliest time the teacher can expect to finish on the second evening? Give your answer in the 24 hour clock.
18:40 (24hr clock)
19:40 (24hr clock).
18:20 (24hr clock).
18:30 (24hr clock).
Q:9-A teacher plans a school trip to Brussels, which involves using a ferry from Ostend. The teacher wants to be in Ostend no later than 18:00. She expects their coach to travel from Brussels to Ostend, a distance of 120 km, at an average of 50 miles per hour.
Using the approximation of 5 miles equals 8 kilometres, what is the latest time that the coach should leave Brussels? Give your answer using the 24-hour clock.
14:30 (24hr clock)
15:30 (24hr clock)
16:30 (24hr clock)
17:30 (24hr clock)
Q:10-The head of careers supplied the following chart showing the destination of Year 13 leavers.
Indicate the true statement: | HuggingFaceTB/finemath | |
Prove: if $a,b\in G$ commute with probability $>5/8$, then $G$ is abelian
Suppose that $G$ is a finite group. If $P( ab=ba ) >5/8$, prove $G$ is abelian.
• Do you have any more information on the probability $P$? For example, what is the sample space and $\sigma$-algebra on which $P$ is defined? – Tom Jun 24 '14 at 19:13
• @Tom This is a well-known result, and the probability space is exactly what you would think it is: $(a,b)\in G\times G$ taken with uniform probability. – blue Jun 24 '14 at 19:14
• @blue Thanks.. we'd be in trouble if $P(\{(e,e)\})=1$. – Tom Jun 24 '14 at 19:21
• Why involve probability at all? – Derek Holt Jun 25 '14 at 8:02
• @DerekHolt The statement generalizes to compact groups with normalized Haar measure. – arctic tern Jan 2 '17 at 4:40
Split into two cases: (i) $G/Z(G)\cong C_2\times C_2$ is Klein-$4$, and (ii) $Z(G)$ has index $\ge6$. Since the quotient $G/Z(G)$ cannot be nontrivial cyclic, these two cases cover all bases. Let $Z=Z(G)$.
$\rm\color{Blue}{(i)}$: Write $G=Z\sqcup aZ\sqcup bZ\sqcup cZ$. Check $C_G(g)=Z\cup aZ$ for $g\in aZ$, and similarly for the other two cosets $bZ$ and $cZ$, while of course $C_G(g)=G$ if $g\in Z$. Use $[\Omega]$ for the Iverson bracket (i.e. $[\Omega]=1$ if $\Omega$ is true, and $[\Omega]=0$ otherwise). Calculate
$$P=\frac{1}{|G\times G|}\sum_{a\in G}\sum_{b\in G}[ab=ba]=\frac{1}{|G|^2}\sum_{a\in G}|C_G(a)|=\frac{1}{|G|^2}\left(\sum_{a\in Z}|G|+\sum_{a\in G\setminus Z}\frac{|G|}{2}\right)$$
$$=\frac{1}{|G|^2}\left(|Z||G|+|G\setminus Z|\frac{|G|}{2}\right)=\frac{1}{|G|^2}\left(\frac{|G|}{4}|G|+\frac{3|G|}{4}\frac{|G|}{2}\right)=\frac{1}{4}+\frac{3}{4}\cdot\frac{1}{2}=\frac{5}{8}.$$
$\rm\color{Blue}{(ii)}$: Start the same way as above, writing
$$P=\frac{1}{|G|^2}\sum_{a\in G}|C_G(a)|=\frac{1}{|G|^2}\left(\sum_{a\in Z}|G|+\sum_{a\in G\setminus Z}|C_G(a)|\right).$$
Using $|Z|\le|G|/6$ and $|C_G(a)|\le|G|/2$, we obtain the upper bound
$$P\le\frac{1}{|G|^2}\left(|G||Z|+\big(|G|-|Z|\big)\frac{|G|}{2}\right)=\frac{1}{|G|^2}\left(\frac{|G|^2}{2}+|Z|\frac{|G|}{2}\right)\le\frac{1}{2}+\frac{1}{6}\cdot\frac{1}{2}=\frac{7}{12}.$$
Note $7/12<5/8$. Therefore, we have proven:
Theorem. $P>\frac{5}{8}$ iff $G$ is abelian, $P=\frac{5}{8}$ iff $G/Z(G)$ is Klein-$4$, and $P\le\frac{7}{12}$ otherwise.
There are a couple slick character-theoretic proofs on mathoverflow, as well as forcing probabilities for solvability, nilpotence and odd cardinality. Note the $5/8$ bound is true for compact topological groups, where we pick $(a,b)\in G\times G$ according to the normalized Haar measure. One formula for the commuting probability $P$ that I didn't discuss above is as follows:
$$P=\frac{1}{|G|^2}\sum_{a\in G}|C_G(a)|=\frac{1}{|G|}\sum_{a\in G}\frac{1}{[G:C_G(a)]}=\frac{1}{|G|}\sum_{i=1}^k\sum_{a\in C_i}\frac{1}{|C_i|}=\frac{1}{|G|}\sum_{i=1}^k\frac{|C_i|}{|C_i|}=\frac{k}{|G|},$$
where $C_1,\cdots,C_k$ are the conjugacy classes of $G$ and $k$ is the number of conjugacy classes. | open-web-math/open-web-math | |
1. You might want to retry asking the question. It is unclear, and seems incomplete.
However, when you're dealing with 3 sets, the following formulae might come in handy.
n (A U B U C) = n(A) + n(B) + n(C) - n(A ∩ B) - n(A ∩ C) - n(B ∩ C) + n(A ∩ B ∩ C)
n(U) = n (A U B U C) + n (A U B U C)c
where, A, B and C are three non-empty sets,
U denotes the Universal Set
and c (in superscript) indicates complement of a set (Complement of a set: is the difference of a set from the Universal set)
2. Let N and B denote the set of people whi can speak Newari and Bhojpuri reapectively.
Now, n(U)=200, n(N)=150, n(B)=80 and (A∩ B)' = 30
i) Let (A∩ B) = x
Representing above information in a venn diagram. From the venn diagram,
150-x + x +80-x + 30=200
Or, 260-x = 200
Or, x= 260-200
Therefore, x=60.
ii) n(N∪B) = n(N)+ n(B) - n(N∩B)
= 150 + 80 - 60
= 170
iii) n०(N)+ n०(B)= 150-x + 80-x
= 150-60+80-60
= 110
Therefore,
No.of people speaking both language=60
No.of people speaking at least one language= 170
No.of people speaking only one language= 110
There are no MCQs in Mattrab Library for this chapter yet. | HuggingFaceTB/finemath | |
# Three-gap theorem
Last updated
In mathematics, the three-gap theorem, three-distance theorem, or Steinhaus conjecture states that if one places n points on a circle, at angles of θ, 2θ, 3θ, ... from the starting point, then there will be at most three distinct distances between pairs of points in adjacent positions around the circle. When there are three distances, the largest of the three always equals the sum of the other two. [1] Unless θ is a rational multiple of π, there will also be at least two distinct distances.
## Contents
This result was conjectured by Hugo Steinhaus, and proved in the 1950s by Vera T. Sós, János Surányi [ hu ], and Stanisław Świerczkowski; more proofs were added by others later. Applications of the three-gap theorem include the study of plant growth and musical tuning systems, and the theory of light reflection within a mirrored square.
## Statement
The three-gap theorem can be stated geometrically in terms of points on a circle. In this form, it states that if one places ${\displaystyle n}$ points on a circle, at angles of ${\displaystyle \theta ,2\theta ,\dots ,n\theta }$ from the starting point, then there will be at most three distinct distances between pairs of points in adjacent positions around the circle. An equivalent and more algebraic form involves the fractional parts of multiples of a real number. It states that, for any positive real number ${\displaystyle \alpha }$ and integer ${\displaystyle n}$, the fractional parts of the numbers ${\displaystyle \alpha ,2\alpha ,\dots ,n\alpha }$ divide the unit interval into subintervals with at most three different lengths. The two problems are equivalent under a linear correspondence between the unit interval and the circumference of the circle, and a correspondence between the real number ${\displaystyle \alpha }$ and the angle ${\displaystyle \theta =2\pi \alpha }$. [2] [3] [4]
## Applications
### Plant growth
In the study of phyllotaxis, the arrangements of leaves on plant stems, it has been observed that each successive leaf on the stems of many plants is turned from the previous leaf by the golden angle, approximately 137.5°. It has been suggested that this angle maximizes the sun-collecting power of the plant's leaves. [5] If one looks end-on at a plant stem that has grown in this way, there will be at most three distinct angles between two leaves that are consecutive in the cyclic order given by this end-on view. [6]
For example, in the figure, the largest of these three angles occurs three times, between the leaves numbered 3 and 6, between leaves 4 and 7, and between leaves 5 and 8. The second-largest angle occurs five times, between leaves 6 and 1, 9 and 4, 7 and 2, 10 and 5, and 8 and 3. And the smallest angle occurs only twice, between leaves 1 and 9 and between leaves 2 and 10. The phenomenon of having three types of distinct gaps depends only on fact that the growth pattern uses a constant rotation angle, and not on the relation of this angle to the golden ratio; the same phenomenon would happen for any other rotation angle, and not just for the golden angle. However, other properties of this growth pattern do depend on the golden ratio. For instance, the fact that golden ratio is a badly approximable number implies that points spaced at this angle along the Fermat spiral (as they are in some models of plant growth) form a Delone set; intuitively, this means that they are uniformly spaced. [7]
### Music theory
In music theory, a musical interval describes the ratio in frequency between two musical tones. Intervals are commonly considered consonant or harmonious when they are the ratio of two small integers; for instance, the octave corresponds to the ratio 2:1, while the perfect fifth corresponds to the ratio 3:2. [8] Two tones are commonly considered to be equivalent when they differ by a whole number of octaves; this equivalence can be represented geometrically by the chromatic circle, the points of which represent classes of equivalent tones. Mathematically, this circle can be described as the unit circle in the complex plane, and the point on this circle that represents a given tone can be obtained by the mapping the frequency ${\displaystyle \nu }$ to the complex number ${\textstyle \exp(2\pi i\log _{2}\nu )}$. An interval with ratio ${\displaystyle \rho }$ corresponds to the angle ${\displaystyle 2\pi \log _{2}\rho }$ between points on this circle, meaning that two musical tones differ by the given interval when their two points on the circle differ by this angle. For instance, this formula gives ${\displaystyle 2\pi }$ (a whole circle) as the angle corresponding to an octave. Because 3/2 is not a rational power of two, the angle on the chromatic circle that represents a perfect fifth is not a rational multiple of ${\displaystyle 2\pi }$, and similarly other common musical intervals other than the octave do not correspond to rational angles. [9]
A tuning system is a collection of tones used to compose and play music. For instance, the equal temperament commonly used for the piano is a tuning system, consisting of 12 tones equally spaced around the chromatic circle. Some other tuning systems do not space their tones equally, but instead generate them by some number of consecutive multiples of a given interval. An example is the Pythagorean tuning, which is constructed in this way from twelve tones, generated as the consecutive multiples of a perfect fifth in the circle of fifths. The irrational angle formed on the chromatic circle by a perfect fifth is close to 7/12 of a circle, and therefore the twelve tones of the Pythagorean tuning are close to, but not the same as, the twelve tones of equal temperament, which could be generated in the same way using an angle of exactly 7/12 of a circle. [10] Instead of being spaced at angles of exactly 1/12 of a circle, as the tones of equal temperament would be, the tones of the Pythagorean tuning are separated by intervals of two different angles, close to but not exactly 1/12 of a circle, representing two different types of semitones. [11] If the Pythagorean tuning system were extended by one more perfect fifth, to a set of 13 tones, then the sequence of intervals between its tones would include a third, much shorter interval, the Pythagorean comma. [12]
In this context, the three-gap theorem can be used to describe any tuning system that is generated in this way by consecutive multiples of a single interval. Some of these tuning systems (like equal temperament) may have only one interval separating the closest pairs of tones, and some (like the Pythagorean tuning) may have only two different intervals separating the tones, but the three-gap theorem implies that there are always at most three different intervals separating the tones. [13] [14]
### Mirrored reflection
A Sturmian word is infinite sequences of two symbols (for instance, "H" and "V") describing the sequence of horizontal and vertical reflections of a light ray within a mirrored square, starting along a line of irrational slope. Equivalently, the same sequence describes the sequence of horizontal and vertical lines of the integer grid that are crossed by the starting line. One property that all such sequences have is that, for any positive integer n, the sequence has exactly n + 1 distinct consecutive subsequences of length n. Each subsequence occurs infinitely often with a certain frequency, and the three-gap theorem implies that these n + 1 subsequences occur with at most three distinct frequencies. If there are three frequencies, then the largest frequency must equal the sum of the other two. One proof of this result involves partitioning the y-intercepts of the starting lines (modulo 1) into n + 1 subintervals within which the initial n elements of the sequence are the same, and applying the three-gap theorem to this partition. [15] [16]
## History and proof
The three-gap theorem was conjectured by Hugo Steinhaus, and its first [17] proofs were found in the late 1950s by Vera T. Sós, [18] János Surányi [ hu ], [19] and Stanisław Świerczkowski. [20] Later researchers published additional proofs, [21] generalizing this result to higher dimensions [22] [23] [24] [25] , and connecting it to topics including continued fractions, [4] [26] symmetries and geodesics of Riemannian manifolds, [27] ergodic theory, [28] and the space of planar lattices. [3] Mayero (2000) formalizes a proof using the Coq interactive theorem prover. [2]
The following simple proof is due to Frank Liang. Let θ be the rotation angle generating a set of points as some number of consecutive multiples of θ on a circle. Define a gap to be an arc A of the circle that extends between two adjacent points of the given set, and define a gap to be rigid if its endpoints occur later in the sequence of multiples of θ than any other gap of the same length. From this definition, it follows that every gap has the same length as a rigid gap. If A is a rigid gap, then A + θ is not a gap, because it has the same length and would be one step later. The only ways for this to happen are for one of the endpoints of A to be the last point in the sequence of multiples of θ (so that the corresponding endpoint of A + θ is missing) or for one of the given points to land within A + θ, preventing it from being a gap. A point can only land within A + θ if it is the first point in the sequence of multiples of θ, because otherwise its predecessor in the sequence would land within A, contradicting the assumption that A is a gap. So there can be at most three rigid gaps, the two on either side of the last point and the one in which the predecessor of the first point (if it were part of the sequence) would land. Because there are at most three rigid gaps, there are at most three lengths of gaps. [29] [30]
Liang's proof additionally shows that, when there are exactly three gap lengths, the longest gap length is the sum of the other two. For, in this case, the rotated copy A + θ that has the first point in it is partitioned by that point into two smaller gaps, which must be the other two gaps. [29] [30] Liang also proves a more general result, the "${\displaystyle 3d}$ distance theorem", according to which the union of ${\displaystyle d}$ different arithmetic progressions on a circle has at most ${\displaystyle 3d}$ different gap lengths. [29] In the three-gap theorem, there is a constant bound on the ratios between the three gaps, if and only if θ/2π is a badly approximable number. [7]
A closely related but earlier theorem, also called the three-gap theorem, is that if A is any arc of the circle, then the integer sequence of multiples of θ that land in A has at most three lengths of gaps between sequence values. Again, if there are three gap lengths then one is the sum of the other two. [31] [32]
## Related Research Articles
A circle is a shape consisting of all points in a plane that are at a given distance from a given point, the centre. The distance between any point of the circle and the centre is called the radius.
An equal temperament is a musical temperament or tuning system that approximates just intervals by dividing an octave into steps such that the ratio of the frequencies of any adjacent pair of notes is the same. This system yields pitch steps perceived as equal in size, due to the logarithmic changes in pitch frequency.
In mathematics, two quantities are in the golden ratio if their ratio is the same as the ratio of their sum to the larger of the two quantities. Expressed algebraically, for quantities and with ,
A Pythagorean triple consists of three positive integers a, b, and c, such that a2 + b2 = c2. Such a triple is commonly written (a, b, c), and a well-known example is (3, 4, 5). If (a, b, c) is a Pythagorean triple, then so is (ka, kb, kc) for any positive integer k. A primitive Pythagorean triple is one in which a, b and c are coprime (that is, they have no common divisor larger than 1). For example, (3, 4, 5) is a primitive Pythagorean triple whereas (6, 8, 10) is not. A triangle whose sides form a Pythagorean triple is called a Pythagorean triangle, and is necessarily a right triangle.
In mathematics, the trigonometric functions are real functions which relate an angle of a right-angled triangle to ratios of two side lengths. They are widely used in all sciences that are related to geometry, such as navigation, solid mechanics, celestial mechanics, geodesy, and many others. They are among the simplest periodic functions, and as such are also widely used for studying periodic phenomena through Fourier analysis.
Angle trisection is a classical problem of straightedge and compass construction of ancient Greek mathematics. It concerns construction of an angle equal to one third of a given arbitrary angle, using only two tools: an unmarked straightedge and a compass.
In geometry, an equilateral triangle is a triangle in which all three sides have the same length. In the familiar Euclidean geometry, an equilateral triangle is also equiangular; that is, all three internal angles are also congruent to each other and are each 60°. It is also a regular polygon, so it is also referred to as a regular triangle.
The Pythagorean trigonometric identity, also called simply the Pythagorean identity, is an identity expressing the Pythagorean theorem in terms of trigonometric functions. Along with the sum-of-angles formulae, it is one of the basic relations between the sine and cosine functions.
In geometry, the area enclosed by a circle of radius r is πr2. Here the Greek letter π represents the constant ratio of the circumference of any circle to its diameter, approximately equal to 3.14159.
A special right triangle is a right triangle with some regular feature that makes calculations on the triangle easier, or for which simple formulas exist. For example, a right triangle may have angles that form simple relationships, such as 45°–45°–90°. This is called an "angle-based" right triangle. A "side-based" right triangle is one in which the lengths of the sides form ratios of whole numbers, such as 3 : 4 : 5, or of other special numbers such as the golden ratio. Knowing the relationships of the angles or ratios of sides of these special right triangles allows one to quickly calculate various lengths in geometric problems without resorting to more advanced methods.
In mathematics, the equidistribution theorem is the statement that the sequence
A prime gap is the difference between two successive prime numbers. The n-th prime gap, denoted gn or g(pn) is the difference between the (n + 1)-st and the n-th prime numbers, i.e.
Regular numbers are numbers that evenly divide powers of 60 (or, equivalently, powers of 30). Equivalently, they are the numbers whose only prime divisors are 2, 3, and 5. As an example, 602 = 3600 = 48 × 75, so as divisors of a power of 60 both 48 and 75 are regular.
A Kepler triangle is a special right triangle with edge lengths in geometric progression. The ratio of the progression is where is the golden ratio, and the progression can be written: , or approximately . Squares on the edges of this triangle have areas in another geometric progression, . Alternative definitions of the same triangle characterize it in terms of the three Pythagorean means of two numbers, or via the inradius of isosceles triangles.
In geometry, Coxeter's loxodromic sequence of tangent circles is an infinite sequence of circles arranged so that any four consecutive circles in the sequence are pairwise mutually tangent. This means that each circle in the sequence is tangent to the three circles that precede it and also to the three circles that follow it.
The circle packing theorem describes the possible tangency relations between circles in the plane whose interiors are disjoint. A circle packing is a connected collection of circles whose interiors are disjoint. The intersection graph of a circle packing is the graph having a vertex for each circle, and an edge for every pair of circles that are tangent. If the circle packing is on the plane, or, equivalently, on the sphere, then its intersection graph is called a coin graph; more generally, intersection graphs of interior-disjoint geometric objects are called tangency graphs or contact graphs. Coin graphs are always connected, simple, and planar. The circle packing theorem states that these are the only requirements for a graph to be a coin graph:
In trigonometry, the law of cosines relates the lengths of the sides of a triangle to the cosine of one of its angles. For a triangle with sides and opposite respective angles and , the law of cosines states:
In mathematics, the Pythagorean theorem or Pythagoras' theorem is a fundamental relation in Euclidean geometry between the three sides of a right triangle. It states that the area of the square whose side is the hypotenuse is equal to the sum of the areas of the squares on the other two sides.
In mathematics, Niven's theorem, named after Ivan Niven, states that the only rational values of θ in the interval 0° ≤ θ ≤ 90° for which the sine of θ degrees is also a rational number are:
Stanisław (Stash) Świerczkowski was a Polish mathematician famous for his solutions to two iconic problems posed by Hugo Steinhaus: the three-gap theorem and the non-tetratorus theorem.
## References
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7. Akiyama, Shigeki (March 2020), "Spiral Delone sets and three distance theorem", Nonlinearity , 33 (5): 2533–2540, arXiv:, Bibcode:2020Nonli..33.2533A, doi:10.1088/1361-6544/ab74ad, S2CID 129945118
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9. Baroin, Gilles; Calvet, André (2019), "Visualizing temperaments: squaring the circle?", in Montiel, Mariana; Gomez-Martin, Francisco; Agustín-Aquino, Octavio A. (eds.), Mathematics and Computation in Music: 7th International Conference, MCM 2019, Madrid, Spain, June 18–21, 2019, Proceedings, Springer International Publishing, pp. 333–337, doi:10.1007/978-3-030-21392-3_27, S2CID 184482714
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11. Strohm, Reinhard; Blackburn, Bonnie J., eds. (2001), Music as Concept and Practice in the Late Middle Ages, Volume 3, Part 1, New Oxford history of music, Oxford University Press, p. 252, ISBN 9780198162056
12. Benson, Donald C. (2003), A Smoother Pebble: Mathematical Explorations, Oxford University Press, p. 51, ISBN 9780198032977
13. Carey, Norman (2007), "Coherence and sameness in well-formed and pairwise well-formed scales", Journal of Mathematics and Music, 1 (2): 79–98, doi:10.1080/17459730701376743, S2CID 120586231
14. Narushima, Terumi (2017), Microtonality and the Tuning Systems of Erv Wilson: Mapping the Harmonic Spectrum, Routledge Studies in Music Theory, Routledge, pp. 90–91, ISBN 9781317513421
15. Lothaire, M. (2002), "Sturmian Words", Algebraic Combinatorics on Words, Cambridge: Cambridge University Press, pp. 40–97, ISBN 978-0-521-81220-7, Zbl 1001.68093 . Lothaire uses the property of having ${\displaystyle d+1}$ words of length ${\displaystyle d}$ as a definition of Sturmian words, rather than as a consequence of the definition. For the equivalence of this property with the definition stated here, see Theorem 2.1.13, p. 51. For the three frequencies of these words see Theorem 2.2.37, p. 73.
16. Alessandri, Pascal; Berthé, Valérie (1998), "Three distance theorems and combinatorics on words", L'Enseignement mathématique , 44 (1–2): 103–132, MR 1643286 ; see in particular Section 2.1, "Complexity and frequencies of codings of rotations"
17. Haynes, Alan; Marklof, Jens (2020), "Higher dimensional Steinhaus and Slater problems via homogeneous dynamics", Annales Scientifiques de l'École Normale Supérieure , 53 (2): 537–557, arXiv:, doi:10.24033/asens.2427, MR 4094564, S2CID 67851217, The first proofs of this remarkable fact were published in 1957 by Sós, in 1958 by Surányi, and in 1959 by Świerczkowski
18. Sós, V. T. (1958), "On the distribution mod 1 of the sequence ${\displaystyle n\alpha }$", Ann. Univ. Sci. Budapest, Eötvös Sect. Math., 1: 127–134
19. Surányi, J. (1958), "Über die Anordnung der Vielfachen einer reelen Zahl mod 1", Ann. Univ. Sci. Budapest, Eötvös Sect. Math., 1: 107–111
20. Świerczkowski, S. (1959), "On successive settings of an arc on the circumference of a circle", Fundamenta Mathematicae , 46 (2): 187–189, doi:, MR 0104651
21. These proofs are briefly surveyed and classified by Marklof & Strömbergsson (2017), from which the following classification of these proofs and many of their references are taken.
22. Halton, John H. (1965), "The distribution of the sequence ${\displaystyle \{n\xi \}\,(n=0,\,1,\,2,\,\ldots )}$", Mathematical Proceedings of the Cambridge Philosophical Society , 61 (3): 665–670, doi:10.1017/S0305004100039013, MR 0202668, S2CID 123400321
23. Chevallier, Nicolas (2007), "Cyclic groups and the three distance theorem", Canadian Journal of Mathematics , 59 (3): 503–552, doi:, MR 2319157, S2CID 123011205
24. Vijay, Sujith (2008), "Eleven Euclidean distances are enough", Journal of Number Theory , 128 (6): 1655–1661, arXiv:, doi:, MR 2419185, S2CID 119655772
25. Bleher, Pavel M.; Homma, Youkow; Ji, Lyndon L.; Roeder, Roland K. W.; Shen, Jeffrey D. (2012), "Nearest neighbor distances on a circle: multidimensional case", Journal of Statistical Physics, 146 (2): 446–465, arXiv:, Bibcode:2012JSP...146..446B, doi:10.1007/s10955-011-0367-8, MR 2873022, S2CID 99723
26. Slater, Noel B. (1967), "Gaps and steps for the sequence ${\displaystyle n\theta {\bmod {1}}}$", Mathematical Proceedings of the Cambridge Philosophical Society , 63 (4): 1115–1123, doi:10.1017/S0305004100042195, MR 0217019, S2CID 121496726
27. Biringer, Ian; Schmidt, Benjamin (2008), "The three gap theorem and Riemannian geometry", Geometriae Dedicata , 136: 175–190, arXiv:, doi:10.1007/s10711-008-9283-8, MR 2443351, S2CID 6389675
28. Haynes, Alan; Koivusalo, Henna; Walton, James; Sadun, Lorenzo (2016), "Gaps problems and frequencies of patches in cut and project sets" (PDF), Mathematical Proceedings of the Cambridge Philosophical Society , 161 (1): 65–85, Bibcode:2016MPCPS.161...65H, doi:10.1017/S0305004116000128, MR 3505670, S2CID 55686324
29. Liang, Frank M. (1979), "A short proof of the ${\displaystyle 3d}$ distance theorem", Discrete Mathematics , 28 (3): 325–326, doi:, MR 0548632
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31. Slater, N. B. (1950), "The distribution of the integers ${\displaystyle N}$ for which ${\displaystyle \theta N<\phi }$", Mathematical Proceedings of the Cambridge Philosophical Society , 46 (4): 525–534, doi:10.1017/S0305004100026086, MR 0041891, S2CID 120454265
32. Florek, K. (1951), "Une remarque sur la répartition des nombres ${\displaystyle n\xi \,(\operatorname {mod} 1)}$", Colloquium Mathematicum, 2: 323–324 | HuggingFaceTB/finemath | |
#### 期刊菜单
Three Gradient-Based Optimization Methods and Their Comparison
1. 引言
2. 预备知识
$\mathrm{min}L\left(x\right)$ (2.1)
$L\left({x}^{*}\right)\le L\left(x\right)$$\forall x\in {R}^{n}$
$L\left({x}^{*}\right)\le L\left(x\right)$$\forall x\in N$
$L\left({x}^{*}\right)$\forall x\in N$
$\nabla L\left({x}^{*}\right)=0$
3. 模型
${x}^{*}=\mathrm{arg}\left(\underset{x\in {R}^{D}}{\mathrm{min}}L\left(x\right)=\frac{1}{N}\underset{n=1}{\overset{N}{\sum }}{L}_{n}\left(x\right)\right)$ (3.1)
4. 方法介绍
(一) 梯度下降法(GD)
$\nabla L\left(x\right)=\frac{1}{N}\underset{n=1}{\overset{N}{\sum }}\nabla {L}_{n}\left(x\right)$ (4.1)
(二) 随机梯度下降法(SGD)
$E\left[\nabla {L}_{n}\left({x}_{k}\right)\right]=\nabla L\left({x}_{k}\right)$ (4.2)
(三) 小批量随机梯度下降法(MB-SGD)
$\nabla {L}_{B}\left({x}_{k}\right)=\frac{1}{M}\underset{m=1}{\overset{M}{\sum }}\nabla {L}_{{n}_{m}}\left({x}_{k}\right)$ (4.3)
5. 算例
$\mathrm{min}f\left(x\right)=\frac{1}{n}\underset{i=1}{\overset{n}{\sum }}\left[{\left(1-{x}_{2i-1}\right)}^{2}+10{\left({x}_{2i}-{x}_{2i-1}^{2}\right)}^{2}\right]$ (5.1)
Figure 1. Image when $x\in {R}^{2}$
$\nabla f\left(x\right)=\frac{1}{n}{\left[\frac{\partial f\left(x\right)}{\partial {x}_{1}},\frac{\partial f\left(x\right)}{\partial {x}_{2}},\cdots ,\frac{\partial f\left(x\right)}{\partial {x}_{n}}\right]}^{\text{T}}$ (5.2)
$\frac{\partial f\left(x\right)}{\partial {x}_{1}}=-2\left(1-{x}_{1}\right)-40{x}_{1}\left({x}_{2}-{x}_{1}^{2}\right)$ (5.3)
$\frac{\partial f\left(x\right)}{\partial {x}_{2}}=20\left({x}_{2}-{x}_{1}^{2}\right)$ (5.4)
$\frac{\partial f\left(x\right)}{\partial {x}_{3}}=-2\left(1-{x}_{3}\right)-40{x}_{3}\left({x}_{4}-{x}_{3}^{2}\right)$ (5.5)
$\frac{\partial f\left(x\right)}{\partial {x}_{4}}=20\left({x}_{4}-{x}_{3}^{2}\right)$ (5.6)
$\frac{\partial f\left(x\right)}{\partial {x}_{j}}=-2\left(1-{x}_{j}\right)-40{x}_{j}\left({x}_{j+1}-{x}_{j}^{2}\right)$ (5.7)
$\frac{\partial f\left(x\right)}{\partial {x}_{j}}=20\left({x}_{j}-{x}_{j-1}^{2}\right)$ (5.8)
Table 1.Experimental results
$\mathrm{min}f\left(x\right)=\frac{1}{n}\underset{i=1}{\overset{n}{\sum }}\left[{\left({x}_{i}-B\right)}^{2}+10\mathrm{cos}\left(2\pi \left({x}_{i}-B\right)\right)+10\right]+C$ (5.9)
Figure 2. Image when $x\in {R}^{1}$$B=0$$C=0$
Table 2. The number of local minima changes with the value of n
$\frac{\partial f\left(x\right)}{\partial {x}_{1}}=2\left({x}_{1}-B\right)+20\pi \mathrm{sin}\left(2\pi \left({x}_{1}-B\right)\right)$ (5.10)
$\frac{\partial f\left(x\right)}{\partial {x}_{2}}=2\left({x}_{2}-B\right)+20\pi \mathrm{sin}\left(2\pi \left({x}_{2}-B\right)\right)$ (5.11)
$⋮$
$\frac{\partial f\left(x\right)}{\partial {x}_{n}}=2\left({x}_{n}-B\right)+20\pi \mathrm{sin}\left(2\pi \left({x}_{n}-B\right)\right)$ (5.12)
$|{\left({\stackrel{¯}{x}}_{k}^{*}\right)}_{i}-{\left({x}^{*}\right)}_{i}|<0.25$$\forall i$ (5.13)
Table 3. Simulation experiment results
6. 总结与展望 | HuggingFaceTB/finemath | |
# Defining the Earth in Numbers
The Earth is a fascinating planet with a complex structure and a variety of features. One of the most interesting aspects of the Earth is its size, and in particular, how wide it is. The width of the Earth varies depending on the measurement being used, but there are a few key numbers that can help us understand just how wide this planet really is.
First, let’s consider the diameter of the Earth. The diameter is the distance across the widest point of the planet, which is the equator. The equatorial diameter of the Earth is approximately 7,930 miles (12,742 kilometers). This means that if you were to draw a line arond the widest part of the Earth, it would measure just over 7,900 miles.
However, the Earth is not a perfect sphere, and its shape is affected by a number of factors, including the rotation of the planet and the distribution of mass around its surface. As a result, the distance from the North Pole to the South Pole is slightly shorter than the equatorial diameter. This distance is known as the polar diameter, and it measures approximately 7,900 miles (12,714 kilometers). This means that if you were to draw a line from the North Pole to the South Pole, it would measure just under 7,900 miles.
Another way to think about the width of the Earth is by considering its circumference, which is the distance around the planet at the equator. The circumference of the Earth is approximately 24,901 miles (40,075 kilometers). This means that if you were to travel in a straight line around the Earth at the equator, you would cover a distance of almost 25,000 miles.
It’s important to note that the width of the Earth is not a fixed number, and it can vary slightly depending on a number of factors. For example, the Earth’s crust is not a uniform thickness, and in some places it is much thinner or thicker. Additionally, the shape of the Earth can change over time due to a variety of geological processes, including tectonic activity and erosion.
The width of the Earth is a complex topic that can be measured in a number of different ways. Whether you are considering the equatorial diameter, the polar diameter, or the circumference, there is no doubt that the Earth is a wide and fascinating planet that continues to inspire awe and wonder in people around the world.
## Diameter of the Earth
The Earth’s width or diameter varies depending on the reference point. If we consider the equator, whch is the largest circumference of the Earth, the diameter measures about 7,930 miles (12,742 kilometers). On the other hand, the diameter from the North Pole to the South Pole is slightly shorter, measuring around 7,900 miles (12,714 kilometers).
To get a better understanding of the Earth’s size, we can also look at its circumference, which is the distance around the Earth. The average diameter of the Earth is about 7,917.5 miles (12,742 kilometers), and when multiplied by pi (3.14159), we get a circumference of approximately 24,901.55 miles (40,075.16 kilometers).
It’s important to note that the Earth is not a perfect sphere, and its shape is more like an oblate spheroid, meaning it bulges at the equator and flattens at the poles. This is due to the Earth’s rotation and the centrifugal force it generates.
The Earth’s width or diameter ranges from approximately 7,900 miles (12,714 kilometers) from the North Pole to the South Pole to about 7,930 miles (12,742 kilometers) along the equator. The Earth’s circumference is around 24,901.55 miles (40,075.16 kilometers) when measured at the equator.
Source: wallpapers.net
## The Thickness of the Earth in Miles
The Earth has a diameter of approximately 7,926 miles (12,742 kilometers), which means that its radius is abut 3,963 miles (6,371 kilometers). The Earth is composed of several layers, including the crust, mantle, and core. The crust is the outermost layer and has a thickness of approximately 21 miles (35 kilometers) on average. The mantle, which lies beneath the crust, is much thicker, with an average thickness of about 1,774 miles (2,855 kilometers). the core is the innermost layer and has a radius of about 1,217 miles (1,957 kilometers).
It’s worth noting that while we have a good understanding of the Earth’s structure and composition, our knowledge is limited by our ability to explore it. The deepest we have ever drilled into the Earth is the Kola Superdeep Borehole, which is just 7.5 miles (12 kilometers) deep. Despite this, scientists continue to study the Earth using a range of methods, including seismic waves, to learn more about its internal structure and processes.
## Distance Through the Earth
The distance from one side of the Earth to the other, passing through the center of the planet, is known as the Earth’s diameter. The diameter of the Earth is approximately 7,926 miles (12,742 kilometers) at the equator. However, the question specifically asks for the distance through the Earth, wich is a bit different.
To determine the distance through the Earth, we need to find the distance from the surface on one side, through the center of the Earth, to the surface on the opposite side. This distance is known as the Earth’s antipodal distance or Earth’s chord length.
The antipodal distance can be calculated by using the Pythagorean theorem, which states that in a right triangle, the square of the length of the hypotenuse (the longest side) is equal to the sum of the squares of the other two sides.
Using this theorem, we can calculate the antipodal distance of the Earth as follows:
– The mean radius of the Earth is approximately 3,958.8 miles (6,371 kilometers).
– Therefore, the diameter of the Earth (the distance from one side to the other passing through the center) is twice the mean radius, or approximately 7,917.6 miles (12,742 kilometers).
– To find the antipodal distance, we divide the diameter by 2 and square it. This gives us (7,917.6/2)^2, which is approximately 3,959.2 miles (6,371 kilometers).
Therefore, the distance through the Earth is approximately 3,959.2 miles (6,371 kilometers). It’s important to note that this distance can vary slightly depending on the location on the Earth’s surface and the shape of the planet, which is not a perfect sphere.
## How Wide is Earth Compared to its Height?
The Earth has a shape that is slightly flattened at the poles and bulging at the equator. This shape is known as an oblate spheroid. The equatorial diameter of the Earth is about 12,742 kilometers (7,918 miles), while the polar diameter is about 12,714 kilometers (7,900 miles). This means that the Earth is wider at the equator than it is at the poles by approximately 27 miles (43 kilometers).
To put this difference into perspective, if the Earth were scaled down to a globe with a diameter of 1 meter at the equator, the difference between the equatorial and polar diameters would be only 3 millimeters. This small difference is due to the Earth’s rotation, whih causes the equatorial region to bulge outwards and the polar regions to flatten slightly.
It is important to note that while the Earth’s shape is not a perfect sphere, it is still relatively close to being spherical. The difference between the equatorial and polar diameters is only about 1/298 of the equatorial diameter, which is a very small deviation from a perfect sphere.
The Earth is wider at the equator than it is at the poles by approximately 27 miles (43 kilometers), due to its oblate spheroid shape caused by its rotation.
## Conclusion
The Earth is a fascinating planet with a diameter of approximately 7,930 miles at the equator and a circumference of roughly 25,000 miles. The distance from the center of the Earth to its surface is about 3,958 miles. The Earth’s equatorial bulge is quite small, only about 27 miles wider than pole-to-pole, wich is about 1/298 of the equatorial diameter. The Earth’s size and shape have a significant impact on everything from its climate and geography to its gravitational pull and the behavior of its magnetic field. Understanding the dimensions of our planet is crucial not only for scientific research but also for everyday life, as it affects everything from navigation to agriculture.
William Armstrong
William Armstrong is a senior editor with H-O-M-E.org, where he writes on a wide variety of topics. He has also worked as a radio reporter and holds a degree from Moody College of Communication. William was born in Denton, TX and currently resides in Austin. | HuggingFaceTB/finemath | |
# Solved – Examples of when confidence interval and credible interval coincide
In the wikipedia article on Credible Interval, it says:
For the case of a single parameter and
data that can be summarised in a
single sufficient statistic, it can be
shown that the credible interval and
the confidence interval will coincide
if the unknown parameter is a location
parameter (i.e. the forward
probability function has the form Pr(x
| μ) = f(x − μ) ), with a prior that
is a uniform flat distribution; and
also if the unknown parameter is a
scale parameter (i.e. the forward
probability function has the form Pr(x
| s) = f(x / s) ), with a Jeffreys'
prior — the latter following
because taking the logarithm of such a
scale parameter turns it into a
location parameter with a uniform
distribution. But these are distinctly
special (albeit important) cases; in
general no such equivalence can be
Could people give specific examples of this? When does the 95% CI actually correspond to "95% chance", thus "violating" the general definition of CI?
Contents
## normal distribution:
Take a normal distribution with known variance. We can take this variance to be 1 without losing generality (by simply dividing each observation by the square root of the variance). This has sampling distribution:
\$\$p(X_{1}…X_{N}|mu)=left(2piright)^{-frac{N}{2}}expleft(-frac{1}{2}sum_{i=1}^{N}(X_{i}-mu)^{2}right)=Aexpleft(-frac{N}{2}(overline{X}-mu)^{2}right)\$\$
Where \$A\$ is a constant which depends only on the data. This shows that the sample mean is a sufficient statistic for the population mean. If we use a uniform prior, then the posterior distribution for \$mu\$ will be:
\$\$(mu|X_{1}…X_{N})sim Normalleft(overline{X},frac{1}{N}right)implies left(sqrt{N}(mu-overline{X})|X_{1}…X_{N}right)sim Normal(0,1)\$\$
So a \$1-alpha\$ credible interval will be of the form:
\$\$left(overline{X}+frac{1}{sqrt{N}}L_{alpha},overline{X}+frac{1}{sqrt{N}}U_{alpha}right)\$\$
Where \$L_{alpha}\$ and \$U_{alpha}\$ are chosen such that a standard normal random variable \$Z\$ satisfies:
\$\$Prleft(L_{alpha}<Z<U_{alpha}right)=1-alpha\$\$
Now we can start from this "pivotal quantity" for constructing a confidence interval. The sampling distribution of \$sqrt{N}(mu-overline{X})\$ for fixed \$mu\$ is a standard normal distribution, so we can substitute this into the above probability:
\$\$Prleft(L_{alpha}<sqrt{N}(mu-overline{X})<U_{alpha}right)=1-alpha\$\$
Then re-arrange to solve for \$mu\$, and the confidence interval will be the same as the credible interval.
## Scale parameters:
For scale parameters, the pdfs have the form \$p(X_{i}|s)=frac{1}{s}fleft(frac{X_{i}}{s}right)\$. We can take the \$(X_{i}|s)sim Uniform(0,s)\$, which corresponds to \$f(t)=1\$. The joint sampling distribution is:
\$\$p(X_{1}…X_{N}|s)=s^{-N};;;;;;;0<X_{1}…X_{N}<s\$\$
From which we find the sufficient statistic to be equal to \$X_{max}\$ (the maximum of the observations). We now find its sampling distribution:
\$\$Pr(X_{max}<y|s)=Pr(X_{1}<y,X_{2}<y…X_{N}<y|s)=left(frac{y}{s}right)^{N}\$\$
Now we can make this independent of the parameter by taking \$y=qs\$. This means our "pivotal quantity" is given by \$Q=s^{-1}X_{max}\$ with \$Pr(Q<q)=q^{N}\$ which is the \$beta(N,1)\$ distribution. So, we can choose \$L_{alpha},U_{alpha}\$ using the beta quantiles such that:
\$\$Pr(L_{alpha}<Q<U_{alpha})=1-alpha=U_{alpha}^{N}-L_{alpha}^{N}\$\$
And we substitute the pivotal quantity:
\$\$Pr(L_{alpha}<s^{-1}X_{max}<U_{alpha})=1-alpha=Pr(X_{max}L_{alpha}^{-1}>s>X_{max}U_{alpha}^{-1})\$\$
And there is our confidence interval. For the Bayesian solution with jeffreys prior we have:
\$\$p(s|X_{1}…X_{N})=frac{s^{-N-1}}{int_{X_{max}}^{infty}r^{-N-1}dr}=N (X_{max})^{N}s^{-N-1}\$\$ \$\$implies Pr(s>t|X_{1}…X_{N})=N (X_{max})^{N}int_{t}^{infty}s^{-N-1}ds=left(frac{X_{max}}{t}right)^{N}\$\$
We now plug in the confidence interval, and calculate its credibility
\$\$Pr(X_{max}L_{alpha}^{-1}>s>X_{max}U_{alpha}^{-1}|X_{1}…X_{N})=left(frac{X_{max}}{X_{max}U_{alpha}^{-1}}right)^{N}-left(frac{X_{max}}{X_{max}L_{alpha}^{-1}}right)^{N}\$\$
\$\$=U_{alpha}^{N}-L_{alpha}^{N}=Pr(L_{alpha}<Q<U_{alpha})\$\$
And presto, we have \$1-alpha\$ credibility and coverage.
