Topic : Simplification
Problem : Simplify (c + 9)(c³ - 3c² - 7c)
Solution :
(c + 9)(c³ - 3c² - 7c)
= c^4 - 3c³ - 7c² + 9c³ - 27c² - 63c
= c^4 + 6c³ - 34c² - 63c
Sunday, April 5, 2009
Tuesday, March 31, 2009
Solving quadratic equation
Problem on how to solve a quadratic equation:
Know more about quadratic equations http://en.wikipedia.org/wiki/Quadratic_equation
Question:
(x + 2) / (x - 3) = 42 (x + 7)
Solution:
(x - 5) (x - 28) = 0
x -5 = 0
OR
x - 28 = 0
Know more about quadratic equations http://en.wikipedia.org/wiki/Quadratic_equation
Question:
(x + 2) / (x - 3) = 42 (x + 7)
Solution:
(x +2 ) (x + 7) = 42 (x - 3)
x² + 7x + 2x + 14 = 42x - 126
x² + 9x + 14 = 42x - 126
x² + 9x + 14 - 42x + 126 = 0
x² - 33x + 140 = 0
x² - 28x - 5x + 140 = 0
Taking out common factors
x(x - 28) -5(x - 28) = 0
(x - 5) (x - 28) = 0
x -5 = 0
OR
x - 28 = 0
Either
x = 5
OR
x = 28
For more help on solving quadratic equations.
Monday, March 30, 2009
Question to Find Summation and Draw Box Plot for all Quartiles
Topic : Box Plot
Question : For given amount 95000, 0 and - 65000 , Function P(Xi) 0.60, 0.20 and 0.20 respectively draw a neat Box Plot and represent quartiles.
Solution :
Tuesday, March 24, 2009
Question on Permutation of Zeros in factorial 500
Topic : Permutation
Question : How many zeros are at the end of factorial 500?
Solution :
Highest power of a prime p that divides n!
Let n be a natural number, and let p be a prime. Then the exponent of the highest power of p that divides n! is equal to
N = [n/p] + [n/p2] + [n/p3] + ..... where [x] denotes the greatest integer less than or equal to x.
The highest power of 2 which divides 500! is = [500/2] + [500/2^2] +[500/2^3] + [500/2^4] +[500/2^5] + [500/2^6] +[500/2^7] +[500/2^8] +[500/2^9]
= 250+125+62+31+15+7+3+1 = 494.
The highest power of 5 which divides 500! is = [500/5] + [500/5^2] +[500/5^3] + [500/5^4]=100+20+4 =124
hence the highest power of 10= 5*2 which divides 500! is 124
Hence the number of zeros in 500! is 124.
Another method to solve the same problem.
you get a zero at the end of a product if one of the factors contains a prime factor of 5 and another of the factors contains a prime factor of 2. The prime factors of 2 and 5 multiplied together make 10 and produce a 0 at the end of the product.
If you want to find the number of 0's at the end of 500! in base 10, you need to find the total number of factors of 5 and the total number of factors of 2 in all the numbers from 1 to 500; the smaller of those two numbers will be the number of 0's at the end of 500!
Here is how you would determine the number of 0's at the end of 500!
1) 1 out of every 5 numbers 1 to 500 inclusive (100 of them) contains at least one factor of 5.
2) Of the 100 numbers 1 to 500 inclusive that contain at least one factor of 5, 1 out of every 5 (20of them) contains a second factor of 5.
3) Of the 20 numbers 1 to 500 inclusive that contain a second factor of 5, 1 out of every 5 (4 of them) contains a third factor of 5.
The total number of factors of 5 contained in the numbers 1 to 500 inclusive is then
100 + 20 + 4 =124
in short :
500/5 = 100
100/5 = 20
20/5 = 4
4/5 = 0
----
124
( Write only the quotient)
similarly, The total number of factors of 2contained in the numbers 1 to 500 inclusive is then
500/2 = 250
250/2 = 125
125/2 = 62
62/2 = 31
31/2 = 15
15/2 = 7
7/2 = 3
3/2 = 1
1/2 = 0
-----
494
so there are 124 factors of 5 , 494 factors of 2 and we get 124 combinations of 5x2 ( match the 124 5's with the 124 2's -- no more 5's left -so we get only 124 combinations of 2 x5 ).thats why we take the smaller of the above two calculations.
(however, 500/2= 250 > 124 so we need not perform the other calculations.)
and the answer is 124 0's at the end of 500 !
Question : How many zeros are at the end of factorial 500?
Solution :
Highest power of a prime p that divides n!
Let n be a natural number, and let p be a prime. Then the exponent of the highest power of p that divides n! is equal to
N = [n/p] + [n/p2] + [n/p3] + ..... where [x] denotes the greatest integer less than or equal to x.
The highest power of 2 which divides 500! is = [500/2] + [500/2^2] +[500/2^3] + [500/2^4] +[500/2^5] + [500/2^6] +[500/2^7] +[500/2^8] +[500/2^9]
= 250+125+62+31+15+7+3+1 = 494.
The highest power of 5 which divides 500! is = [500/5] + [500/5^2] +[500/5^3] + [500/5^4]=100+20+4 =124
hence the highest power of 10= 5*2 which divides 500! is 124
Hence the number of zeros in 500! is 124.
Another method to solve the same problem.
you get a zero at the end of a product if one of the factors contains a prime factor of 5 and another of the factors contains a prime factor of 2. The prime factors of 2 and 5 multiplied together make 10 and produce a 0 at the end of the product.
If you want to find the number of 0's at the end of 500! in base 10, you need to find the total number of factors of 5 and the total number of factors of 2 in all the numbers from 1 to 500; the smaller of those two numbers will be the number of 0's at the end of 500!
Here is how you would determine the number of 0's at the end of 500!
1) 1 out of every 5 numbers 1 to 500 inclusive (100 of them) contains at least one factor of 5.
2) Of the 100 numbers 1 to 500 inclusive that contain at least one factor of 5, 1 out of every 5 (20of them) contains a second factor of 5.
3) Of the 20 numbers 1 to 500 inclusive that contain a second factor of 5, 1 out of every 5 (4 of them) contains a third factor of 5.
The total number of factors of 5 contained in the numbers 1 to 500 inclusive is then
100 + 20 + 4 =124
in short :
500/5 = 100
100/5 = 20
20/5 = 4
4/5 = 0
----
124
( Write only the quotient)
similarly, The total number of factors of 2contained in the numbers 1 to 500 inclusive is then
500/2 = 250
250/2 = 125
125/2 = 62
62/2 = 31
31/2 = 15
15/2 = 7
7/2 = 3
3/2 = 1
1/2 = 0
-----
494
so there are 124 factors of 5 , 494 factors of 2 and we get 124 combinations of 5x2 ( match the 124 5's with the 124 2's -- no more 5's left -so we get only 124 combinations of 2 x5 ).thats why we take the smaller of the above two calculations.
(however, 500/2= 250 > 124 so we need not perform the other calculations.)
and the answer is 124 0's at the end of 500 !
Thursday, March 19, 2009
Solved Problem on Partial Differentiation
Topic : Partial Differentiation
![f(x,y) = xe^{x^2y}\\\frac{\partial f}{\partial x}=? And \frac{\partial f}{\partial y}=?\\\frac{\partial f}{\partial x}_{(1,2)}=? And \frac{\partial f}{\partial y}_{(1,ln 2)}=?\\ f(x,y) = xe^{x^2y}<br />\\Consider \frac{\partial e^{x^2y}}{\partial x}\\\frac