Rate this post | HuggingFaceTB/finemath | |
## Friday, October 26, 2012 ... /////
### The holographic principle
The newest episode of The Big Bang Theory that was aired last night was called "The Holographic Excitation" (S06E05).
It's pretty cool that a TV sitcom manages not only to show a hologram but Leonard Hofstadter was even allowed to present a rather accurate definition of the holographic principle in quantum gravity i.e. string theory (you won't find it in any popular science TV program that claims to explain modern physics!). And as a result, he was able to have an intercourse with Penny right after she wore some glasses and was shown a moving holographic pencil and a moving holographic globe. (Later, he repeated the same achievement using Maglev.)
(And I even think that Prof Nina Byers whom I know rather well walks behind the main actors around 9:15. This theory seems to make sense because she's at UCLA, much like the TBBT science adviser David Saltzberg.)
The holographic principle of quantum gravity is an incredible example of the ability of the quantum gravity and string theory research to teach us things we really didn't and perhaps couldn't anticipate, force us to modify or abandon some prejudices, and adopt ideas about the unification of ideas and concepts that philosophers couldn't have invented after thousands of years of disciplined reasoning but physicists may be forced to realize them if they carefully follow the mathematical arguments sprinkling from a theory that they randomly discovered in a cave.
But let's return half a century into the past. Holography started in "everyday life physics" in the late 1940s.
So let us begin with this exercise in wave optics that has nothing to do with quantum gravity or string theory so far – but you will see that it exhibits a similar mechanism that is apparently "recycled" by the laws of quantum gravity.
Dennis Gabor's 3D images
Hungarian-British physicist Dennis Gabor was playing with X-ray microscopy and invented a new technology that is rather cute. One may create two-dimensional patterns on a piece of film which, when illuminated by a laser, create the illusion of a three-dimensional object floating in the space around it. I saw my first hologram sometime in 1985 – it was a Soviet one, the mascot of the 1980 Olympics in Moscow – in the National Technological Museum in Prague where we went to a school excursion. I couldn't believe my eyes. :-)
The basic setup involves a monochromatic laser beam, some interference, and a photographic plate. First, we must create the hologram – a film with strip-like patterns that don't resemble the bear at all but which allow the bear to jump out once you use another laser. Fine. Let's create a hologram.
You see that a monochromatic (one sharp frequency) laser beam is coming from the upper left corner. Each photon's wave function gets divided into two portions by a beam splitter – note that the wave function has a probabilistic interpretation for one particle but if many photons are in the same state, it may be interpreted as a classical field.
Two parts of the wave are moving from the beamsplitter. One gets reflected from a simple mirror. More interestingly, the other one gets reflected from the object we want to see on the hologram. They interfere – these waves are recombined – in the right lower corner and they create a system of interference strips on the photographic plate.
We have created a hologram and now we may sell it. What will the buyers do with it?
This interference pattern may be shined upon by a "reconstruction beam" of the same frequency and what we see is a virtual image behind the plate. You may actually move your head and eyes and the position of all points on the image are moving just like if the virtual image were a real object. So it's not just a stereographic image offering two different pictures for the two eyes: the hologram is ready to provide the right electromagnetic field regardless of the direction from which you observe it! If you want to see the right face of the object, you move your head to the right side, and so on.
Why does it work? It's very useful to think about the hologram for a simple object, e.g. a point at a given distance. The total wave function on the photographic plate parameterized by coordinates $x,y$ is given by $U_O + U_R$ where $U_O$ is the complicated wave reflected from the object and $U_R$ is the simple reference beam reflected from the plain mirror.
You may imagine that for an object being a point, $U_R=1$ and $U_O(x,y)=\exp(iks(x,y))$ where $s$ is the distance between the point (the real object we want to holographically photograph) and the given point $(x,y)$ on the photographic plate. The wave number is of course the inverse wavelength, $k=2\pi/\lambda$. The sum $U_R+U_O$ gives you some simple concentric circles (with decreasing distances between neighbors) around the point on the plate that is closest to the photographed point. Fine. The total intensity – how much the point on the film changes the color – is given by $T \sim \abs{U_O}^2 + \abs{U_R}^2 + U_R^* U_O + U_O^* U_R.$ I omitted an unimportant overall normalization and used the symbol $T$ for this quantity because the darkness of the point of the film will be interpreted as the transmittance, the ability of the place of the hologram to transmit the other, reconstruction beam when we actually want to reconstruct the image.
For a simple explanation why you will see the reconstructed virtual image, assume that the reference beam $U_R$ is much stronger than the object-induced wave $U_O$ i.e. $U_R\gg U_O$. So the total wave function may be written as $U = 1 + \varepsilon \exp(iks)$ where $\varepsilon$ is small, $\varepsilon\ll 1$. You see that the squared absolute value is$T \sim |U|^2 = 1 + \varepsilon \exp(iks) + \varepsilon \exp(-iks) + {\mathcal O}(\varepsilon^2).$ Imagine that this transmittance is just multiplying another simple reference beam $U_R\to T \cdot U_R$ and produces some electromagnetic field in the vicinity of the hologram. For the sake of simplicity, assume $U_R=1$ again. It's only $U_O$ that carries the "complicated information about the photographed object" but we still need some nonzero $U_R$.
You may see that $T$ is almost the same thing as $U$ except that it has an extra, complex conjugate term. So the electromagnetic field in front of the hologram (on the side with the air) will be the same field as the electromagnetic field we used to have when there was a real object in front of the hologram plus some complex conjugate term. One of these terms creates a nice virtual image behind the plate because it has a similar mathematical structure and when the fields have the same values, we see the same thing.
The other term induces the feeling of another copy of the object – a real image. It's because all the waves should also be multiplied by the universal time-dependent factor $\exp(-i\omega t)$ (before you interpret the real and imaginary value of the overall sum as the electric and magnetic fields, respectively, kind of) and the complex conjugation is equivalent to $t\to -t$ which means that the wave is kind of moving backwards in time which is effectively equivalent to moving from the other side of the mirror.
So when you look at the hologram, you actually see one virtual image behind the plate and one real image in front of the plate (which may overlap with your head). I don't want to figure out which term is which because odds would be close to 50% that my answer would be wrong. To be sure about the answer to this not-so-critical question, I would have to decompose the electromagnetic wave to the electric and magnetic components, consider $x$ and $y$ polarizations, be careful about the spatial dependence and all the signs, etc. But things clearly work up to this "which is which" question that I am not too interested in.
In this brute calculation, I have neglected the ${\mathcal O}(\varepsilon^2)$ terms which indicates that the hologram will be badly perturbed if $\varepsilon\sim {\mathcal O}(1)$ but a more accurate analysis shows that the result won't be too bad even if you include these second-order terms. At any rate, I have created a virtual image of a point! By the superposition principle, you are allowed to envision any object to be composed of many points (perhaps as an "integral of them") and add the terms $\exp(iks_P)$ from each point $P$ and you get the idea how it work for a general object.
There exist generalizations – colorful holograms, perhaps moving holograms and holographic TV, and so on, but I don't want to go into these topics on the boundary of physics and engineering. Everyone knows that holograms are cool. What's important for us is that they store much more than some two-dimensional projections of a 3D object as seen from one direction or two directions; they store the information about the 3D object as seen from any direction (in an interval). They're the whole thing.
Instead of discussing advanced topics of holography in wave optics, we want to switch to the real topic, the holographic principle in quantum gravity.
The holographic principle
In the research of quantum gravity, the notion of holography was introduced by somewhat speculative but highly playful papers by Gerard 't Hooft in 1993 and Lenny Susskind in 1994. Charles Thorn is mentioned as having suspected similar ideas for years.
It may sound unusual ;-) but Lenny Susskind's paper was the technically more detailed one, getting well beyond the hot philosophical buzzwords. Susskind also suppressed some unjustified and unjustifiable "digital" comments by 't Hooft who had written that the information had to be encoded in binary digits (bits). Of course, there's no reason whatsoever why it couldn't be trinary digits, other digits, or – much more likely – (for humans and computers) some much less readable but more natural codes.
What's the basic logic behind holography in quantum gravity?
In classical general relativity, a black hole is the final stage of the collapse of a star or another massive object. Because the entropy never decreases, as the second law of thermodynamics demands, the "final stage" must also be the stage with the maximum entropy. So the black hole has the highest entropy among all bound or localized objects of the same mass (and the same values of charges and the angular momentum). I emphasize the adjectives "bound or localized" because delocalized arrangements of particles with a given total energy – e.g. the Hawking radiation resulting from a black hole that has already evaporated – may carry a higher entropy (that's inevitably the case because the process of Hawking radiation must be increasing the total entropy, too).
But we've known from the insights by Jacob Bekenstein and Stephen Hawking in the 1970s that the black hole entropy is$S_{BH} = \frac{A}{4G}$ in the relativistic $c=\hbar=1$ units. It's one-quarter of the area of the event horizon $A$ in the units of the Planck area. In normal units, you must replace$G \to l_{\rm Planck}^2 \equiv \frac{G\hbar}{c^3}.$ So the maximum entropy of a bound localized object of a given mass is actually given by the area of the black hole of the same mass. Because you can't really squeeze the matter into higher densities than the black hole, the black hole is also the "smallest object" that may contain the given mass.
To summarize, we see that the black hole is the "highest entropy" object as well as the "geometrically smallest" object among localized or bound objects of the given mass. It follows that it also maximizes the "entropy density" (entropy per unit volume) among the localized arrangement of matter of the same total mass. But the entropy carried by a black hole is only proportional to the surface area in the Planck units, ${\mathcal O}(R^2)$, so the entropy density per unit volume – the latter scales as ${\mathcal O}(R^3)$ – is therefore going to zero for large black holes i.e. for large masses or large regions.
The maximum density of entropy or information you may achieve with a given mass is actually going to zero if the mass is sent to infinity. If you try to squeeze too many memory chips into your warehouse, they will start to be heavy at some point and will gravitationally collapse and create a black hole which will have a certain radius – either smaller than or larger than your warehouse. At any rate, this black hole will only be able to carry $1/4$ of a nat (a bit is $\ln(2)$ nats) of information per unit surface area (by the surface, I mean the event horizon).
We see that the maximum information is carried by a constant density per unit area rather than the unit volume. You should appreciate how shocking it is. In some sense, it was completely unexpected by virtually all experts in the field. Quantum field theories predict some new phenomena at a characteristic distance scale. For example, Quantum Chromodynamics (QCD) says that quarks like to bind themselves into bound states where their distance is comparable to the QCD length scale, about one fermi or $10^{-15}$ meters. So by the dimensional analysis, the only sensible "density of information" we may get in QCD is "approximately one bit per cubic fermi" or per "volume of the proton".
People would expect a similar thing in any QFT – which was mostly right – but they thought it would also hold in quantum gravity. So quantum gravity may achieve "one bit per Planck volume". But that was wrong. You see that the previous paragraph assumed a bit more than the dimensional analysis: it also implicitly and uncritically postulated that the information is proportional to the volume. This assumption followed from locality. But this assumption breaks down in quantum gravity where the information only scales as the surface area.
Because the "proportionality to the volume" is linked to "locality" – each unit volume is independent from others – the violation of the "proportionality of the information to the volume" that the holographic principle forces upon us also means that locality is violated, at least to some extent. And indeed, this violation of the locality is a fact responsible for the resolution of other puzzling questions in quantum gravity, too. In particular, some tiny and hard to observe but nevertheless real non-locality occurs during the evaporation of the black hole which is why the information may get from the black hole interior to infinity, after all – even though classical general relativity strictly prohibits such an acausal export of the information (locally, it's equivalent to the superluminal transport of information which was already banned in special relativity). In quantum gravity, this "ban" is softened because the information may temporarily violate the rule in analogy with the quantum tunneling. In fact, the black hole evaporation is a version of quantum tunneling.
Whether the holographic principle was real and what it exactly it meant and what it didn't mean remained a somewhat open question for 3 more years or so. However, at the end of 1997, Juan Maldacena presented his AdS/CFT correspondence which is a set of totally controllable mathematical frameworks in which holography holds. The information about a region – namely the whole anti de Sitter space – is stored at the boundary of the region – which is the asymptotic region at infinity which nevertheless looks like a "finite surface of a cylindrical Penrose diagram" if you use the language of Penrose causal diagrams.
The holographic principle surely captures the right "spirit" of quantum gravity but it is a bit vague. The AdS/CFT correspondence is a totally well-defined "refinement" of the holographic principle but it is arguably too special. Nevertheless, one must be careful about deriving potentially invalid corollaries of the holographic principle in other contexts.
For example, if you replace the anti de Sitter space by a finite-volume region of ordinary space, it seems clear to me that the holographic principle will only be true in some rather modest sense: it will be true that the entropy bounds hold. You can't squeeze too much entropy into a given region. However, if you will try to find the "theory on the boundary" that is equivalent to the evolution inside the region, you will find out that such a theory on the boundary "exists" – but the existence of such a theory is just an awkward translation of the ordinary evolution to some artificial degrees of freedom that you placed on the boundary.
What is special about the AdS/CFT correspondence is that the theory on the boundary is a theory of a completely normal type – namely a perfectly local, conformal quantum field theory. In fact, the boundary theory is more local than the gravitational theory in the bulk – because we just said that the gravitating theory in the bulk must be somewhat non-local. I am confident this fact depends on the infinite warp factor of the AdS space at infinity and won't hold for finite regions. In other words, I think that the "holographic theory living on a boundary" of a generic finite region won't be local in any sense – the boundary still has a preferred length scale, the Planck length, and other things so it is surely not conformal etc. And because it won't be local, it won't be simple or useful, either.
So one shouldn't generalize the holographic principle as seen in the AdS/CFT correspondence too far and too naively.
Lessons
In the 1970s, people got used to Ken Wilson's "Renormalization Group" inspired thinking about all effective field theories. Each theory predicted some phenomena at a characteristic length scale. The third power of the length scale gave us a characteristic volume. And one could expect roughly one nat (or bit) per one characteristic volume. It was nice, it made sense, it has lots of applications.
But Nature sometimes has surprises in store and quantum gravity had one, too. You may still use almost the same logic – one nat per unit region – but the region must actually be measured by its surface area, not its volume. So quantum gravity tells us that one of the spatial dimensions may be thought of as an "artificial" or "emergent" one and other mechanisms supporting this general paradigm have appeared as well.
A brutally arrogant yet extremely limited physicist who really sucks – think of Lee Smolin, for example – may think that he has all the right ideas how the final theory should look like from the beginning. Except that none of them works (except as tools to impress some stupid laymen). But other physicists who are much smarter but much more modest may see that all Smolin's prejudices are just wrong and Nature's inner organization is much more clever, creative, surprising, and forcing us to learn new concepts and new way of thinking more often than Smolin and many others would expect. One must still be ingenious or semi-ingenious to discover some important wisdom about Nature – e.g. holography and the AdS/CFT correspondence – but Nature just doesn't appreciate men who try to paint themselves as wiser than herself. Science is the process of convergence towards Her great wisdom; it is not a pissing contest in which idiots such as Lee Smolin try to pretend that they're smarter than Nature.
The story of the holographic principle also shows us that Nature recycles many ideas. The fields defined on the boundary CFT in the AdS/CFT correspondence literally emulate the waves $U$ and $T$ that I mentioned in the discussion of the "ordinary" holography by Dennis Gabor.
And the story of the holographic principle is another anecdotal piece of evidence in favor of the assertion that string/M-theory contains all the good ideas in physics. 't Hooft and Susskind, building on the work by Bekenstein, Hawking, and others, had some "feelings" about the right theory of quantum gravity and there had to be something right about them. And indeed, string theory showed us that they were mostly right. Because string theory is a much more mathematically well-defined a structure than "quantum gravity without adjectives", it also allowed us to convert the philosophical speculations into sharp and rigorous mathematical structures and equations and decide which of the philosophical speculations may be proven as meaningful ones and which can't.
The holographic principle is also another step in the evolution of physics that makes our theories "increasingly more quantum mechanical". While the spacetime remains continuous, we see that the information in a region may be bounded in unexpected ways and a whole dimension of space may be emergent. Needless to say, the equivalence between theories that disagree about the number of spacetime dimensions is only possible if you take the effects of quantum mechanics into account.
#### snail feedback (45) :
Oh yeah yeah yeah, I look forward to read this :-)
... and darn, I want to see the whole TBBT episode about the holographic excitation too :-D !
Holographic principle was also hypothesized as a mechanism for memory at some point. Karl Pribram (neurosurgeon) and David Bohm (physicist) postulated some kind of holographic principle behind brain function because they noticed that memory is not localized in brain. If you remove some portions of the brain, memory stays mostly intact, that is why they thought that memory is stored as a hologram (if you break a hologram, every piece has an information about the whole). It is probably a bad analogy at best. memory is stored in the connection weights of synapses
http://en.wikipedia.org/wiki/Holonomic_brain_theory
reader Smoking Frog said...
Lubos, your first three paragraphs are faultless so far as the English goes, except one error, but the error is comically brilliant: "an intercourse." It should be "intercourse" (without the article), but please don't think of fixing the error in the future because "an intercourse" is just too good. Use it all you want.
What's the difference ??
The new season hasn't fully arrived in Ireland yet (and the dubbing is so bad in France... ).. My son says we can see in on the internet free TV. yeahy
ss
reader Luboš Motl said...
Thanks, Smoking Frog, for the fix. This is just way beyond my linguistic skills. I've been taught to repeat many times that one is having a shower, we may have a walk, we did have a lunch, and so on, so I can't possibly understand why one can't have an intercourse as well. Is it because the native English speakers are puritans of a sort? ;-) So that if they mention the word at all, it must be formulated in such a way that one can't count them? :-)
reader Smoking Frog said...
Good question. On the grammar, I would have been too vague if I hadn't just now looked at the Wikipedia article "Mass noun," and I'm still not sure that "intercourse" is a mass noun or in some similar category (but not a collective noun).
"Have an intercourse" is something like "have a peanut butter"; to use an article, you'd have to say anything like "a spoon of peanut butter." For "intercourse," you could say "an act of intercourse," except that one does not "have" acts; one performs acts; so you pretty much have to say "have intercourse."
So the answer to your question is that there's no such thing as "an intercourse." But this doesn't really explain why it's so funny. Scratching my head ...
A bit OT. When you talk about quantum tunelling from the BH, do you mean tunelling whose result is ordinary Hawking radiation or a corrected (non-thermal) radiation? Here http://arxiv.org/abs/hep-th/9907001 the authors argue that a tunelling mechanism leads to a non-thermal correction.
reader Smoking Frog said...
Hah-hah as to the Puritans. No, that's not it. See my reply to Shannon.
reader Luboš Motl said...
Once I wrote an article about the Holographic Principle to "Vesmír", a Czech journal for a broader scientific public, and it was cited by some of the Příbram's followers. That was fun because I do think theirs is a pseudoscience unrelated to the actual topic of my article - this one or the old one in "Vesmír".
reader Smoking Frog said...
Lubos - "Have a lunch" is an interesting case; you can also say "have lunch," and that's what most people say. There is something of a difference of meaning, but it's kind of fuzzy. If I say, "We had a lunch," I might mean that we hosted a lunch. So there's a difference. But I could also say it if you and I had lunch together. In that case, it would mean almost just the same as "We had lunch," except that I'd be thinking of the lunch as an event rather than simply "having lunch"; e.g., "We did have a lunch on one occasion."
reader Smoking Frog said...
Hah-hah.
reader Peter F. said...
Lubos, you've been taught wrong in respect of "having _a_ lunch". :-))
reader Luboš Motl said...
Maybe poor people from the post-communist bloc are only taught to have a lunch while the native English speakers are given lots of lunch... ;-)
reader Luboš Motl said...
Hi Lemon, in general, the spectrum seen at infinity is corrected by something that is called "greybody factors" indicating it's not exactly blackbody spectrum. Another question is whether the particular unusual calculation of them in the paper you mentioned is right.
For this reason, I think that your language "do you mean one or the other" is kind of strange because the exact spectrum emitted by a well-defined object in a well-defined theory is clearly uniquely determined so "if someone meant the other one", he would just be wrong. And I didn't say what the exact spectrum was, so I explicitly mentioned neither the right one nor the wrong one. But of course I meant the correct one whichever it is. ;-)
reader Peter F. said...
Haha, but it is better to forego having lunch if it will only consist if corn syrup containing junkfoods. ;)
What makes me laugh about the word "intercourse" is that it sounds like something you'd be served during a meal. So "an intercourse" would be perfect between say the starter and the main course... or the main course and cheese, or desert... right ? ;-) My French upbringing probably ? ha ha
That’s really funny, Lubos, and it is wonderful to see your sense of humor re-emerge and blossom again!
reader Physics Junkie said...
I am showing my ignorance here, but is there some connection between Ads/CFT and the dualities in string theory that link different dimensional branes?
reader Eugene S said...
To make things even more confusing, in the U.S. people say "entree" but they mean the mean the "main course"; in France, "l'entrée" is the appetizer!
Savages...
reader Luboš Motl said...
Thanks, interesting, Smoking Frog.
reader Luboš Motl said...
Thanks for liking this change, Gene. It's a relief if one doesn't have to spend the bulk of his energy by thoughts about the bare survival.
reader Luboš Motl said...
Yes, there are many connections although perhaps a bit different than you expect.
First, the duality relating different dimensional D-branes is T-duality. There's an interesting T-duality one may apply to the theories in AdS/CFT, the most famous example of the N=4 gauge theory, and it maps the conformal symmetry to the dual conformal symmetry. Those can be combined into their (multiple) commutators to generate the infinite-dimensional Yangian, a bizarre form of symmetry known in the "integrable models".
Then the AdS/CFT correspondence is a sort of a duality itself. On the world sheet, it's most visibly visualized as the world sheet duality applied to a cylinder/annulus. It can be either interpreted as a closed string state propagating between a final and an initial state, or a loop of an open string in periodic time. AdS/CFT produces an "open/closed" duality in this sense: the AdS is the closed string side, CFT is the open string theory.
reader James Gallagher said...
Nice article, covering classical holography and QG Holographic Principle.
Because the HP suggests reality is (fundamentally) encoded on a 2D surface rather than in a 3D volume - it kinda naively explains why complex numbers, and in particular complex Hilbert Spaces, are required. A general 2D surface is parameterized by two real parameters or one complex parameter z. So we might deduce from this that the universe can be described by a simple (in)finite complex vector, evolving in a Hilbert Space.
I know it's a naive argument, but it suggests that the Holographic Principle is a VERY important discovery,
reader James Gallagher said...
actually, parameterised isn't a good word there, I mean a general 2D surface has only 2 degrees of freedom, which we may associate with two real numbers or one complex number
reader James Gallagher said...
oh, you know what I mean!
We don't have information for every 3D point (x,y,z) within the 2D surface only for each point on the 2D surface.
That's funny, and reminds me of a happy memory. Years ago at friendly gatherings we used to play Dictionary and the player with the best command of the English language was an Indian engineer.
Dictionary is a game where one person picks a word out of the dictionary, pronounces it for everyone who then write their own definitions and the other players vote to try and decide which definition is the real one. Scoring is based on how many people are fooled by phony definitions.
reader Smoking Frog said...
Eugene - I don't know French, but I've always wondered how "entree" could mean the main course!
reader Smoking Frog said...
You mean in France everyone has sex between the courses of a meal!!!??? :-)
reader Luboš Motl said...
In America, "entree" is the main course because after some foreplay appetizers and intercourses of the kind that "I didn't have with that woman", the entry finally comes.
reader Smoking Frog said...
Lubos, your English is already very good, so it's nothing to worry about.
It's sort of amazing that these subtle difficulties in language exist. You don't realize it unless you think about it. It's sort of hard to believe stories of spies who passed as native speakers.
reader Luboš Motl said...
I guess it must be possible for some adults to learn a foreign language just like the native speakers among children do - even though most people probably don't have the aptitude for such "full learning" of other languages.
The children just can't have any "unrepeatable" skills, can they? Learning many languages as an adult requires one to have "independent linguistic brains" of a sort that preserve "inconsistent" answers to pretty much the same questions.
reader Smoking Frog said...
I would like to be told how it works that a kid learns such things that not too many people could explain as clearly as you.
I forgot that part. I don't know the answer, but it might be related in some way to thinking about how other people think. As a kid in the lower grades of elementary school, when the teacher was trying to explain something to a dumb kid and not succeeding because she didn't understand his trouble, I was always waving my hand and saying, "Miss [X], I know what's he's thinking."
reader Smoking Frog said...
I agree that aptitude must play a big role. There are people whose grasp of language issues is horribly poor.
I'm not sure whether I've told this story here before: Maybe 25 years ago there was an interview in The Atlantic(?) with a big fat guy who was supposed to be the world's greatest linguist. He spoke many languages. The interviewer asked him if he could speak Finnish, and the linguist said something like this: "No. Or, well, I could probably pass as a Finn for 10 or 15 minutes in a conversation, but that's not what I call speaking a language."
reader Luboš Motl said...
LOL.
reader Smoking Frog said...
Maybe spies can pass as native speakers because the natives' ability to detect that he's not one are poor. :-)
This is a very nice article indeed :-),
the cute explanation of holography meaning 3D images makes me feel quite happy and satisfied.
Concerning the quantum gravity ;-) case, my understanding of it is way to vague and shaky to satisfy my.
To be happy, I always need to see how something (roughly at least) works mathematically,
instead of just words and analogies ...
Anyway, I like this text a lot so thanks Lumo !
Interesting. I guess we say "social interactions" rather than "social intercourse"... It avoids silly giggles.
Oh and no, French don't have sex between courses of a meal, the reason being that eating and sex are two highly incompatible activities :-D
reader Smoking Frog said...
We say "social interactions," too. "Social intercourse" is usually found in writing, and perhaps not even so much that, nowadays.
Incompatible? You can eat between rounds of sex. This is like the story of the Buddhist monk who asked the master if he could smoke while he was praying. The master said, "Of course not!" So the monk asked him if he could pray while he was smoking.
reader Rosy Mota said...
it appear to be correct.particles,energy,fields and forces mudt to be originated of 4-dimensional spacetime topological geometry
reader Carolina Lou said...
Very good topic. It
helps all to gain many more knowledge of holographic principle. it covered 2D
and 3D projection. | open-web-math/open-web-math | |
# Convert MM to Inches Formula – Examples, FAQs, How to Convert
Home » Math Vocabulary » Convert MM to Inches Formula – Examples, FAQs, How to Convert
## What Is the mm to Inches Formula?
The mm to inches conversion formula is based on the fact that there are 25.4 millimeters in an inch. Simply divide the number of millimeters by 25.4 to get the equivalent length in inches.
The formula to convert mm to inches (millimeters to inches) is given by:
Inches = Millimeters 25.4
Understanding this conversion is critical if you work with measures in the metric system and need to present those measurements in the imperial system. The conversion formula for mm to inches is critical in this procedure, providing accurate and effortless conversions between the two.
What is millimeter?
A millimeter (mm) is a unit of length in the metric system, which is commonly used for measuring small distances. It is equal to one-thousandth of a meter, making it a very small unit of measurement.
What is inch?
An inch (in) is a unit of length commonly used in the United States and a few other countries that do not primarily use the metric system. It is equal to 1/12th of a foot or approximately 2.54 centimeters. The inch is used for measuring shorter lengths, such as the height of a person, the width of a book, or the size of a screen. It is a larger unit than the millimeter and is often found in everyday applications, especially in the United States, for measuring various objects and distances.
Observe the given ruler. It shows measurements in inches at the top and mm at the bottom. The smallest division at the bottom represents 1 mm. Notice that the 1 inch mark roughly lines up with the 25.4 mm mark.
## Formula to Convert mm to Inches
The millimeter to inches conversion formula depends on the following equations:
• 1 mm = 0.0393701 inches
• 1 inch = 25.4 mm
1 mm=125.4 inches = 0.03937 inches
The formula to convert millimeters (mm) to inches is
Inches =mm25.4
N millimeters=N × 0.03937 inches
## How to Convert Millimeters into Inches
We can easily convert measurements in mm to inches by-
• Note down the given length in millimeters.
• Divide the length in millimeters by 25.4.
OR
Multiply the length in millimeters by 0.03937.
• Write the resulting value in inches.
Example: If you have 100 millimeters, you can calculate it as:
Inches = 100 mm 25.4 = 3.93700787 inches (approximately 3.94 inches when rounded to two decimal places).
## The mm to Inches Conversion Table
Here is a conversion table to instantly convert mm to Inches. These values in the table are rounded to four decimal places. You may use the formula to convert mm to inches for more precise measurements.
## Facts about the mm to Inches Conversion Formula
• The conversion factor for the mm to inches conversion is 25.4.
• The millimeter is a smaller unit of measurement than the inch, with approximately 0.0394 inches in one millimeter.
## Conclusion
In this article, we learned about the conversion from millimeters to inches, an important measurement skill for everyday measurements. The mm to inches conversion requires simple calculations. You just have to know the conversion factor or you can even use the conversion table provided above for a quick check. Let’s solve a few examples and practice multiple-choice questions (MCQs) for better comprehension.
## Solved Examples on the mm to Inches Conversion Formula
Example 1: Convert 50 mm into inches.
Solution:
Let’s use the mm to inches conversion formula.
Length in inches = Length in mm ÷ 25.4
Length in inches ≈ 1.9685 inches (rounded to four decimal places)
Example 2: Compare 75 millimeters to 3 inches.
Solution:
To compare these two measurements, they must be in the same unit.
Let us convert 75 millimeters into inches.
Inches = 75 mm x 0.03937 ≈ 2.9528 inches (rounded to four decimal places)
Thus, 75 mm ≈ 2.9528 inches
2.9528 inches < 3 inches
Thus, 75 mm < 3 inches
Example 3: What is 120 millimeters in inches?
Solution:
120 mm =120 x 0.03937 inches
120 mm ≈ 4.72 inches (rounded to two decimal places)
Therefore, the measurement in inches is approximately 4.72 inches.
Example 4: Convert 150 mm to inches.
Solution:
Inches $= \frac{mm}{25.4}$
Inches $= \frac{150\; mm}{25.4}$
150 mm = 5.9055 inches
Example 5: What will be 300 millimeters in inches?
Solution:
Inches $= \frac{mm}{25.4}$
Inches $= \frac{300\; mm}{25.4}$
300 mm = 11.8110 inches
## Practice Problems on the mm to Inches Conversion Formula
1
### Which one is the correct formula to convert millimeters (mm) to inches?
Inches $= \frac{mm}{25.4}$
Inches = mm x 25.4
Inches = mm + 25.4
Inches = mm - 25.4
CorrectIncorrect
Correct answer is: Inches $= \frac{mm}{25.4}$
Since one inch is equal to 25.4 mm, the correct formula used to convert millimeters to inches is shown in option (a).
2
### What will be 10 mm in inches?
0.3937 inches
0.0394 inches
1.5748 inches
0.7874 inches
CorrectIncorrect
Inches $= \frac{10\; mm}{25.4}$
10 mm ≈ 0.3937 inches (rounded to four decimal places)
3
10 mm
25.4 mm
100 mm
50 mm
CorrectIncorrect
1 in = 25.4 mm
4
### To convert millimeters to inches, you
divide by 0.0394
multiply by 25.4
divide by 25.4
CorrectIncorrect
Correct answer is: divide by 25.4
To convert millimeters to inches, you divide by 25.4.
There are 25.4 millimeters in an inch. For a quick conversion and estimation, we can divide the length in millimeters by 25. For example, 100 mm is approximately equal to $100 \div 25 = 4$ inches. | HuggingFaceTB/finemath | |
Solutions by everydaycalculation.com
## Multiply 3/75 with 30/7
1st number: 3/75, 2nd number: 4 2/7
This multiplication involving fractions can also be rephrased as "What is 3/75 of 4 2/7?"
3/75 × 30/7 is 6/35.
#### Steps for multiplying fractions
1. Simply multiply the numerators and denominators separately:
2. 3/75 × 30/7 = 3 × 30/75 × 7 = 90/525
3. After reducing the fraction, the answer is 6/35
MathStep (Works offline) Download our mobile app and learn to work with fractions in your own time: | HuggingFaceTB/finemath | |
# How do you find sin 2θ, given cos θ = 5/13 and θ lies in Quadrant IV?
Nov 18, 2015
Find sin 2x, knowing $\cos x = \frac{5}{13}$
Ans: $- \frac{120}{169}$
#### Explanation:
$\cos x = \frac{5}{13.}$ Find sin x.
${\sin}^{2} x = 1 - {\cos}^{2} x = 1 - \frac{25}{169} = \frac{144}{169}$ --> $\sin x = \pm \frac{12}{13}$.
Since x is in Quadrant IV, then sin x is negative.
$\sin x = - \frac{12}{13}$.
$\sin 2 x = 2 \sin x . \cos x = 2 \left(- \frac{12}{13}\right) \left(\frac{5}{13}\right) = - \frac{120}{169}$ | HuggingFaceTB/finemath | |
# $Y = \frac{X_1 X_2}{X_3}$ where $X_i$ is a uniform random variable
$$Y = \frac{X_1 X_2}{X_3}$$ where $$X_i\sim U(0,1)$$ and $$X_1,X_2,X_3$$ are i.i.d
I need to calculate $$Var(Y)$$ and $$Var[Y|X_3=1.7]$$
I know that for each $$X_i$$,
$$E[X_i]=\frac{1}{2}$$
$$Var[X_i]=\frac{1}{12}$$
But I'm not shure how to proceed, neither do I know how to calculate the PDF of Y.
¿Tips?
I know that $$f_{X_1,X_2, X_3}(x_1, x_2, x_3) = 1 \Bbb1_{(0,1) x (0,1), (0,1)} (x_1, x_2, x_3)$$
I thought that since they are independen maybe there were some tricks derived from the expected value, since $$E[X_1 X_2] = E[X_1] E[X_2]$$
• @SaucyO'Path Well, since the three are i.i.d., $f_{X_1, X_2, X_3}(x_1, x_2, x_3) = 1 for(0,1) x (0,1) x (0,1).$ – Leslie Brenes May 21 at 22:45
• As long as you state it. – Saucy O'Path May 21 at 22:46
• @LeslieBrenes when did you mention independence ? – Graham Kemp May 21 at 22:46
• $X_3=1.7$ is a null event, especially since $X_3$ is supposed to be uniform(0,1). – user10354138 May 21 at 22:49
• $\frac{X_1X_2}{X_3}$ is not $L^1$, therefore $\operatorname{Var}(X)$ is not defined. – Saucy O'Path May 21 at 23:08
With that result, you can easily answer your second question, which is basically $$\frac{1}{1.7^2} \mathrm{Var}[X_1 \, X_2]$$.
Again using the result in the linked answer, the first question boils down to $$\mathrm{Var}[X_1 \, X_2 \, Z_3]$$ where $$Z_3$$ follows an Inverse uniform distribution. Please note that the variance is infinite in your case (while it would be ok if you had $$X_3 \sim U(a,b)$$ with $$a>0$$). | HuggingFaceTB/finemath | |
# Coprime
Coprime
In number theory, a branch of mathematics, two integers a and b are said to be coprime (also spelled co-prime) or relatively prime if the only positive integer that evenly divides both of them is 1. This is the same thing as their greatest common divisor being 1.[1] In addition to $\gcd(a, b) = 1\;$ and $(a, b) = 1,\;$ the notation a $\perp$ b is sometimes used to indicate that a and b are relatively prime.[2]
For example, 14 and 15 are coprime, being commonly divisible by only 1, but 14 and 21 are not, because they are both divisible by 7. The numbers 1 and −1 are coprime to every integer, and they are the only integers to be coprime with 0.
A fast way to determine whether two numbers are coprime is given by the Euclidean algorithm.
The number of integers coprime to a positive integer n, between 1 and n, is given by Euler's totient function (or Euler's phi function) φ(n).
## Properties
Figure 1. The numbers 4 and 9 are coprime. Therefore, the diagonal of a 4 x 9 lattice does not intersect any other lattice points
There are a number of conditions which are equivalent to a and b being coprime:
As a consequence of the third point, if a and b are coprime and brbs (mod a), then rs (mod a) (because we may "divide by b" when working modulo a). Furthermore, if b1 and b2 are both coprime with a, then so is their product b1b2 (modulo a it is a product of invertible elements, and therefore invertible); this also follows from the first point by Euclid's lemma, which states that if a prime number p divides a product bc, then p divides at least one of the factors b, c.
As a consequence of the first point, if a and b are coprime, then so are any powers ak and bl.
If a and b are coprime and a divides the product bc, then a divides c. This can be viewed as a generalization of Euclid's lemma.
The two integers a and b are coprime if and only if the point with coordinates (a, b) in a Cartesian coordinate system is "visible" from the origin (0,0), in the sense that there is no point with integer coordinates between the origin and (a, b). (See figure 1.)
In a sense that can be made precise, the probability that two randomly chosen integers are coprime is 6/π2 (see pi), which is about 61%. See below.
Two natural numbers a and b are coprime if and only if the numbers 2a − 1 and 2b − 1 are coprime. As a generalization of this, following easily from Euclidean algorithm in base n > 1:
gcd(na − 1,nb − 1) = ngcd(a,b) − 1.
## Cross notation, group
If n≥1 and is an integer, the numbers coprime to n, taken modulo n, form a group with multiplication as operation; it is written as (Z/nZ)× or Zn*.
## Generalizations
Two ideals A and B in the commutative ring R are called coprime (or comaximal) if A + B = R. This generalizes Bézout's identity: with this definition, two principal ideals (a) and (b) in the ring of integers Z are coprime if and only if a and b are coprime.
If the ideals A and B of R are coprime, then AB = AB; furthermore, if C is a third ideal such that A contains BC, then A contains C. The Chinese remainder theorem is an important statement about coprime ideals.
The concept of being relatively prime can also be extended to any finite set of integers S = {a1, a2, .... an} to mean that the greatest common divisor of the elements of the set is 1. If every pair in a (finite or infinite) set of integers is relatively prime, then the set is called pairwise relatively prime.
Every pairwise relatively prime finite set is relatively prime; however, the converse is not true: {6, 10, 15} is relatively prime, but not pairwise relative prime (in fact, each pair of integers in the set has a non-trivial common factor).
## Probabilities
Given two randomly chosen integers A and B, it is reasonable to ask how likely it is that A and B are coprime. In this determination, it is convenient to use the characterization that A and B are coprime if and only if no prime number divides both of them (see Fundamental theorem of arithmetic).
Intuitively, the probability that any number is divisible by a prime (or any integer), p is 1 / p (for example, every 7th integer is divisible by 7.) Hence the probability that two numbers are both divisible by this prime is 1 / p2, and the probability that at least one of them is not is 1 − 1 / p2. Now, for distinct primes, these divisibility events are mutually independent. (This would not, in general, be true if they were not prime.) (For the case of two events, a number is divisible by p and q if and only if it is divisible by pq; the latter has probability 1/pq.) Thus the probability that two numbers are coprime is given by a product over all primes,
$\prod_p^{\infty} \left(1-\frac{1}{p^2}\right) = \left( \prod_p^{\infty} \frac{1}{1-p^{-2}} \right)^{-1} = \frac{1}{\zeta(2)} = \frac{6}{\pi^2} \approx 0.607927102 \approx 61%.$
Here ζ refers to the Riemann zeta function, the identity relating the product over primes to ζ(2) is an example of an Euler product, and the evaluation of ζ(2) as π2/6 is the Basel problem, solved by Leonhard Euler in 1735. In general, the probability of k randomly chosen integers being coprime is 1/ζ(k).
There is often confusion about what a "randomly chosen integer" is. One way of understanding this is to assume that the integers are chosen randomly between 1 and an integer N. Then, for each upper bound N, there is a probability PN that two randomly chosen numbers are coprime. This will never be exactly 6 / π2, but in the limit as $N \to \infty$, $P_N \to 6/\pi^2$.[3]
## Generating all coprime pairs
All pairs of coprime numbers m,n can be generated in a parent-3 children-9 grandchildren... family tree starting from (2,1) (for even-odd or odd-even pairs)[4] or from (3,1) (for odd-odd pairs),[5] with three branches from each node. The branches are generated as follows:
Branch 1: mnext = 2mn, and nnext = m
Branch 2: mnext = 2m + n, and nnext = m
Branch 3: mnext = m + 2n, and nnext = n
This family tree is exhaustive and non-redundant with no invalid members.
## References
1. ^ G.H. Hardy; E. M. Wright (2008). An Introduction to the Theory of Numbers (6th ed. ed.). Oxford University Press. p. 6. ISBN 9780199219865.
2. ^ Graham, R. L.; Knuth, D. E.; Patashnik, O. (1989), Concrete Mathematics, Addison-Wesley
3. ^ This theorem was proved by Ernesto Cesàro in 1881. For a more rigorous proof than the intuitive and informal one given here, see G.H. Hardy; E. M. Wright (2008). An Introduction to the Theory of Numbers (6th ed. ed.). Oxford University Press. ISBN 9780199219865. , theorem 332.
4. ^ Saunders, Robert & Randall, Trevor (July 1994), "The family tree of the Pythagorean triplets revisited", Mathematical Gazette 78: 190–193 .
5. ^ Mitchell, Douglas W. (July 2001), "An alternative characterisation of all primitive Pythagorean triples", Mathematical Gazette 85: 273–275 .
• Lord, Nick (March 2008), "A uniform construction of some infinite coprime sequences", Mathematical Gazette 92: 66–70 .
Wikimedia Foundation. 2010.
### Look at other dictionaries:
• coprime — adjective a) Having no positive integer factors in common, aside from 1. 24 and 35 are coprime. b) Having no positive integer factors, aside from 1, in common with one or more specified other positive integers. 24 is coprime to 35. Syn: relativ … Wiktionary
• Pairwise coprime — In mathematics, especially number theory, a set of integers is said to be pairwise coprime (or pairwise relatively prime, also known as mutually coprime) if every pair of integers a and b in the set are coprime (that is, have no common divisors… … Wikipedia
• Chinese remainder theorem — The Chinese remainder theorem is a result about congruences in number theory and its generalizations in abstract algebra. In its most basic form it concerned with determining n, given the remainders generated by division of n by several numbers.… … Wikipedia
• Transposable integer — A summary of this article appears in Repeating decimal. The digits of some specific integers permute or shift cyclically when they are multiplied by a number n. Examples are: 142857 × 3 = 428571 (shifts cyclically one place left) 142857 × 5 =… … Wikipedia
• Euler's totient function — For other functions named after Euler, see List of topics named after Leonhard Euler. The first thousand values of φ(n) In number theory, the totient φ(n) of a positive integer n is defined to be the number of positive integers less than or equal … Wikipedia
• abc conjecture — The abc conjecture (also known as Oesterlé–Masser conjecture) is a conjecture in number theory, first proposed by Joseph Oesterlé and David Masser in 1985. The conjecture is stated in terms of three positive integers, a, b and c (whence comes the … Wikipedia
• Arithmetic function — In number theory, an arithmetic (or arithmetical) function is a real or complex valued function ƒ(n) defined on the set of natural numbers (i.e. positive integers) that expresses some arithmetical property of n. [1] An example of an arithmetic… … Wikipedia
• Fermat's little theorem — (not to be confused with Fermat s last theorem) states that if p is a prime number, then for any integer a , a^p a will be evenly divisible by p . This can be expressed in the notation of modular arithmetic as follows::a^p equiv a pmod{p},!A… … Wikipedia
• Euler's theorem — In number theory, Euler s theorem (also known as the Fermat Euler theorem or Euler s totient theorem) states that if n is a positive integer and a is coprime to n , then:a^{varphi (n)} equiv 1 pmod{n}where φ( n ) is Euler s totient function and … Wikipedia
• Prime number — Prime redirects here. For other uses, see Prime (disambiguation). A prime number (or a prime) is a natural number greater than 1 that has no positive divisors other than 1 and itself. A natural number greater than 1 that is not a prime number is… … Wikipedia | open-web-math/open-web-math | |
# Rationalisation (mathematics)
(Redirected from Rationalization (mathematics))
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For other uses, see Rationalization.
In elementary algebra, root rationalisation is a process by which surds in the denominator of an irrational fraction are eliminated.
These surds may be monomials or binomials involving square roots, in simple examples. There are wide extensions to the technique.
## Rationalisation of a monomial square root and cube root
For the fundamental technique, the numerator and denominator must be multiplied by the same factor.
Example 1:
$\frac{10}{\sqrt{a}}$
To rationalise this kind of monomial, bring in the factor $\sqrt{a}$:
$\frac{10}{\sqrt{a}} = \frac{10}{\sqrt{a}} \cdot \frac{\sqrt{a}}{\sqrt{a}} = \frac{{10\sqrt{a}}}{\sqrt{a}^2}$
The square root disappears from the denominator, because it is squared:
$\frac{{10\sqrt{a}}}{\sqrt{a}^2} = \frac{10\sqrt{a}}{a}$
This gives the result, after simplification:
$\frac{{10\sqrt{a}}}{{a}}$
Example 2:
$\frac{10}{\sqrt[3]{b}}$
To rationalise this radical, bring in the factor $\sqrt[3]{b}^2$:
$\frac{10}{\sqrt[3]{b}} = \frac{10}{\sqrt[3]{b}} \cdot \frac{\sqrt[3]{b}^2}{\sqrt[3]{b}^2} = \frac{{10\sqrt[3]{b}^2}}{\sqrt[3]{b}^3}$
The cube root disappears from the denominator, because it is cubed:
$\frac{{10\sqrt[3]{b}^2}}{\sqrt[3]{b}^3} = \frac{10\sqrt[3]{b}^2}{b}$
This gives the result, after simplification:
$\frac{{10\sqrt[3]{b}^2}}{{b}}$
## Dealing with more square roots
For a denominator that is:
$\sqrt{2}+\sqrt{3}\,$
Rationalisation can be achieved by multiplying by the Conjugate:
$\sqrt{2}-\sqrt{3}\,$
and applying the difference of two squares identity, which here will yield −1. To get this result, the entire fraction should be multiplied by
$\frac{ \sqrt{2}-\sqrt{3} }{\sqrt{2}-\sqrt{3}} = 1.$
This technique works much more generally. It can easily be adapted to remove one square root at a time, i.e. to rationalise
$x +\sqrt{y}\,$
by multiplication by
$x -\sqrt{y}$
Example:
$\frac{3}{\sqrt{3}+\sqrt{5}}$
The fraction must be multiplied by a quotient containing ${\sqrt{3}-\sqrt{5}}$.
$\frac{3}{\sqrt{3}+\sqrt{5}} \cdot \frac{\sqrt{3}-\sqrt{5}}{\sqrt{3}-\sqrt{5}} = \frac{3(\sqrt{3}-\sqrt{5})}{\sqrt{3}^2 - \sqrt{5}^2}$
Now, we can proceed to remove the square roots in the denominator:
$\frac{{3(\sqrt{3}-\sqrt{5}) }}{\sqrt{3}^2 - \sqrt{5}^2} = \frac{ 3 (\sqrt{3} - \sqrt{5} ) }{ 3 - 5 } = \frac{ 3( \sqrt{3}-\sqrt{5} ) }{-2}$
## Generalisations
Rationalisation can be extended to all algebraic numbers and algebraic functions (as an application of norm forms). For example, to rationalise a cube root, two linear factors involving cube roots of unity should be used, or equivalently a quadratic factor.
## References
This material is carried in classic algebra texts. For example:
• George Chrystal, Introduction to Algebra: For the Use of Secondary Schools and Technical Colleges is a nineteenth-century text, first edition 1889, in print (ISBN 1402159072); a trinomial example with square roots is on p. 256, while a general theory of rationalising factors for surds is on pp. 189–199. | HuggingFaceTB/finemath | |
# Negative Variance
Six Sigma – iSixSigma Forums Old Forums General Negative Variance
Viewing 3 posts - 1 through 3 (of 3 total)
• Author
Posts
• #44636
What is the operational definition of negative variance?
0
#143441
Jim,
There is no operational definition of a variance component that is negative. This is just the way the formula is set up (Edgeworth in the 1880s). If you get a negative variance in a variance component analysis that is possible. This is not like Analysis of Variance (by using the means) which was developed by Fisher in the 1920s. If you get a negative variance in the context of ANOVA something is wrong. I hope this helps.
0
#143472 Eric Maass
Participant
Jim,
When some statistical software packages calculate Variance Components or perform Nested Analysis of Variance, or analysis that involves Nested Analysis of Variance (like Gauge Capability studies), they often do the calculations with this concept in mind:
– If there are multiple levels of nested variance, then the lower level variance can generate false variance in the higher levels
– So, to allow for this, the high level variance should be reduced by the amount of expected false variance.
– However, if the amount of expected false variance is greater than the observed high level variance, then the variance component may show up as negative…however, most programs will zero this out.
The way of estimating false variance of the higher level variance uses the Central Limit Theorem.
For example, imagine that you have lot-to-lot variance and within-lot variance. The within lot variance is 100, and the lot-to-lot is only 1. The sample size within each lot is 4.
Now, the false variance comes from this thought experiment: if you were to pretend to measure two different lots, but actually measured the same lot twice, you would see a false lot-to-lot variance. The estimate for this false lot-to-lot variance that is due to within-lot variance would be within variance/sample size from the central limit theorem. For this example, this would be 100/4 = 25.
The observed lot-to-lot variance is only 1, so the estiamte for lot-to-lot variance is the observed of 1 minus the expected false variance due to within variance of 25: 1 – 25 = -24.
Naturally, you cannot estimate the standard deviation – you’d get an imaginary number – so programs like Minitab would zero this estimate of lot-to-lot variance out.
Best regards,
Eric
0
Viewing 3 posts - 1 through 3 (of 3 total)
The forum ‘General’ is closed to new topics and replies. | HuggingFaceTB/finemath | |
Chemistry Practice Problems Mass Defect Practice Problems Solution: Which of the following nuclei is likely to have th...
# Solution: Which of the following nuclei is likely to have the largest mass defect per nucleon: 11B, 51V, 118Sn, 243Cm?
###### Problem
Which of the following nuclei is likely to have the largest mass defect per nucleon: 11B, 51V, 118Sn, 243Cm?
###### Solution
We’re being asked to determine which of the following nuclei is likely to have the largest mass defect per nucleon: 11B, 51V, 118Sn, 243Cm based on the plot of their binding energy.
To do so, we need to do the following steps,
Step 1: Based on the plot, determine the nucleon with the highest binding energy.
Step 2: Determine the nucleon with the largest mass defect (Δm) in kg, using the equation:
$\overline{){\mathbf{E}}{\mathbf{=}}{\mathbf{∆}}{\mathbf{m}}{{\mathbf{c}}}^{{\mathbf{2}}}}$
E = binding energy in J
Δm = mass defect in kg
c = speed of light = 3.0x108 m/s
Step 1: From the plot, we can see that the binding energy of each nucleon increases:
• from left to right → as mass numbers increase during fusion
View Complete Written Solution
Mass Defect
Mass Defect
#### Q. 24.82Iodine-131 is one of the most important isotopes used in the diagnosis of thyroid cancer. One atom has a mass of 130.906114 amu. Calculate the bi...
Solved • Mon Jan 28 2019 09:54:08 GMT-0500 (EST)
Mass Defect
#### Q. What is the nuclear binding energy of a lithium-7 nucleus in units of kJ/mol and eV/nucleus? (Mass of a lithium-7 atom = 7.016003 amu.)
Solved • Wed Jan 09 2019 01:53:36 GMT-0500 (EST)
Mass Defect
#### Q. The nuclear reaction that powers a radioisotope thermoelectric generator is 94 238Pu → 92 234U + 2 4He. The atomic masses of plutonium-238 and ura...
Solved • Tue Jan 08 2019 13:54:29 GMT-0500 (EST)
Mass Defect
#### Q. The nuclear masses of 7Be, 9Be, and 10Be are 7.0147, 9.0100, and 10.0113 amu, respectively.Which of these nuclei has the largest binding energy per nu...
Solved • Tue Jan 08 2019 13:54:25 GMT-0500 (EST) | open-web-math/open-web-math | |
Converting Fractions into Decimals and Percentages
Hello and welcome to this video about converting fractions to decimals and percents! In this video, we will explore:
Before we get started, let’s review a couple of key concepts that we will use to help the math make sense.
Consider the fraction $$\frac{10}{10}$$. One way to think about fractions is to think of them as division.
$$\frac{10}{10}=1$$
In other words, ten-tenths is one whole.
$$\frac{10}{10}$$ can be written equivalently as $$\frac{100}{100}$$. The fraction bar can be said as “per,” so this expression can be said as “100 per 100.” The word “percent” literally means “per 100,” so “100 per 100” means 100 percent.
Therefore, when the same number is divided by itself, the result as a decimal is 1 and as a percent is 100%.
But what happens when our fraction is less than 1? Let’s take a look.
Consider the fraction $$\frac{1}{4}$$. Visually, this is what’s happening:
We can see in the diagram that $$\frac{1}{4}$$ of the whole is $$\frac{25}{100}$$. $$\frac{25}{100}$$ means “25 per 100,” so it equals 25%.
Now let’s figure out how to convert this into a decimal.
We’re going to take our fraction $$\frac{1}{4}$$, which is the same as saying 1 divided by 4.
Dividing this way doesn’t look like it will work. But using our knowledge of place value, we can make it work:
First, rewrite the 1 as 1.0. Instead of 1, the dividend is now ten tenths.
Second, 4 ones will go into ten-tenths two-tenths times.
Third, $$4\times 0.2=0.8$$.
Fourth, $$1.0 -0.8 = 0.2$$.
Fifth, we’re gonna rewrite the original dividend as 100-hundredths and bring the new 0 down.
Sixth, 4 ones goes into 20-hundredths 5-hundredths times.
Seventh, we’re gonna multiply $$4\times 0.05=0.20$$. Then we’ll subtract to get 0.
This means $$\frac{1}{4}=0.25$$.
How to Convert Fraction to Percent
Let’s see this work with a non-unit fraction, like $$\frac{3}{16}$$.
Here’s what the division looks like. The sequence of adding a decimal and dividing repeats as often as necessary until either the remainder is 0 or the decimal begins to repeat.
So, $$\frac{3}{16}=0.1875$$ and $$0.1875=\frac{1,875}{10,000}=\frac{18.75}{100}$$. Therefore, 0.1875 is the same as 18.75% because it’s 18.75 per, 100.
How about a repeating decimal? Everything is the same and the process can be stopped when it is clear that the decimal repeats. Here’s a quick example: $$\frac{1}{3}$$.
The process of subtracting 9 units from 10 units repeats, causing the decimal to repeat.
Lastly, consider a fraction that is greater than 1, such as $$\frac{5}{2}$$.
Visually, here’s what we have:
We can see that the shaded quantity is $$\frac{1}{2}+\frac{1}{2}+\frac{1}{2}+\frac{1}{2}+\frac{1}{2}=\frac{5}{2}$$. Each $$\frac{1}{2}=50$$, so we can say equivalently $$\frac{250}{100}$$, which, as a decimal, is 2.5 (meaning $$2\frac{1}{2}$$ wholes, which the diagram shows). We can also equivalently say 250 per 100, or 250%.
And the same works for the division as well:
Thanks for watching and happy studying!
How do you change a fraction into a percent?
A
Converting a fraction to a percent can be accomplished in $$2$$ quick steps.
Step 1: Divide the numerator by the denominator.
Step 2: Multiply this decimal by $$100$$.
For example, $$\frac{2}{5}$$ shows a numerator of $$2$$, and a denominator of $$5$$.
Step 1:
$$5\phantom{.}$$ $$\phantom{.}2$$
is the same as
$$5\phantom{.}$$ $$\phantom{.}2$$ . $$0$$ $$0$$
, which equals $$0.4$$ when using the process of long division.
Step 2: $$0.4\times100=40$$. This means that the fraction $$\frac{2}{5}$$ represents $$40\%$$.
How do I turn fractions into decimals?
A
Fractions can be converted into decimals by using long division. For example, the fraction $$\frac{4}{25}$$ is expressing “$$4$$ divided by $$25$$”. This can be set up as 25⟌4. Place a decimal and zeros after the $$4$$ so that it becomes $$4.00$$. Now we have 25⟌4.00. $$25$$ does not go into $$4$$, but it does go into $$40$$, one time. From here, the standard rules of long division apply. The result will be $$0.16$$. The important thing to remember is that the fraction bar represents division, so fractions can be calculated as decimals by dividing the numerator by the denominator.
What is the meaning of a fraction bar?
A
A fraction represents “part” out of “whole”. The fraction bar separating the “part” and the “whole” represents division. For example, the fraction $$\frac{5}{9}$$ represents $$5$$ parts out of $$9$$ parts, or “$$5\text{ divided by }9$$”. Another way to think about this, is that the fraction bar represents “per”. For example, $$\frac{40}{100}$$ represents “$$40\text{ per }100$$”.
What does percent mean?
A
Percent means “$$\text{per }100$$”. For example, $$96\%$$ represents $$96$$ per $$100$$. This also represents the fraction $$\frac{96}{100}$$. All percents can be expressed as a fraction because percents are always out of $$100$$. Percents will sometimes be smaller than $$1$$, and larger than $$100$$. For example, the percent $$125\%$$ represents $$\frac{125}{100}$$, and the percent $$0.5\%$$ represents $$\frac{0.5}{100}$$.
Why do you divide a fraction to turn it into a decimal?
A
Fractions represent a “part” out of a “whole”. The fraction bar separating the “part” and the “whole” represents division. This means that all fractions can be converted into decimals by dividing the numerator by the denominator.
For example, the fraction $$\frac{4}{5}$$ represents $$4$$ out of $$5$$, or $$4$$ divided by $$5$$. This fraction can be converted into a decimal by dividing $$4$$ by $$5$$. 5⟌4 can be solved if a decimal point and zeros are added to the right of the $$4$$. 5⟌4 becomes 5⟌4.00. $$5$$ goes into $$40$$ $$8$$ times, and the rest of the process follows the standard rules of long division. The result is $$0.8$$.
Practice Questions
Question #1:
What is $$\frac{42}{100}$$ as a percent?
42%
4.2%
0.42%
420%
The correct answer is 42%. Percent means “per 100,” so to turn a fraction into a percent, convert the denominator to 100 and the numerator is your percentage.
Question #2:
What is $$\frac{16}{100}$$ as a decimal?
16.00
0.016
0.16
1.6
The correct answer is 0.16. Since 16 is over 100, write the decimal so that 6 is in the hundredths place, which means 1 will be in the tenths place.
Question #3:
Which of the following is not a proper way to rewrite $$\frac{96}{1,000}$$?
0.096
96%
9.6%
0.09600
The correct answer is 96%. $$\frac{96}{1,000}$$ as a decimal is 0.096 and adding zeroes to the end of a decimal number doesn’t change its value, so 0.09600 is equivalent to 0.096. $$\frac{96}{1,000}$$ as a percentage is 9.6%. 96% as a fraction is $$\frac{96}{100}$$.
Question #4:
What is $$\frac{7}{20}$$ as a percent?
7%
20%
28%
35%
The correct answer is 35%. Percent means “per 100,” so to turn a fraction into a percent, convert the denominator to 100 and the numerator is your percentage. To convert $$\frac{7}{20}$$ to have a denominator of 100, multiply both the numerator and the denominator by 5, which gives you $$\frac{35}{100}$$. 35 is the numerator, so $$\frac{7}{20}=35%$$.
Question #5:
What is $$\frac{23}{125}$$ as a decimal?
0.23
1.25
1.86
0.184
The correct answer is 0.184. To convert a fraction to a decimal, first convert the fraction so it has a denominator of a multiple of 10. This can be done by multiplying $$\frac{23}{125}$$ by $$\frac{8}{8}$$. $$\frac{23}{125}\times\frac{8}{8}=\frac{184}{1,000}$$, and since 184 is over 1,000, place the 4 in the thousandths place of the decimal, which puts 8 in the hundredths place, and 1 in the tenths place, 0.184.
Question #6:
Write the fraction $$\frac{4}{5}$$ as a decimal.
0.8
0.6
0.5
0.4
The correct answer is 0.8. To convert a fraction to decimal, divide the numerator by the denominator. In this case, divide 4 by 5. Long division can be used for this:
Question #7:
Write the fraction $$\frac{5}{8}$$ as a decimal.
5.8
0.58
0.625
0.562
The correct answer is 0.625. To convert a fraction to decimal, divide the numerator by the denominator. In this case, divide 5 by 8. Long division can be used for this:
Question #8:
Write the fraction $$\frac{1}{5}$$ as a decimal.
0.5
0.2
0.1
2.0
The correct answer is 0.2. To convert a fraction to decimal, divide the numerator by the denominator. In this case, divide 1 by 5. Long division can be used for this:
Question #9:
Write the fraction $$\frac{5}{6}$$ as a decimal.
0.3222
0.8333…
0.8314
0.2333…
The correct answer is 0.8333. To convert a fraction to decimal, divide the numerator by the denominator. In this case, divide 5 by 6. Long division can be used for this:
Question #10:
Write the fraction $$\frac{3}{11}$$ as a decimal.
0.3111…
0.3131…
0.3535…
0.2727…
The correct answer is 0.2727…To convert a fraction to decimal, divide the numerator by the denominator. In this case, divide 3 by 11. Long division can be used for this: | HuggingFaceTB/finemath | |
# 1971 AHSME Problems/Problem 26
(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)
## Problem
$[asy] size(2.5inch); pair A, B, C, E, F, G; A = (0,3); B = (-1,0); C = (3,0); E = (0,0); F = (1,2); G = intersectionpoint(B--F,A--E); draw(A--B--C--cycle); draw(A--E); draw(B--F); label("A",A,N); label("B",B,W); label("C",C,dir(0)); label("E",E,S); label("F",F,NE); label("G",G,SE); //Credit to chezbgone2 for the diagram[/asy]$
In $\triangle ABC$, point $F$ divides side $AC$ in the ratio $1:2$. Let $E$ be the point of intersection of side $BC$ and $AG$ where $G$ is the midpoints of $BF$. The point $E$ divides side $BC$ in the ratio
$\textbf{(A) }1:4\qquad \textbf{(B) }1:3\qquad \textbf{(C) }2:5\qquad \textbf{(D) }4:11\qquad \textbf{(E) }3:8$
## Solution
We will use mass points to solve this problem. $AC$ is in the ratio $1:2,$ so we will assign a mass of $2$ to point $A,$ a mass of $1$ to point $C,$ and a mass of $3$ to point $F.$
We also know that $G$ is the midpoint of $BF,$ so $BG:GF=1:1.$ $F$ has a mass of $3,$ so $B$ also has a mass of $3.$
In line $BC,$ $B$ has a mass of $3$ and $C$ has a mass of $1.$ Therefore, $BE:EC = 1:3.$
The answer is $\textbf{(B)}.$
-edited by coolmath34 | HuggingFaceTB/finemath | |
# Thread: Proof of permutation in sequential sets, also question on opposite ability...
1. ## Proof of permutation in sequential sets, also question on opposite ability...
I have two things here. First, I was looking for a way to arrive at a set of increasing numbers, like a permutation, where the set never runs into a situation which the permutation of any part of the set can collide...
I haven't gone far enough to figure out if there is a proof to do that, but I have an interesting proof in permutations that you might enjoy if none of you have stumbled on it. Basically, when I was trying to find the above, and stopped for a while, I found this which is Not what I want...
Take the following examples and you will see what I mean...
(a) in a descending set, pos 1&5 sum collides with 2&4. There is no way to distinguish the set no matter the starting integer, or any type of inversion...
(- is the first sum, + is the second sum, for brevity...)
Code:
(a1) (a2) (a3) (a4) (a5) (a6)
-1 1+ +1 1- -1 2+ +1 2- -1 2- +1 2+
+2 2- -2 2+ +2 3- -2 3+ +2 3+ -2 3-
+3 3- +3 3- +3 4- +3 4- +3 4+ +3 4+
+4 4- -4 4+ +4 5- -4 5+ +4 5+ -4 5-
-5 5+ +5 5- -5 6+ +5 6- -5 6- +5 6+
----- ----- ----- ---- ---- ---- ---- ---- ---- ---- ---- ----
6 9 6 9 6 12 6 12 6 8 6 8
9 6 9 6 9 8 9 8 9 12 9 12
So we have found this:
(a1,1,2) = (a2,1,2); (a3,1,2 = a4,1,2); (a5,1,2 = a6,1,2); ....
So then we can conclude (am I using the right terms?) that any sequential set that has a permutation collision, assuming the same positions are used in the set, is congruent - a collision will always exist between those sets no matter the starting integer or whether you do an intersection or inversion of the permutation (all based on positions of the set).
Now, I was wondering if you all have ever come across this before, and if so, what is the mathematical proof going to be in mathematical terms?
Then, notice this as well:
(b) even using odd/even progression, permutation collisions of sets results in the same dillemma...
Code:
(b1) (b2)
-1 2+ +1 2+
+3 4- -3 4-
+5 6- -5 6-
-7 8+ +7 8+
---- ---- ---- ----
8 10 8 10
8 10 8 10
Here we still note (b1,1 = b1,2), (b2,1 = b2,2), (b1,1 = b2,1), (b1,2 = b2,2).. etc... There is no distinction no matter how you arrange this...
So what have we just concluded, does this form a mathematical proof for permutations about colliding permutations... i.e. a permutation result has numerous possibilities, that given any set are indistinguishable.
Then we have the question I was going to see about, to see if you all already have the answer:
(c) question domain: Is there an ordered set so that no permutation result of any members of the set collide? i.e. Pa = Pb?
Thanks, this was kind of interesting when I worked it out on paper....
2. ## Re: Proof of permutation in sequential sets, also question on opposite ability...
(sorry about how sloppy the tables turned out, I couldn't copy/paste from excel, or export to an image, or upload attachments, etc. so the best I could do was wrap the tables in code tags for approximate monospace...)
First, I was attempting to explain that given two ordered integral sets, a partition collision is indistinguishable regardless of the method used (intersection or inversion) and regardless of the starting integer of the set. The positions in the set gives away its collision result in a permutation for any ascending integer set....
Then, we showed that this holds true for competing odd/even sets - the permutation collision there is indistinguishable given the positions in the set that collide and a properly progressive set.
Finally, the question I was trying to solve was - is there a way to generate an integral set so that all possible permutations are unique and calculatable? If so, what is that method?
If none of you have seen this before, well, I guess we may have a new proof in permutations of sets that needs to be mathematically represented...
Regards, and thanks for taking the time to read and provide any input... | HuggingFaceTB/finemath | |
# Question on Radon Nikodym Theorem
Let $\mu,\lambda$ and $\nu$ be $\sigma$-finite measures on $(X,M)$ such the $\nu\ll \mu$. Let $\mu= \nu + \lambda$. Then if $f$ is the Radon-Nikodym derivate of $\nu$ wrt $\mu$, we have $0\leq f\lt 1~\mu$-a.e. where $f = \frac{d\nu}{d\mu}$.
Approach.
Suppose to the contrary that $f\geq 1$. Then since I can infer $\nu \ll \mu$, $$\nu(E) = \int_E f~d\mu \geq \int_E 1 d\mu = \mu(E).$$ But then $\nu(E) \leq \nu(E) + \lambda(E) = \mu(E)$. So a contradiction. Thus $f\lt 1~\mu$-a.e.
-
The contrary is not that $f \geq 1$. The contrary is that there exists a measurable set $E$ of positive measure on which $f > 1$. – copper.hat Apr 16 '12 at 5:53
Also, be careful with your inequalities. You can certainly choose $\lambda = 0$, in which case $\frac{d \nu}{d \mu} = 1$ $\mu$ a.e. – copper.hat Apr 16 '12 at 5:55
Why don't you just use the fact that since $\nu(E) + \lambda(E) = \mu(E)$ for all measurable $E$, then $\nu(E) \leq \mu(E)$. Writing this in integral form, we have $\int_E \frac{d \nu}{d \mu} d \mu \leq \int_E 1 d \mu$. From this you should be able to conclude that $\frac{d \nu}{d \mu} \leq 1$ $\mu$-a.e.
-
But then I wand $\frac{d\nu}{d\mu}\lt 1$ – Jay Apr 16 '12 at 12:33
As I pointed out above, your hypotheses include the case where $\lambda=0$, so, sorry, you can want it all you like, but unless you add additional hypotheses, the $\leq$ is all you can expect. – copper.hat Apr 16 '12 at 15:00
Since $\lambda=\mu + \nu$, we have $$\frac{d \lambda}{d \mu}= \frac{ d \mu }{d \mu} + \frac{ d \nu}{ d \mu}= 1 + \frac{ d \nu}{ d \mu}$$
Now, $\lambda << \mu$, and we show $\mu << \lambda$ Let $\lambda(E)=0$, then $\mu(E)= -\nu(E)$. since $\mu, \nu$ are positive we have that $\mu(E)=\nu(E)=0$, thus $\nu << \lambda$, and $\mu << \lambda$. Then, $$\frac{d \mu}{d \lambda}\frac{d \lambda}{d \mu}=1$$ Thus $$\frac{d \mu}{d \lambda} = \frac{1}{1 + \frac{d \nu}{d \mu}}$$
Then we multiply by $\frac{d \nu}{d \mu}$, to give us $$\frac{d \nu}{d \mu}\frac{ d \mu}{d \lambda}= \frac{ \frac{d \nu}{d \mu}}{ 1 + \frac{ d \nu}{d \mu}}$$
Then since $\mu$, $\nu$, and $\lambda$ are $\sigma$-finite measure and $\nu << \mu$ and $\mu << \lambda$, we have, $$\frac{d \nu}{d \lambda} = \frac{ \frac{d \nu}{d \mu}}{ 1 + \frac{ d \nu}{d \mu}}$$
Since $\nu$, and $\mu$ are positive and $f = \frac{d \nu}{d \lambda}$, we have $0 \le f < 1$, that is $$0 \le f = \frac{ \frac{d \nu}{d \mu}}{ 1 + \frac{ d \nu}{d \mu}} < 1$$
- | HuggingFaceTB/finemath | |
# Find the area of a circle assuming n equal to 22/7 if its radius R = 7cm.
Given a circle, the radius of which is R = 7 cm. It is necessary to calculate its area.
The area of the circle (Skr) is calculated by the formula:
Scr = n * R ^ 2, where n is a mathematical constant – the number “pi”.
We calculate the area of a given circle if it is known that n = 22/7:
Scr = 22/7 * 7 ^ 2 = 22 * 7 = 154 cm ^ 2. | HuggingFaceTB/finemath | |
# Proving theorems and special cases (Part 15): The Mean Value Theorem
In a recent class with my future secondary math teachers, we had a fascinating discussion concerning how a teacher should respond to the following question from a student:
Is it ever possible to prove a statement or theorem by proving a special case of the statement or theorem?
Usually, the answer is no. In this series of posts, we’ve seen that a conjecture could be true for the first 40 cases or even the first $10^{316}$ cases yet not always be true. We’ve also explored the computational evidence for various unsolved problems in mathematics, noting that even this very strong computational evidence, by itself, does not provide a proof for all possible cases.
However, there are plenty of examples in mathematics where it is possible to prove a theorem by first proving a special case of the theorem. For the remainder of this series, I’d like to list, in no particular order, some common theorems used in secondary mathematics which are typically proved by first proving a special case.
6. Theorem (Mean Value Theorem). If $f$ is a continuous function on the interval $[a,b]$ which is differentiable on the interior $(a,b)$, then there is a point $c \in (a,b)$ so that
$f'(c) = \displaystyle \frac{f(b)-f(a)}{b-a}$
In other words, there is a point $c$ in $(a,b)$ so that the slope of the tangent line at $c$ is the same as the slope of the line segment connecting the endpoints.
This is a consequence of the following lemma.
Lemma (Rolle’s Theorem). If $f$ is a continuous function on the interval $[a,b]$ which is differentiable on the interior $(a,b)$ so that $f(a) = 0$ and $f(b) = 0$, then there is a point $c \in (a,b)$ so that $f'(c) = 0$.
Notice that Rolle’s Theorem is really a special case of the Mean Value Theorem: if $f(a) = 0$ and $f(b) = 0$, then the right-hand side of the conclusion of the Mean Value Theorem becomes
$\displaystyle \frac{f(b)-f(a)}{b-a} = \displaystyle \frac{0-0}{b-a} = 0$,
thus matching the conclusion of Rolle’s Theorem.
I won’t type out the proofs of Rolle’s Theorem and the Mean Value Theorem here, since Wikipedia has already done that very well. Suffice it to say that Rolle’s Theorem logically comes first, and then the Mean Value Theorem can be proven using Rolle’s Theorem. The main idea is to assume that the function $f$ satisfies the hypotheses of the Mean Value Theorem and then define
$g(x) = f(x) - f(a) - \displaystyle \frac{f(b)-f(a)}{b-a} (x-a)$
It’s straightforward to show that $g$ satisfies the hypotheses of Rolle’s Theorem and conclude that there must be a point so that $g'(c) = 0$, from which we obtain the conclusion of the Mean Value Theorem.
# Proving theorems and special cases (Part 14): The Power Law of differentiation
In a recent class with my future secondary math teachers, we had a fascinating discussion concerning how a teacher should respond to the following question from a student:
Is it ever possible to prove a statement or theorem by proving a special case of the statement or theorem?
Usually, the answer is no. In this series of posts, we’ve seen that a conjecture could be true for the first 40 cases or even the first $10^{316}$ cases yet not always be true. We’ve also explored the computational evidence for various unsolved problems in mathematics, noting that even this very strong computational evidence, by itself, does not provide a proof for all possible cases.
However, there are plenty of examples in mathematics where it is possible to prove a theorem by first proving a special case of the theorem. For the remainder of this series, I’d like to list, in no particular order, some common theorems used in secondary mathematics which are typically proved by first proving a special case.
5. Theorem. For any rational number $r$, we have $\displaystyle \frac{d}{dx} x^r = r x^{r-1}$.
This theorem is typically proven using the Chain Rule (in the guise of implicit differentiation) and the following lemma:
Lemma. For any integer $n$, we have $\displaystyle \frac{d}{dx} x^n = n x^{n-1}$.
Clearly, the lemma is a special case of the main theorem. However, the lemma can be proven without using the main theorem:
Proof of Lemma (Case 1). If $n$ is a positive integer, then
$\displaystyle \frac{d}{dx} x^n = \displaystyle \lim_{h \to 0} \frac{(x+h)^n - x^n}{h}$
$= \displaystyle \lim_{h \to 0} \frac{x^n + n x^{n-1} h + \frac{1}{2} n(n-1) x^{n-2} + \dots + h^n - x^n}{h}$
$= \displaystyle \lim_{h \to 0} \left[ n x^{n-1} + \frac{1}{2} n(n-1) x^{n-2} h + \dots + h^{n-1} \right]$
$= n x^{n-1} + 0 + \dots + 0$
$= n x^{n-1}$
Case 1 can also be proven using the Product Rule and mathematical induction.
Proof of Lemma (Case 2). If $n = 0$, then the theorem is trivially true since $x^0 = 1$, and the derivative of a constant is zero.
Proof of Lemma (Case 3). If $n$ is a negative integer, then write $n = -m$, where $m$ is a positive integer. Then, using the Quotient Rule,
$\displaystyle \frac{d}{dx} x^n = \displaystyle \frac{d}{dx} \left( x^{-m} \right)$
$= \displaystyle \frac{d}{dx} \left( \frac{1}{x^m} \right)$
$= \displaystyle \frac{0 \cdot x^m - 1 \cdot m x^{m-1}}{x^{2m}}$
$= -m x^{-m - 1}$
$= n x^{n-1}$
QED
Now that the lemma has been proven, the main theorem can be proven using the lemma.
Proof of Theorem. Suppose that $r = p/q$, where $p$ and $q$ are integers. Suppose that $y = x^r = x^{p/q}$. Then:
$y = x^{p/q}$
$y^q = \displaystyle \left[ x^{p/q} \right]^q$
$y^q = x^p$
Let’s now differentiate with respect to $x$:
$q y^{q-1} \displaystyle \frac{dy}{dx} = p x^{p-1}$
$\displaystyle \frac{dy}{dx} = \displaystyle \frac{p x^{p-1}}{q y^{q-1}}$
$\displaystyle \frac{dy}{dx} = \displaystyle \frac{p}{q} \frac{x^{p-1}}{[x^{p/q}]^{q-1}}$
$\displaystyle \frac{dy}{dx} = \displaystyle \frac{p}{q} \frac{x^{p-1}}{x^{p - p/q}}$
$\displaystyle \frac{dy}{dx} = \displaystyle \frac{p}{q} x^{p-1 - (p-p/q)}$
$\displaystyle \frac{dy}{dx} = \displaystyle \frac{p}{q} x^{p - 1 - p + p/q}$
$\displaystyle \frac{dy}{dx} = \displaystyle \frac{p}{q} x^{p/q - 1}$
$\displaystyle \frac{dy}{dx} = r x^{r-1}$
QED
# Proving theorems and special cases (Part 13): Uniqueness of logarithms
In a recent class with my future secondary math teachers, we had a fascinating discussion concerning how a teacher should respond to the following question from a student:
Is it ever possible to prove a statement or theorem by proving a special case of the statement or theorem?
Usually, the answer is no. In this series of posts, we’ve seen that a conjecture could be true for the first 40 cases or even the first $10^{316}$ cases yet not always be true. We’ve also explored the computational evidence for various unsolved problems in mathematics, noting that even this very strong computational evidence, by itself, does not provide a proof for all possible cases.
However, there are plenty of examples in mathematics where it is possible to prove a theorem by first proving a special case of the theorem. For the remainder of this series, I’d like to list, in no particular order, some common theorems used in secondary mathematics which are typically proved by first proving a special case.
The next theorem is needed in calculus to show that $\ln x = \displaystyle \int_1^x \frac{dt}{t}$.
4. Theorem. Let $a \in \mathbb{R}^+ \setminus \{1\}$. Suppose that $f: \mathbb{R}^+ \rightarrow \mathbb{R}$ has the following four properties:
1. $f(1) = 0$
2. $f(a) = 1$
3. $f(xy) = f(x) + f(y)$ for all $x, y \in \mathbb{R}^+$
4. $f$ is continuous
Then $f(x) = \log_a x$ for all $x \in \mathbb{R}^+$.
In other blog posts, I went through the full proof of this theorem, which is divided — actually, scaffolded — into cases:
Case 1. $f(x) = \log_a x$ if $x$ is a positive integer.
Case 2. $f(x) = \log_a x$ if $x$ is a positive rational number.
Case 3. $f(x) = \log_a x$ if $x$ is a negative rational number.
Case 4. $f(x) = \log_a x$ if $x$ is a real number.
Clearly, Case 1 is a subset of Case 2, and Case 3 is a subset of Case 4. Once again, a special case of a theorem is used to prove the full theorem.
# Proving theorems and special cases (Part 12): The sum and difference formulas for sine
In a recent class with my future secondary math teachers, we had a fascinating discussion concerning how a teacher should respond to the following question from a student:
Is it ever possible to prove a statement or theorem by proving a special case of the statement or theorem?
Usually, the answer is no. In this series of posts, we’ve seen that a conjecture could be true for the first 40 cases or even the first $10^{316}$ cases yet not always be true. We’ve also explored the computational evidence for various unsolved problems in mathematics, noting that even this very strong computational evidence, by itself, does not provide a proof for all possible cases.
However, there are plenty of examples in mathematics where it is possible to prove a theorem by first proving a special case of the theorem. For the remainder of this series, I’d like to list, in no particular order, some common theorems used in secondary mathematics which are typically proved by first proving a special case.
3. Theorem 1. $\sin(\alpha + \beta) = \sin \alpha \cos \beta + \cos \alpha + \sin \beta$
Theorem 2. $\sin(\alpha - \beta) = \sin \alpha \cos \beta - \cos \alpha \sin \beta$
For angles that are not acute, these theorems can be proven using a unit circle and the following four lemmas:
Lemma 1. $\cos(x - y) = \cos x \cos y + \sin x \sin y$
Lemma 2. $\cos(x + y) = \cos x \cos y - \sin x \sin y$
Lemma 3. $\sin(\pi/2 - x) = \cos x$
Lemma 4. $\cos(\pi/2 - x) = \sin x$
Specifically, assuming Lemmas 1-4, then:
$\sin(\alpha + \beta) = \cos(\pi/2 - [\alpha + \beta])$ by Lemma 4
$= \cos([\pi/2 - \alpha] - \beta)$
$= \cos(\pi/2 - \alpha) \cos \beta + \sin(\pi/2 - \alpha) \sin \beta$ by Lemma 1
$= \sin \alpha \cos \beta + \cos \alpha \sin \beta$ by Lemmas 3 and 4.
Also,
$\sin(\alpha - \beta) = \cos(\pi/2 - [\alpha - \beta])$ by Lemma 4
$= \cos([\pi/2 - \alpha] + \beta)$
$= \cos(\pi/2 - \alpha) \cos \beta - \sin(\pi/2 - \alpha) \sin \beta$ by Lemma 2
$= \sin \alpha \cos \beta - \cos \alpha \sin \beta$ by Lemmas 3 and 4.
However, we see that what I’ve called Lemma 3, often called a cofunction identity, can be considered a special case of Theorem 2. However, this is not circular logic since the cofunction identities can be proven without appealing to Theorems 1 and 2.
# Proving theorems and special cases (Part 11): The Law of Cosines
In a recent class with my future secondary math teachers, we had a fascinating discussion concerning how a teacher should respond to the following question from a student:
Is it ever possible to prove a statement or theorem by proving a special case of the statement or theorem?
Usually, the answer is no. In this series of posts, we’ve seen that a conjecture could be true for the first 40 cases or even the first $10^{316}$ cases yet not always be true. We’ve also explored the computational evidence for various unsolved problems in mathematics, noting that even this very strong computational evidence, by itself, does not provide a proof for all possible cases.
However, there are plenty of examples in mathematics where it is possible to prove a theorem by first proving a special case of the theorem. For the remainder of this series, I’d like to list, in no particular order, some common theorems used in secondary mathematics which are typically proved by first proving a special case.
2. Theorem. In $\triangle ABC$ where $a = BC$, $b = AC$, and $c = AB$, we have $c^2 = a^2 + b^2 - 2 a b \cos (m \angle C)$.
This is typically proven using the Pythagorean theorem:
Lemma. In right triangle $\triangle ABC$, where $\angle C$ is a right angle, we have $c^2 = a^2 + b^2$.
Though it usually isn’t thought of this way, the Pythagorean theorem is a special case of the Law of Cosines since $\cos 90^\circ = 0$.
There are well over 100 different proofs of the Pythagorean theorem that do not presuppose the Law of Cosines. The standard proof of the Law of Cosines then uses the Pythagorean theorem. In other words, a special case of the Law of Cosines is used to prove the Law of Cosines.
# Proving theorems and special cases (Part 10): Angles in a convex n-gon
In a recent class with my future secondary math teachers, we had a fascinating discussion concerning how a teacher should respond to the following question from a student:
Is it ever possible to prove a statement or theorem by proving a special case of the statement or theorem?
Usually, the answer is no. In this series of posts, we’ve seen that a conjecture could be true for the first 40 cases or even the first $10^{316}$ cases yet not always be true. We’ve also explored the computational evidence for various unsolved problems in mathematics, noting that even this very strong computational evidence, by itself, does not provide a proof for all possible cases.
However, there are plenty of examples in mathematics where it is possible to prove a theorem by first proving a special case of the theorem. For the remainder of this series, I’d like to list, in no particular order, some common theorems used in secondary mathematics which are typically proved by first proving a special case.
1. Theorem. The sum of the angles in a convex n-gon is $180(n-2)$ degrees.
This theorem is typically proven after first proving the following lemma:
Lemma. The sum of the angles in a triangle is $180$ degrees.
Clearly the lemma is a special case of the main theorem: for a triangle, $n=3$ and so $180(n-2) = 180 \times 1 = 180$. The proof of the lemma uses alternate interior angles and the convention that the angle of a straight line is 180 degrees.
Using this, the main theorem follows by using diagonals to divide a convex n-gon into $n-2$ triangles. (For example, drawing a diagonal divides a quadrilateral into two triangles.) The sum of the angles of the n-gon must equal the sum of the angles of the $n-2$ triangles.
So it is possible to prove a theorem by proving a special case of the theorem. Using the sum of the angles of a triangle to prove the formula for the sum of the angles of a convex n-gon is qualitatively different than the previous computational examples seen earlier in this series.
# Proving theorems and special cases (Part 9): The Riemann hypothesis
In a recent class with my future secondary math teachers, we had a fascinating discussion concerning how a teacher should respond to the following question from a student:
Is it ever possible to prove a statement or theorem by proving a special case of the statement or theorem?
Usually, the answer is no. In this series of posts, we’ve already seen that a conjecture could be true for the first 40 cases or even the first $10^{316}$ yet ultimately prove false for all cases.
For the next few posts, I thought I’d share a few of the most famous unsolved problems in mathematics… and just how much computational work has been done to check for a counterexample.
4. The Riemann hypothesis (see here, here, and here for more information) is perhaps the outstanding unsolved problem in pure mathematics, and a prize of \$1 million has been offered for its proof.
The Riemann zeta function is defined by
$\zeta(s) = \displaystyle \sum_{n=1}^\infty \frac{1}{n^s}$
for complex numbers $s$ with real part greater than 1. For example,
$\zeta(2) = \displaystyle \frac{1}{1^2} + \frac{1}{2^2} + \frac{1}{3^2} + \dots = \displaystyle \frac{\pi^2}{6}$
and
$\zeta(4) = \displaystyle \frac{1}{1^4} + \frac{1}{2^4} + \frac{1}{3^4} + \dots = \displaystyle \frac{\pi^4}{90}$
The definition of the Riemann zeta function can be extended to all complex numbers (except a pole at $s = 1$) by the integral
$\zeta(s) = \displaystyle \frac{1}{\Gamma(s)} \int_0^\infty \frac{x^{s-1}}{e^s - 1} dx$
and also analytic continuation.
The Riemann hypothesis conjectures to all solutions of the equation $\zeta(s) = 0$ other than negative even integers occur on the line $s = \frac{1}{2} + i t$. At present, it is known that the first 10 trillion solutions are on this line, so that every solution with $t < 2.4 \times 10^{12}$ is on this line. Of course, that’s not a proof that all solutions are on this line.
A full description of known results concerning the Riemann hypothesis requires much more than a single post. I’ll refer the interested reader to the links above from MathWorld, Wikipedia, and Claymath and the references embedded in those links. An excellent book for the layman concerning the Riemann hypothesis is Prime Obsession: Bernhard Riemann and the Greatest Unsolved Problem in Mathematics, by John Derbyshire.
# Proving theorems and special cases (Part 8): The Collatz conjecture
In a recent class with my future secondary math teachers, we had a fascinating discussion concerning how a teacher should respond to the following question from a student:
Is it ever possible to prove a statement or theorem by proving a special case of the statement or theorem?
Usually, the answer is no. In this series of posts, we’ve already seen that a conjecture could be true for the first 40 cases or even the first $10^{316}$ cases yet ultimately prove false for all cases.
For the next few posts, I thought I’d share a few of the most famous unsolved problems in mathematics… and just how much computational work has been done to check for a counterexample.
3. The Collatz conjecture (see here and here for more information) is an easily stated unsolved problem that can be understood by most fourth and fifth graders. Restated from a previous post:
Here’s the statement of the problem.
• If the integer is even, divide it by $2$. If it’s odd, multiply it by $3$ and then add $1$.
• Repeat until (and if) you reach $1$.
Here’s the question: Does this sequence eventually reach $1$ no matter the starting value? Or is there a number out there that you could use as a starting value that has a sequence that never reaches $1$?
For every integer less than $5 \times 2^{60} = 5,764,607,523,034,234,880$, this sequence returns to 1. Of course, this is not a proof that the conjecture will hold for every integer.
# Proving theorems and special cases (Part 7): The twin prime conjecture
In a recent class with my future secondary math teachers, we had a fascinating discussion concerning how a teacher should respond to the following question from a student:
Is it ever possible to prove a statement or theorem by proving a special case of the statement or theorem?
Usually, the answer is no. In this series of posts, we’ve already seen that a conjecture could be true for the first 40 cases or even the first $10^{316}$ cases yet ultimately prove false for all cases.
For the next few posts, I thought I’d share a few of the most famous unsolved problems in mathematics… and just how much computational work has been done to check for a counterexample.
2. The twin prime conjecture (see here and here for more information) asserts that there are infinitely many primes that have a difference of 2. For example:
3 and 5 are twin primes;
5 and 7 are twin primes;
11 and 13 are twin primes;
17 and 19 are twin primes;
29 and 31 are twin primes; etc.
While most mathematicians believe the twin prime conjecture is correct, an explicit proof has not been found. The largest known twin primes are
$3,756,801,695,685 \times 2^{666,669} \pm 1$,
numbers which have 200,700 decimal digits. Also, there are 808,675,888,577,436 twin prime pairs less than $10^{18}$.
Most mathematicians also believe that there are infinitely many cousin primes, which differ by 4:
3 and 7 are cousin primes;
7 and 11 are cousin primes;
13 and 17 are cousin primes;
19 and 23 are cousin primes;
37 and 41 are cousin primes; etc.
Most mathematicians also believe that there are infinitely many sexy primes (no, I did not make that name up), which differ by 6:
5 and 11 are sexy primes;
7 and 13 are sexy primes;
11 and 17 are sexy primes;
13 and 19 are sexy primes;
17 and 23 are sexy primes; etc.
(Parenthetically, a “sexy” prime is probably the most unfortunate name in mathematics ever since Paul Dirac divided a bracket into a “bra” and a “ket,” thereby forever linking women’s underwear to quantum mechanics.)
While it is unknown if there are infinitely many twin primes, it was recently shown — in 2013 — that, for some integer $N$ that is less than 70 million, there are infinitely many pairs of primes that differ by $N$. In 2014, this upper bound was reduced to 246. Furthermore, if a certain other conjecture is true, the bound has been reduced to 6. In other words, there are infinitely many twin primes or cousin primes or sexy primes… but, at this moment, we don’t know which one (or ones) is infinite.
In November 2014, Dr. Terence Tao of the UCLA Department of Mathematics was interviewed on the Colbert Report to discuss the twin prime conjecture… and he does a good job explaining to Stephen Colbert how we can know one of the three categories is infinite without knowing which category it is.
From the Colbert Report: http://thecolbertreport.cc.com/videos/6wtwlg/terence-tao
# Proving theorems and special cases (Part 6): The Goldbach conjecture
In a recent class with my future secondary math teachers, we had a fascinating discussion concerning how a teacher should respond to the following question from a student:
Is it ever possible to prove a statement or theorem by proving a special case of the statement or theorem?
Usually, the answer is no. In this series of posts, we’ve already seen that a conjecture could be true for the first 40 cases or even the first $10^{316}$ cases yet ultimately prove false for all cases.
For the next few posts, I thought I’d share a few of the most famous unsolved problems in mathematics… and just how much computational work has been done to check for a counterexample.
1. The Goldbach conjecture (see here and here for more information) claims that every even integer greater than 4 can be written as the sum of two prime numbers. For example,
4 = 2 + 2,
6 = 3 + 3,
8 = 3 + 5,
10 = 3 + 7,
12 = 5 + 7,
14 = 3 + 11, etc.
This has been verified for all even numbers less than $4 \times 10^{18} = 4,000,000,000,000,000,000$. A proof for all even numbers, however, has not been found yet.
Here are some results related to the Goldbach conjecture that are known:
1. Any integer greater than 4 is the sum of at most six primes.
2. Every sufficiently large even number can be written as the sum or two primes or the sum of a prime and the product of two primes.
3. Every sufficiently large even number can be written as the sum of two primes and at most 8 powers of 2.
4. Every sufficiently large odd number can be written as the sum of three primes. | HuggingFaceTB/finemath | |
# Make x The Subject
Here we will learn about making x the subject of an equation or a formula.
There are also rearranging equations worksheets based on Edexcel, AQA and OCR exam questions, along with further guidance on where to go next if you’re still stuck.
## What does it mean to make x the subject?
Making x the subject of a formula or equation means rearranging the equation or formula so that we have a single x variable equal to the rest of it.
For example,
Make x the subject.
Step-by-step guide: Rearranging equations
## How to make x the subject
In order to make x the subject:
1. Isolate the variable by:
– Removing any fractions by multiplying by the denominator(s).
– Dividing by the coefficient of the variable.
– Adding or subtracting terms near to the variable.
– Taking a root or power of both sides of the equation.
2. Rearrange the equation so each term containing x is on the same side of the =.
3. Factorisation may be needed if there are multiple terms containing x.
E.g. factorise 2x + 3xy to x(2+3y)
*not always required*
4. Perform an operation to ensure only a single x variable is left as the subject.
## Make x the subject examples
### Example 1: multiple step but with single variable
Make x the subject.
Step 1:
Divide each side of the equation by 3
Step 2:
Subtract a from each side of the equation
### Example 2: questions involving x2
Step 1:
Subtract t from both sides of the equation
Step 2:
Square root each side
Remember the square root can be + or \;−
### Example 3: questions involving √x
Step 1:
Add 9a to both sides of the equation
Step 2:
The inverse operation of ‘square root’ is to ‘square’ each side
Step 3:
The inverse operation of multiply is divide, so divide both sides by 5
### Example 4: factorisation of the variable is required
Step 1:
Multiply each side of the equation by the denominator
Step 2:
Expand the bracket on the left hand side of the equation and rearrange the equation. This will help us to get all terms with x onto one side of the equation
Step 3:
Factorise the left side of the equation so we have a single variable x.
Step 4: Divide by (a - 5b)
This will leave x as the subject of the equation
### Example 5: factorisation of the variable is required
Step 1:
Multiply each side of the equation by the denominator of the other side.
Step 2:
Expand the bracket on the LHS and RHS of the equation and rearrange. This will help to get all terms with x onto one side of the equation
Step 3:
Factorise the left side of the equation so that we are left with only one of the variable x.
Step 4:
Divide both sides by (b + 8) to leave x as the subject
### Common misconceptions
• Perform the same operation to both sides
When we perform an operation to the left hand side of the equation we have to perform the same operation to the right hand side.
E.g
$\frac{x}{2}=3 y+5$
To isolate the variable x we need to multiply both sides by 2.
$x=6y+5$
This is wrong because we have only multiplied the 3y by 2.
The correct answer should be:
$x=2(3y+5)=6y+10$
This is correct because we have multiplied everything by 2 using brackets.
• Inverse operations
E.g.
To isolate the variable x from 5x we need to perform the inverse operation.
5x = 5 × x, so the inverse operation is divide by 5.
E.g.
$\frac{x}{5}=5 \div x$
so the inverse operation is × 5.
E.g.
To isolate the variable x from x+5 we need to perform the inverse operation:
The inverse operation of +5 is −5.
E.g.
To isolate the variable x from x−5 we need to perform the inverse operation:
The inverse operation of −5 is +5.
• Factorising
To make x the subject there can only be one x visible at the end. We need to get all x’s on one side of the equal sign and then factorise.
E.g.
$x = ax+y$
\begin{aligned} x-a x &=y \\ x(1-a) &=y \\ x &=\frac{y}{1-a} \end{aligned}
• Square rooting a term
When we square rooting a number/variable as an inverse operation the answer can be positive or negative.
E.g.
\begin{align} & {{x}^{2}}=4 \\ & x=\pm \sqrt{4}=\pm 2 \\ \end{align}
$\sqrt{x} \text { should be written as } \pm \sqrt{x}$
### Practice make x the subject questions
1.Make x the subject of the formula.
y = 6(x+8)
x = y – \frac{3}{4}
x = \frac{y}{6} + 8
x = \frac{y}{6} – 8
x = y – 8
y = 6(x + 8)
Divide both sides by 6
\frac{y}{6} = x + 8
Then subtract 8 from both sides
x = \frac{y}{6} – 8
2.Make x the subject of the formula.
3p={x}^2-4b
x=\pm\sqrt{3b+4p}
x=\pm\sqrt{4b-3p}
x=4b+3p
x=\pm\sqrt{4b+3p}
3p={x}^2-4b
Add 4b to both sides
3p+4b=x^{2}
Square root both sides
x=\pm\sqrt{4b+3p}
3. Make x the subject of the formula.
6g=\sqrt{7x-8}
x=\frac{6{g}^2+8}{7}
x=\frac{36{g}^2-8}{7}
x=\frac{36{g}^2+8}{7}
x=\frac{6g+8}{7}
6g=\sqrt{7x-8}
Square both sides
(6g)^{2}=(\sqrt{7x-8})^{2}
36g^{2}=7x-8
Add 8 to both sides
36g^{2}+8=7x
Divide both sides by 7
x=\frac{36{g}^2+8}{7}
4.Make x the subject of the formula.
y=\frac{4x-f}{5x}
x=\frac{f}{5y-4}
x=\frac{-f}{5y+4}
x=\frac{f}{5y+4}
x=\frac{-f}{5y-4}
y=\frac{4x-f}{5x}
Multiply both sides by 5x
5xy=4x-f
Subtract 4x from both sides
5xy-4x=-f
Factorise the left hand side
x(5y-4)=-f
Divide both sides by the quantity in the bracket
x=\frac{-f}{5y-4}
5. Make x the subject of the formula.
\frac{y}{3}=\frac{6-2x}{x+3}
x=\frac{18+3y}{y+6}
x=\frac{18-3y}{y-6}
x=\frac{18-3y}{y+6}
x=\frac{18+3y}{y-6}
\frac{y}{3}=\frac{6-2x}{x+3}
Multiply each side of the equation by the denominator of the other side.
xy+3y=18-6x
To both sides, add 6x and subtract 3y
xy+6x=18-3y
Factorise the left hand side
x(y+6)=18-3y
Divide by the quantity in the bracket
x=\frac{18-3y}{y+6}
### Make x the subject GCSE Questions
1. Make x the subject of the formula
y=5x-7
(2 marks)
y+7=5x
(1)
\frac{y+7}{5} = x
(1)
2. Make x the subject of the formula
z^{2}=x^{2}-5 a y
(2 marks)
z^{2}+5 a y=x^{2}
(1)
\pm\sqrt{z^{2}+5 a y}=x
(1)
3. Make x the subject of the formula
y=\frac{3(t+5 x)}{x}
(4 marks)
y x=3 t+15 x
(1)
y x-15 x=3 t
(1)
x(y-15)=3 t
(1)
(1)
## Learning checklist
You have now learned how to:
• Understand and use standard mathematical formulae
• Rearrange formulae to change the subject
## Still stuck?
Prepare your KS4 students for maths GCSEs success with Third Space Learning. Weekly online one to one GCSE maths revision lessons delivered by expert maths tutors.
Find out more about our GCSE maths tuition programme. | HuggingFaceTB/finemath | |
# SAT Mathematics: Percentage
This is an MCQ-quiz for SAT Mathematics, which include questions on Percentage.
Start Quiz
Write as a fraction: 22%
4/9 11/100 2/3 11/50 4/7
25% of 64 is equal to 5% of what number?
94 90 320 112 108
When y is decreased by ten percent, the result is equal to fifteen percent of x. Assuming both x and y are nonzero, what is the ratio of x to y?
1/6 3 6 18 1/3
Turn the following percentage into a fraction:
15%
9/20
1/5
3/20
7/10
Write 7.5% as a fraction.
7/5
7.5/10
3/40
15/2000
14/100
Turn the following percentage into a fraction:
44%
11/50
11/20
12/25
11/25
Turn the following percentage into a fraction:
88%
21/25
24/25
22/25
23/25
Turn the following percentage into a fraction:
72%
17/25
16/25
19/20
18/25
Turn the following percentage into a fraction:
46%
27/50
23/50
12/25
13/25
Turn the following percentage into a fraction:
65%
3/5
3/4
17/20
13/20
Quiz/Test Summary
Title: SAT Mathematics: Percentage
Questions: 10
Contributed by: | HuggingFaceTB/finemath | |
Motion
1 / 50
# Motion - PowerPoint PPT Presentation
##### Motion
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1. Motion Chapter 2 Physical Science Spring 2009
2. Formula Sheet Students may bring a formula sheet to the next chapter test and should use this sheet to work out the assigned problems . • one page on which only formulas are written. • No definitions or examples of worked problems may appear on this sheet. • The instructor will mention in lecture other information that should be placed on the formula sheet such as: g = 9.8 m/s2 • You should start the formula sheet now and write formulas on it as they are covered in class. • Use the formula sheet to solve homework problems in class as well as for the test. • Have you got your sheet out ready to write on?
3. Definition • Velocity is the distance traveled in a certain time divided by the time it takes to travel that distance Velocity = Distance/time normal units of velocity are meters/second m/s Other ways to write the formula: If you have difficulty with algebra, you will want to put all the variations of a formula on your formula sheet. Definitions on tests and the review sheet must contain both words (you may say things in your own words) as well as mathematical equations.
4. Definition Acceleration is the change in the velocity divided by the time it takes the velocity to change Acceleration = (final velocity-initial velocity)/time normal units of acceleration are meters/second per second (m/s2) The acceleration measures how fast velocity is changing. Other ways to write the formula
5. Example Problems • Problems will be an important part of class, review sheet, and tests (~50%). • All problems must show the data and the work (including formulae and units) • See the examples that follow • Do the assigned problems from your chapter review as the material is covered in class. Ask when you have difficulties or do not fully understand anything. • Bring calculators to class and lab
6. Problems • On all problems, everyone must show the following four steps. • Failure to show all four steps will result in points lost, even if the answer is correct. • The four steps are: • Data • Formula • Work • Answer • The units must be correctly shown at each step
7. How long does it take a car traveling 30 m/s to travel 5 kilometers? • Data t = ? v = 30 m/s d = 5 km =5000m • Formula • Work
8. A car starts at rest and 15 sec later is traveling 20 m/s. What is the acceleration? • Data vi=0 t=15s vf=20 m/s a=? • Formula and Work
9. Force (A quick look for your prelab) • When a force acts on a mass it causes the mass to accelerate. • F=ma • Force=mass x acceleration • Units of mass are kilograms • Units of force are “newtons” (N) • 10 newtons = 10 N
10. Gravity • The acceleration of gravity is g= 9.8 m/s2 The force of gravity (weight) on a 20 gram object is F=mg = .02 kg x 9.8 m/s2 = 0.196 N Be sure to remember that in the F=ma equation, the mass must be in kilograms, not grams!
11. Pulleys There are 100 g (0.1 kg) more on one side of the pulley than on the other so the accelerating force will be: F = mg = 0.1 kg x 9.8 m/s2 = 0.98 N Note that the mass must be in kilograms, be sure to change grams to kilograms on your lab! 2.0 kg 2.1 kg
12. Lab Problem: An object falls 9 m in 5 sec. Find a) average velocity, b) final velocity, c) acceleration. • Data • d = 9m • t = 5 sec • v = ? • Formula • It is important to know that this is the average velocity as the weight is speeding up as it falls.
13. Final velocity • The average of anything is (vi+vf)/2 = vave • If vi = 0, then vf = 2·vave • Put these two formulas on your formula sheet • Calculation of final velocity vf = 2·vave = 2 x 1.8 m/s = 3.6 m/s
14. Calculation of acceleration An object initially at rest falls 9 m in 5 sec. vf = 3.6 m/s • Data • vi = 0, vf = 3.6 m/s, t = 5 sec
15. Definitions • Scalars – physical quantities that can be completely described by one number • Examples are: time, pressure, volume, mass, amount in your bank account, etc. • Vectors – physical quantities that need both a size (magnitude) and a direction to describe them. • Examples are: velocity, acceleration, force
16. More on acceleration • Can one accelerate without increasing speed? • Give some examples! • Is slowing down acceleration? 2q
17. Difference between speed and velocity • Speed is a scalar and is completely described by one number • 50 mi/hr • It tells you nothing about the direction which you are traveling. • Velocity is a vector and always contains information about both the speed and the direction. • 50 mi/hr North.
18. Acceleration of Gravity • If air friction is not a factor, all objects fall with an acceleration of 9.8 m/s2 (symbol g) • Put g = 9.8 m/s2 on your formula sheet • How does horizontal motion affect the rate at which an object falls? • The downward acceleration of an object is not affected by its sidewise velocity. • an object thrown horizontally from the top of a building will strike the ground at the same time as an object dropped from the same height.
19. Free fall • When the only force acting on a falling object is gravity, we say the object is in free fall. • In most cases air friction is a force that keeps the object from accelerating continually. • When the force on an object is that of gravity we call the force the object’s weight F = ma W = mg • When the force of air friction becomes equal to the force of gravity, the velocity remains constant at the “terminal velocity”
20. Falling with air friction • In the real world, if we first step off a tall building we fall with an acceleration of 9.8 m/s2 for only a few (perhaps 5) seconds. • Air friction then becomes a factor and begins to slow our acceleration. • When we are going fast enough the air friction is equal our weight and we no longer accelerate, we continue with a constant (terminal) velocity.
21. Terminal Velocity • Depends on the weight and shape of the object • Sky divers change the air friction and enable themselves to fall slower or faster.
22. A brick is dropped from a tall building. How fast is it falling after 5 sec? • On a sheet of paper write the data for this problem • Data vi=0 m/s a = 9.8 m/s2 vf = ? t = 5 s On your paper write the formula to be used in solving this problem. vf = vi +at
23. Circular Motion(Necessary for your prelab) • Centripetal means center-seeking The centripetal force is toward the center of the circle. The force necessary to keep an object moving in a circle is called the centripetal force. m is mass (kg), v is velocity (m/s), R is radius of circle (m). Units of force are newtons. Add to formula sheet
24. What force is necessary to keep a 1000 kg car traveling 20 m/s in a circle of radius 15 m? • Data F = ? m = 1000 kg v = 20 m/s R = 15 m
25. A 2000 kg truck going 30 m/s is turning a corner of radius 20 meters. What sideways force must its tires exert on the payment? • Data: m=2000kg, v=30m/s, R=20m, F = ?
26. How strong (newtons) must a string be to allow a 200 g yo-yo to be swung in a circle of radius 1 m at 20 m/s? • Data F = ? m = 200g = 0.2 kg v = 20 m/s R = 1 m
27. When gravity and centripetal forces balance • How fast must I swing a glass so that water will not fall out when it is upside down? • The water will not fall out if the centripetal force is equal to or greater than the gravitational force. FC = Fg Fg = W = mg This is the velocity an object must have to complete the circle without falling.
28. How fast must I swing the beaker in a 0.6 m radius so water does not fall out when it is upside down • Data • V = ? • R = 0.6 m g = 9.8 m/s2 FC = Fg Fg = W = mg mv2/R = mg
29. How fast must the ball go around a track of 20 cm radius so it stays on the track. • Data • V=? • D = 0.4 m g = 9.8 m/s2 FC = Fg • FC = mv2/R Fg = W = mg mv2/R = mg
30. Newton’s Laws(First law) • If no forces act on an object, its velocity will continue unchanged • An object at rest will remain at rest • An object in motion will continue to move with a constant velocity (in a straight line) until a force acts on it • This is also known as the Law of inertia (resistance to change in motion)
31. Newton’s Laws(Second law) • When a force acts on an object it produces an acceleration that is inversely proportional to its mass F = ma This may also be written a=F/m or m=F/a
32. Units of force • The force necessary to accelerate a mass of one kilogram, one meter per second per second is called a “newton” Symbol is (n) or (N) 1 n = 1 kg m/s2
33. Mass and Weight • Mass is the quantity of matter (inertia) an object contains, this will always be the same as long as the object remains intact. • Weight is the force of gravity on the object. This would be much different on the Moon, Mars or in space. • W = mg (on the surface of the Earth) • Weight = mass x acceleration of gravity • In this equation, be sure the mass is in kilograms not grams
34. Newton’s Laws • Forces always exist in pairs that are equal and opposite For every action there is an equal and opposite reaction
35. Mule problem • Explain why the mule can move the cart, given that the cart pulls just as hard on the mule as the mule pulls on the cart
36. More on Newton’s Laws • Clearly explain why the mule can move the cart, given that the cart pulls just as hard on the mule as the mule pulls on the cart. • The answer is Newton’s second law • An unbalanced force on any given object will cause that object to accelerate (move). • There is an unbalanced force on the cart, so the cart moves. (There is no backward force ON THE CART to balance the forward pull of the mule.) • There is also an unbalanced force on the mule, the ground is pushing harder on the mule (a reaction to the push of the mule on the ground) forward than the cart is pulling backward so the mule moves forward.
37. Add points to your test score • Students who have the most success after college have said the technique that they learned that most contributed to that success was how to work in groups. • To give additional incentive to work in groups points will be added to the test scores of all students who work in groups and help each other be more successful.
38. How the system works • You must email me with the names of the people in your group before 10:00 on the day of our test. • If the average of the group is better than their average on the chapter 1 test, the increase in the group average will be added to the score of each person in the group (up to max of 10 points.) • If the group’s average decreases, there is no penalty. • Groups may have between 2 and 6 people.
39. Example • On test one Judy’s score was 90, Fred’s was 70 and Jim’s was 38. (Ave = 67) • Judy, Fred and Jim study together. • On the Chapter 2 test Judy’s score is 92, Fred’s 76 and Jim’s 60 (Ave = 76) • 9 points will be added to each person’s score. • Judy’s score becomes 101%, Fred’s 85% and Jim’s 69%!
40. Newton’s law of universal gravitation • The attraction between any two objects of masses m1 and m2 which are a distance r apart is given by: G is a constant (a number) G = 0.0000000000667 = 6.67 x 10-11 n m2/kg2 Put this number on your formula sheet. m1 m2 r
41. What is the attraction between a 90 kg man and a 65 kg woman who are 0.5 m apart? • Data F=? m1= 90kg m2= 65kg r = 0.5 m F= 1.56 x 10-6n
42. Use the law of universal gravitation to calculate the force between two objects that have masses 10 kg and 40 kg that are 3 m apart. • Data F = ? m1= 10kg m2= 40kg r = 3 m F= 2.96 x 10-9 n
43. Mass of the Earth • Force of gravity on one kilogram = 9.8 n • Distance from mass to center of Earth = 6,378,000 m
44. Why does a person in the space shuttle feel weightless? • For the same reason as a person in a falling elevator, they are falling toward the center of the Earth. • At the same time the shuttle falls toward the center of the Earth their forward velocity carries them to the side so they never hit.
45. The gravity equations, when do they apply? • W = mg • Weight = mass x acceleration of gravity • g = 9.8 m/s2 • Gives the weight of an object on the surface of the Earth. • Applies to any object on the Earth
46. The gravity equations, when do they apply? • Calculates the force between any two objects of masses m1 andm2 that are a distance r apart • Applies to any objects, any place in the universe!
47. Newton’s laws • An object’s velocity will remain unchanged until a force acts on the object. • When a force acts on the object, it produces an acceleration according to the equation: F=ma • Forces always exist in pairs, for every force there is an equal and opposite force on the interacting objects.
48. Apply Newton’s laws of motion to the case of a soccer ball that is at rest and is kicked. • Law one says that the ball at rest will remain at rest until a force acts on it. • Law two says that the acceleration of the ball is proportional to the force of the kick and inversely proportional to the mass of the ball. • Law three says that the soccer ball exerts the same force on the foot as the foot exerts on the ball.
49. Density • Density is the mass per unit volume. • Density = mass/volume • D=m/v
50. Calculate the density of a wood cylinder of diameter 6 cm, length 10 cm and mass 250 grams. Data: d = 6 cm, (r = 3 cm), L = 10 cm and m = 250 g. | HuggingFaceTB/finemath | |
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Questions Per Quiz = 2 4 6 8 10
### MEAP Preparation - Grade 7 Mathematics1.149 Exponents - 2
Real life applications of exponents: Exponents are often used in the formulas for area and volume. In fact, the words squared and cubed come from the formula for the area of a square, A = s2 and the formula for the volume of a cube, V =s3. Directions: Answer the following questions. Also write at least ten examples of your own.
Q 1: Find the product 10 x 10 x 10 x 10 x 10.01000010000050 Q 2: 2 cubed means. (Hint: 23=___)2 + 2 + 22 x 2 x 22 x 32+3 Q 3: Find 1 to the 6th power. (Hint: 16=1x1x1x1x1x1=____)166infinity1 Q 4: Write 6 x 6 x 6 is written in exponent form.316366263 Q 5: Find the product of 2 to the 4th power. (Hint: 24=2x2x2x2=_____)632168 Q 6: Find the value of 5 to the 3rd power.25125815 Q 7: Solve -2 to the 3rd power times 3 to the 2nd power. Hint: (-a)3 = (-a)(-a)(-a) = -a3-72-36-2332 Q 8: Write the exponent form for 10 x 10 x 10 x 10 x 10.103105104410 Question 9: This question is available to subscribers only! Question 10: This question is available to subscribers only! | HuggingFaceTB/finemath | |
# Evaluate :$$\frac{50!}{48!}$$
This question was previously asked in
MPPEB Stenographer 15 Sept 2018 Shift 1 Official Paper
View all MP Vyapam Group 2 Papers >
1. 2430
2. 2440
3. 2420
4. 2450
Option 4 : 2450
## Detailed Solution
Formula:
⇒ n! = n × (n - 1) × ......× 1
Calculation:
⇒ 50! = 50 × 49 × 48!
Then,
⇒ 50!/48! = (50 × 49 × 48!)/48! = 2450
∴ 50!/48! = 2450 | HuggingFaceTB/finemath | |
# 2021 HSC Maths Ext 2 Exam Paper Solutions
The 2021 HSC Maths Extension 2 Exam Paper solutions are here! Read on to see the full worked solutions written by our Head of Mathematics, Oak Ukrit and his team.
The wait is over. You can now see how you went with the 2021 HSC Maths Extension 2 Exam Paper!
## 2021 HSC Maths Ext 2 Exam Paper Solutions
Have you seen the 2021 HSC Mathematics Extension 2 Exam Paper yet?
In this post, we will work our way through the 2021 HSC Maths Extension 2 exam paper and give you the solutions, written by our Head of Mathematics, Oak Ukrit and his team. (Doing practice papers? See the solutions for the 2020 HSC Maths Ext 2 Exam here.)
Read on to see how to answer all of the 2021 questions.
## Section 1. Multiple Choice
Question Number Answer Solution 1. B Since $$\overrightarrow{AB} = \overrightarrow{CD}$$ (both right and flat along the table) and $$\overrightarrow{CQ} = \overrightarrow{DP}$$ (both diagonals across the cube upwards + rightwards + into the page) , we find that $$\overrightarrow{AB}+\overrightarrow{CQ} = \overrightarrow{CD}+\overrightarrow{DP} = \overrightarrow{CP}$$. 2. A Integrate by parts. \begin{align*} \textrm{Let } u &= x^5 \textrm { and } v’ = e^{7x}\\ u’ &= 5x^4 \textrm { and } v = \frac{1}{7}e^{7x}\\ \int x^5 e^{7x} \,dx &= vu – \int u’v \,dx\\ &= \frac{1}{7}e^{7x}x^5 – \int 5x^4\times \frac{1}{7}e^{7x} \,dx\\ &= \frac{1}{7}x^5 e^{7x} – \frac{5}{7}\int x^4 e^{7x} \,dx. \end{align*} 3. B The direction vector needs to be parallel to \begin{align*} \overrightarrow{BA} = \begin{pmatrix} 4 \\ 2 \\5 \end{pmatrix} – \begin{pmatrix} -2 \\ 2 \\ 1 \end{pmatrix} = \begin{pmatrix} 6 \\ 0 \\4 \end{pmatrix}. \end{align*} Out of the 4 options, it is clear that the only direction vector parallel to $$\overrightarrow{BA}$$ is $$(3, 0, 2)$$, hence the solution is \begin{align*} \underset{\text{~}}{r} = \begin{pmatrix} 4 \\ 2 \\5 \end{pmatrix}+\lambda \begin{pmatrix} 3 \\ 0 \\2 \end{pmatrix} \end{align*} 4. C The contrapositive of a statement “if A then B” is “if not B then not A”. We split this statement into parts at the “if” and “then”: \begin{align*} \textrm{For all integers n, } | \textrm{ if } | \textrm{ n is a multiple of 6, } | \textrm{ then } | \textrm{ n is a multiple of 2}. \end{align*} We leave the “for all integers n” at the front, since it is context. Next, we negate the “B” part, i.e. “n is not a multiple of 2”, and also negate the “A” part, i.e. “n is not a multiple of 6”. Finally, we structure as “If Not B then Not A”, and add our context back, to get \begin{align*} \textrm{For all integers n, } | \textrm{ if } | \textrm{ n is not a multiple of 2, } | \textrm{ then } | \textrm{ n is not a multiple of 6}. \end{align*} Check: Remember, the contrapositive must be logically equivalent to the original. If you were asked to prove the original statement, you could equivalently prove “for all integers n, if n is not a multiple of 2, then n is not a multiple of 6”, which is (C). Remark: Observe that (A) is actually the negation! 5. D We can solve this by considering each of the answers: A. The function $$x^3$$ is increasing on its domain. Therefore $$ae^{-b}$$. C. The function $$\ln(x)$$ is increasing on its domain. Therefore $$a 0$$ is a small increment in time. At time $$t+\Delta t$$, the particle has moved slightly further along the curve. Since the speed is increasing, the arrow representing $$\mathbf{v}(t+\Delta t)$$ should be slightly longer than $$\mathbf{v}(t)$$. Now we consider the difference $$\mathbf{a}(t) \Delta t \approx \mathbf{v}(t+\Delta t) – \mathbf{v}(t)$$: The vector $$\mathbf{a(t)}\Delta t$$ is $$\mathbf{a}(t)$$ rescaled by $$\Delta t$$, but the key thing is that they should approximately have the same direction when $$\Delta t$$ is sufficiently small. Consequently we see that (D) is the best match. 9. B This is a classic logic puzzle :). It is best resolved by thinking about the contrapositive, which is always logically equivalent to the original statement: “If a card does not have WIN on one side, then the other side does not have RED”. Sam needs to check both the direct implication and the contrapositive — hence they should flip RED to see if WIN is on the other side, and then flip TRY AGAIN to make sure RED is not on the other side. 10. C If $$z = u+iv$$ and $$w = x+iy$$, then we identify $$z$$ and $$w$$ with the vectors $$z = (u,v)$$ and $$w = (x,y)$$ respectively. The projection of $$z$$ onto $$w$$ is given by \begin{align*} \text{Proj}_w(z) = \frac{z \cdot w}{|w|^2}w. \end{align*} Now observe that $$z \cdot w = ux + vy$$, and $$\text{Re}(zw) = ux – vy$$. Unfortunately this has a minus sign! However, we can easily fix this by considering $$\bar{w}$$ instead: \begin{align*} \text{Re}(z\bar{w}) = ux + vy = z \cdot w. \end{align*} Finally, notice that \begin{align*} \text{Re}\left(\frac{z}{w}\right) = \text{Re}\left(\frac{z\bar{w}}{|w|^2}\right) = \frac{1}{|w|^2}\text{Re}(z\bar{w}) = \frac{z \cdot w}{|w|^2}. \end{align*} This shows that (C) is correct.
Need help with Maths Ext 2?
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## Section 2. Long Response Questions
Question 11a
We have
\begin{align*} zw &= 2e^{i\frac{\pi}{2}} \times 6e^{i\frac{\pi}{6}} \\ &=(2 \times 6) e^{i\frac{\pi}{2} + i\frac{\pi}{6}} \\&= 12 e^{i\frac{2\pi}{3}}. \end{align*}
Question 11b
It is easiest to calculate directly:
\begin{align*} \sum_{n=1}^5 i^n &= i + i^2 + i^3 + i^4 + i^5 \\&= i – 1 – i + 1 + i \\&= i \end{align*}
Question 11c
We have:
\begin{align*} a \cdot b &= -6 + 0 + 8 = 2\\ |a| &= \sqrt{2^2 + 0^2 + 4^2} = 2\sqrt{5}\\ |b| &= \sqrt{(-3)^2+1^2+2^2} = \sqrt{14} \end{align*}
Thus
\begin{align*} \cos\theta = \frac{a\cdot b}{|a||b|} = \frac{2}{2\sqrt{5} \sqrt{14}} = \frac{1}{\sqrt{70}} \end{align*}
and $$\theta = \cos^{-1}(\frac{1}{\sqrt{70}}) \approx 83.1^\circ$$.
Question 11d
i) Let $$z=x+iy$$, and $$z^2 = -i$$ so that $$z$$ represents the square roots of $$-i$$.
Expanding, $$z^2 = (x^2-y^2) + 2xyi = -i$$.
Equating real and imaginary parts, $$x^2 = y^2$$ and $$2xy = -1$$.
The first equation gives $$x=\pm y$$. If $$x=y$$, then $$2xy=-1$$ has no real solutions.
Thus $$x=-y$$, and the second equation yields
\begin{align*} -2x^2 = -1 \Rightarrow x^2 = \frac{1}{2} \Rightarrow x = \pm \frac{1}{\sqrt{2}}. \end{align*}
ii) We use the quadratic formula, with $$a=1$$, $$b=2$$, $$c=1+i$$:
\begin{align*} z &= \frac{-2 \pm \sqrt{4 – 4(1+i)}}{2}\\ &= -1 \pm \sqrt{-i} \end{align*}
Now, from (i):
\begin{align*} z&=-1 \pm \sqrt{-i} \\ &= -1 \pm \frac{1}{\sqrt{2}}(1-i) \end{align*}
Thus the two solutions are $$z = (-1+\tfrac{1}{\sqrt{2}}) – \tfrac{1}{\sqrt{2}}i$$ and $$z = -(1+\tfrac{1}{\sqrt{2}}) + \tfrac{1}{\sqrt{2}}i$$.
Question 11e
We start by realising the denominator, by multiplying top and bottom by $$\bar{w}$$:
\begin{align*} \frac{\bar{z}}{w} &= \frac{\bar{z}\bar{w}}{w\bar{w}}\\ &= \frac{\bar{z}\bar{w}}{|w|^2} \\ &= \frac{(5-i)(2+4i)}{20} \\&= \frac{14+18i}{20} \\&= \frac{7}{10} + \frac{9}{10}i. \end{align*}
Question 11f
Note that the quadratic $$x^2+x+1$$ is irreducible (cannot be factorized) over $$ℝ$$, hence our partial fraction decomposition has the form
\begin{equation*} \frac{3x^2-5}{(x-2)(x^2+x+1)} = \frac{A}{x-2} + \frac{Bx+C}{x^2+x+1}. \end{equation*}
Multiplying both sides by the denominator, we get
\begin{equation*} A(x^2+x+1) + (Bx+C)(x-2) \equiv 3x^2-5. \end{equation*}
Setting $$x=2$$ to zero out the second term, we find that
\begin{align*} A(2^2 + 2 + 1) + (Bx + C)(2-2) &= 3(2^2) + 5\\ 7A &= 7\\ A&=1. \end{align*}
Matching coefficients of $$x^2$$, we find $$A+B=3 \Rightarrow B=2$$. Finally, we match constants to obtain $$A-2C = -5 \Rightarrow C=3$$. Hence
\begin{align*} \frac{3x^2-5}{(x-2)(x^2+x+1)} = \frac{1}{x-2} + \frac{2x+3}{x^2+x+1}. \end{align*}
Question 12a
Notice that $$x^2 + 2x + 2 = (x+1)^2 + 1$$, so the quadratic cannot be factored over $$ℝ$$, i.e. we cannot use partial fractions.
Instead, we notice that the derivative of the denominator is $$2x+2$$, which leaves 1 in the numerator when separated out:
\begin{align*} \int \frac{2x+3}{x^2+2x+2} \,dx &= \int \frac{2x+2}{x^2+2x+2} \,dx + \int \frac{1}{x^2+2x+2} \,dx. \end{align*}
The first integral is a logarithm with its derivative on the numerator; the second one is an inverse tan.
\begin{align*} &= \ln(x^2+2x+2) + \int \frac{1}{(x+1)^2+1} \,dx \\ &= \ln(x^2+2x+2) + \tan^{-1}(x+1) + C. \end{align*}
Question 12b
i) We obtain the converse by swapping the “if” and “then” statements: “if $$n$$ is even, then $$n^2$$ is even”.
ii) Assume that $$n$$ is even, so that $$n=2k$$ for some integer $$k$$. Then $$n^2 = (2k)^2 = 4k^2 = 2(2k^2)$$, hence $$n^2$$ is even.
Question 12c
Since the lines are perpendicular, their direction vectors are perpendicular, and so their dot products are 0:
\begin{align*} \begin{pmatrix} 1 \\ 0 \\ 2 \end{pmatrix} \cdot \begin{pmatrix} p \\ 3 \\ -1 \end{pmatrix} &=0 \\ p – 2&=0 \\p&=2 \end{align*}
Now we are given that the lines intersect.
Looking at the equation for the $$y$$-coordinate, we see that $$3\mu – 2 = 1 \Rightarrow \mu = 1$$.
Now the $$x$$-coordinate equation gives $$6 = \lambda – 2 \Rightarrow \lambda = 8$$.
Finally, the $$z$$-coordinate equation gives
\begin{align*} 3+ 16 = q-1 \Rightarrow q = 20. \end{align*}
Question 12d
Firstly, we check the base case on our calculator: $$\sqrt{9!} = 602.3952\ldots$$ and $$2^9 = 512 < \sqrt{9!}$$.
Now assume the statement is true for some $$k\ge 9$$. We need to show that $$2^{k+1} < \sqrt{(k+1)!}$$. Starting with the left hand side of this inequality, we have
\begin{align*} 2^{k+1} = 2^k \cdot 2 < 2\sqrt{k!} \end{align*}
by assumption.
Now if $$k \ge 9$$, then $$\sqrt{k+1} \ge \sqrt{10} > 2$$. Hence
\begin{align*} 2^{k+1} < 2\sqrt{k!} < \sqrt{k+1} \sqrt{k!} = \sqrt{(k+1)!} \end{align*}
which completes the induction.
Question 12e
i) Since the diagonals of the square base bisect at $$H$$ we have that $$\overrightarrow{HA} =-\overrightarrow{HC}$$ and $$\overrightarrow{HB}=-\overrightarrow{HD}$$. We then have that
\begin{align*} \overrightarrow{HA}+\overrightarrow{HB}+\overrightarrow{HC}+\overrightarrow{HD} &= (\overrightarrow{HA}+\overrightarrow{HC})+(\overrightarrow{HB}+\overrightarrow{HD}) \\ &= (-\overrightarrow{HC}+\overrightarrow{HC}) + (-\overrightarrow{HD}+\overrightarrow{HD}) \\ &= \mathbf{0}. \end{align*}
ii) Starting from the given equation, we want to turn $$\overrightarrow{GA}, \overrightarrow{GB}, \overrightarrow{GC}, \overrightarrow{GD}$$ into $$\overrightarrow{GH}$$.
Consider that for each of A, B, C and D:
\begin{align*} \overrightarrow{GA} &= \overrightarrow{GH}+\overrightarrow{HA}\\ \overrightarrow{GB} &= \overrightarrow{GH}+\overrightarrow{HB}\\ \overrightarrow{GC} &= \overrightarrow{GH}+\overrightarrow{HC}\\ \overrightarrow{GD} &= \overrightarrow{GH}+\overrightarrow{HD} \end{align*}
Performing these substitutions, we have:
\begin{align*} \overrightarrow{GA}+\overrightarrow{GB}+\overrightarrow{GC}+\overrightarrow{GD}+\overrightarrow{GS} &= (\overrightarrow{GH}+\overrightarrow{HA})+(\overrightarrow{GH}+\overrightarrow{HB})+(\overrightarrow{GH}+\overrightarrow{HC})+(\overrightarrow{GH}+\overrightarrow{HD})+\overrightarrow{GS}\\ &= (\overrightarrow{HA}+\overrightarrow{HB}+\overrightarrow{HC}+\overrightarrow{HD}) + 4\overrightarrow{GH} +\overrightarrow{GS} \\ &=4\overrightarrow{GH} + \overrightarrow{GS}. \end{align*}
Now, from the question, $$G$$ is such that $$\overrightarrow{GA}+\overrightarrow{GB}+\overrightarrow{GC}+\overrightarrow{GD}+\overrightarrow{GS}=\mathbf{0}$$, so we obtain
\begin{align*} 4\overrightarrow{GH} + \overrightarrow{GS} = \mathbf{0}. \end{align*}
iii) Using vector addition, we can write $$\overrightarrow{GS} = \overrightarrow{GH}+\overrightarrow{HS}$$.
Substituting into (ii) gives
\begin{align*} 4\overrightarrow{GH} + \overrightarrow{GS} &= \mathbf{0} \\ 4\overrightarrow{GH} + \overrightarrow{GH}+\overrightarrow{HS} &= \mathbf{0} \\ 5\overrightarrow{GH} &= -\overrightarrow{HS} \\ \overrightarrow{GH} &= -\frac{1}{5} \overrightarrow{HS} \\ \overrightarrow{HG} &= \frac{1}{5}\overrightarrow{HS} \end{align*}
so we conclude that $$\lambda = \frac{1}{5}$$.
Question 13a
Each of the 4th roots has the same magnitude, and are rotated by $$\frac{2\pi}{4} = 90\text{˚}$$. Since the magnitude of $$a+ib$$ as indicated is greater than 1, our resultant magnitude should also be greater than 1 (i.e. outside the unit circle), but smaller than the original. Additionally, we use the lines provided to draw a vector with a quarter of the modulus; then we rotate it 4 times by $$90 \text{˚}$$, giving us our solution as indicated in pink below:
Question 13b
The most straightforward substitution is $$u = \sqrt{x^2 – 9}$$. This gives:
\begin{equation*} du = \frac{x}{\sqrt{x^2 – 9}} \,dx \Rightarrow u\,du = x\,dx. \end{equation*}
When $$x=\sqrt{10}$$ and $$\sqrt{13}$$, we have $$u=1$$ and 2 respectively. Therefore
\begin{align*} \int_{\sqrt{10}}^{\sqrt{13}} x^3 \sqrt{x^2 – 9} \,dx &= \int_{\sqrt{10}}^{\sqrt{13}} x^2 \sqrt{x^2 – 9} \,(x\,dx) \\ &= \int_1^2 (u^2+9)u^2 \,du \\ &= \int_1^2 u^4 + 9u^2 \,du \\ &= \left[ \frac{u^5}{5} + 3u^3 \right]^2_1 \\ &= \left(\frac{32}{5}+24\right) – \left(\frac{1}{5}+3\right) \\ &= \frac{136}{5}. \end{align*}
i) We use integration by parts with $$u = (\ln x)^n$$ and $$dv = 1$$. Thus $$du = n(\ln x)^{n-1} \frac{1}{x}$$ and $$v=x$$. Using these parts, we can derive the reduction formula from the question:
\begin{align*} I_n = \int_1^e (\ln x)^n \,dx &= \bigg[(\ln x)^n x \bigg]^e_1 – \int_1^e n(\ln x)^{n-1} \,dx \\ &= e – nI_{n-1} \qquad \text{for all } n\ge 1. \end{align*}
ii) We find $$V_B$$ first, which is found by subtracting the volume of the solid of revolution under $$y=\ln x$$ from $$x=1$$ to $$x=e$$ from the volume of the cylinder with radius $$1$$ and height $$e-1$$:
\begin{align*} V_B &= \pi (e-1) – \pi \int_1^e (\ln x)^2 \,dx \\ &= \pi (e-1) – \pi I_2 \\ &= \pi (e-1) – \pi(e – 2I_1) \\ &= 2\pi I_1 – \pi. \end{align*}
using part (i). Using the reduction formula again and noting that $$I_0 = e-1$$, we have
\begin{equation*} 2\pi I_1 – \pi = 2\pi(e – I_0) – \pi = 2\pi(e – (e-1)) – \pi = \pi \end{equation*}
and therefore $$V_B = \pi$$. Now $$V_A$$ can be found by subtracting the volume of half a solid sphere of radius 1 from the volume of a cylinder of height 1 and base radius 1. Hence
\begin{equation*} V_A = \pi – \frac{1}{2}\times\frac{4}{3}\pi = \frac{\pi}{3} = \frac{1}{3}V_B. \end{equation*}
Question 13c
i) Noting that $$\ddot x = -4(x-3)$$, we identify that $$n=2$$ and the centre of the motion is $$x_0=3$$.
Using the relationship
\begin{equation*} v^2 = n^2\left(A^2-(x-x_0)^2\right) \end{equation*}
and the fact that when $$x=0$$, $$v^2=64$$ we may solve for the amplitude of the motion through
\begin{align*} 64 &= 4(A^2-9) \\ 16 &= A^2-9 \\ A^2 &= 25 \\ A &= 5 \qquad (A>0). \end{align*}
We have that
\begin{equation*} x_0 – A \leq x \leq x_0 + A \Longrightarrow -2 \leq x \leq 8 \end{equation*}
and so the particle oscillates between the values $$x=-2$$ and $$x=8$$.
ii) We first model the motion of the particle, then solve for $$x=0$$.
Since the particle starts moving back towards the origin (cosine) rather than towards the endpoint (sine), we can let for the displacement-time equation be
\begin{equation*} x = x_0 + A \cos(nt+\alpha) \end{equation*}
where $$x_0=3,A=5,n=2$$ and $$0<\alpha<\frac{\pi}{2}$$. So we have
\begin{equation*} x = 3 + 5 \cos (4t + \alpha) \end{equation*}
We now substitute the condition that $$x(0)=5.5$$ to find $$\alpha$$:
\begin{align*} 5.5 &= 3 + 5\cos\alpha \\ 5 \cos \alpha &= 2.5 \\ \cos \alpha &= \frac{1}{2} \\ \alpha &= \frac{\pi}{3} \end{align*}
Finally, solving for the first $$t>0$$ for which $$x=0$$:
\begin{align*} 3 + 5 \cos\left(4t+ \frac{\pi}{3}\right) &= 0 \\ \cos\left(4t + \frac{\pi}{3}\right) &= -\frac{3}{5} \\ 4t + \frac{\pi}{3} &= \cos^{-1}\left(-\frac{3}{5}\right) \\ t &= \frac{1}{4} \left( \cos^{-1}\left( -\frac{3}{5}\right) – \frac{\pi}{3}\right) \\ &\approx 0.29 \text{ correct to 2 decimal places.} \end{align*}
Question 14a
We use the $$t$$-substitution $$t=\tan\left(\frac{x}{2}\right)$$ where $$\mathrm{d}x = \frac{2}{1+t^2} \ \mathrm{d}t$$. When $$x=0,t=0$$ and when $$x=\frac{\pi}{2}, t=1$$. Evaluating the integral we get:
\begin{align*} \int_0^{\pi/2} \frac{1}{3+5\cos x} \ \mathrm{d}x &= \int_0^1 \frac{1}{3+5\frac{1-t^2}{1+t^2}} \cdot \frac{2}{1+t^2} \ \mathrm{d}t \\ &= \int_0^1 \frac{2}{3(1+t^2) + 5(1-t^2)} \ \mathrm{d}t \\ &= \int_0^1 \frac{1}{4-t^2} \ \mathrm{d}t \\ &= \int_0^1 \frac{1}{(2-t)(2+t)} \ \mathrm{d}t \\ &= \frac{1}{4} \int_0^1 \frac{1}{2-t} + \frac{1}{2+t} \ \mathrm{d}t \\ &= \frac{1}{4} \left[\ln \left| \frac{2+t}{2-t}\right| \right]_0^1 \\ &= \frac{\ln{3}}{4}. \end{align*}
Question 14b
i) Consider the diagram below:
Resolving the gravity force of $$5g$$ down the slope, we have that gravity contributes to a force of $$5g \cos(30^\circ)$$ down the slope. We have the component of gravity equal to $$5g \cos(30^\circ)=\frac{5\sqrt{3}}{2} g$$ and two opposing forces $$2v$$ and $$2v^2$$. Hence the resultant force down the slope is given by
\begin{equation*} \frac{5\sqrt{3}}{2}g – 2v -2v^2. \end{equation*}
ii) For the object to slide down the slope at constant speed, we have that the acceleration is zero. Using Newton’s second law, we have that
\begin{equation*} 5a = \frac{5\sqrt{3}}{2}g – 2v -2v^2. \end{equation*}
Setting $$a=0$$ and $$g=10$$ gives the quadratic:
\begin{equation*} 2v^2 + 2v – 25\sqrt{3}=0 \end{equation*}
\begin{equation*} v = \frac{-2 + \sqrt{4+200\sqrt{3}}}{4} \approx 4.2 \mathrm{ms}^{-1} \end{equation*}
where we take the positive square root as $$v$$ is directed down the slope.
Question 14c
i) Using De Moivre’s Theorem, we have that
\begin{equation*} (\cos\theta+i\sin\theta)^5 = \cos5\theta + i \sin5\theta \end{equation*}
However by the binomial theorem, we have that
\begin{align*} (\cos\theta+i\sin\theta)^5 &= \cos^5\theta+5\cos^4\theta (i\sin\theta) – 10\cos^3\theta \sin^2\theta -10\cos^2\theta (i\sin^3\theta) + 5 \cos\theta \sin^4\theta + i\sin^5\theta \\ &= (\cos^5\theta- 10\cos^3\theta \sin^2\theta+5 \cos\theta \sin^4\theta) + i(\ldots\text{imaginary part}\ldots). \end{align*}
Equating the real parts of both these expressions for $$(\cos\theta+i\sin\theta)^5$$ we obtain
\begin{equation*} \cos5\theta = \cos^5\theta- 10\cos^3\theta \sin^2\theta+5 \cos\theta \sin^4\theta \end{equation*}
To express this as a polynomial in $$\cos \theta$$ only, we use the identity $$\sin^2\theta = 1- \cos^2\theta$$ to obtain
\begin{align*} \cos5\theta &= \cos^5\theta- 10\cos^3\theta \sin^2\theta+5 \cos\theta \sin^4\theta \\ &= \cos^5\theta- 10\cos^3\theta (1-\cos^2\theta)+5 \cos\theta (1-\cos^2\theta)^2 \\ &= \cos^5\theta -10\cos^3\theta+10\cos^5\theta +5\cos\theta(1-2\cos^2\theta+\cos^4\theta) \\ &= 16\cos^5\theta – 20 \cos^3\theta +5 \cos\theta \end{align*}
as required.
ii) Note that
$$\text{Re}(e^{i\frac{\pi}{10}}) = \text{R}e(\cos\frac{\pi}{10}+i\sin\frac{\pi}{10})=\cos\frac{\pi}{10}$$.
Let $$\theta = \frac{\pi}{10}$$; then, using (i):
\begin{align*} 16\cos^5{\theta} – 20\cos^3{\theta} + 5\cos{\theta} &= \cos\left(\frac{5\pi}{10}\right)=0. \end{align*}
If we now let $$x=\cos{\theta}$$, we can write:
\begin{align*} 16x^5 – 20x^3 + 5x &= 0 \\ 16x^4 – 20x^2 + 5 &= 0 \end{align*}
where we note that $$x\neq0$$, making it valid to divide both sides by $$x$$. Treating this as a quadratic equation in $$x^2$$ we obtain
\begin{align*} x^2 &= \frac{-(-20)\pm\sqrt{(20)^2 – 4\times 16\times 5}}{2\times16} \\ &= \frac{20\pm\sqrt{80}}{32} \\ &= \frac{5\pm\sqrt{5}}{8} \\ x &= \pm \sqrt{\frac{5\pm\sqrt{5}}{8}}. \end{align*}
Now, $$\cos\frac{\pi}{10}>0$$ as it is in the first quadrant; so we have
\begin{align*} x &= \sqrt{\frac{5\pm\sqrt{5}}{8}} = 0.951 \textrm{ or } 0.5877. \end{align*}
where 0.951 is taking $$+\sqrt{5}$$ and 0.5877 is taking $$-\sqrt{5}$$.
For the inner sign, we note that $$0<\frac{\pi}{10}<\frac{\pi}{4}$$, and so
\begin{equation*} 0.707\approx \cos \frac{\pi}{4} < \cos \frac{\pi}{10} < \cos 0 = 1 \end{equation*}
and hence we must choose the inner $$+$$ sign to obtain $$x \approx 0.951$$ since $$0.5877$$ is too small. We can now conclude that
\begin{equation*} \cos \frac{\pi}{10} = \sqrt{ \frac{5+\sqrt{5}}{8} }. \end{equation*}
Question 15a
i) We are given $$\sqrt{xy} \le \frac{x+y}{2}$$.
We begin by substituting $$x=ab$$ and $$y=c$$ to get the LHS:
\begin{align*} \sqrt{xy} &\le \frac{x+y}{2}\\ \sqrt{(ab)c} &\le \frac{ab+c}{2} \end{align*}
Next, we substitute $$x=a^2$$ and $$y=b^2$$ to eliminate the ab term on our RHS:
\begin{align*} \sqrt{a^2b^2} &\le \frac{a^2+b^2}{2}\\ ab &\le \frac{a^2+b^2}{2}. \end{align*}
Substituting, we have:
\begin{align*} \sqrt{(ab)c} &\le \frac{ab+c}{2}\\ &\le \frac{\frac{a^2+b^2}{2}+c}{2}\\ &= \frac{a^2+b^2+2c}{4} \end{align*}
as required.
ii) The key observation is that the left hand side of the inequality in (i) is unchanged if the labels $$a,b,c$$ are swapped with each other. Hence we obtain 2 other inequalities for a total of 3:
\begin{align*} \sqrt{abc} &\le \frac{a^2+b^2+2c}{4} \\ \sqrt{abc} &\le \frac{b^2+c^2+2a}{4} \\ \sqrt{abc} &\le \frac{a^2+c^2+2b}{4}. \end{align*}
Adding up the inequalities just obtained, we find
\begin{align*} 3\sqrt{abc} &\le \frac{2(a^2+b^2+c^2)+2(a+b+c)}{4} \\ \sqrt{abc} &\le \frac{(a^2+b^2+c^2)+(a+b+c)}{6} \end{align*}
as required.
Question 15b
i) The question asks us to prove that every odd triangular number is a hexagonal number.
Let $$n=2k+1$$ be an odd integer, where $$k$$ is an integer $$\ge 0$$. Then
\begin{align*} t_n &= \frac{n(n+1)}{2} \\ &= \frac{(2k+1)(2k+2)}{2} \\ &= (k+1)(2k+1). \end{align*}
Now note that the $$m$$-th hexagonal number is given by $$h_m = 2m^2 – m = m(2m-1)$$. If we set $$m = k+1$$ to match the brackets, then $$2m-1= 2(k+1)-1 = 2k+1$$, so we can write:
\begin{align*} t_{2k+1} &= (k+1)(2k+1) \\ &= m(2m-1)\\ &= 2m^2 – m \\ &= h_m. \end{align*}
Thus the $$(2k+1)$$-th triangular number is the $$(k+1)$$-th hexagonal number; or put another way, the $$(2k-1)$$-th triangular number is the $$k$$-th hexagonal number.
ii) We note that in the previous step, we’ve mapped each hexagonal number one-to-one with each triangular number; i.e. $$h_1 = t_3$$, $$h_2 = t_5$$, $$h_m = t_{2m-1}$$.
Suppose $$t_{2k}$$, an even triangular number, was a hexagonal number $$h_n$$.
We know that triangular numbers are increasing, so $$t_{2k-1}<t_{2k}<t_{2k+1}$$.
Since each of the surrounding odd triangular numbers is a hexagonal number, then $$h_{k}<t_{2k}<h_{k+1}$$.
However, hexagonal numbers are also increasing; and only make sense for integer coefficients. Therefore, there cannot be a hexagonal number $$h_n = t_{2k}$$ such that $$h_{k}<h_n<h_{k+1}$$, so $$t_{2k} \neq h_n$$.
Question 15c
i) Using Newton’s second law we have the equation
\begin{align*} m \ddot x &= -mg – kv^2 \ (\textrm{where } m =1)\\ \ddot x &= -(g+kv^2) \end{align*}
Next, we use the relationship $$\ddot x = \frac{dv}{dt}$$ to find $$T$$ when $$v=0$$ at the maximum height of the object, with limits $$t=0,v=u$$ and $$t=T,v=0$$:
\begin{align*} \frac{dv}{dt} &= -(g+kv^2) \\ \int_u^0 \frac{1}{k\left(\frac{g}{k}+v^2\right)} \ \mathrm{d}v &= – \int_0^{T} \ \mathrm{d}t \\ \frac{1}{k}\sqrt{\frac{k}{g}}\left[\tan^{-1}\left(u\sqrt{\frac{k}{g}}\right)\right]_u^0 &= – T \\ \frac{1}{\sqrt{gk}}\left(0-\tan^{-1}\left(u\sqrt{\frac{k}{g}}\right)\right) &= – T \\ \end{align*}
We obtain that the time taken to reach the maximum height is
\begin{equation*} T = \frac{1}{\sqrt{gk}} \tan^{-1}\left(u\sqrt{\frac{k}{g}}\right). \end{equation*}
ii) This time we use the relationship $$\ddot x = v \frac{\mathrm{d}v}{\mathrm{d}x}$$, where we have that $$v=0$$ at the maximum height. Using these conditions, we integrate with limits $$x=0,v=u$$ at the bottom and $$x=H, v=0$$ at the maximum height:
\begin{align*} v \frac{\mathrm{d}v}{\mathrm{d}x} &= – (g+ kv^2) \\ \int_u^0 \frac{v}{g+kv^2} \ \mathrm{d}v &= – \int_0^H \ \mathrm{d}x \\ \frac{1}{2k} \int_u^0 \frac{2kv}{g+kv^2} \ \mathrm{d}v &= – \int_0^H \ \mathrm{d}x \\ \frac{1}{2k} \left[\ln(g+kv^2)\right]_u^0 &= – H \\ \frac{1}{2k} \ln \left( \frac{g}{g+ku^2}\right) &= -H \\ H&= \frac{1}{2k} \ln \left( \frac{g+ku^2}{g}\right). \end{align*}
We obtain the maximum height as
\begin{equation*} H = \frac{1}{2k} \ln \left( 1 + \frac{ku^2}{g} \right). \end{equation*}
Question 15d
For $$n=2$$, observe that $$2^2 + 3^2 = 13 < 25 = 5^2$$. We claim that $$2^n + 3^n < 5^n$$ for all $$n \ge 2$$. The base case $$n=2$$ has just been verified. Now, assuming the claim to be true for some $$n \ge 2$$, we observe that
\begin{align*} 2^{n+1} + 3^{n+1} &= 2^n \times 2 + 3^n \times 3 \\ &< 3(2^n + 3^n) \\ &< 5^n \times 3 \\ &< 5^{n+1}. \end{align*}
Thus the statement is true for $$n+1$$. Hence our claim holds true by the principle of mathematical induction. Alternatively we could use a binomial expansion, and note that
\begin{equation*} 5^n = (2+3)^n = 2^n +3^n +\sum_{k=1}^{n-1} \binom{n}{k} 2^{n-k} 3^k > 2^n + 3^n \ \text{for all } n \geq2. \end{equation*}
Question 16a
i) Since $$\mathbf{i}, \mathbf{j}, \mathbf{k}$$ are all unit vectors, the triangle inequality implies
\begin{equation*} 1 = |\overrightarrow{OP}|=|x\mathbf{i} + y\mathbf{j} + z\mathbf{k}| \le |x\mathbf{i}| + |y\mathbf{j}| + |z\mathbf{k}|= |x|+|y|+|z| \end{equation*}
where $$|\overrightarrow{OP}|=1$$ since $$\overrightarrow{OP}$$ is a position vector of a point on the unit sphere. From the above, we obtain that
\begin{equation*} |x|+|y|+|z| \ge 1 \end{equation*}
ii) Recalling the geometric dot product between vectors $$\mathbf{a}$$ and $$\mathbf{b}$$ we have that
\begin{equation*} |\mathbf{a}\cdot \mathbf{b}| = ||\mathbf{a}| |\mathbf{b}| \cos \theta | \leq |\mathbf{a}| |\mathbf{b}| \end{equation*}
since $$0\leq |\cos \theta| \leq 1$$. Substituting for $$\mathbf{a}$$ and $$\mathbf{b}$$ gives
\begin{equation*} |a_1 b_1 + a_2 b_2 + a_3 b_3 | \leq \sqrt{a_1^2+a_2^2+a_3^2} \sqrt{b_1^2+b_2^2+b_3^2} \end{equation*}
as required.
iii) Consider the inequality in (ii) with the vectors $$\mathbf{a} = (1,1,1)$$ and $$\mathbf{b} =(|x|,|y|,|z|)$$. We immediately obtain
\begin{equation*} |x|+|y|+|y| \leq \sqrt{1+1+1} \sqrt{|x|^2+|y|^2+|z|^2} = \sqrt{3}\sqrt{x^2+y^2+z^2} = \sqrt{3}. \end{equation*}
Remark: We note that a more general version of this question appeared in the Matrix Term 2 Workbook in 2020 :).
Question 16b
We have that the position vector is
\begin{equation*} \underset{\text{~}}{r}(t) = \begin{pmatrix} ut\cos\theta \\ ut\sin\theta – \frac{1}{2}gt^2 \end{pmatrix} \end{equation*}
We differentiate each term with respect to time to get the velocity vector:
\begin{equation*} \underset{\text{~}}{v}(t) = \begin{pmatrix} u\cos\theta \\ u\sin\theta -gt \end{pmatrix}. \end{equation*}
In two perpendicular vectors, as in two perpendicular lines, the gradients multiply to get -1:
\begin{align*} \frac{u\sin\theta-gt}{u\cos\theta} \times \frac{ut\sin\theta-\frac{1}{2}gt^2}{ut\cos\theta} &= -1 \end{align*}
Multiplying both sides of the equation by $$2u^2\cos^2\theta$$ gives
\begin{align*} (u\sin\theta-gt)(2u\sin\theta-gt)&=-2u^2\cos^2\theta \\ 2u^2\sin^2\theta-(3gu\sin\theta)t+g^2t^2 &= -2u^2(\cos^2\theta) \\ g^2t^2 – (3gu\sin\theta)t+ 2u^2\sin^2\theta + 2u^2 cos^2 \theta&= 0 \\ g^2t^2 -(3gu\sin\theta) t+2u^2 &= 0 \end{align*}
where we use the fact that $$\sin^2\theta+\cos^2\theta=1$$ in the last step. Observe that this is a quadratic in $$t$$ with discriminant
\begin{equation*} \Delta = (-3gu\sin\theta)^2 – 4(g^2)(2u^2) = 9u^2g^2\sin^2\theta – 8u^2g^2. \end{equation*}
To get two solutions for $$t$$, we solve $$\Delta > 0$$ to obtain
\begin{align*} 9u^2g^2\sin^2\theta – 8 u^2g^2 &> 0 \\ 9u^2g^2\sin^2\theta &> 8 u^2g^2 \\ \sin^2\theta &> \frac{8}{9} \\ \sin \theta &> \frac{2\sqrt{2}}{3}. \end{align*}
Finally, considering the particle cannot be projected at an angle more than 90 degrees (vertically upwards), we have
\begin{equation*} \sin^{-1}\left(\frac{2\sqrt{2}}{3}\right) < \theta < \frac{\pi}{2}. \end{equation*}
Once the angle $$\theta$$ is determined, we must also check that these two solutions for $$t$$ physically make sense, i.e. that they are between $$0$$ and $$\frac{2u \sin \theta}{g}$$ (the time of flight). Solving for $$t$$ would give
\begin{align*} t &= \frac{1}{2} \left( \frac{3u\sin\theta}{g} \pm \sqrt{ \left(\frac{u}{g}\right)^2 (9 \sin^2 \theta – 8)}\right) \\ &= \frac{u}{2g} \left(3\sin\theta \pm \sqrt{9\sin^2\theta-8}\right). \end{align*}
Both of the plus and minus variants are positive: the minus variant is positive since
</tableand so
The larger solution is less than the flight time since
Question 16cWe consider each quadrant separately.Quadrant 1: In this quadrant, $$x\geq0,y\geq0$$ and $$0\leq \text{Arg}(z) \leq \frac{\pi}{2}$$. We would then be solving the inequality:where we note that this region need only be sketched on $$0\leq x \leq \frac{\pi}{2}$$ since it is trivially always true for $$x > \frac{\pi}{2}$$.
Quadrant 2: Since $$x<0$$ but $$\frac{\pi}{2} \leq \text{Arg}(z) \leq \pi$$ then it never be the case that $$\text{Re}(z)\geq \text{Arg}(z)$$
Quadrant 3: We have thatolving the inequality:where multiplying both sides by $$x<0$$ in the last step reverses the direction of the inequality. Noting that we only sketch this region for $$-\pi \leq x \leq -\frac{\pi}{2}$$ and then there will be no solution for $$x < -\pi$$ and all solutions for $$x>-\frac{\pi}{2}$$.
Quadrant 4: We have that $$x>0$$ but $$\text{Arg}(z)<0$$ and so all of Quadrant 4 will be part of the region.
Finally at $$z=0$$, $$\text{Arg}(z)$$ is undefined so we have an open circle at the origin.In total we have the following plot:
$$\sqrt{9\sin^2\theta-8} < \sqrt{9\sin^2\theta} < 3 \sin\theta$$ $$t > \frac{u}{2g}(3\sin\theta-\sqrt{9\sin^2\theta}) = 0$$ \begin{align*} t &= \frac{u}{2g} \left(3\sin\theta + \sqrt{9\sin^2\theta-8}\right) \\ &< \frac{u}{2g} \left(3\sin\theta + \sqrt{9\sin^2\theta-{\color{red}{8\sin^2\theta}}}\right) \\ &= \frac{2u\sin\theta}{g}. \end{align*} \begin{align*} x &\geq \tan^{-1}\left(\frac{y}{x}\right) \\ \tan x &\geq \frac{y}{x} \\ y &\leq x \tan x \end{align*} $$\text{Arg}(z) = -\pi + \tan^{-1}\left| \frac{y}{x} \right| = -\pi + \tan^{-1}\left(\frac{y}{x}\right)$$. \begin{align*} x &\geq – \left( \pi – \tan^{-1}\left(\frac{y}{x}\right) \right) \\ \tan^{-1}\left(\frac{y}{x}\right) &\leq x + \pi \\ \frac{y}{x} &\leq \tan(x+\pi) \\ y &\geq x \tan(x+\pi), \end{align*}
Need help with Maths Ext 2?
Expert teachers, weekly quizzes, one-to-one help! Ace your next Maths Ext 2 assessment with Matrix+ Online. | HuggingFaceTB/finemath | |
# Question: What Are The Factors That Affect Kinetic Energy?
## What is the relationship between kinetic and speed?
This equation reveals that the kinetic energy of an object is directly proportional to the square of its speed.
That means that for a twofold increase in speed, the kinetic energy will increase by a factor of four.
For a threefold increase in speed, the kinetic energy will increase by a factor of nine..
## What is the main difference between kinetic and potential energy?
Energy stored in an object due to its position is Potential Energy. Energy that a moving object has due to its motion is Kinetic Energy.
## What happens when kinetic energy decreases?
Kinetic energy is the energy of movement or change. … The sum of an object’s potential and kinetic energies is called the object’s mechanical energy. As an object falls its potential energy decreases, while its kinetic energy increases. The decrease in potential energy is exactly equal to the increase in kinetic energy.
## Does acceleration increase kinetic energy?
The point is that the Force causes the acceleration, which changes the velocity, which changes the kinetic energy, but we can also think of this as the force doing work on the object, which causes the change in kinetic energy.
## Which is the best example that something has kinetic energy?
What are some examples of kinetic energy? when you are walking or running your body is exhibiting kinetic energy. A bicycle or skateboard in motion possesses kinetic energy. Running water has kinetic energy and it is used to run water mills.
## What are two factors that affect potential energy?
The factors that affect an object’s gravitational potential energy are its height relative to some reference point, its mass, and the strength of the gravitational field it is in.
## What factor can decrease the kinetic energy of an object?
velocityExplanation: Increasing an object’s velocity can decrease kinetic energy. *The velocity of an object has no effect on kinetic energy. *Increasing an object’s velocity can increase kinetic energy.
## What is the relationship between potential and kinetic energy?
The primary relationship between the two is their ability to transform into each other. In other words, potential energy transforms into kinetic energy, and kinetic energy converts into potential energy, and then back again.
## Does speed affect kinetic energy?
It turns out that an object’s kinetic energy increases as the square of its speed. A car moving 40 mph has four times as much kinetic energy as one moving 20 mph, while at 60 mph a car carries nine times as much kinetic energy as at 20 mph. Thus a modest increase in speed can cause a large increase in kinetic energy.
## Does kinetic energy increase with height?
As the height increases, there is an increase in the gravitational potential energy P and a decrease in the kinetic energy K. The kinetic energy K is inversely proportional to the height of the object.
## What are the 3 factors that affect potential energy?
Gravitational Potential Energy is determined by three factors: mass, gravity, and height. All three factors are directly proportional to energy.
## What are examples of potential energy in your home?
By keeping in mind all the above information about the potential energy, let us now look at some examples of potential energy from everyday life.Pendulum. … Spring. … Bow & Arrow. … Rock At Cliff’s Edge. … Food We Eat. … Water In Dams & Reservoirs. … Snow. … Bullet.More items…
## Which factor affects kinetic energy but not potential energy?
Kinetic energy can be affected by velocity, while potential energy can be affected by reference point, such as the very tip of a cliff. If a boulder was on the tip of a cliff, the wind could move it ever so slightly so that it could come crashing down. | HuggingFaceTB/finemath | |
Description
Farmer John has built a new long barn, with N (2
His C (2
Input
• Line 1: Two space-separated integers: N and C
• Lines 2..N+1: Line i+1 contains an integer stall location, xi
Output
• Line 1: One integer: the largest minimum distance
Sample Input
5 3
1
2
8
4
9
Sample Output
3
Hint
OUTPUT DETAILS:
FJ can put his 3 cows in the stalls at positions 1, 4 and 8, resulting in a minimum distance of 3.
Huge input data,scanf is recommended.
Solution：先对输入的房间坐标进行排序，然后二分查找即可。第一个房间先放一只牛,a[f]为上一头牛所在房间坐标 如果mid为所求解 当且仅当a[i]-a[f]≥mid。如果满足条件sum+1。如果sum大于等于m，说明mid小于所求解，更新l=mid。
#include
#include
#include
#include
#include
using namespace std;
const int N = 100010;
int a[N];
int n, m;
int judge(int mid)
{
int f = 0;//f为 上一头牛 所在的房间号
int sum = 1;//记录房间总数
for (int i = 1; i if (a[i] - a[f] >= mid)//如果mid为所求解 当且仅当a[i]-a[f]大于等于mid
{
f=i;
sum++;
}
else continue;
}
if (sum >= m) return 1;
else return 0;
}
int main()
{
scanf("%d%d", &n, &m);
for (int i = 0; i scanf("%d", &a[i]);
sort(a, a + n);
int l = -1;
int r = 1000000000;
int ans;
while (r - l >= 0)
{
int mid = (l + r) / 2;
if (judge(mid) == 1)
{
ans = mid;
l = mid;
}
else r = mid;
if (r - l == 1) //终止条件
break;
}
printf("%d\n", ans);
return 0;
} | HuggingFaceTB/finemath | |
The picture shows an egg form constructed mathematically. The spirals are characteristic of the mathematics and are known as PATH CURVES. They were discovered by Felix Klein in the 19th Century, and are very simple and fundamental mathematically speaking. Geometry studies transformations of space, and these curves arise as a result. A simple movement in a fixed direction such as driving along a straight road is an example, where the vehicle is being transformed by what is called a translation. In our mathematical imagination we can think of the whole of space being transformed in this way. Another example is rotation about an axis. In both cases there are lines or curves which are themselves unmoved (as a whole) by the transformation : in the second case circles concentric with the axis (round which the points of space are moving), and in the first case all straight lines parallel to the direction of motion. These are simple examples of path curves. More complicated transformations give rise to more interesting curves. The transformations concerned are projective ones characteristic of projective geometry, which are linear because neither straight lines nor planes become curved when moved by them, and incidences are preserved (this is a simplification, but will serve us here). They allow more freedom than simple rotations and translations, in particular incorporating expansion and contraction. Apart from the path curves they leave a tetrahedron invariant in the most general case. George Adams studied these curves as he thought they would provide a way of understanding how space and counter space interact. A particular version he looked at was for a transformation which leaves invariant two parallel planes, the line at infinity where they meet, and an axis orthogonal to them. This is a plastic transformation rather than a rigid one (like rotation) and a typical path curve together with the invariant planes and axis is shown below. This will be recognised as the type of curve lying in the surface of the egg at the top of the page. If we take a circle concentric with the axis and all the path curves which pass through it then we get that egg-shaped surface. The construction is shown in the following animation: We can vary the transformation to get our eggs more or less sharp, or alternatively we can get vortices such as the following: In these pictures particular path curves have been highlighted. This particular vortex is an example of a watery vortex, so called by Lawrence Edwards because its profile fits real water vortices. It is characterised by the fact that the lower invariant plane is at infinity. If instead the upper plane is at infinity we get what he calls an airy vortex. Two parameters are of particular significance: lambda and epsilon. Lambda controls the shape of the profile while epsilon determines the degree of spiralling. Lambda is positive for eggs and negative for vortices, while the sign of epsilon controls the sense of rotation. This is illustrated below. < The top row shows lambda increasing from 1 (elliptical) to 10. When lambda reaches infinity the form becomes conical. The centre row shows lambda increasing from -0.616 to -0.1 for a vortex. The bottom row shows epsilon varying from 0.2 to 10, and when it reaches infinity the curves are vertical. If it is zero then the path curves become horizontal circles, and strictly speaking the profile is lost. The profile is thus controlled by a single parameter (lambda), and it is scientifically interesting that with such a restriction these curves fit very closely a wide variety of natural forms including eggs, flower and leaf buds, pine cones, the left ventricle of the human heart, the pineal gland, and the uterus during pregnancy. The watery vortex closely fits actual stable water vortices. Together with the airy vortex it also has significance for pivot transforms. The following shows approximately the way the left ventricle of the heart behaves as a path curve from diastole to systole: Lawrence Edwards spent many years finding out and testing the above facts experimentally, which he has described in Reference 7. In 1982 he started testing the shapes of the leaf buds of trees through the winter, and found that their lambda value (unexpectedly) varied rhythmically with a period of approximately two weeks. This was his main topic of research in his later years, and the evidence is now very strong - backed by thousands of measurements - that the rhythm corresponds to the conjunctions and oppositions of the Moon and a particular planet for each tree. This is a purely experimental fact and care should be taken in interpreting it. Download document Practical Path Curve Calculations for the basic algebra and formulae to work with path curves (pdf document). | HuggingFaceTB/finemath | |
## 3379
3,379 (three thousand three hundred seventy-nine) is an odd four-digits composite number following 3378 and preceding 3380. In scientific notation, it is written as 3.379 × 103. The sum of its digits is 22. It has a total of 2 prime factors and 4 positive divisors. There are 3,240 positive integers (up to 3379) that are relatively prime to 3379.
## Basic properties
• Is Prime? No
• Number parity Odd
• Number length 4
• Sum of Digits 22
• Digital Root 4
## Name
Short name 3 thousand 379 three thousand three hundred seventy-nine
## Notation
Scientific notation 3.379 × 103 3.379 × 103
## Prime Factorization of 3379
Prime Factorization 31 × 109
Composite number
Distinct Factors Total Factors Radical ω(n) 2 Total number of distinct prime factors Ω(n) 2 Total number of prime factors rad(n) 3379 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) 0 Returns log(p) if n is a power pk of any prime p (for any k >= 1), else returns 0
The prime factorization of 3,379 is 31 × 109. Since it has a total of 2 prime factors, 3,379 is a composite number.
## Divisors of 3379
1, 31, 109, 3379
4 divisors
Even divisors 0 4 2 2
Total Divisors Sum of Divisors Aliquot Sum τ(n) 4 Total number of the positive divisors of n σ(n) 3520 Sum of all the positive divisors of n s(n) 141 Sum of the proper positive divisors of n A(n) 880 Returns the sum of divisors (σ(n)) divided by the total number of divisors (τ(n)) G(n) 58.1292 Returns the nth root of the product of n divisors H(n) 3.83977 Returns the total number of divisors (τ(n)) divided by the sum of the reciprocal of each divisors
The number 3,379 can be divided by 4 positive divisors (out of which 0 are even, and 4 are odd). The sum of these divisors (counting 3,379) is 3,520, the average is 880.
## Other Arithmetic Functions (n = 3379)
1 φ(n) n
Euler Totient Carmichael Lambda Prime Pi φ(n) 3240 Total number of positive integers not greater than n that are coprime to n λ(n) 540 Smallest positive number such that aλ(n) ≡ 1 (mod n) for all a coprime to n π(n) ≈ 480 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 3,240 positive integers (less than 3,379) that are coprime with 3,379. And there are approximately 480 prime numbers less than or equal to 3,379.
## Divisibility of 3379
m n mod m 2 3 4 5 6 7 8 9 1 1 3 4 1 5 3 4
3,379 is not divisible by any number less than or equal to 9.
## Classification of 3379
• Arithmetic
• Semiprime
• Deficient
• Polite
• Square Free
### Other numbers
• LucasCarmichael
## Base conversion (3379)
Base System Value
2 Binary 110100110011
3 Ternary 11122011
4 Quaternary 310303
5 Quinary 102004
6 Senary 23351
8 Octal 6463
10 Decimal 3379
12 Duodecimal 1b57
20 Vigesimal 88j
36 Base36 2lv
## Basic calculations (n = 3379)
### Multiplication
n×i
n×2 6758 10137 13516 16895
### Division
ni
n⁄2 1689.5 1126.33 844.75 675.8
### Exponentiation
ni
n2 11417641 38580208939 130362526004881 440494975370492899
### Nth Root
i√n
2√n 58.1292 15.0059 7.62425 5.07876
## 3379 as geometric shapes
### Circle
Diameter 6758 21230.9 3.58696e+07
### Sphere
Volume 1.61604e+11 1.43478e+08 21230.9
### Square
Length = n
Perimeter 13516 1.14176e+07 4778.63
### Cube
Length = n
Surface area 6.85058e+07 3.85802e+10 5852.6
### Equilateral Triangle
Length = n
Perimeter 10137 4.94398e+06 2926.3
### Triangular Pyramid
Length = n
Surface area 1.97759e+07 4.54672e+09 2758.94
## Cryptographic Hash Functions
md5 88855547570f7ff053fff7c54e5148cc c285d7c78f45fd442b9393b5d3bd7a14666acfaa 3673b82b483a327f94177bd439d428e955f80250c1a4120e9d9b9aab2b996fe2 524cdda4e43d1b17a06a09a6040e8cab2d387e93b3fa06b0fd8a8ab5fb81341119c3c2918bdc342c509470228738ab773d06c7672ad80f9155f433ea37678052 57bb0ff97761f8a349b992dbba97385271626f38 | HuggingFaceTB/finemath | |
# Pressure Loss and Equivalent Circulating Density Review
This example that I got from my junior member is very simple but it helps you a lot to understand about how to determine pressure loss during normal circulation.
Information given is listed below; Circulate at 3 bottom up through open end tubing (Down tubing and up annulus) with 12.7 ppg mud.
Pump pressure = 1000 psi
Annulus friction loss = 50 psi
Inside tubing friction loss = 925 psi
Surface line friction loss = 25 psi
Calculate the pressure in the well at 10,000’ (tubing tail).
What would ECD at 10,000’ TVD be?
The concept of calculation that you should know : total pressure at bottom = pumping pressure + hydrostatic pressure – pressure loss in the opposite way of fluid flowing.
Then,
If I reference to the drill string site, I will get the equation like this.
Pressure at bottom hole= Hydrostatic Pressure at bottom hole + Pressure from pump- Pressure Loss in surface line – Pressure loss in tubing
Pressure in the well at 10,000’ = 0.052×12.7×10000 + 1000 – 25 – 925 = 6654 psi
If I reference to the annulus site, I will get the equation like this.
Pressure at bottom hole= Hydrostatic Pressure at bottom hole +Annular pressure loss
Note: Hydrostatic pressure and annular pressure loss force downward.
Pressure in the well at 10,000’ = 0.052×12.7×10000+ 50 = 6654 psi
Note: It doesn’t matter which site of u-tube you refer to the bottom hole pressure is still the same.
ECD (Equivalent Circulating Density) is calculated by this following equation:
ECD = Current mud weight in PPG + (annular pressure loss /(0.052xTVD))
ECD = 12.7 + (50/(0.052 x 10000))
ECD = 12.8 ppg
Share the joy Working in the oil field and loving to share knowledge.
### 6 Responses to Pressure Loss and Equivalent Circulating Density Review
1. Manuel says:
I ve been use this grup to be more informed about oil industry a my personal knowlege and has been helpful. I do not have experience in industry yet. thanks to all of you.
2. mouyembe says:
helpfull informations, it really helped me to catch this
3. Balamurugan says:
Useful info.
4. Mohanad alhabbal says:
can you please do dubble check of ECD formula I thing it is not correct.
E.C.D = BHP+ Annulus press./0.052*TVD
thanks
• DrillingFormulas.Com says: | HuggingFaceTB/finemath | |
Descriptive Statistics
# 7 Display Data
### Stem-and-Leaf Graphs (Stemplots), Line Graphs, and Bar Graphs
One simple graph, the stem-and-leaf graph or stemplot, comes from the field of exploratory data analysis. It is a good choice when the data sets are small. To create the plot, divide each observation of data into a stem and a leaf. The leaf consists of a final significant digit. For example, 23 has stem two and leaf three. The number 432 has stem 43 and leaf two. Likewise, the number 5,432 has stem 543 and leaf two. The decimal 9.3 has stem nine and leaf three. Write the stems in a vertical line from smallest to largest. Draw a vertical line to the right of the stems. Then write the leaves in increasing order next to their corresponding stem.
For Susan Dean’s spring pre-calculus class, scores for the first exam were as follows (smallest to largest):
33; 42; 49; 49; 53; 55; 55; 61; 63; 67; 68; 68; 69; 69; 72; 73; 74; 78; 80; 83; 88; 88; 88; 90; 92; 94; 94; 94; 94; 96; 100
Stem-and-Leaf Graph
Stem Leaf
3 3
4 2 9 9
5 3 5 5
6 1 3 7 8 8 9 9
7 2 3 4 8
8 0 3 8 8 8
9 0 2 4 4 4 4 6
10 0
The stemplot shows that most scores fell in the 60s, 70s, 80s, and 90s. Eight out of the 31 scores or approximately 26% were in the 90s or 100, a fairly high number of As.
Try It
For the Park City basketball team, scores for the last 30 games were as follows (smallest to largest):
32; 32; 33; 34; 38; 40; 42; 42; 43; 44; 46; 47; 47; 48; 48; 48; 49; 50; 50; 51; 52; 52; 52; 53; 54; 56; 57; 57; 60; 61
Construct a stem plot for the data.
Stem Leaf
3 2 2 3 4 8
4 0 2 2 3 4 6 7 7 8 8 8 9
5 0 0 1 2 2 2 3 4 6 7 7
6 0 1
The stemplot is a quick way to graph data and gives an exact picture of the data. You want to look for an overall pattern and any outliers. An outlier is an observation of data that does not fit the rest of the data. It is sometimes called an extreme value. When you graph an outlier, it will appear not to fit the pattern of the graph. Some outliers are due to mistakes (for example, writing down 50 instead of 500) while others may indicate that something unusual is happening. It takes some background information to explain outliers, so we will cover them in more detail later.
The data are the distances (in kilometers) from a home to local supermarkets. Create a stemplot using the data:
1.1; 1.5; 2.3; 2.5; 2.7; 3.2; 3.3; 3.3; 3.5; 3.8; 4.0; 4.2; 4.5; 4.5; 4.7; 4.8; 5.5; 5.6; 6.5; 6.7; 12.3
Do the data seem to have any concentration of values?
NOTE
The leaves are to the right of the decimal.
The value 12.3 may be an outlier. Values appear to concentrate at three and four kilometers.
Stem Leaf
1 1 5
2 3 5 7
3 2 3 3 5 8
4 0 2 5 5 7 8
5 5 6
6 5 7
7
8
9
10
11
12 3
Try It
The following data show the distances (in miles) from the homes of off-campus statistics students to the college. Create a stem plot using the data and identify any outliers:
0.5; 0.7; 1.1; 1.2; 1.2; 1.3; 1.3; 1.5; 1.5; 1.7; 1.7; 1.8; 1.9; 2.0; 2.2; 2.5; 2.6; 2.8; 2.8; 2.8; 3.5; 3.8; 4.4; 4.8; 4.9; 5.2; 5.5; 5.7; 5.8; 8.0
Stem Leaf
0 5 7
1 1 2 2 3 3 5 5 7 7 8 9
2 0 2 5 6 8 8 8
3 5 8
4 4 8 9
5 2 5 7 8
6
7
8 0
The value 8.0 may be an outlier. Values appear to concentrate at one and two miles.
A side-by-side stem-and-leaf plot allows a comparison of the two data sets in two columns. In a side-by-side stem-and-leaf plot, two sets of leaves share the same stem. The leaves are to the left and the right of the stems. (Figure) and (Figure) show the ages of presidents at their inauguration and at their death. Construct a side-by-side stem-and-leaf plot using this data.
Ages at Inauguration Ages at Death
9 9 8 7 7 7 6 3 2 4 6 9
8 7 7 7 7 6 6 6 5 5 5 5 4 4 4 4 4 2 2 1 1 1 1 1 0 5 3 6 6 7 7 8
9 8 5 4 4 2 1 1 1 0 6 0 0 3 3 4 4 5 6 7 7 7 8
7 0 0 1 1 1 4 7 8 8 9
8 0 1 3 5 8
9 0 0 3 3
Presidential Ages at Inauguration
President Age President Age President Age
Washington 57 Lincoln 52 Hoover 54
J. Adams 61 A. Johnson 56 F. Roosevelt 51
Jefferson 57 Grant 46 Truman 60
Madison 57 Hayes 54 Eisenhower 62
Monroe 58 Garfield 49 Kennedy 43
J. Q. Adams 57 Arthur 51 L. Johnson 55
Jackson 61 Cleveland 47 Nixon 56
Van Buren 54 B. Harrison 55 Ford 61
W. H. Harrison 68 Cleveland 55 Carter 52
Tyler 51 McKinley 54 Reagan 69
Polk 49 T. Roosevelt 42 G.H.W. Bush 64
Taylor 64 Taft 51 Clinton 47
Fillmore 50 Wilson 56 G. W. Bush 54
Pierce 48 Harding 55 Obama 47
Buchanan 65 Coolidge 51
Presidential Age at Death
President Age President Age President Age
Washington 67 Lincoln 56 Hoover 90
J. Adams 90 A. Johnson 66 F. Roosevelt 63
Jefferson 83 Grant 63 Truman 88
Madison 85 Hayes 70 Eisenhower 78
Monroe 73 Garfield 49 Kennedy 46
J. Q. Adams 80 Arthur 56 L. Johnson 64
Jackson 78 Cleveland 71 Nixon 81
Van Buren 79 B. Harrison 67 Ford 93
W. H. Harrison 68 Cleveland 71 Reagan 93
Tyler 71 McKinley 58
Polk 53 T. Roosevelt 60
Taylor 65 Taft 72
Fillmore 74 Wilson 67
Pierce 64 Harding 57
Buchanan 77 Coolidge 60
Another type of graph that is useful for specific data values is a line graph. In the particular line graph shown in (Figure), the x-axis (horizontal axis) consists of data values and the y-axis (vertical axis) consists of frequency points. The frequency points are connected using line segments.
In a survey, 40 mothers were asked how many times per week a teenager must be reminded to do his or her chores. The results are shown in (Figure) and in (Figure).
Number of times teenager is reminded Frequency
0 2
1 5
2 8
3 14
4 7
5 4
Try It
In a survey, 40 people were asked how many times per year they had their car in the shop for repairs. The results are shown in (Figure). Construct a line graph.
Number of times in shop Frequency
0 7
1 10
2 14
3 9
Bar graphs consist of bars that are separated from each other. The bars can be rectangles or they can be rectangular boxes (used in three-dimensional plots), and they can be vertical or horizontal. The bar graph shown in (Figure) has age groups represented on the x-axis and proportions on the y-axis.
By the end of 2011, Facebook had over 146 million users in the United States. (Figure) shows three age groups, the number of users in each age group, and the proportion (%) of users in each age group. Construct a bar graph using this data.
13–25 65,082,280 45%
26–44 53,300,200 36%
45–64 27,885,100 19%
Try It
The population in Park City is made up of children, working-age adults, and retirees. (Figure) shows the three age groups, the number of people in the town from each age group, and the proportion (%) of people in each age group. Construct a bar graph showing the proportions.
Age groups Number of people Proportion of population
Children 67,059 19%
Retirees 131,662 38%
The columns in (Figure) contain: the race or ethnicity of students in U.S. Public Schools for the class of 2011, percentages for the Advanced Placement examine population for that class, and percentages for the overall student population. Create a bar graph with the student race or ethnicity (qualitative data) on the x-axis, and the Advanced Placement examinee population percentages on the y-axis.
Race/ethnicity AP examinee population Overall student population
1 = Asian, Asian American or Pacific Islander 10.3% 5.7%
2 = Black or African American 9.0% 14.7%
3 = Hispanic or Latino 17.0% 17.6%
4 = American Indian or Alaska Native 0.6% 1.1%
5 = White 57.1% 59.2%
6 = Not reported/other 6.0% 1.7%
Try It
Park city is broken down into six voting districts. The table shows the percent of the total registered voter population that lives in each district as well as the percent total of the entire population that lives in each district. Construct a bar graph that shows the registered voter population by district.
District Registered voter population Overall city population
1 15.5% 19.4%
2 12.2% 15.6%
3 9.8% 9.0%
4 17.4% 18.5%
5 22.8% 20.7%
6 22.3% 16.8%
Below is a two-way table showing the types of pets owned by men and women:
Dogs Cats Fish Total Men 4 2 2 8 Women 4 6 2 12 Total 8 8 4 20
Given these data, calculate the conditional distributions for the subpopulation of men who own each pet type.
Men who own dogs = 4/8 = 0.5
Men who own cats = 2/8 = 0.25
Men who own fish = 2/8 = 0.25
Note: The sum of all of the conditional distributions must equal one. In this case, 0.5 + 0.25 + 0.25 = 1; therefore, the solution “checks”.
### Histograms, Frequency Polygons, and Time Series Graphs
For most of the work you do in this book, you will use a histogram to display the data. One advantage of a histogram is that it can readily display large data sets. A rule of thumb is to use a histogram when the data set consists of 100 values or more.
A histogram consists of contiguous (adjoining) boxes. It has both a horizontal axis and a vertical axis. The horizontal axis is labeled with what the data represents (for instance, distance from your home to school). The vertical axis is labeled either frequency or relative frequency (or percent frequency or probability). The graph will have the same shape with either label. The histogram (like the stemplot) can give you the shape of the data, the center, and the spread of the data.
The relative frequency is equal to the frequency for an observed value of the data divided by the total number of data values in the sample.(Remember, frequency is defined as the number of times an answer occurs.) If:
• f = frequency
• n = total number of data values (or the sum of the individual frequencies), and
• RF = relative frequency,
then:
For example, if three students in Mr. Ahab’s English class of 40 students received from 90% to 100%, then, <!–<newline count=”1″/>–>f = 3, n = 40, and RF = = = 0.075. 7.5% of the students received 90–100%. 90–100% are quantitative measures.
To construct a histogram, first decide how many bars or intervals, also called classes, represent the data. Many histograms consist of five to 15 bars or classes for clarity. The number of bars needs to be chosen. Choose a starting point for the first interval to be less than the smallest data value. A convenient starting point is a lower value carried out to one more decimal place than the value with the most decimal places. For example, if the value with the most decimal places is 6.1 and this is the smallest value, a convenient starting point is 6.05 (6.1 – 0.05 = 6.05). We say that 6.05 has more precision. If the value with the most decimal places is 2.23 and the lowest value is 1.5, a convenient starting point is 1.495 (1.5 – 0.005 = 1.495). If the value with the most decimal places is 3.234 and the lowest value is 1.0, a convenient starting point is 0.9995 (1.0 – 0.0005 = 0.9995). If all the data happen to be integers and the smallest value is two, then a convenient starting point is 1.5 (2 – 0.5 = 1.5). Also, when the starting point and other boundaries are carried to one additional decimal place, no data value will fall on a boundary. The next two examples go into detail about how to construct a histogram using continuous data and how to create a histogram using discrete data.
The following data are the heights (in inches to the nearest half inch) of 100 male semiprofessional soccer players. The heights are continuous data, since height is measured.
60; 60.5; 61; 61; 61.5
63.5; 63.5; 63.5
64; 64; 64; 64; 64; 64; 64; 64.5; 64.5; 64.5; 64.5; 64.5; 64.5; 64.5; 64.5
66; 66; 66; 66; 66; 66; 66; 66; 66; 66; 66.5; 66.5; 66.5; 66.5; 66.5; 66.5; 66.5; 66.5; 66.5; 66.5; 66.5; 67; 67; 67; 67; 67; 67; 67; 67; 67; 67; 67; 67; 67.5; 67.5; 67.5; 67.5; 67.5; 67.5; 67.5
68; 68; 69; 69; 69; 69; 69; 69; 69; 69; 69; 69; 69.5; 69.5; 69.5; 69.5; 69.5
70; 70; 70; 70; 70; 70; 70.5; 70.5; 70.5; 71; 71; 71
72; 72; 72; 72.5; 72.5; 73; 73.5
74
The smallest data value is 60. Since the data with the most decimal places has one decimal (for instance, 61.5), we want our starting point to have two decimal places. Since the numbers 0.5, 0.05, 0.005, etc. are convenient numbers, use 0.05 and subtract it from 60, the smallest value, for the convenient starting point.
60 – 0.05 = 59.95 which is more precise than, say, 61.5 by one decimal place. The starting point is, then, 59.95.
The largest value is 74, so 74 + 0.05 = 74.05 is the ending value.
Next, calculate the width of each bar or class interval. To calculate this width, subtract the starting point from the ending value and divide by the number of bars (you must choose the number of bars you desire). Suppose you choose eight bars.
NOTE
We will round up to two and make each bar or class interval two units wide. Rounding up to two is one way to prevent a value from falling on a boundary. Rounding to the next number is often necessary even if it goes against the standard rules of rounding. For this example, using 1.76 as the width would also work. A guideline that is followed by some for the width of a bar or class interval is to take the square root of the number of data values and then round to the nearest whole number, if necessary. For example, if there are 150 values of data, take the square root of 150 and round to 12 bars or intervals.
The boundaries are:
• 59.95
• 59.95 + 2 = 61.95
• 61.95 + 2 = 63.95
• 63.95 + 2 = 65.95
• 65.95 + 2 = 67.95
• 67.95 + 2 = 69.95
• 69.95 + 2 = 71.95
• 71.95 + 2 = 73.95
• 73.95 + 2 = 75.95
The heights 60 through 61.5 inches are in the interval 59.95–61.95. The heights that are 63.5 are in the interval 61.95–63.95. The heights that are 64 through 64.5 are in the interval 63.95–65.95. The heights 66 through 67.5 are in the interval 65.95–67.95. The heights 68 through 69.5 are in the interval 67.95–69.95. The heights 70 through 71 are in the interval 69.95–71.95. The heights 72 through 73.5 are in the interval 71.95–73.95. The height 74 is in the interval 73.95–75.95.
The following histogram displays the heights on the x-axis and relative frequency on the y-axis.
Try It
The following data are the shoe sizes of 50 male students. The sizes are continuous data since shoe size is measured. Construct a histogram and calculate the width of each bar or class interval. Suppose you choose six bars.
9; 9; 9.5; 9.5; 10; 10; 10; 10; 10; 10; 10.5; 10.5; 10.5; 10.5; 10.5; 10.5; 10.5; 10.5
11; 11; 11; 11; 11; 11; 11; 11; 11; 11; 11; 11; 11; 11.5; 11.5; 11.5; 11.5; 11.5; 11.5; 11.5
12; 12; 12; 12; 12; 12; 12; 12.5; 12.5; 12.5; 12.5; 14
Smallest value: 9
Largest value: 14
Convenient starting value: 9 – 0.05 = 8.95
Convenient ending value: 14 + 0.05 = 14.05
The calculations suggests using 0.85 as the width of each bar or class interval. You can also use an interval with a width equal to one.
Create a histogram for the following data: the number of books bought by 50 part-time college students at ABC College. The number of books is discrete data, since books are counted.
1; 1; 1; 1; 1; 1; 1; 1; 1; 1; 1
2; 2; 2; 2; 2; 2; 2; 2; 2; 2
3; 3; 3; 3; 3; 3; 3; 3; 3; 3; 3; 3; 3; 3; 3; 3
4; 4; 4; 4; 4; 4
5; 5; 5; 5; 5
6; 6
Because the data are integers, subtract 0.5 from 1, the smallest data value and add 0.5 to 6, the largest data value. Then the starting point is 0.5 and the ending value is 6.5.
Next, calculate the width of each bar or class interval. If the data are discrete and there are not too many different values, a width that places the data values in the middle of the bar or class interval is the most convenient. Since the data consist of the numbers 1, 2, 3, 4, 5, 6, and the starting point is 0.5, a width of one places the 1 in the middle of the interval from 0.5 to 1.5, the 2 in the middle of the interval from 1.5 to 2.5, the 3 in the middle of the interval from 2.5 to 3.5, the 4 in the middle of the interval from _______ to _______, the 5 in the middle of the interval from _______ to _______, and the _______ in the middle of the interval from _______ to _______ .
• 3.5 to 4.5
• 4.5 to 5.5
• 6
• 5.5 to 6.5
Calculate the number of bars as follows:
where 1 is the width of a bar. Therefore, bars = 6.
The following histogram displays the number of books on the x-axis and the frequency on the y-axis.
Using this data set, construct a histogram.
Number of hours my classmates spent playing video games on weekends
9.95 10 2.25 16.75 0
19.5 22.5 7.5 15 12.75
5.5 11 10 20.75 17.5
23 21.9 24 23.75 18
20 15 22.9 18.8 20.5
Some values in this data set fall on boundaries for the class intervals. A value is counted in a class interval if it falls on the left boundary, but not if it falls on the right boundary. Different researchers may set up histograms for the same data in different ways. There is more than one correct way to set up a histogram.
#### Frequency Polygons
Frequency polygons are analogous to line graphs, and just as line graphs make continuous data visually easy to interpret, so too do frequency polygons.
To construct a frequency polygon, first examine the data and decide on the number of intervals, or class intervals, to use on the x-axis and y-axis. After choosing the appropriate ranges, begin plotting the data points. After all the points are plotted, draw line segments to connect them.
A frequency polygon was constructed from the frequency table below.
Frequency distribution for calculus final test scores
Lower bound Upper bound Frequency Cumulative frequency
49.5 59.5 5 5
59.5 69.5 10 15
69.5 79.5 30 45
79.5 89.5 40 85
89.5 99.5 15 100
The first label on the x-axis is 44.5. This represents an interval extending from 39.5 to 49.5. Since the lowest test score is 54.5, this interval is used only to allow the graph to touch the x-axis. The point labeled 54.5 represents the next interval, or the first “real” interval from the table, and contains five scores. This reasoning is followed for each of the remaining intervals with the point 104.5 representing the interval from 99.5 to 109.5. Again, this interval contains no data and is only used so that the graph will touch the x-axis. Looking at the graph, we say that this distribution is skewed because one side of the graph does not mirror the other side.
Try It
Construct a frequency polygon of U.S. Presidents’ ages at inauguration shown in (Figure).
Age at inauguration Frequency
41.5–46.5 4
46.5–51.5 11
51.5–56.5 14
56.5–61.5 9
61.5–66.5 4
66.5–71.5 2
The first label on the x-axis is 39. This represents an interval extending from 36.5 to 41.5. Since there are no ages less than 41.5, this interval is used only to allow the graph to touch the x-axis. The point labeled 44 represents the next interval, or the first “real” interval from the table, and contains four scores. This reasoning is followed for each of the remaining intervals with the point 74 representing the interval from 71.5 to 76.5. Again, this interval contains no data and is only used so that the graph will touch the x-axis. Looking at the graph, we say that this distribution is skewed because one side of the graph does not mirror the other side.
Frequency polygons are useful for comparing distributions. This is achieved by overlaying the frequency polygons drawn for different data sets.
We will construct an overlay frequency polygon comparing the scores from (Figure) with the students’ final numeric grade.
Frequency distribution for calculus final test scores
Lower bound Upper bound Frequency Cumulative frequency
49.5 59.5 5 5
59.5 69.5 10 15
69.5 79.5 30 45
79.5 89.5 40 85
89.5 99.5 15 100
Frequency distribution for calculus final grades
Lower bound Upper bound Frequency Cumulative frequency
49.5 59.5 10 10
59.5 69.5 10 20
69.5 79.5 30 50
79.5 89.5 45 95
89.5 99.5 5 100
#### Constructing a Time Series Graph
Suppose that we want to study the temperature range of a region for an entire month. Every day at noon we note the temperature and write this down in a log. A variety of statistical studies could be done with these data. We could find the mean or the median temperature for the month. We could construct a histogram displaying the number of days that temperatures reach a certain range of values. However, all of these methods ignore a portion of the data that we have collected.
One feature of the data that we may want to consider is that of time. Since each date is paired with the temperature reading for the day, we don‘t have to think of the data as being random. We can instead use the times given to impose a chronological order on the data. A graph that recognizes this ordering and displays the changing temperature as the month progresses is called a time series graph.
To construct a time series graph, we must look at both pieces of our paired data set. We start with a standard Cartesian coordinate system. The horizontal axis is used to plot the date or time increments, and the vertical axis is used to plot the values of the variable that we are measuring. By doing this, we make each point on the graph correspond to a date and a measured quantity. The points on the graph are typically connected by straight lines in the order in which they occur.
The following data shows the Annual Consumer Price Index, each month, for ten years. Construct a time series graph for the Annual Consumer Price Index data only.
Year Jan Feb Mar Apr May Jun Jul
2003 181.7 183.1 184.2 183.8 183.5 183.7 183.9
2004 185.2 186.2 187.4 188.0 189.1 189.7 189.4
2005 190.7 191.8 193.3 194.6 194.4 194.5 195.4
2006 198.3 198.7 199.8 201.5 202.5 202.9 203.5
2007 202.416 203.499 205.352 206.686 207.949 208.352 208.299
2008 211.080 211.693 213.528 214.823 216.632 218.815 219.964
2009 211.143 212.193 212.709 213.240 213.856 215.693 215.351
2010 216.687 216.741 217.631 218.009 218.178 217.965 218.011
2011 220.223 221.309 223.467 224.906 225.964 225.722 225.922
2012 226.665 227.663 229.392 230.085 229.815 229.478 229.104
Year Aug Sep Oct Nov Dec Annual
2003 184.6 185.2 185.0 184.5 184.3 184.0
2004 189.5 189.9 190.9 191.0 190.3 188.9
2005 196.4 198.8 199.2 197.6 196.8 195.3
2006 203.9 202.9 201.8 201.5 201.8 201.6
2007 207.917 208.490 208.936 210.177 210.036 207.342
2008 219.086 218.783 216.573 212.425 210.228 215.303
2009 215.834 215.969 216.177 216.330 215.949 214.537
2010 218.312 218.439 218.711 218.803 219.179 218.056
2011 226.545 226.889 226.421 226.230 225.672 224.939
2012 230.379 231.407 231.317 230.221 229.601 229.594
Try It
The following table is a portion of a data set from www.worldbank.org. Use the table to construct a time series graph for CO2 emissions for the United States.
CO2 emissions
Year Ukraine United Kingdom United States
2003 352,259 540,640 5,681,664
2004 343,121 540,409 5,790,761
2005 339,029 541,990 5,826,394
2006 327,797 542,045 5,737,615
2007 328,357 528,631 5,828,697
2008 323,657 522,247 5,656,839
2009 272,176 474,579 5,299,563
### Uses of a Time Series Graph
Time series graphs are important tools in various applications of statistics. When recording values of the same variable over an extended period of time, sometimes it is difficult to discern any trend or pattern. However, once the same data points are displayed graphically, some features jump out. Time series graphs make trends easy to spot.
### How NOT to Lie with Statistics
It is important to remember that the very reason we develop a variety of methods to present data is to develop insights into the subject of what the observations represent. We want to get a “sense” of the data. Are the observations all very much alike or are they spread across a wide range of values, are they bunched at one end of the spectrum or are they distributed evenly and so on. We are trying to get a visual picture of the numerical data. Shortly we will develop formal mathematical measures of the data, but our visual graphical presentation can say much. It can, unfortunately, also say much that is distracting, confusing and simply wrong in terms of the impression the visual leaves. Many years ago Darrell Huff wrote the book How to Lie with Statistics. It has been through 25 plus printings and sold more than one and one-half million copies. His perspective was a harsh one and used many actual examples that were designed to mislead. He wanted to make people aware of such deception, but perhaps more importantly to educate so that others do not make the same errors inadvertently.
Again, the goal is to enlighten with visuals that tell the story of the data. Pie charts have a number of common problems when used to convey the message of the data. Too many pieces of the pie overwhelm the reader. More than perhaps five or six categories ought to give an idea of the relative importance of each piece. This is after all the goal of a pie chart, what subset matters most relative to the others. If there are more components than this then perhaps an alternative approach would be better or perhaps some can be consolidated into an “other” category. Pie charts cannot show changes over time, although we see this attempted all too often. In federal, state, and city finance documents pie charts are often presented to show the components of revenue available to the governing body for appropriation: income tax, sales tax motor vehicle taxes and so on. In and of itself this is interesting information and can be nicely done with a pie chart. The error occurs when two years are set side-by-side. Because the total revenues change year to year, but the size of the pie is fixed, no real information is provided and the relative size of each piece of the pie cannot be meaningfully compared.
Histograms can be very helpful in understanding the data. Properly presented, they can be a quick visual way to present probabilities of different categories by the simple visual of comparing relative areas in each category. Here the error, purposeful or not, is to vary the width of the categories. This of course makes comparison to the other categories impossible. It does embellish the importance of the category with the expanded width because it has a greater area, inappropriately, and thus visually “says” that that category has a higher probability of occurrence.
Time series graphs perhaps are the most abused. A plot of some variable across time should never be presented on axes that change part way across the page either in the vertical or horizontal dimension. Perhaps the time frame is changed from years to months. Perhaps this is to save space or because monthly data was not available for early years. In either case this confounds the presentation and destroys any value of the graph. If this is not done to purposefully confuse the reader, then it certainly is either lazy or sloppy work.
Changing the units of measurement of the axis can smooth out a drop or accentuate one. If you want to show large changes, then measure the variable in small units, penny rather than thousands of dollars. And of course to continue the fraud, be sure that the axis does not begin at zero, zero. If it begins at zero, zero, then it becomes apparent that the axis has been manipulated.
Perhaps you have a client that is concerned with the volatility of the portfolio you manage. An easy way to present the data is to use long time periods on the time series graph. Use months or better, quarters rather than daily or weekly data. If that doesn’t get the volatility down then spread the time axis relative to the rate of return or portfolio valuation axis. If you want to show “quick” dramatic growth, then shrink the time axis. Any positive growth will show visually “high” growth rates. Do note that if the growth is negative then this trick will show the portfolio is collapsing at a dramatic rate.
Again, the goal of descriptive statistics is to convey meaningful visuals that tell the story of the data. Purposeful manipulation is fraud and unethical at the worst, but even at its best, making these type of errors will lead to confusion on the part of the analysis.
### References
Burbary, Ken. Facebook Demographics Revisited – 2001 Statistics, 2011. Available online at http://www.kenburbary.com/2011/03/facebook-demographics-revisited-2011-statistics-2/ (accessed August 21, 2013).
“9th Annual AP Report to the Nation.” CollegeBoard, 2013. Available online at http://apreport.collegeboard.org/goals-and-findings/promoting-equity (accessed September 13, 2013).
“Overweight and Obesity: Adult Obesity Facts.” Centers for Disease Control and Prevention. Available online at http://www.cdc.gov/obesity/data/adult.html (accessed September 13, 2013).
Data on annual homicides in Detroit, 1961–73, from Gunst & Mason’s book ‘Regression Analysis and its Application’, Marcel Dekker
“Timeline: Guide to the U.S. Presidents: Information on every president’s birthplace, political party, term of office, and more.” Scholastic, 2013. Available online at http://www.scholastic.com/teachers/article/timeline-guide-us-presidents (accessed April 3, 2013).
“Presidents.” Fact Monster. Pearson Education, 2007. Available online at http://www.factmonster.com/ipka/A0194030.html (accessed April 3, 2013).
“Food Security Statistics.” Food and Agriculture Organization of the United Nations. Available online at http://www.fao.org/economic/ess/ess-fs/en/ (accessed April 3, 2013).
“Consumer Price Index.” United States Department of Labor: Bureau of Labor Statistics. Available online at http://data.bls.gov/pdq/SurveyOutputServlet (accessed April 3, 2013).
“CO2 emissions (kt).” The World Bank, 2013. Available online at http://databank.worldbank.org/data/home.aspx (accessed April 3, 2013).
“Births Time Series Data.” General Register Office For Scotland, 2013. Available online at http://www.gro-scotland.gov.uk/statistics/theme/vital-events/births/time-series.html (accessed April 3, 2013).
“Demographics: Children under the age of 5 years underweight.” Indexmundi. Available online at http://www.indexmundi.com/g/r.aspx?t=50&v=2224&aml=en (accessed April 3, 2013).
Gunst, Richard, Robert Mason. Regression Analysis and Its Application: A Data-Oriented Approach. CRC Press: 1980.
“Overweight and Obesity: Adult Obesity Facts.” Centers for Disease Control and Prevention. Available online at http://www.cdc.gov/obesity/data/adult.html (accessed September 13, 2013).
### Chapter Review
A stem-and-leaf plot is a way to plot data and look at the distribution. In a stem-and-leaf plot, all data values within a class are visible. The advantage in a stem-and-leaf plot is that all values are listed, unlike a histogram, which gives classes of data values. A line graph is often used to represent a set of data values in which a quantity varies with time. These graphs are useful for finding trends. That is, finding a general pattern in data sets including temperature, sales, employment, company profit or cost over a period of time. A bar graph is a chart that uses either horizontal or vertical bars to show comparisons among categories. One axis of the chart shows the specific categories being compared, and the other axis represents a discrete value. Some bar graphs present bars clustered in groups of more than one (grouped bar graphs), and others show the bars divided into subparts to show cumulative effect (stacked bar graphs). Bar graphs are especially useful when categorical data is being used.
A histogram is a graphic version of a frequency distribution. The graph consists of bars of equal width drawn adjacent to each other. The horizontal scale represents classes of quantitative data values and the vertical scale represents frequencies. The heights of the bars correspond to frequency values. Histograms are typically used for large, continuous, quantitative data sets. A frequency polygon can also be used when graphing large data sets with data points that repeat. The data usually goes on y-axis with the frequency being graphed on the x-axis. Time series graphs can be helpful when looking at large amounts of data for one variable over a period of time.
### For the next three exercises, use the data to construct a line graph.In a survey, 40 people were asked how many times they visited a store before making a major purchase. The results are shown in (Figure).Number of times in storeFrequency142103164654
In a survey, several people were asked how many years it has been since they purchased a mattress. The results are shown in (Figure).
Years since last purchase Frequency
0 2
1 8
2 13
3 22
4 16
5 9
Several children were asked how many TV shows they watch each day. The results of the survey are shown in (Figure).
Number of TV shows Frequency
0 12
1 18
2 36
3 7
4 2
The students in Ms. Ramirez’s math class have birthdays in each of the four seasons. (Figure) shows the four seasons, the number of students who have birthdays in each season, and the percentage (%) of students in each group. Construct a bar graph showing the number of students.
Seasons Number of students Proportion of population
Spring 8 24%
Summer 9 26%
Autumn 11 32%
Winter 6 18%
Using the data from Mrs. Ramirez’s math class supplied in (Figure), construct a bar graph showing the percentages.
David County has six high schools. Each school sent students to participate in a county-wide science competition. (Figure) shows the percentage breakdown of competitors from each school, and the percentage of the entire student population of the county that goes to each school. Construct a bar graph that shows the population percentage of competitors from each school.
High school Science competition population Overall student population
Alabaster 28.9% 8.6%
Concordia 7.6% 23.2%
Genoa 12.1% 15.0%
Mocksville 18.5% 14.3%
Tynneson 24.2% 10.1%
West End 8.7% 28.8%
Use the data from the David County science competition supplied in (Figure). Construct a bar graph that shows the county-wide population percentage of students at each school.
Sixty-five randomly selected car salespersons were asked the number of cars they generally sell in one week. Fourteen people answered that they generally sell three cars; nineteen generally sell four cars; twelve generally sell five cars; nine generally sell six cars; eleven generally sell seven cars. Complete the table.
Data value (# cars) Frequency Relative frequency Cumulative relative frequency
What does the frequency column in (Figure) sum to? Why?
65
What does the relative frequency column in (Figure) sum to? Why?
What is the difference between relative frequency and frequency for each data value in (Figure)?
The relative frequency shows the proportion of data points that have each value. The frequency tells the number of data points that have each value.
What is the difference between cumulative relative frequency and relative frequency for each data value?
To construct the histogram for the data in (Figure), determine appropriate minimum and maximum x and y values and the scaling. Sketch the histogram. Label the horizontal and vertical axes with words. Include numerical scaling.
Answers will vary. One possible histogram is shown:
Construct a frequency polygon for the following:
1. Pulse rates for women Frequency
60–69 12
70–79 14
80–89 11
90–99 1
100–109 1
110–119 0
120–129 1
2. Actual speed in a 30 MPH zone Frequency
42–45 25
46–49 14
50–53 7
54–57 3
58–61 1
3. Tar (mg) in nonfiltered cigarettes Frequency
10–13 1
14–17 0
18–21 15
22–25 7
26–29 2
Construct a frequency polygon from the frequency distribution for the 50 highest ranked countries for depth of hunger.
Depth of hunger Frequency
230–259 21
260–289 13
290–319 5
320–349 7
350–379 1
380–409 1
410–439 1
Find the midpoint for each class. These will be graphed on the x-axis. The frequency values will be graphed on the y-axis values.
Use the two frequency tables to compare the life expectancy of men and women from 20 randomly selected countries. Include an overlayed frequency polygon and discuss the shapes of the distributions, the center, the spread, and any outliers. What can we conclude about the life expectancy of women compared to men?
Life expectancy at birth – women Frequency
49–55 3
56–62 3
63–69 1
70–76 3
77–83 8
84–90 2
Life expectancy at birth – men Frequency
49–55 3
56–62 3
63–69 1
70–76 1
77–83 7
84–90 5
Construct a times series graph for (a) the number of male births, (b) the number of female births, and (c) the total number of births.
Sex/Year 1855 1856 1857 1858 1859 1860 1861 Female 45,545 49,582 50,257 50,324 51,915 51,220 52,403 Male 47,804 52,239 53,158 53,694 54,628 54,409 54,606 Total 93,349 101,821 103,415 104,018 106,543 105,629 107,009
Sex/Year 1862 1863 1864 1865 1866 1867 1868 1869 Female 51,812 53,115 54,959 54,850 55,307 55,527 56,292 55,033 Male 55,257 56,226 57,374 58,220 58,360 58,517 59,222 58,321 Total 107,069 109,341 112,333 113,070 113,667 114,044 115,514 113,354
Sex/Year 1870 1871 1872 1873 1874 1875 Female 56,431 56,099 57,472 58,233 60,109 60,146 Male 58,959 60,029 61,293 61,467 63,602 63,432 Total 115,390 116,128 118,765 119,700 123,711 123,578
The following data sets list full time police per 100,000 citizens along with homicides per 100,000 citizens for the city of Detroit, Michigan during the period from 1961 to 1973.
Year 1961 1962 1963 1964 1965 1966 1967 Police 260.35 269.8 272.04 272.96 272.51 261.34 268.89 Homicides 8.6 8.9 8.52 8.89 13.07 14.57 21.36
Year 1968 1969 1970 1971 1972 1973 Police 295.99 319.87 341.43 356.59 376.69 390.19 Homicides 28.03 31.49 37.39 46.26 47.24 52.33
1. Construct a double time series graph using a common x-axis for both sets of data.
2. Which variable increased the fastest? Explain.
3. Did Detroit’s increase in police officers have an impact on the murder rate? Explain.
### Homework
(Figure) contains the 2010 obesity rates in U.S. states and Washington, DC.
State Percent (%) State Percent (%) State Percent (%)
Alabama 32.2 Kentucky 31.3 North Dakota 27.2
Alaska 24.5 Louisiana 31.0 Ohio 29.2
Arizona 24.3 Maine 26.8 Oklahoma 30.4
Arkansas 30.1 Maryland 27.1 Oregon 26.8
California 24.0 Massachusetts 23.0 Pennsylvania 28.6
Colorado 21.0 Michigan 30.9 Rhode Island 25.5
Connecticut 22.5 Minnesota 24.8 South Carolina 31.5
Delaware 28.0 Mississippi 34.0 South Dakota 27.3
Washington, DC 22.2 Missouri 30.5 Tennessee 30.8
Florida 26.6 Montana 23.0 Texas 31.0
Georgia 29.6 Nebraska 26.9 Utah 22.5
Hawaii 22.7 Nevada 22.4 Vermont 23.2
Idaho 26.5 New Hampshire 25.0 Virginia 26.0
Illinois 28.2 New Jersey 23.8 Washington 25.5
Indiana 29.6 New Mexico 25.1 West Virginia 32.5
Iowa 28.4 New York 23.9 Wisconsin 26.3
Kansas 29.4 North Carolina 27.8 Wyoming 25.1
1. Use a random number generator to randomly pick eight states. Construct a bar graph of the obesity rates of those eight states.
2. Construct a bar graph for all the states beginning with the letter “A.”
3. Construct a bar graph for all the states beginning with the letter “M.”
1. Example solution for using the random number generator for the TI-84+ to generate a simple random sample of 8 states. Instructions are as follows.
• Number the entries in the table 1–51 (Includes Washington, DC; Numbered vertically)
• Press MATH
• Arrow over to PRB
• Press 5:randInt(
• Enter 51,1,8)
Eight numbers are generated (use the right arrow key to scroll through the numbers). The numbers correspond to the numbered states (for this example: {47 21 9 23 51 13 25 4}. If any numbers are repeated, generate a different number by using 5:randInt(51,1)). Here, the states (and Washington DC) are {Arkansas, Washington DC, Idaho, Maryland, Michigan, Mississippi, Virginia, Wyoming}.
Corresponding percents are {30.1, 22.2, 26.5, 27.1, 30.9, 34.0, 26.0, 25.1}.
Suppose that three book publishers were interested in the number of fiction paperbacks adult consumers purchase per month. Each publisher conducted a survey. In the survey, adult consumers were asked the number of fiction paperbacks they had purchased the previous month. The results are as follows:
Publisher A
# of books Freq. Rel. freq.
0 10
1 12
2 16
3 12
4 8
5 6
6 2
8 2
Publisher B
# of books Freq. Rel. freq.
0 18
1 24
2 24
3 22
4 15
5 10
7 5
9 1
Publisher C
# of books Freq. Rel. freq.
0–1 20
2–3 35
4–5 12
6–7 2
8–9 1
1. Find the relative frequencies for each survey. Write them in the charts.
2. Use the frequency column to construct a histogram for each publisher’s survey. For Publishers A and B, make bar widths of one. For Publisher C, make bar widths of two.
3. In complete sentences, give two reasons why the graphs for Publishers A and B are not identical.
4. Would you have expected the graph for Publisher C to look like the other two graphs? Why or why not?
5. Make new histograms for Publisher A and Publisher B. This time, make bar widths of two.
6. Now, compare the graph for Publisher C to the new graphs for Publishers A and B. Are the graphs more similar or more different? Explain your answer.
Often, cruise ships conduct all on-board transactions, with the exception of gambling, on a cashless basis. At the end of the cruise, guests pay one bill that covers all onboard transactions. Suppose that 60 single travelers and 70 couples were surveyed as to their on-board bills for a seven-day cruise from Los Angeles to the Mexican Riviera. Following is a summary of the bills for each group.
Singles
Amount(?) Frequency Rel. frequency
51–100 5
101–150 10
151–200 15
201–250 15
251–300 10
301–350 5
Couples
Amount(?) Frequency Rel. frequency
100–150 5
201–250 5
251–300 5
301–350 5
351–400 10
401–450 10
451–500 10
501–550 10
551–600 5
601–650 5
1. Fill in the relative frequency for each group.
2. Construct a histogram for the singles group. Scale the x-axis by ?50 widths. Use relative frequency on the y-axis.
3. Construct a histogram for the couples group. Scale the x-axis by ?50 widths. Use relative frequency on the y-axis.
4. Compare the two graphs:
1. List two similarities between the graphs.
2. List two differences between the graphs.
3. Overall, are the graphs more similar or different?
5. Construct a new graph for the couples by hand. Since each couple is paying for two individuals, instead of scaling the x-axis by ?50, scale it by ?100. Use relative frequency on the y-axis.
6. Compare the graph for the singles with the new graph for the couples:
1. List two similarities between the graphs.
2. Overall, are the graphs more similar or different?
7. How did scaling the couples graph differently change the way you compared it to the singles graph?
8. Based on the graphs, do you think that individuals spend the same amount, more or less, as singles as they do person by person as a couple? Explain why in one or two complete sentences.
Singles
Amount(?) Frequency Relative frequency
51–100 5 0.08
101–150 10 0.17
151–200 15 0.25
201–250 15 0.25
251–300 10 0.17
301–350 5 0.08
Couples
Amount(?) Frequency Relative frequency
100–150 5 0.07
201–250 5 0.07
251–300 5 0.07
301–350 5 0.07
351–400 10 0.14
401–450 10 0.14
451–500 10 0.14
501–550 10 0.14
551–600 5 0.07
601–650 5 0.07
1. See (Figure) and (Figure).
2. In the following histogram data values that fall on the right boundary are counted in the class interval, while values that fall on the left boundary are not counted (with the exception of the first interval where both boundary values are included).
3. In the following histogram, the data values that fall on the right boundary are counted in the class interval, while values that fall on the left boundary are not counted (with the exception of the first interval where values on both boundaries are included).
4. Compare the two graphs:
• Both graphs have a single peak.
• Both graphs use class intervals with width equal to ?50.
• The couples graph has a class interval with no values.
• It takes almost twice as many class intervals to display the data for couples.
3. Answers may vary. Possible answers include: The graphs are more similar than different because the overall patterns for the graphs are the same.
5. Check student’s solution.
6. Compare the graph for the Singles with the new graph for the Couples:
• Both graphs have a single peak.
• Both graphs display 6 class intervals.
• Both graphs show the same general pattern.
1. Answers may vary. Possible answers include: Although the width of the class intervals for couples is double that of the class intervals for singles, the graphs are more similar than they are different.
7. Answers may vary. Possible answers include: You are able to compare the graphs interval by interval. It is easier to compare the overall patterns with the new scale on the Couples graph. Because a couple represents two individuals, the new scale leads to a more accurate comparison.
8. Answers may vary. Possible answers include: Based on the histograms, it seems that spending does not vary much from singles to individuals who are part of a couple. The overall patterns are the same. The range of spending for couples is approximately double the range for individuals.
Twenty-five randomly selected students were asked the number of movies they watched the previous week. The results are as follows.
# of movies Frequency Relative frequency Cumulative relative frequency
0 5
1 9
2 6
3 4
4 1
1. Construct a histogram of the data.
2. Complete the columns of the chart.
Use the following information to answer the next two exercises: Suppose one hundred eleven people who shopped in a special t-shirt store were asked the number of t-shirts they own costing more than ?19 each.
The percentage of people who own at most three t-shirts costing more than ?19 each is approximately:
1. 21
2. 59
3. 41
4. Cannot be determined
c
If the data were collected by asking the first 111 people who entered the store, then the type of sampling is:
1. cluster
2. simple random
3. stratified
4. convenience
Following are the 2010 obesity rates by U.S. states and Washington, DC.
State Percent (%) State Percent (%) State Percent (%)
Alabama 32.2 Kentucky 31.3 North Dakota 27.2
Alaska 24.5 Louisiana 31.0 Ohio 29.2
Arizona 24.3 Maine 26.8 Oklahoma 30.4
Arkansas 30.1 Maryland 27.1 Oregon 26.8
California 24.0 Massachusetts 23.0 Pennsylvania 28.6
Colorado 21.0 Michigan 30.9 Rhode Island 25.5
Connecticut 22.5 Minnesota 24.8 South Carolina 31.5
Delaware 28.0 Mississippi 34.0 South Dakota 27.3
Washington, DC 22.2 Missouri 30.5 Tennessee 30.8
Florida 26.6 Montana 23.0 Texas 31.0
Georgia 29.6 Nebraska 26.9 Utah 22.5
Hawaii 22.7 Nevada 22.4 Vermont 23.2
Idaho 26.5 New Hampshire 25.0 Virginia 26.0
Illinois 28.2 New Jersey 23.8 Washington 25.5
Indiana 29.6 New Mexico 25.1 West Virginia 32.5
Iowa 28.4 New York 23.9 Wisconsin 26.3
Kansas 29.4 North Carolina 27.8 Wyoming 25.1
Construct a bar graph of obesity rates of your state and the four states closest to your state. Hint: Label the x-axis with the states.
### Key Terms
Frequency
the number of times a value of the data occurs
Histogram
a graphical representation in xy form of the distribution of data in a data set; x represents the data and y represents the frequency, or relative frequency. The graph consists of contiguous rectangles.
Relative Frequency
the ratio of the number of times a value of the data occurs in the set of all outcomes to the number of all outcomes | HuggingFaceTB/finemath | |
# Search by Topic
#### Resources tagged with Generalising similar to Counting Triangles:
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### Chess
##### Age 11 to 14 Challenge Level:
What would be the smallest number of moves needed to move a Knight from a chess set from one corner to the opposite corner of a 99 by 99 square board?
### Christmas Chocolates
##### Age 11 to 14 Challenge Level:
How could Penny, Tom and Matthew work out how many chocolates there are in different sized boxes?
### Cubes Within Cubes Revisited
##### Age 11 to 14 Challenge Level:
Imagine starting with one yellow cube and covering it all over with a single layer of red cubes, and then covering that cube with a layer of blue cubes. How many red and blue cubes would you need?
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##### Age 11 to 14 Challenge Level:
Rectangles are considered different if they vary in size or have different locations. How many different rectangles can be drawn on a chessboard?
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##### Age 11 to 14 Challenge Level:
Can you dissect a square into: 4, 7, 10, 13... other squares? 6, 9, 12, 15... other squares? 8, 11, 14... other squares?
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##### Age 11 to 14 Challenge Level:
A 2 by 3 rectangle contains 8 squares and a 3 by 4 rectangle contains 20 squares. What size rectangle(s) contain(s) exactly 100 squares? Can you find them all?
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##### Age 11 to 14 Challenge Level:
If you can copy a network without lifting your pen off the paper and without drawing any line twice, then it is traversable. Decide which of these diagrams are traversable.
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##### Age 11 to 14 Challenge Level:
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### Dotty Triangles
##### Age 11 to 14 Challenge Level:
Imagine an infinitely large sheet of square dotty paper on which you can draw triangles of any size you wish (providing each vertex is on a dot). What areas is it/is it not possible to draw?
### Picturing Square Numbers
##### Age 11 to 14 Challenge Level:
Square numbers can be represented as the sum of consecutive odd numbers. What is the sum of 1 + 3 + ..... + 149 + 151 + 153?
### Route to Infinity
##### Age 11 to 14 Challenge Level:
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##### Age 11 to 14 Challenge Level:
Draw a square. A second square of the same size slides around the first always maintaining contact and keeping the same orientation. How far does the dot travel?
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##### Age 11 to 14 Challenge Level:
Triangular numbers can be represented by a triangular array of squares. What do you notice about the sum of identical triangle numbers?
### Shear Magic
##### Age 11 to 14 Challenge Level:
What are the areas of these triangles? What do you notice? Can you generalise to other "families" of triangles?
### Frogs
##### Age 11 to 14 Challenge Level:
How many moves does it take to swap over some red and blue frogs? Do you have a method?
### Card Trick 2
##### Age 11 to 14 Challenge Level:
Can you explain how this card trick works?
### Number Pyramids
##### Age 11 to 14 Challenge Level:
Try entering different sets of numbers in the number pyramids. How does the total at the top change?
### Konigsberg Plus
##### Age 11 to 14 Challenge Level:
Euler discussed whether or not it was possible to stroll around Koenigsberg crossing each of its seven bridges exactly once. Experiment with different numbers of islands and bridges.
### Jam
##### Age 14 to 16 Challenge Level:
A game for 2 players
### Cunning Card Trick
##### Age 11 to 14 Challenge Level:
Delight your friends with this cunning trick! Can you explain how it works?
### A Tilted Square
##### Age 14 to 16 Challenge Level:
The opposite vertices of a square have coordinates (a,b) and (c,d). What are the coordinates of the other vertices?
### Mirror, Mirror...
##### Age 11 to 14 Challenge Level:
Explore the effect of reflecting in two parallel mirror lines.
### ...on the Wall
##### Age 11 to 14 Challenge Level:
Explore the effect of reflecting in two intersecting mirror lines.
### Have You Got It?
##### Age 11 to 14 Challenge Level:
Can you explain the strategy for winning this game with any target?
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##### Age 5 to 14 Challenge Level:
Nim-7 game for an adult and child. Who will be the one to take the last counter?
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To avoid losing think of another very well known game where the patterns of play are similar.
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##### Age 11 to 16 Challenge Level:
The aim of the game is to slide the green square from the top right hand corner to the bottom left hand corner in the least number of moves.
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##### Age 11 to 14 Challenge Level:
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A three digit number abc is always divisible by 7 when 2a+3b+c is divisible by 7. Why?
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##### Age 7 to 14 Challenge Level:
Got It game for an adult and child. How can you play so that you know you will always win?
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Can you find sets of sloping lines that enclose a square?
### Mini-max
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Consider all two digit numbers (10, 11, . . . ,99). In writing down all these numbers, which digits occur least often, and which occur most often ? What about three digit numbers, four digit numbers. . . .
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##### Age 11 to 14 Challenge Level:
The sum of the numbers 4 and 1 [1/3] is the same as the product of 4 and 1 [1/3]; that is to say 4 + 1 [1/3] = 4 × 1 [1/3]. What other numbers have the sum equal to the product and can this be so for. . . .
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Find some examples of pairs of numbers such that their sum is a factor of their product. eg. 4 + 12 = 16 and 4 × 12 = 48 and 16 is a factor of 48.
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##### Age 11 to 14
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### Partitioning Revisited
##### Age 11 to 14 Challenge Level:
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##### Age 7 to 16
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##### Age 14 to 16 Challenge Level:
Use the animation to help you work out how many lines are needed to draw mystic roses of different sizes. | HuggingFaceTB/finemath | |
# Figure 7.3
## GUIDE: Mathematics of the Discrete Fourier Transform (DFT) - Julius O. Smith III. Figure 7.3
It appears that you are using AdBlocking software. The cost of running this website is covered by advertisements. If you like it please feel free to a small amount of money to secure the future of this website. NOTE: THIS DOCUMENT IS OBSOLETE, PLEASE CHECK THE NEW VERSION: "Mathematics of the Discrete Fourier Transform (DFT), with Audio Applications --- Second Edition", by Julius O. Smith III, W3K Publishing, 2007, ISBN 978-0-9745607-4-8. - Copyright © 2017-09-28 by Julius O. Smith III - Center for Computer Research in Music and Acoustics (CCRMA), Stanford University
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## Figure 7.3
Below is the Matlab for Fig. 7.3:
```% Parameters (sampling rate = 1)
N = 16; % DFT length
k = N/4; % bin where DFT filter is centered
wk = 2*pi*k/N; % normalized radian center-frequency for DFT_k()
wStep = 2*pi/N;
w = [0:wStep:2*pi - wStep]; % DFT frequency grid
interp = 10;
N2 = interp*N; % Denser grid showing “arbitrary” frequencies
w2Step = 2*pi/N2;
w2 = [0:w2Step:2pi - w2Step]; % Extra dense frequency grid
X = (1 - exp(j(w2-wk)N)) ./ (1 - exp(j(w2-wk)));
% slightly offset to avoid divide by zero at wk
X(1+k*interp) = N; % Fix divide-by-zero point (overwrite “NaN”)
% Plot spectral magnitude
clf;
magX = abs(X);
magXd = magX(1:interp:N2); % DFT frequencies only
subplot(2,1,1);
plot(w2,magX,‘-’); hold on; grid;
plot(w,magXd,’*‘); % Show DFT sample points
title(‘DFT Amplitude Response at k=N/4’);
xlabel(‘Normalized Radian Frequency (radians per sample)’);
ylabel(‘Magnitude (Linear)’);
text(-1,20,‘a)’);
% Same thing on a dB scale
magXdb = 20*log10(magX); % Spectral magnitude in dB
% Since the zeros go to minus infinity, clip at -60 dB:
magXdb = max(magXdb,-60ones(1,N2));
magXddb = magXdb(1:interp:N2); % DFT frequencies only
subplot(2,1,2);
hold off; plot(w2,magXdb,‘-’); hold on; plot(w,magXddb,’’); grid;
xlabel(‘Normalized Radian Frequency (radians per sample)’);
ylabel(‘Magnitude (dB)’);
text(-1,40,‘b)’);
print -deps ‘https://ccrma.stanford.edu/~jos//eps/dftfilter.eps';
hold off;```
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Convert to a + ib form
Chapter 4 Class 11 Complex Numbers
Concept wise
Learn in your speed, with individual attention - Teachoo Maths 1-on-1 Class
Transcript
Example 4 Express ( (3 ) "+ " ( 2) ) (2 3 ) in the form of a + ib ( 3 "+ " ( 2) ) (2 3 ) = ( 3 " + " ( 1 2) ) (2 3 ) = (" " 3 " + " ( 1 ) (2 ) ) (2 3 ) = ( 3 " + i" (2 )) (2 3 ) = ( 3 " " ) (2 3 ) "+ i" (2 ) (2 3 ) = 3 (2 3 ) 3 ( ) + i 2 (2 3 ) + i 2 "(" ) = 3 2 + 3i + 2 i (2 3) i 2 2 = 6 + (3"i" ) + 2"i" (2 3) ( 1 ) 2 = 6 + (3"i" ) + 2 (2 3) + 2 = ( 6+ (2 ) ) + 3 +2 (2 3) = ( 6+ (2 ) ) + 3 (1+2 (2 )) | HuggingFaceTB/finemath | |
# 1710119 (number)
1,710,119 (one million seven hundred ten thousand one hundred nineteen) is an odd seven-digits composite number following 1710118 and preceding 1710120. In scientific notation, it is written as 1.710119 × 106. The sum of its digits is 20. It has a total of 2 prime factors and 4 positive divisors. There are 1,635,744 positive integers (up to 1710119) that are relatively prime to 1710119.
## Basic properties
• Is Prime? No
• Number parity Odd
• Number length 7
• Sum of Digits 20
• Digital Root 2
## Name
Short name 1 million 710 thousand 119 one million seven hundred ten thousand one hundred nineteen
## Notation
Scientific notation 1.710119 × 106 1.710119 × 106
## Prime Factorization of 1710119
Prime Factorization 23 × 74353
Composite number
Distinct Factors Total Factors Radical ω(n) 2 Total number of distinct prime factors Ω(n) 2 Total number of prime factors rad(n) 1710119 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) 0 Returns log(p) if n is a power pk of any prime p (for any k >= 1), else returns 0
The prime factorization of 1,710,119 is 23 × 74353. Since it has a total of 2 prime factors, 1,710,119 is a composite number.
## Divisors of 1710119
4 divisors
Even divisors 0 4 2 2
Total Divisors Sum of Divisors Aliquot Sum τ(n) 4 Total number of the positive divisors of n σ(n) 1.7845e+06 Sum of all the positive divisors of n s(n) 74377 Sum of the proper positive divisors of n A(n) 446124 Returns the sum of divisors (σ(n)) divided by the total number of divisors (τ(n)) G(n) 1307.72 Returns the nth root of the product of n divisors H(n) 3.83328 Returns the total number of divisors (τ(n)) divided by the sum of the reciprocal of each divisors
The number 1,710,119 can be divided by 4 positive divisors (out of which 0 are even, and 4 are odd). The sum of these divisors (counting 1,710,119) is 1,784,496, the average is 446,124.
## Other Arithmetic Functions (n = 1710119)
1 φ(n) n
Euler Totient Carmichael Lambda Prime Pi φ(n) 1635744 Total number of positive integers not greater than n that are coprime to n λ(n) 817872 Smallest positive number such that aλ(n) ≡ 1 (mod n) for all a coprime to n π(n) ≈ 128563 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 1,635,744 positive integers (less than 1,710,119) that are coprime with 1,710,119. And there are approximately 128,563 prime numbers less than or equal to 1,710,119.
## Divisibility of 1710119
m n mod m 2 3 4 5 6 7 8 9 1 2 3 4 5 5 7 2
1,710,119 is not divisible by any number less than or equal to 9.
## Classification of 1710119
• Arithmetic
• Semiprime
• Deficient
• Polite
• Square Free
### Other numbers
• LucasCarmichael
## Base conversion (1710119)
Base System Value
2 Binary 110100001100000100111
3 Ternary 10012212211202
4 Quaternary 12201200213
5 Quinary 414210434
6 Senary 100353115
8 Octal 6414047
10 Decimal 1710119
12 Duodecimal 6a579b
36 Base36 10njb
## Basic calculations (n = 1710119)
### Multiplication
n×y
n×2 3420238 5130357 6840476 8550595
### Division
n÷y
n÷2 855060 570040 427530 342024
### Exponentiation
ny
n2 2924506994161 5001254976347615159 8552741158896607288093921 14626205157911107158907888086599
### Nth Root
y√n
2√n 1307.72 119.585 36.1623 17.6443
## 1710119 as geometric shapes
### Circle
Diameter 3.42024e+06 1.0745e+07 9.18761e+12
### Sphere
Volume 2.09492e+19 3.67504e+13 1.0745e+07
### Square
Length = n
Perimeter 6.84048e+06 2.92451e+12 2.41847e+06
### Cube
Length = n
Surface area 1.7547e+13 5.00125e+18 2.96201e+06
### Equilateral Triangle
Length = n
Perimeter 5.13036e+06 1.26635e+12 1.48101e+06
### Triangular Pyramid
Length = n
Surface area 5.06539e+12 5.89404e+17 1.39631e+06
## Cryptographic Hash Functions
md5 53b30d25e3f2b0b91bc46afcf352d503 ffeec180afbdf8f70a54272765ce078a5c77c0c2 6fca85b36b6244b2068a0572b911eb1cbada549725fe6b3cc9d08623e43015ab 2d6dd264054c47c07c7481edd3404fb17f65076fe9bd64c308e8d03d30bb4daf9da124e52ba9951a6ecfb8ac2ecb1003b7d0ee8bcffa7815b5740fa121e79e1a 106736f48922a4af9c5f4d5fc5f7a7e5d3f9e911 | HuggingFaceTB/finemath | |
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# Simple Interest Questions Worksheet (CAT, SSC, Bank, Competitive) Exams(function() { var po = document.createElement('script'); po.type = 'text/javascript'; po.async = true; po.src = 'https://apis.google.com/js/plusone.js'; var s = document.getElementsByTagName('script')[0]; s.parentNode.insertBefore(po, s); })();
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## Simple Interest Questions Worksheet
Candidates who are preparing for any Competitive exams like CAT, SSC, Bank, railway etc, they can acquire Simple Interest Questions Worksheet from this page for the better preparation of arithmetic aptitude section. Examines here we are settled the Questions and answers of Simple Interest. So, Candidates are advised to do practice of these Questions and train your brain to solve the any question of Simple Interest within 1-2 minute minimum.
## Simple Interest Questions And Answers
Q1- Sachin borrows Rs. 5000 for 2 years at 4% p.a. simple interest. He immediately lends money to Rahul at 25/4% p.a. for 2 years. Find the gain of one year by Sachin.
Q2- A man took some money for borrowed, for 3 years the total will be Rs.4000 and 5 years it will be Rs.5000/-. Then how much amount he borrowed?
Q3- A person took some amount with some interest for 2 years, but increase the interest for 1%, he paid Rs.120/- extra, then how much amount he took?
Q4- If A lends Rs. 3500 to B at 10% p.a. and B lends the same sum to C at 11.5% p.a., then the gain of B (in Rs.) in a period of 3 years is
Q5- Sahil took a loan for 6 years at the rate of 5% per annum on Simple Interest, If the total interest paid was Rs. 1230, the principal was
Q6- There was simple interest of Rs. 4016.25 on a principal amount at the rate of 9%p.a. in 5 years. Find the principal amount
Q7- Find the rate at Simple interest, at which a sum becomes four times of itself in 15 years.
Q8- At 5% per annum simple interest, Rahul borrowed Rs. 500. What amount will he pay to clear the debt after 4 years?
Q9- If a sum of money doubles itself in 8 years at simple interest, the ratepercent per annum is
Q10- A sum of Rs 12,500 amounts to Rs. 15,500 in the 4 years at the rate of simple interest. Find the rate percent
Q11- What is the present worth of Rs. 132 due in 2 years at 5% simple interest per annum?
Q12- Reema took a loan of Rs 1200 with simple interest for as many years as the rate of interest. If she paid Rs. 432 as interest at the end of the loan period, what was the rate of interest?
Q13- The simple interest on a certain sum of money at the rate of 5% p.a. for 8 years is Rs. 840. At what rate of intrest the same amount of interest can be received on the same sum after 5 years.
Q14- A man took a loan at rate of 12% per annum simple interest. After 3 years he had to pay 5400 interest. The principal amount borrowed by him was.
Q15- A sum of money at simple interest amounts to Rs. 2240 in 2 years and to Rs. 2600 in 5 years. What is the principal amount?
16. At what rate percent per annum will the simple interest on a sum of money be 2/5 of the amount in 10 years?
Q17- Rs. 800 becomes Rs. 956 in 3 years at a certain rate of simple interest. If the rate of interest is increased by 4%, what amount will Rs. 800 become in 3 years.
Q18- In how many years Rs 150 will produce the same interest at 8% as Rs. 800 produce in 3 years at 9/2%
Q19- What will the ratio of simple interest earned by certain amount at the same rate of interest for 6 years and that for 9 years?
Q20- A financier claims to be lending money at simple interest, but he includes the interest every six months for calculating the principal. If he is charging an interest of 10%, the effective rate of interest becomes.
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# Cube 5
The content area of one cube wall is 32 square centimeters. Determine the length of its edges, its surface and volume.
Correct result:
a = 5.657 cm
S = 192 cm2
V = 181.024 cm3
#### Solution:
$S_{1}=32 \ \\ a=\sqrt{ S_{1} }=\sqrt{ 32 }=4 \ \sqrt{ 2 }=5.657 \ \text{cm}$
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# Free Rational Numbers 02 Practice Test - 7th grade
38 to get 516?
A.
1216
B.
1316
C.
1116
D.
1416
#### SOLUTION
Solution : C
Let the number to be added be x.
38+x=516
x=516+38=516+616=1116
In the given fig. locate the position where the ball lies on the number line.
A.
15
B.
25
C.
35
D.
45
#### SOLUTION
Solution : C
The space between 0 and 1 is divided into 10 equal parts. The ball lies in sixth place.
Therefore, the location ball lies in is, 610=35
Find the value of
163÷43.
A.
1
B.
3
C.
4
D.
2
#### SOLUTION
Solution : C
163÷43=163×34=4
Compare the rational numbers and choose the correct option.
3926 ___64
A.
>
B.
<
C.
=
D. None of the above
#### SOLUTION
Solution : C
Reducing 3926 to its lowest form , we get,
39÷1326÷1332( 13 is the common factor)
Multiplying numerator and denominator by 2 we get
3×22×2 = 64
Hence, 3926 is also equal to 64
So, both the fraction are equal
Hence, 3926=64
Which of the following is a rational number equal to 47 with 16 as the numerator?
A.
1621
B.
1628
C.
1615
D.
1620
#### SOLUTION
Solution : B
Consider the given rational number 47.
To make the numerator as 16, we have to multiply it by 4.
Multiplying both numerator and denominator by 4, we get
4× 47× 4=1628
Therefore, 1628 is an equivalent rational number of 47 with numerator equal to 16.
Add 213 to the reciprocal of 2612.
A.
57
B.
56
C.
67
D.
413
#### SOLUTION
Solution : D
Reciprocal of 2612 is 1226.
213+1226 [Taking LCM]
=4261226=41226=826=413
Find the sum of the rational numbers
719 and 257.
A.
2157
B.
2357
C.
2247
D.
3247
#### SOLUTION
Solution : B
719+257=(7×3)(19×3)+257
=2157+257=2357
102 in simplest form is ___
___
#### SOLUTION
Solution :
102=2×52
=5
413>0. True or False?
A.
True
B.
False
#### SOLUTION
Solution : A
Any positive number is greater than 0.
432436
A.
True
B.
False
#### SOLUTION
Solution : B
A negative number cannot be greater than a positive number.
so,43<2436 | HuggingFaceTB/finemath | |
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In the unordered pair, order doesn't matter and those two pairs are the same. For example, if x=2, then f(2)=6 for the point (2,6). Ordered Pair. ☼ Ordered Pair : A pair of two elements a and b listed in a specific order, is called an ordered pair ( a, b ). While the two elements of an ordered pair (a, b) need not be distinct, modern authors only call {a, b} an unordered pair if a ≠ b. In the previous examples, we substituted the $x\text{- and }y\text{-values}$ of a given ordered pair to determine whether or not it was a solution to a linear equation. The important thing to remember when considering if an equatin or set of ordered pairs is a function is that for a function, any value of x can produce no more than one value of f(x) . For example, (4, 7) is an ordered-pair number; the order is designated by the first element 4 and the second element 7. Print Ordered Pair: Definition & Examples Worksheet 1. Free vista key download Messages. In mathematics, an ordered pair (a, b) is a pair of mathematical objects.In the ordered pair (a, b), the object a is called the first entry, and the object b the second entry of the pair.Alternatively, the objects are called the first and second coordinates, or the left and right projections of the ordered pair. Identify the x-value in the ordered pair and plug it into the equation. a set having two elements a and b with no particular relation between them. Ordered Pairs There is an important branch of mathematics, known as coordinate geometry. Solving systems of linear equations in two variables. Download Image. Coordinate system and ordered … This code is not a good example as any Map implementation will behave as your sample code. Practical example. Ordered-Pair Numbers. (Note that in higher math, an ordered pair doesn't have to be an ordered pair of numbers; you can have ordered pairs of sets, functions, or even ordered pairs!) more ... Two numbers written in a certain order. Usually, Ordered Pair is used to locate a point in the Coordinate System. Ordered pair definition, two quantities written in such a way as to indicate that one quantity precedes or is to be considered before the other, as (3, 4) indicates the Cartesian coordinates of … The range will be any five integers between -10 and 10. Question 393212: Provide an example of at least five ordered pairs that do not model a function. Sepulchral. Wettest. The ordered pair ( 5 , 2 ) works, since 2 = 5 − 3 . • the second term is … The ordered pairs of the function would be represented by the points (x, f(x) ). The y-value (the second number in the ordered pair) is … Graph the ordered pair (π, 0).We don't know all the digits of π, but we know it's a little more than 3.Therefore, we draw the point in approximately the right place and then label it: When the points are not exact and not labeled, we can't do what we did with integer coordinates and go backwards from the graph to the ordered pair. sorted, ordered or not. 39. An ordered pair tells you the location of a point by relating the point’s location along the x-axis (the first value of the ordered pair) and along the y-axis (the second value of the ordered pair). An object or a point is defined by means of coordinates in the space. So they're telling us that we have an x … To figure out if an ordered pair is a solution to an equation, you could perform a test. One way is to choose a value for … Rainy Days :: Red Hunter Boots & Plaid Blanket Scarf Why Elton John's Drummer, Nigel Olsson, Is A True Gentleman English Jubilee Orpington Chickens And Hatching Eggs For Sale Bolt-On Toe/Heel Supports(pair). The vertical axis here is the y-axis. (1, 3) is the same as (3, 1) x = 3. y = 3. x is between 1 and 3. Ordered Pair - Displaying top 8 worksheets found for this concept.. The x-value (the first number in the ordered pair) is the distance left or right from the center. Note (i) ( 2, 3 ) We were looking for an equivalent class for pair in Java but Pair class did not come into existence till Java 7. An ordered-pair number is a pair of numbers that go together. In C++, we have std::pair in the utility library which is of immense use if we want to keep a pair of values together. But how do we find the ordered pairs if they are not given? I will illustrate the difference for pairs of numbers. So (12,5) is 12 units along, and 5 units up. It studies about the position of objects in space and problems based on that. 72. Select the TRUE statement about the ordered pair (1, 3). Sliding pair, turning pair and screw pair are form of lower pair. In mathematics, an unordered pair or pair set is a set of the form {a, b}, i.e. Ordered Pairs - Sample Math Practice Problems The math problems below can be generated by MathScore.com, a math practice program for schools and individual families. Ordered pair: definition & examples video & lesson transcript. If you are talking about unordered pairs of numbers, then (13, 16) = (16, 13). ordered pair • a pair of numbers where order is important, for example, (4, 6) is different to (6, 4). I already ordered your present. This week, we will present a discussion of functions of one real variable, illustrated by some economic examples. Ordered pair examples math. How to graph an ordered pair algebra 1. 2. Plotting a point (ordered pair) (video) | khan academy. This first quadrant of the xy-plane exercise has novelty in spades for children in grade 4, grade 5, grade 6. : You are free: to share – to copy, distribute and transmit the work; to remix – to adapt the work; Under the following conditions: attribution – You must give appropriate credit, provide a link to the license, and indicate if changes were made. 65. Your example must not be the same as those of other students or the textbook. If you are talking about ordered pairs of numbers, then (13, 16) $\neq$ (16, 13). Relations and functions chilimath. Stress's. Ordered Pair. So this is a coordinate plane right over here. References to complexity and mode refer to the overall difficulty of the problems as they appear in the main program. ... For example, the language of economic analysis is full of terms like demand and supply functions, cost functions, production functions, consumption functions, and so on. Write the point located at each ordered pair in part A, and represent the location of the given points using the ordered pairs in part B. Plotting Points on the 1st Quadrant. Hippies. If we are ordered to die, we must die. 38. 15 In an ordered pair, such as (x, y), the first value is called the x-coordinate and the second value is the y-coordinate. Algebra examples | functions | finding ordered pair solutions. 36. A quadrant is set up in such a way that the positive numbers are to the right and top, while the negative numbers are to the left and bottom. 31. And the convention, when we get an ordered pair like this, is that the first coordinate is the x-coordinate, and the second coordinate is the y-coordinate. A Function as an Ordered Pair of Elements. In the ordered pair, order matters: (a,b) is different from (b,a). 25. Cartesian Product and Ordered pairs Examples. When you simplify, if the y-value you get is the same as the y-value in the ordered pair, then that ordered pair is indeed a solution to the equation. Ordered sentence examples "Lisa," he ordered in a calm voice. In contrast, an ordered pair (a, b) has a as its first element and b as its second element.. Precalculus examples | linear equations | finding ordered pair. Ordered pair ~ a maths dictionary for kids quick reference by. Ordered pair ~ a maths dictionary for kids quick reference by. Example of a non functional ordered pair with intergers between 0. • the first term is called the abscissa and denotes the horizontal position. Psychologists Ordered pair wikipedia. Snowmobiles. JavaFX 2.2 has the javafx.util.Pair class which can be used to store a pair. ordered pair Two ordered pairs (a, b) and (c, d) are equal if and only if a = c and b = d. For example the ordered pair (1, 2) is not equal to the ordered pair … Kinematic Pair Based on contact between two link or element (types of contact) Lower pair. Dismantled. Ordered Pair (3, 4) is not equal to the Ordered Pair (4, 3). 56. 25. If objects are represented by x and y, then we write the ordered pair as (x, y). Ordered pair wikipedia. Coordinate system and ordered pairs (pre-algebra, introducing. By ordered pair, it is meant that two elements taken from each set are written in particular order. Ordered pairs are a crucial part of graphing, but you need to know how to identify the coordinates in an ordered pair if you're going to plot it on a coordinate plane. • coordinates are written as ordered pairs of numbers or letters and numbers. This file is licensed under the Creative Commons Attribution-Share Alike 3.0 Unported license. The lower pair is a surface contact between joint of to links or element is called lower pair. Further Exploration. In an alphabet of n letters, there are n^2 ordered pairs (choose the first letter n ways; choose the second letter n … In this tutorial, you'll see how to identify the y-coordinate in an ordered pair! Usually written in parentheses like this: (12,5) Which can be used to show the position on a graph, where the "x" (horizontal) value is first, and the "y" (vertical) value is second. So, if a ≠ b , ordered pairs (a,b) and (b,a) are distinct. The horizontal axis here, this is the x-axis. The domain will be any five integers between 0 and 20. – Peter Lawrey Mar 19 '09 at 20:42. We were ordered to be at the place before nine, but we haven't got halfway. Some of the worksheets for this concept are Ordered pairs, Ordered pairs, 2006 ordered pairs work and activity, Math 6 notes the coordinate system, Ordered pairs chess fsjis, Lesson 14 ordered pairs, Showing route positive s1, 3 points in the coordinate. The numbers are written within a set of parentheses and separated by a comma. Ordered Pair - symbol description, layout, design and history from Symbols.com Definition (ordered pair): An ordered pair is a pair of objects with an order associated with them. In Order Pair, we have a pair of integers in which the first one is called abscissa and the second one is called the ordinate. She ordered me up for a job, but I.m considering not going back. 2. Example: Find an ordered pair which is a solution of the equation y = x − 3 , and plot the point on the coordinate plane. Example 1: To take an example, let us take P as the set of grades in … Moderately 74hc32 datasheet. In spades for children in grade 4, 3 ) a as its first element ordered pair example as... Usually, ordered pairs ( pre-algebra, introducing that two elements taken from each set written... Layout, design and history from Symbols.com Download Image to choose a value for … pair... This week, we must die see how to identify the y-coordinate in an pair. Between 0 and 20 | linear equations | finding ordered pair - Displaying top 8 worksheets found for this..... Week, we must die -10 and 10 distance left or right the! To links or element is called lower pair is used to locate a point ( 2,6.!, design and history from Symbols.com Download Image this is a solution to an,! X … ordered pair ( 4, 3 ) ordered pair ~ a maths dictionary for kids quick by... How do we find the ordered pair ~ a maths dictionary for kids quick reference by you are about... Or right from the center contact ) lower pair is used to locate a point is by... Definition & examples Worksheet 1 that two elements taken from each set are written in certain! Before nine, but I.m considering not going back dictionary for kids quick reference by tutorial, you perform! Pair, it is meant that two elements a and b with no particular relation between them for … pair... Is meant that two elements a and b with no particular relation between them having two elements and., we must die contact ) lower pair, an unordered pair or pair set is set. Each set are written as ordered pairs ( pre-algebra, introducing elements taken from each set are written ordered... This tutorial, you could perform a test to choose a value for … ordered pair ( 1 3... Be used to store a pair elements taken from each set are within! To an equation, you 'll see how to identify the x-value ( the second in. You could perform a test f ( x, y ) the function would be represented by points. The function would be represented by the points ( x, f ( 2, 3 ) 2 works! Into the equation pairs There is an important branch of mathematics, an pair! By the points ( x, f ( 2 ) works, since 2 = −! With an order associated with them linear equations | finding ordered pair is a surface contact between joint to. Choose a value for … ordered pair, turning pair and plug it into the equation (,! The distance left or right from the center same as those of other or! Pairs are the same, grade 6 'll see how to identify the y-coordinate in ordered. By means of coordinates in the coordinate System and ordered pairs There is an important branch of mathematics known! Called ordered pair example pair see how to identify the y-coordinate in an ordered pair ) ( video ) khan! Are ordered to be at the place before nine, but we have n't got halfway do model! Grade 5, 2 ) =6 for the point ( ordered pair ( 5, grade,! You are talking about unordered pairs of numbers that go together you 'll see to. Mode refer to the overall difficulty of the function would be represented by and! Figure out if an ordered pair ( types of contact ) lower pair objects with an order with... Difficulty of the problems as they appear in the ordered pair solutions from the center to links element. Those of other students or the textbook between -10 and 10 they appear in the ordered pair a! 13 ) a pair of objects with an order associated with them a discussion functions... Would be represented by the points ( x, f ( 2, 3 ) a as its element... • the first number in the unordered pair, order does n't matter and those two pairs are same... ) lower pair with them they 're telling us that we have x... Nine, ordered pair example we have an x … ordered pair solutions first term is called the abscissa and denotes horizontal. The x-axis coordinates in the ordered pairs that do not model a function if an ordered pair (,... ) ( video ) | khan academy other students or the textbook we find the ordered pair ) 2., illustrated by some economic examples works, since 2 = 5 3... Class for pair in Java but pair class did not come into existence till Java 7 (... Unordered pair or pair set is a set of the function would be represented by and!: an ordered pair and plug it into the equation a function we write the ordered pair based that. Overall difficulty of the form { a, b }, i.e more... two numbers written in certain. Can be used to store a pair of objects with an order associated with them class which can used! Units along, and 5 units up here, this is a set having two elements a and as. Locate a point ordered pair example 2,6 ) called lower pair is 12 units,. For example, if x=2, then f ( 2, 3 ) ordered pair set are written as pairs! The center any five integers between 0 and 20 if x=2, then f x! The points ( x, y ) as any Map implementation will behave as your sample code symbol description layout! To be at the place before nine, but I.m considering not going back represented by the (. Java 7 more... two numbers written in a certain order, if x=2, then (. Linear equations | finding ordered pair ~ a maths dictionary for kids quick reference by how we! Contact ) lower pair spades for children in grade 4, 3 ) as... As ( x, y ) ) = ( 16, 13 ) as its second..! N'T got halfway an unordered pair or pair set is a surface contact between joint of links! Lower pair is used to store a pair of numbers, then we the... There is an important branch of mathematics, an unordered pair, it is meant that two elements taken each! Your example must not be the same as those of other students the... Difficulty of the problems as they appear in the coordinate System and ordered pairs There an..., if x=2, then we write the ordered pair 0 and 20 ( ordered pair, order does matter! The javafx.util.Pair class which can be used to store a pair of numbers that together... Lesson transcript it is meant that two elements a and b with particular... ) lower pair x-value ( the second number in the ordered pair: definition & examples Worksheet.! The function would be represented by the points ( x, y ) the {. Javafx 2.2 has the javafx.util.Pair class which can be used to locate a in... An important branch of mathematics, known as coordinate geometry we must die of numbers, then we the! Ordered pairs that do not model a function telling us that we have an x … ordered pair (,! Choose a value for … ordered pair is a pair of objects in space and problems based on.! N'T matter and those two pairs are the same as those of other or! Going back at the place before nine, but we have an x … ordered )... A comma range will be any five integers between -10 and 10 two pairs are the same for a,. As they appear in the ordered pair is a pair of numbers letters... Before nine, but I.m considering not going back at the place before nine, I.m... The x-axis element ( types of contact ) lower pair or the textbook the. The position of objects in space and problems based on that week, we must die meant two! Did not come into existence till Java 7 up for a job, but I.m considering going! Coordinate geometry good example as any Map implementation will behave as your sample code - Displaying top worksheets... Grade 6 x=2, then we write the ordered pair is used to store a of! For pair in Java but pair class did not come into existence till Java 7 illustrated by economic! Symbols.Com Download Image will illustrate the difference for pairs of numbers or letters and numbers in an pair... Into the equation an important branch of mathematics, known as coordinate geometry ordered-pair is!, f ( x ) ) if objects are represented by the points ( x, f ( )... Kinematic pair based on that ( pre-algebra, introducing be at the place before nine, we... The x-value in the coordinate System element is called lower pair as any Map implementation will behave as your code... Are form of lower pair 2 ) works, since 2 = 5 −.! The overall difficulty of the problems as they appear in the ordered:! For pairs of numbers, then ( 13, 16 ) = ( 16, 13 ) position objects! For pairs of numbers or letters and numbers Java 7 particular relation between them (! Is meant that two elements a ordered pair example b as its first element b! A test point is defined by means of coordinates in the coordinate System and pairs... Licensed under the Creative Commons Attribution-Share Alike 3.0 Unported license the numbers are written within set! Main program 13 ) surface contact between two link or element ( types of )... Means of coordinates in the ordered pairs of numbers or letters and numbers order associated with.... On contact between joint of to links or element is called the abscissa and denotes the axis... | open-web-math/open-web-math | |
# simplify 6 : 2
#### Understand the Problem
The question is asking to simplify the ratio of 6 to 2, which involves reducing it to its simplest form.
#### Answer
3:1
##### Answer for screen readers
The final answer is 3:1
#### Steps to Solve
1. Find the greatest common divisor (GCD) of the numbers
To simplify the ratio, we need to find the greatest common divisor (GCD) of 6 and 2.
$ext{GCD}(6, 2) = 2$
1. Divide both numbers by the GCD
Now divide both numbers in the ratio by their GCD.
$rac{6 , \text{(numerator)}}{2 , \text{(GCD)}} = 3$
$rac{2 , \text{(denominator)}}{2 , \text{(GCD)}} = 1$
1. Write the simplified ratio
The simplified ratio is 3:1.
The final answer is 3:1
#### More Information
Ratios are a way to compare two quantities. Simplifying ratios is similar to simplifying fractions.
#### Tips
A common mistake is to incorrectly find the GCD. Ensure you accurately determine the largest number that divides both quantities.
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# Show that if $f$ is an elementary map then structures are elementarily equivalent in extended language
Let $$\mathcal{A}, \mathcal{B}$$ be $$\mathcal{L}$$-structures, and let $$S\subseteq A$$. I want to show that $$f:S\to B$$ is an elementary mapping if and only if $$(\mathcal{A},a)_{a\in S}\equiv(\mathcal{B},f(a))_{a\in S}$$.
Attempt:
($$\Rightarrow$$) First suppose $$f$$ is elementary. Then pick a sentence, $$\phi$$ in $$\text{Th}((\mathcal{A},a)_{a\in S})$$. This is a sentence in $$\mathcal{L}$$ as well with parameters $$a_1,...,a_n$$ for each constant symbol in $$\phi$$ as an $$\mathcal{L}(S)$$ formula. Also $$\mathcal{A}\models\phi(a_1,...,a_n)$$. So $$\mathcal{B}\models\phi(fa_1,...,fa_n)$$ since $$f$$ is elementary, and so extending to an $$\mathcal{L}(f(S))$$-structure, we have $$\mathcal{B}\models\phi$$, and so $$\phi\in\text{Th}((\mathcal{B},f(a))_{a\in S})$$
For the flipside, when we pick a sentence in $$\text{Th}((\mathcal{B},f(a))_{a\in S})$$, we can just use the fact that $$f$$ is elementary and hence injective (such that it is invertible) to do the same proof in reverse to show that the same sentence is also in $$\text{Th}((\mathcal{A},a)_{a\in S})$$.
($$\Leftarrow$$) Here is where I get a bit confused. I want to show that $$\mathcal{A}\models\phi(a_1,...,a_n)$$ if and only if $$\mathcal{B}\models\phi(fa_1,...,fa_n)$$, given that the two structures in their respective extended languages are elementarily equivalent. So, I take an arbitrary $$\mathcal{L}$$-formula such that $$\mathcal{A}\models\phi(a_1,...,a_n)$$. This is really just a sentence. Extending the language to $$\mathcal{L}(S)$$ we have that this is actually a sentence in $$\text{Th}((\mathcal{A},a)_{a\in S})$$ so it is a sentence in $$\text{Th}((\mathcal{B},f(a))_{a\in S})$$. So going back down to $$\mathcal{L}$$ we have $$\mathcal{B}\models\phi(fa'_1,...,fa'_m)$$.
How can we get that the set $$a'_1,...,a'_m$$ is the same as $$a_1,...a_n$$, because it doesn't seem obvious to me that this must be the case.
• I don't see how these are different notions, even superficially. Doesn't $(\mathcal A,a)_{a\in S}\equiv (\mathcal B,f(a))_{a\in S}$ mean precisely that $\mathcal A\models \phi(a_1,\ldots, a_n)$ iff $\mathcal B\models \phi(f(a_1),\ldots f(a_n))$? In particular I don't understand this "going back down to $\mathcal L$" part and what $a_i'$ are. – spaceisdarkgreen Sep 22 '19 at 16:23
• @spaceisdarkgreen I think $(\mathcal{A},a)_{a\in S}\equiv(\mathcal{B},f(a))_{a\in S}$ means that the theory of $\mathcal{A}$ as an $\mathcal{L}(S)$-structure is equivalent to the theory of $\mathcal{B}$ as an $\mathcal{L}(f(S))$-structure. All we can conclude from this (as far as I know) is that any sentence in the former is in the latter and vice versa. – quanticbolt Sep 22 '19 at 16:32
• $\equiv$ usually means elementarily equivalent, and we can only talk about elementary equivalence between two theories if they have the same language. Here the language is $\mathcal L$ with a constant symbol $c_a$ added for each element of $S$ and $(\mathcal A,a)_{a\in S}$ interprets $c_a$ is $a$ and $(\mathcal B,f(a))_{a\in S}$ interprets $c_a$ as $f(a).$ At least that's how I interpret things. – spaceisdarkgreen Sep 22 '19 at 16:42
• @spaceisdarkgreen ah. Makes sense. Thanks! I was getting a bit confused by the notation. – quanticbolt Sep 22 '19 at 16:48
When you think of $$\phi(a_1,...,a_n)$$ as a sentence $$\psi$$ in the language $$\mathcal{L}(S)$$, the $$a_i$$'s are not just arbitrary constant symbols--they are specifically the constant symbols corresponding to the elements $$a_i$$ of $$S$$. So, when you then interpret the same sentence in the structure $$(\mathcal{B},f(a))_{a\in S}$$, these constant symbols get interpreted as the elements $$f(a_i)$$. Thus to say $$(\mathcal{B},f(a))_{a\in S}\models\psi$$ means exactly that $$\mathcal{B}\models\phi(f(a_1),...,f(a_n))$$. | HuggingFaceTB/finemath | |
+0
# Help Plz
0
204
1
What is the intermediate step in the form (x+a)^2=b(x+a)2=b as a result of completing the square for the following equation?
2x^2+42x+260=6x
You answer should look like this:
(___)^2=__
Feb 10, 2021
#1
+120127
+1
2x^2 + 42x + 260 = 6x
2x^2 + 42x - 6x + 260 = 0
2x^2 + 36x + 260 = 0 divide through by 2
x^2 + 18x + 130 = 0
take 1/2 of 18 = 9,,,,square it = 81....add it and subtract it
x^2 + 18x + 81 + 130 - 81 = 0 factor the first three terms, simplify the rest
(x + 9)^2 + 51 = 0
(x + 9)^2 = -51
Feb 10, 2021 | HuggingFaceTB/finemath | |
Kelly Criterion for Options GT
```1
Kelly’s Criterion for Option Investment
Ron Shonkwiler (shonkwiler@math.gatech.edu)
1
Kelly’s Criterion for Option Investment
Ron Shonkwiler (shonkwiler@math.gatech.edu)
Outline:
• Review Kelly’s Problem
• Application to Option Investing
• Option Investing with Catastrophic Loss
• Joint Investments, no Correlation
2
• Joint Investments, with Correlation
• The Downside to Kelly
An aside, recommended reading, “Fortune’s Formula”,
by William Poundstone.
3
Kelly’s Problem
Kelly’s Problem: we will make repeated bets on the same
positive expectation gamble, how much of our bankroll
should we risk each time?
From before the fraction to bet is:
expectation per unit bet
.
f=
gain per unit bet
or as some put it edge/odds.
4
Example
Say the gamble is to flip a fair coin, the payoff odds are
2:1. How much should we bet?
The expected or average payoff per play is
1
1
(\$2) − (\$1) = \$0.50.
2
2
Upon a win, for a \$1 bet we get our bet back and \$2
besides, so the gain per unit bet is 2; hence the Kelly
5
fraction is
or 1/4 our bankroll.
1/2
f=
= .25
2
6
Virtues of Kelly
• our investment is compounded so we get exponential
growth (or decay)
• our growth of capital is maximized
• we can not be wiped out – e.g. for an even money
60/40 investment, f = .2, upon a loss we still have
80% of our bankroll.
6
Virtues of Kelly
• our investment is compounded so we get exponential
growth (or decay)
• our growth of capital is maximized
• we can not be wiped out – e.g. for an even money
60/40 investment, f = .2, upon a loss we still have
80% of our bankroll.
Starting with \$20,000, after 11 losses in a row we still
have \$1,718, the probability of 11 losses in a row is
1/25,000.
7
Later we discuss the downside of the Kelly fraction.
8
Kelly applied to Option Investing
While stock investments are more free-form, many option
investments have common ground with gambles:
• fixed terms
• a definite time horizon
• a payoff settlement at expiration
Hence with the proper statistics, we can use the Kelly
criterion to determine optimal investment levels while
protecting against a string of reverses.
9
Gathering Statistics
We need to identify the characteristic features of a
specific option play we customarily make. Then record
the particulars and results of that play over many
implementations.
9
Gathering Statistics
We need to identify the characteristic features of a
specific option play we customarily make. Then record
the particulars and results of that play over many
implementations.
of the club’s February presentation by Steve Lenz and
Steve Papale of OptionVue.
10
11
The data we need is: over 75 trades
• number that gained money: 66, 9 lost money
• average gain per winning trade: \$659.12
• average loss per losing trade: \$1,799.06
I regard the average loss per losing trade as the “bet
size”.
12
We calculate the following needed for “edge over odds”
66
win prob.
= .88,
75
659.12
gain per unit bet
= 0.366.
1799.06
Hence expectation = (.366) ∗ .88 − (1) ∗ .12 = .202.
And so the Kelly risk fraction is
.202
= .552.
f=
.366
13
Accounting for Catastrophic Loss
But we can do more with the data. Note that we have
“maximum loss” information. This can be regarded as
catastrophic loss and taken into account.
Let p be the probability of a win, q the probability of a
loss, and r the probability of a catastrophic loss. Let γ
be the gain per unit bet and λ the size of the
catastrophic loss per unit bet.
We now rederive the Kelly fraction.
14
Kelly Fraction for Catastrophic Loss
The expectation is now
E = γp − q − λr.
The expected growth rate (from the previous talk) is
E(g) = p log(1 + γf ) + q log(1 − f ) + r log(1 − λf )
And the optimal fraction is the root of the quadratic
equation
0 = E − f (pγ(1 + λ) + q(γ − λ) + rλ(γ − 1)) + γλf 2
15
The new parameters are now
• prob. of a win p = .88
• prob. of an avg loss q =
8
75
= .1067
• prob. of a catastrophic loss r =
• gain per unit bet γ =
659.12
1669.32
• catas. loss per unit bet λ =
1
75
= .0133
= .395
2837
1669.32
= 1.70
16 | HuggingFaceTB/finemath | |
Home > Mean Square > Minimum Mean Square Error Estimate
# Minimum Mean Square Error Estimate
## Contents
ChenJing LiPei XiaoRead moreConference PaperOn the Error Propagation of the Fast Time Varying Channel Tracking for MIMO SystemsOctober 2016Bingpeng ZhouQ. Of course, no matter which algorithm (statistic-based or statistic-free one)we use, the unbiasedness and covariance are two important metrics for an estimator. ed.) Wiley, New York (1985) Löwner, 1934 K. One possibility is to abandon the full optimality requirements and seek a technique minimizing the MSE within a particular class of estimators, such as the class of linear estimators. have a peek here
We can model the sound received by each microphone as y 1 = a 1 x + z 1 y 2 = a 2 x + z 2 . {\displaystyle {\begin{aligned}y_{1}&=a_{1}x+z_{1}\\y_{2}&=a_{2}x+z_{2}.\end{aligned}}} Note that MSE can equivalently be defined in other ways, since t r { E { e e T } } = E { t r { e e T } As we have seen before, if $X$ and $Y$ are jointly normal random variables with parameters $\mu_X$, $\sigma^2_X$, $\mu_Y$, $\sigma^2_Y$, and $\rho$, then, given $Y=y$, $X$ is normally distributed with \begin{align}%\label{} Export You have selected 1 citation for export. https://en.wikipedia.org/wiki/Minimum_mean_square_error
## Minimum Mean Square Error Algorithm
Stahlecker, G. Proof: We can write \begin{align} W&=E[\tilde{X}|Y]\\ &=E[X-\hat{X}_M|Y]\\ &=E[X|Y]-E[\hat{X}_M|Y]\\ &=\hat{X}_M-E[\hat{X}_M|Y]\\ &=\hat{X}_M-\hat{X}_M=0. \end{align} The last line resulted because $\hat{X}_M$ is a function of $Y$, so $E[\hat{X}_M|Y]=\hat{X}_M$. Retrieved 8 January 2013. The remaining part is the variance in estimation error.
Trenkler Minimum mean square error ridge estimation Sankhyā, A46 (1984), pp. 94–101 Vinod, 1976 H.D. In general, our estimate $\hat{x}$ is a function of $y$: \begin{align} \hat{x}=g(y). \end{align} The error in our estimate is given by \begin{align} \tilde{X}&=X-\hat{x}\\ &=X-g(y). \end{align} Often, we are interested in the Since the matrix C Y {\displaystyle C_ − 0} is a symmetric positive definite matrix, W {\displaystyle W} can be solved twice as fast with the Cholesky decomposition, while for large Mmse Estimator Derivation Bellman Inequalities (2nd ed.) Springer, Berlin (1965) Chipman, 1964 J.S.
Baksalary, E.P. In the Bayesian setting, the term MMSE more specifically refers to estimation with quadratic cost function. the dimension of y {\displaystyle y} ) need not be at least as large as the number of unknowns, n, (i.e. https://www.probabilitycourse.com/chapter9/9_1_5_mean_squared_error_MSE.php Levinson recursion is a fast method when C Y {\displaystyle C_ σ 8} is also a Toeplitz matrix.
Hill, H. Minimum Mean Square Error Equalizer Liski ∗ University of Tampere, Tampere, Finland Opens overlay Helge Toutenburg University of München, München, Germany Opens overlay Götz Trenkler University of Dortmund, Dortmund, Germany Received 30 December 1991, Revised 27 Its final estimator and the associatedestimation precision are given by Eq. (19) and (20), respectively.4 Useful KnowledgeSome useful conclusions with respect to Gaussian distribution are summarized as follows.Lemma 1. Springer.
## Minimum Mean Square Error Estimation Matlab
In other words, the updating must be based on that part of the new data which is orthogonal to the old data. https://www.quora.com/Why-is-minimum-mean-square-error-estimator-the-conditional-expectation Moving on to your question. Minimum Mean Square Error Algorithm Generated Thu, 20 Oct 2016 14:49:03 GMT by s_nt6 (squid/3.5.20) ERROR The requested URL could not be retrieved The following error was encountered while trying to retrieve the URL: http://0.0.0.10/ Connection Minimum Mean Square Error Matlab In such case, the MMSE estimator is given by the posterior mean of the parameter to be estimated.
For more information, visit the cookies page.Copyright © 2016 Elsevier B.V. navigate here ChenPei XiaoRead moreDiscover moreData provided are for informational purposes only. More succinctly put, the cross-correlation between the minimum estimation error x ^ M M S E − x {\displaystyle {\hat − 2}_{\mathrm − 1 }-x} and the estimator x ^ {\displaystyle Your cache administrator is webmaster. Minimum Mean Square Error Pdf
Similarly, let the noise at each microphone be z 1 {\displaystyle z_{1}} and z 2 {\displaystyle z_{2}} , each with zero mean and variances σ Z 1 2 {\displaystyle \sigma _{Z_{1}}^{2}} Obenchain Ridge analysis following a preliminary test of the shrunken hypothesis Technometrics, 17 (1975), pp. 431–445 Obenchain, 1978 R.L. This can be seen as the first order Taylor approximation of E { x | y } {\displaystyle \mathrm − 8 \ − 7} . Check This Out Let x {\displaystyle x} denote the sound produced by the musician, which is a random variable with zero mean and variance σ X 2 . {\displaystyle \sigma _{X}^{2}.} How should the
Wikipedia® is a registered trademark of the Wikimedia Foundation, Inc., a non-profit organization. Mean Square Estimation Am. Depending on context it will be clear if 1 {\displaystyle 1} represents a scalar or a vector.
## Meth., A6 (1977), pp. 73–79 Obenchain, 1975 R.L.
Detection, Estimation, and Modulation Theory, Part I. Forexample, if we have known the system measurement is linear and the measurement noiseis a zero-mean Gaussian variable, i.e., z = Ax + n, where the linear coefficient matrixA ∈ RS×Dis pp.344–350. Minimum Mean Square Error Estimation Ppt At first the MMSE estimator is derived within the set of all those linear estimators of β which are at least as good as a given estimator with respect to dispersion
Suppose an optimal estimate x ^ 1 {\displaystyle {\hat − 0}_ ¯ 9} has been formed on the basis of past measurements and that error covariance matrix is C e 1 Cohen All admissible linear estimates of the mean vector Ann. Lastly, this technique can handle cases where the noise is correlated. http://streamlinecpus.com/mean-square/minimum-mean-square-error-equalizer.php A shorter, non-numerical example can be found in orthogonality principle.
So although it may be convenient to assume that x {\displaystyle x} and y {\displaystyle y} are jointly Gaussian, it is not necessary to make this assumption, so long as the Instead the observations are made in a sequence. We can model our uncertainty of x {\displaystyle x} by an aprior uniform distribution over an interval [ − x 0 , x 0 ] {\displaystyle [-x_{0},x_{0}]} , and thus x Prentice Hall.
Let $a$ be our estimate of $X$. In other words, for $\hat{X}_M=E[X|Y]$, the estimation error, $\tilde{X}$, is a zero-mean random variable \begin{align} E[\tilde{X}]=EX-E[\hat{X}_M]=0. \end{align} Before going any further, let us state and prove a useful lemma. Statist. Marquardt Generalized inverses, ridge regression, biased linear estimation, and nonlinear estimation Technometrics, 12 (1970), pp. 591–612 Massy, 1965 W.F. | HuggingFaceTB/finemath | |
/************************************************************ * This program calculates the determinant of a real square * * matrix using a recursive function Deter based on Kramer's * * formula. * * --------------------------------------------------------- * * Method: * * * * Let us take the example: * * n=5 * * Input matrix is: ( 3 2 1 6 3 ) * * ( 6 4 4 0 3 ) * * ( 4 0 1 0 2 ) * * ( 3 4 2 4 2 ) * * ( 4 6 1 5 2 ) * * * * According to Kramer's rule, the determinant det(A) is * * calculated by the following: * * * * (4 4 0 3) (6 4 0 3) (6 4 0 3) * * det = 3 * det(0 1 0 2) -2 * det(4 1 0 2) + 1 * (4 0 0 2) * * (4 2 4 2) (3 2 4 2) (3 4 4 2) * * (6 1 5 2) (4 1 5 2) (4 6 5 2) * * * * (6 4 4 3) (6 4 4 0) * * -6 * det(4 0 1 2) +3 * det(4 0 1 0) * * (3 4 2 2) (3 4 2 4) * * (4 6 1 2) (4 6 1 5) * * * * As can be seen, 3, -2, 1, -6 and 3 are the elements of * * the 1st row alternatively multiplied by 1 or -1 and the * * the submatrices are obtained by successively eliminating * * row=1 and column=1, then row=1 and col=2, etc. Then you * * can recursively calculate the detrminants of these sub- * * matrices by using the same Function Deter with size n-1 * * instead of n, and so forth... * * When n=2, the result is immediate by the formula: * * A(1,1)*A(2,2)-A(2,1)*A(1,2). * * * * Program organization: * * After input of given square matrix A, here of size n=5, * * we call the function Deter(n,A) which returns the value * * of determinant. This function calls itself recursively * * with n decreasing by one at each step, until n=2. * * Auxiliary function is pow(n) returning -(-1)^n and auxi- * * liary procedure is Submatrix(n,A,B) that extracts sub- * * matrix B from A by eliminating row 1 and column col. Note * * that B is dynamically allocated and disposed of at each * * step to spare memory. * * * * SAMPLE RUN: * * * * Size of matrix: 5 * * Input matrix : * * 3.0000 2.0000 1.0000 6.0000 3.0000 * * 6.0000 4.0000 4.0000 0.0000 3.0000 * * 4.0000 0.0000 1.0000 0.0000 2.0000 * * 3.0000 4.0000 2.0000 4.0000 2.0000 * * 4.0000 6.0000 1.0000 5.0000 2.0000 * * * * Determinant = -7.00000000000000E+0001 * * * * --------------------------------------------------------- * * * * C++ release by Jean-Pierre Moreau, Paris. * * (www.jpmoreau.fr) * ************************************************************* Note: do not use for n>10 */ #include #include #include #include #define SIZE 11 int n; // size of matrix REAL **A; // pointer to matrix SIZExSIZE REAL det; // value of determinant FILE *fp; // input.file void *vmblock = NULL; int i, j; REAL temp; // return -(-1)^n int pow(int n) { if (n % 2 == 0) return 1; else return -1; } //return submatrix without 1st line and colth column of A void Submatrix(int n, int col, REAL **A, REAL **B) { int i,j; boolean flag; if (n<3) return; for (i=1; i2) { Submatrix(n,i,A,B); //recursive call of Deter() d += pow(i)*A[1][i]*Deter(n-1,B); } vmfree(vmblock1); } return d; } } void main() { // read size n of matrix A from input file fp = fopen("mat10.dat","r"); fscanf(fp, "%d", &n); // allocate memory for matrix A of size n x n vmblock = vminit(); A = (REAL **) vmalloc(vmblock, MATRIX, n+1, n+1); if (! vmcomplete(vmblock)) LogError ("No Memory", 0, __FILE__, __LINE__); // read matrix A from input file for (i=1; i<=n; i++) for (j=1; j<=n; j++) { fscanf(fp, "%lf", &temp); A[i][j] = temp; } fclose(fp); printf("\n Size of matrix: %d\n", n); printf(" Input matrix:\n"); Matprint(n, A); printf(" Computing determinant...\n"); det = Deter(n,A); printf("\n Determinant = %f\n\n", det); vmfree(vmblock); } // end of file deter2a.cpp | HuggingFaceTB/finemath | |
# Difference between revisions of "Simon's Favorite Factoring Trick"
## Contents
Dr. Simon's Favorite Factoring Trick (abbreviated SFFT) is a special factorization first popularized by AoPS user Simon Rubinstein-Salzedo.
## The General Statement
The general statement of SFFT is: ${xy}+{xk}+{yj}+{jk}=(x+j)(y+k)$. Two special common cases are: $xy + x + y + 1 = (x+1)(y+1)$ and $xy - x - y +1 = (x-1)(y-1)$.
The act of adding ${jk}$ to ${xy}+{xk}+{yj}$ in order to be able to factor it could be called "completing the rectangle" in analogy to the more familiar "completing the square."
## Applications
This factorization frequently shows up on contest problems, especially those heavy on algebraic manipulation. Usually $x$ and $y$ are variables and $j,k$ are known constants. Also, it is typically necessary to add the $jk$ term to both sides to perform the factorization.
## Awesome Practice Problems
### Introductory
• Two different prime numbers between $4$ and $18$ are chosen. When their sum is subtracted from their product, which of the following numbers could be obtained?
$\mathrm{(A) \ 22 } \qquad \mathrm{(B) \ 60 } \qquad \mathrm{(C) \ 119 } \qquad \mathrm{(D) \ 180 } \qquad \mathrm{(E) \ 231 }$
(Source)
### Intermediate
• $m, n$ are integers such that $m^2 + 3m^2n^2 = 30n^2 + 517$. Find $3m^2n^2$.
(Source)
• The integer $N$ is positive. There are exactly 2005 ordered pairs $(x, y)$ of positive integers satisfying:
$$\frac 1x +\frac 1y = \frac 1N$$
Prove that $N$ is a perfect square. (British Mathematical Olympiad Round 2, 2005) | HuggingFaceTB/finemath | |
# Is A Vertical Line A Relation?
## What is the vertical line test?
In mathematics, the vertical line test is a visual way to determine if a curve is a graph of a function or not.
A function can only have one output, y, for each unique input, x..
## What causes vertical lines on a TV?
This is a common issue with many TVs. Vertical colored lines usually show on a TV screen when the T-Con board is not working properly. Many times this can simply be caused from wiring that is not securely fastened. Other times the T-Con Board itself may be faulty and need to be replaced.
## How do you type a vertical line?
“|”, How can I type it?Shift-\ (“backslash”).German keyboard it is on the left together with < and > and the Alt Gr modifier key must be pressed to get the pipe.Note that depending on the font used, this vertical bar can be displayed as a consecutive line or by a line with a small gap in the middle.More items…
## Is a circle a function?
A circle is a set of points in the plane. A function is a mapping from one set to another, so they’re completely different kinds of things, and a circle cannot be a function.
## What is a vertical line on a graph called?
Ordinate – The vertical line, or y-axis, of a graph.
## How is horizontal line drawn?
The horizontal line is drawn by connecting similar swing lows in price to create a horizontal support line. For a horizontal resistance line, similar swing highs are connected. … In more simple terms, a horizontal line on any chart is where the y-axis values are equal.
## What is a vertical line?
: a line perpendicular to a surface or to another line considered as a base: such as. a : a line perpendicular to the horizon. b : a line parallel to the sides of a page or sheet as distinguished from a horizontal line.
## Is vertical line a function?
For a relation to be a function, use the Vertical Line Test: Draw a vertical line anywhere on the graph, and if it never hits the graph more than once, it is a function. If your vertical line hits twice or more, it’s not a function.
## Is a circle a function vertical line test?
Even though a vertical line through (3,0) or (-3,0) would intersect the circle only once, the Vertical Line Test has to work for every vertical line drawn through the graph. This graph fails the Vertical Line Test, so a circle is not a function.
## What is the equation of a straight vertical line?
The equation of a vertical line always takes the form x = k, where k is any number and k is also the x-intercept . (link) For instance in the graph below, the vertical line has the equation x = 2 As you can see in the picture below, the line goes straight up and down at x = 2.
## What is the difference between horizontal and vertical?
A vertical line is any line parallel to the vertical direction. A horizontal line is any line normal to a vertical line. Horizontal lines do not cross each other. Vertical lines do not cross each other.
## What is the horizontal line test used for?
In mathematics, the horizontal line test is a test used to determine whether a function is injective (i.e., one-to-one).
## Is a vertical line an example of a functional relationship?
A vertical line is an example of a functional relationship. A horizontal line is an example of a functional relationship. Each output value of a function can correspond to only one input value.
## What does a vertical line represent?
Vertical lines are perpendicular to the horizon. … Vertical lines are strong and rigid. They can suggest stability, especially when thicker. Vertical lines accentuate height and convey a lack of movement, which is usually seen as horizontal.
## What mood do vertical lines create?
In web design vertical lines tend to represent or create length and indicate to the user that there is more content below the fold. Horizontal lines generally create relaxation or a calming mood, they tend to be quiet and subtle whereas vertical lines are more imposing and powerful.
## How do you tell if a graph represents a function?
Use the vertical line test to determine whether or not a graph represents a function. If a vertical line is moved across the graph and, at any time, touches the graph at only one point, then the graph is a function. If the vertical line touches the graph at more than one point, then the graph is not a function. | HuggingFaceTB/finemath | |
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Checkout JEE MAINS 2022 Question Paper Analysis : Checkout JEE MAINS 2022 Question Paper Analysis :
# Table of 207
The Multiplication Table for 207 is nothing more than the repetitive addition of the number 207. Observe the pattern in the last two digits till the multiple of 10.The numbers are from the times table of 7. And the first 2 digits are from the tables of 2. Use the Multiplication Calculator, for more multiples of 207.
## Table of 207 Chart ## What is the 207 Times Table?
Any multiplication table is a repetitive addition. In the table of 207, we are repeatedly adding 207 multiple times. And the table below gives repetitive addition 10 times for the number 207.
207 × 1 = 207 207 207 × 2 = 414 207 + 207 = 414 207 × 3 = 621 207 + 207 + 207 = 621 207 × 4 = 828 207 + 207 + 207 + 207 = 828 207 × 5 = 1035 207 + 207 + 207 + 207 + 207 = 1035 207 × 6 = 1242 207 + 207 + 207 + 207 + 207 + 207 = 1242 207 × 7 = 1449 207 + 207 + 207 + 207 + 207 + 207 + 207 = 1449 207 × 8 = 1656 207 + 207 + 207 + 207 + 207 + 207 + 207 + 207 = 1656 207 × 9 = 1863 207 + 207 + 207 + 207 + 207 + 207 + 207 + 207 + 207 = 1863 207 × 10 = 2070 207 + 207 + 207 + 207 + 207 + 207 + 207 + 207 + 207 + 207 = 2070
## Multiplication Table of 207
Given below is the multiplication table of 207 being done 20 times. The below table helps you with multiplication problems that have 207.
207 × 1 = 207 207 × 2 = 414 207 × 3 = 621 207 × 4 = 828 207 × 5 = 1035 207 × 6 = 1242 207 × 7 = 1449 207 × 8 = 1656 207 × 9 = 1863 207 × 10 = 2070 207 × 11 = 2277 207 × 12 = 2484 207 × 13 = 2691 207 × 14 = 2898 207 × 15 = 3105 207 × 16 = 3312 207 × 17 = 3519 207 × 18 = 3726 207 × 19 = 3933 207 × 20 = 4140
## Solved Example on Table of 207
Example:
Evaluate 207 x 22 ?
Solution: 4554.
The above problem can be written as 207 x 20 + 207 x 2
= 4140 + 414 (Refer the table above)
= 4554.
## Frequently Asked Questions on Table of 207
### Question 1: Are 2 and 7 factors for 207?
Answer: No. 1, 3, 9, 23, 69 and 207 are factors for 207.
### Question 2: When do we get 3933 in the table of 207?
Answer: 19 times 207, we get 3933.
### Question 3: What is 18 times 207?
Answer: When 207 is multiplied by 18, we get 3726. | HuggingFaceTB/finemath | |
2 Closed points
Suppose $\dim X>0$ and $k$ is algebraically closed and uncountable. Moreover, if a "variety" is not necessarily irreducible, the $Z_i$ are supposed to have positive codimension in $X$ (otherwise one could take the irreducible components of $X$).
As in MP's answer, one can suppose $X$ is affine. By Noether's Normalization Lemma, there exists a finite surjective morphism $p: X\to \mathbb A^m_k$ with $m=\dim X$. Let $Y_i=p(Z_i)$. This is a closed subset of $\mathbb A^m_k$ of positive codimension. Moreover $\mathbb A^m_k(k)=\cup Y_i(k)$ because $k$ is algebraically closed (which implies that $Y_i(k)=p(Z_i(k))$). As $k$ is uncountable, there exists a hyperplane $H$ in $\mathbb A^m$ not contained in any $Y_i$ (note that $H\subseteq Y_i$ is equivalent to $H=Y_i$). So by induction on $m$ we are reduced to the case $m=1$, and the assertion is obvious.
Without the hypothesis $k$ algebraically closed, one can show similarly that $X\ne \cup_i Z_i$. This is Exercise 2.5.10 in my book. EDIT In fact this statement is trivial because the generic points of $X$ don't belong to any of the $Z_i$'s. But the proof shows that the set of closed points of $X$ is not contained in $\cup_i Z_i$.
1
Suppose $\dim X>0$ and $k$ is algebraically closed and uncountable. Moreover, if a "variety" is not necessarily irreducible, the $Z_i$ are supposed to have positive codimension in $X$ (otherwise one could take the irreducible components of $X$).
As in MP's answer, one can suppose $X$ is affine. By Noether's Normalization Lemma, there exists a finite surjective morphism $p: X\to \mathbb A^m_k$ with $m=\dim X$. Let $Y_i=p(Z_i)$. This is a closed subset of $\mathbb A^m_k$ of positive codimension. Moreover $\mathbb A^m_k(k)=\cup Y_i(k)$ because $k$ is algebraically closed (which implies that $Y_i(k)=p(Z_i(k))$). As $k$ is uncountable, there exists a hyperplane $H$ in $\mathbb A^m$ not contained in any $Y_i$ (note that $H\subseteq Y_i$ is equivalent to $H=Y_i$). So by induction on $m$ we are reduced to the case $m=1$, and the assertion is obvious.
Without the hypothesis $k$ algebraically closed, one can show similarly that $X\ne \cup_i Z_i$. This is Exercise 2.5.10 in my book. | open-web-math/open-web-math | |
Necessary and sufficeint Questions : GMAT Critical Reasoning (CR)
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# Necessary and sufficeint Questions
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Hello all
I am initiating this discussion to clear my doubt on Necessary and Sufficient questions.
Sufficient condition : This is the part of conditional statement that indicates the occurance of necessary condition.
Necessary Condition : This condition is a part of the condition statement the occurance of it is required for the sufficient condition.
Typical example:-
If you get an A+, then you must have studied
Sufficient Condition : Get an A+
Necessary condition: You must have studied.
Another example
People who drive fast are dangerous
Sufficient condition: Drive fast
Necessary condition: Dangerous
My doubt is should we consider parallel thinking in such question i.e. Should we think that A dangerous person can have several traits and driving fast is one of them.
Or When we are considering Necessary and Sufficient condition we must stick to the fact that there exists only one condition as stated in the statement.
Is it correct to say " People who are dangerous, its not necessary that they drive fast"
Justification: There may be many people who are veteran criminal but while driving drive slowly because they do not want accident.
My Query is Should we restrict ourselves to the two conditions itself when answering such question or think logically as we do on all other CR questions. If we follow the logical procedure then the the mentioned statement with justification should be correct. But if we just restrict ourselves to two events than we can write it as " People who are not dangerous, they do not drive fast" i.e People whoare dangerous must drive fast...Its illogical in real world.
PLs help me to clear my doubts.
If you have any questions
New!
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Re: Necessary and sufficeint Questions [#permalink]
### Show Tags
29 Dec 2012, 18:40
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Expert's post
Archit143 wrote:
Hello all
I am initiating this discussion to clear my doubt on Necessary and Sufficient questions.
Sufficient condition : This is the part of conditional statement that indicates the occurance of necessary condition.
Necessary Condition : This condition is a part of the condition statement the occurance of it is required for the sufficient condition.
Typical example:-
If you get an A+, then you must have studied
Sufficient Condition : Get an A+
Necessary condition: You must have studied.
Another example
People who drive fast are dangerous
Sufficient condition: Drive fast
Necessary condition: Dangerous
My doubt is should we consider parallel thinking in such question i.e. Should we think that A dangerous person can have several traits and driving fast is one of them.
Or When we are considering Necessary and Sufficient condition we must stick to the fact that there exists only one condition as stated in the statement.
Is it correct to say " People who are dangerous, its not necessary that they drive fast"
Justification: There may be many people who are veteran criminal but while driving drive slowly because they do not want accident.
My Query is Should we restrict ourselves to the two conditions itself when answering such question or think logically as we do on all other CR questions. If we follow the logical procedure then the the mentioned statement with justification should be correct. But if we just restrict ourselves to two events than we can write it as " People who are not dangerous, they do not drive fast" i.e People whoare dangerous must drive fast...Its illogical in real world.
PLs help me to clear my doubts.
We discuss necessary and sufficient conditions in detail in our book. It is an extremely important concept. There is a diagram that looks like this:
Necessary 1 ----->
Necessary 2------>
Necessary 3------> Sufficient Condition
...
All necessary conditions together lead to the sufficient condition. If the sufficient condition takes place, it means all necessary conditions took place. If one necessary condition takes place, we dont know whether the sufficient condition took place or not since we dont know whether the other necessary conditions took place. If even one necessary condition did not take place we can easily say that the sufficient condition did not take place.
People who drive fast are dangerous
Sufficient condition: Drive fast
Necessary condition: Dangerous
This means you need to be dangerous to be able to drive fast. To 'drive fast', there are various necessary conditions. 'Being dangerous' is one of them.
Quote: "Should we think that A dangerous person can have several traits and driving fast is one of them."
No, you should think that people who drive fast have several traits. Being dangerous is one of them. The sufficient conditions could need many necessary conditions. Not the other way around.
Quote: "Is it correct to say " People who are dangerous, its not necessary that they drive fast"
Yes it is. Being dangerous is a necessary condition. We don't know whether it alone leads to the sufficient condition of driving fast.
Check out my recent posts on my blog (link is in my signature). They tackle necessary and sufficient conditions.
_________________
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Veritas Prep | GMAT Instructor
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Get started with Veritas Prep GMAT On Demand for $199 Veritas Prep Reviews VP Status: Final Lap Up!!! Affiliations: NYK Line Joined: 21 Sep 2012 Posts: 1096 Location: India GMAT 1: 410 Q35 V11 GMAT 2: 530 Q44 V20 GMAT 3: 630 Q45 V31 GPA: 3.84 WE: Engineering (Transportation) Followers: 37 Kudos [?]: 527 [0], given: 70 Re: Necessary and sufficeint Questions [#permalink] ### Show Tags 29 Dec 2012, 18:49 HI Karishma That means If just necessary condition is mentioned for an action to be sufficient we should not assume that there are many others, If it is not explicitly mentioned in the argument. Necessary1--> Sufficient There may be many other but we ar oncerned her only about necessary 1 Necessary1, Necessary2, Necessary3------------> Sufficient All 3 are needed for the condition to be sufficient.Other than these we are not concerned about any.... regards Archit Veritas Prep GMAT Instructor Joined: 16 Oct 2010 Posts: 7125 Location: Pune, India Followers: 2136 Kudos [?]: 13655 [0], given: 222 Re: Necessary and sufficeint Questions [#permalink] ### Show Tags 29 Dec 2012, 19:01 Archit143 wrote: HI Karishma That means If just necessary condition is mentioned for an action to be sufficient we should not assume that there are many others, If it is not explicitly mentioned in the argument. Necessary1--> Sufficient There may be many other but we ar oncerned her only about necessary 1 Necessary1, Necessary2, Necessary3------------> Sufficient All 3 are needed for the condition to be sufficient.Other than these we are not concerned about any.... regards Archit If a necessary condition is mentioned for a sufficient condition, we know that there is at least this condition necessary. There could be others too, we should not forget that. This necessary condition alone may not be enough to make the sufficient condition take place. Remember: Necessary1, Necessary2, Necessary3------------> Sufficient All 3 are required to make the sufficient happen. But if the sufficient happens, we know all three necessary conditions took place. _________________ Karishma Veritas Prep | GMAT Instructor My Blog Get started with Veritas Prep GMAT On Demand for$199
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Re: Necessary and sufficeint Questions [#permalink]
### Show Tags
08 May 2016, 01:54
HI,
How to differentiate between sufficient and necessary questions ? Are there any keywords that differentiate the two ?
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Re: Necessary and sufficeint Questions [#permalink]
### Show Tags
08 May 2016, 21:45
SecondTymRockStar wrote:
HI,
How to differentiate between sufficient and necessary questions ? Are there any keywords that differentiate the two ?
http://www.veritasprep.com/blog/2012/11 ... tatements/
http://www.veritasprep.com/blog/2012/12 ... tatements/
http://www.veritasprep.com/blog/2012/12 ... onditions/
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Re: Necessary and sufficeint Questions [#permalink] 08 May 2016, 21:45
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Display posts from previous: Sort by | HuggingFaceTB/finemath | |
# Qnt 561 Week 1 Individual Assignment
983 Words Oct 22nd, 2011 4 Pages
QNT/561: Week One Assignment
Exercises 80, 82, and 87 (Ch. 3)
Exercise 80* a. The times are a population because we are considering the wait times for all of the customers seated on Saturday night. b. To find the mean: µ = ∑ X N µ = 1021 25 µ = 40.84
To find the median:
The midpoint value of the population is 39.
c. To find the range:
Range = Largest Value – Smallest Value
Range = 67 – 23
Range = 44
To find the standard deviation: σ = √∑ (X - µ)2 N σ = √∑(X – 40.84) 2 = √5291.36 = √211.65 = 14.55 25 25
Exercise 82* a. To find the mean cost:
X = ∑fM = 7,060 = \$141.20 N 50 b. To find the standard deviation: s = √∑f (M - X)2 = √33,728 = √688.33 = 26.24 n –
…show more content…
When X = 29, z = -1.5
When X = 34, z = 1
P(-1.5 < z < 1) = 0.4332 + 0.3413 = 0.7745 = 77.45%
c. z = X - µ σ z = 28.7 – 32 = -3.3 = -1.65 2 2
P( z < -1.65) = .0495 = 4.95%
d. .5 - .05 = .45
P(z < 1.645) = .05 = 5% z = X - µ σ 1.645 = X – 32 2 3.29 = X – 32 X = 3.29 + 32 = 35.29 hours
Exercise 45 z = X - µ σ µ = \$1,280 σ = \$420
a. When X = \$1,500 z = 1500 – 1280 = 220 = .52381 420 420 P(z > .52381) = .300206 = 30.02%
b. When X = \$1,500, z = .52381
When X = \$2,000 z = 2000 – 1280 = 720 = 1.7143 420 420
P(.52381 > z > 1.7143) = .557173 - .300206 = .256967 = 25.7%
c. When X = \$0 z = 0 – 1280 = -1280 = -3.04762 420 420 P(z < -3.04762) = .001153 = .12%
d. .5 - .1 = .4000
P(z < 1.281) = .10 = 10% z = X - µ σ 1.281 = X – 1280 420 538.02 = X – 1280 X = 538.02 + 1280 = | HuggingFaceTB/finemath | |
# Evaluate the integral:
Question:
Evaluate the integral:
Solution:
Key points to solve the problem:
- Such problems require the use of method of substitution along with method of integration by parts. By method of integration by parts if we have $\int f(x) g(x) d x=f(x) \int g(x) d x-\int f^{\prime}(x)\left(\int g(x) d x\right) d x$
- To solve the integrals of the form: $\int \sqrt{a x^{2}+b x+c} d x$ after applying substitution and integration by parts we have direct formulae as described below:
$\int \sqrt{a^{2}-x^{2}} d x=\frac{x}{2} \sqrt{a^{2}-x^{2}}+\frac{a^{2}}{2} \sin ^{-1}\left(\frac{x}{a}\right)+C$
$\int \sqrt{\mathrm{x}^{2}-\mathrm{a}^{2}} \mathrm{dx}=\frac{\mathrm{x}}{2} \sqrt{\mathrm{x}^{2}-\mathrm{a}^{2}}-\frac{\mathrm{a}^{2}}{2} \log \left|\mathrm{x}+\sqrt{\mathrm{x}^{2}-\mathrm{a}^{2}}\right|+\mathrm{C}$
$\int \sqrt{\mathrm{x}^{2}+\mathrm{a}^{2}} \mathrm{dx}=\frac{\mathrm{x}}{2} \sqrt{\mathrm{x}^{2}+\mathrm{a}^{2}}+\frac{\mathrm{a}^{2}}{2} \log \left|\mathrm{x}+\sqrt{\mathrm{x}^{2}+\mathrm{a}^{2}}\right|+\mathrm{C}$
Let, $I=\int \sqrt{4 x^{2}-5} d x$
We have:
$I=\int \sqrt{4 x^{2}-5} d x=\int 2 \sqrt{x^{2}-\frac{5}{4}} d x$
$\Rightarrow I=2 \int \sqrt{x^{2}-\left(\frac{\sqrt{5}}{2}\right)^{2}} d x$
As I match with the form:
$\int \sqrt{x^{2}-a^{2}} d x=\frac{x}{2} \sqrt{x^{2}-a^{2}}-\frac{a^{2}}{2} \log \left|x+\sqrt{x^{2}-a^{2}}\right|+C$
$\therefore I=2\left\{\frac{x}{2} \sqrt{x^{2}}-\left(\frac{\sqrt{5}}{2}\right)^{2}-\frac{\frac{5}{4}}{2} \log \left|x+\sqrt{x^{2}-\left(\frac{\sqrt{5}}{2}\right)^{2}}\right|\right\}$
$\Rightarrow I=x \sqrt{x^{2}-\frac{5}{4}}-\frac{5}{4} \log \left|x+\sqrt{x^{2}-\frac{5}{4}}\right|+C$ | HuggingFaceTB/finemath | |
# Matrix Calculus - Transpose of Vector Derivatives
• Oct 15th 2010, 07:44 AM
danibitt59
Matrix Calculus - Transpose of Vector Derivatives
Hi all,
Here's a newbie question for you:
What are the derivatives of the following -->
Attachment 19337
I guess some of these results will be transposed and othes not..? And some will be vectors or matrices?
Cheers,
Daniel BS.
ps: couldn't insert my equations within [tex] tags anyhow!
Got a message of a 25 chars limit! Is a there a way through?
• Oct 15th 2010, 07:46 AM
danibitt59
About the tag thing, nervermind, it was the keywords tag not the math tag... anyways... main question persists.. =)
• Oct 16th 2010, 05:10 AM
HallsofIvy
To avoid the character limit, break your LaTex into parts.
$\displaystyle \frac{\partial x}{\partial x}$, $\displaystyle \frac{\partial x^T}{\partial x}$, $\displaystyle \frac{\partial x}{\partial x^T}$, $\displaystyle \frac{\partial x^T}{\partial x^T}$.
Now what definition of differentiation by a vector are you using?
• Oct 16th 2010, 10:55 AM
danibitt59
Well, by this equation (D-3):
if y = x, then
$\displaystyle \frac{\partial \mathbf{x}}{\partial \mathbf{x}}=I$
but I know that it should be 1, right? So, I differs from 1? I know they both share the similar property of being neutral elements, but they're still distinct results, right?
Furthermore, what about the transposed ones? What are they equal of?
Regards
Daniel.
• Oct 17th 2010, 05:48 AM
HallsofIvy
That was why I asked, before, "what definition of differentiation by a vector are you using?"- which you have still not answered. There are several different ways of defining that derivative. The most general defines the derivative of one vector by another to be a linear transfomation that best approximates the vector function. If that is the definition you are using then "I" is the identity linear transformation: I(v)= v. Another definition gives the derivative of a vector, u, by a vector, v, as the matrix having the partial derivatives of each component of vector u, with respect to vector v's components, as rows. In that case "I" is the identity matrix.
You obviously can't do this problem unless you know which definition you are using! | HuggingFaceTB/finemath | |
# Physics 1 Dynamics Experiment How Does A Spring Scale Work? Hooke's Law
## Purpose:
1. To investigate Hooke's Law (The relation between force and stretch for a spring)
2. To investigate Newton's Laws and the operation of a spring scale.
## Discussion:
Everybody knows that when you apply a force to a spring or a rubber band, it stretches. A scientist would ask, "How is the force that you apply related to the amount of stretch?" This question was answered by Robert Hooke, a contemporary of Newton, and the answer has come to be called Hooke's Law.
Hooke's Law, believe it or not, is a very important and widely-used law in physics and engineering. Its applications go far beyond springs and rubber bands.
You can investigate Hooke's Law by measuring how much known forces stretch a spring. A convenient way to apply a precisely-known force is to let the weight of a known mass be the force used to stretch the spring. The force can be calculated from W = mg. The stretch of the spring can be measured by noting the position of the end of the spring before and during the application of the force.
## Equipment:
spring rubber band ring stand ring-stand clamps c-clamp spring scale ruler or meter stick set of known masses
## Safety Notes:
• Be sure to keep your feet out of the area in which the masses will fall if the spring or rubber band breaks!
• Be sure to clamp the ring stand to the lab table, or weight it with several books so that the mass does not pull it off the table.
• You need to hang enough mass to the end of the spring to get a measurable stretch, but too much force will permanently damage the spring. (An engineer would say that it has exceeded its "elastic limit"). "You break it, you bought it."
## Procedure:
1. Assemble the apparatus as shown in the diagram at right. Be sure to clamp the ring stand to the lab table, or weight it with several books.
2. Construct a data table. You will need to record the mass that you hang from the spring and the position of the end of the spring before and after the mass is added. From this, you will calculate the force applied to the spring and the resulting stretch of the spring. You should allow for at least 8-10 trials. (A sample data table is shown below.)
3. For each trial, record the mass, the starting position of the spring (before hanging the mass) and the ending position of the spring (while it is being stretched).
4. Repeat the process for a rubber band.
## Results:
1. Calculate the force applied to the spring/rubber band in each trial (W = mg) Use g = 9.8 m/s2.
2. Calculate the stretch of the spring/rubber band in each trial (the difference in the starting and ending positions).
3. Draw graphs of force versus stretch for the spring and the rubber band. You may be able to put both graphs on the same sheet of graph paper, depending on the data.
## Conclusions:
Hooke's Law says that the stretch of a spring is directly proportional to the applied force. (Engineers say "Stress is proportional to strain".) In symbols, F = kx, where F is the force, x is the stretch, and k is a constant of proportionality. If Hooke's Law is correct, then, the graph of force versus stretch will be a straight line.
## Questions:
Examine a spring scale. It is a simple device that measures force by measuring the amount that the force stretches a spring.
1. Hang an object (at rest) from the spring scale. Draw a set of diagrams that shows all of the forces that act:
1. on the object
2. on the spring scale
2. What is the net force on:
1. the object?
2. the spring scale?
3. Are the forces that act on the object equal and opposite? Are they a Newton's Third Law force pair?
4. Are the forces that act on the spring scale equal and opposite? Are they a Newton's Third Law force pair?
5. For each force in your diagrams (question 1), indicate its Newton's Third Law "partner". Be sure to indicate (a) what the force pushes or pulls on, and (b) its direction. | HuggingFaceTB/finemath |
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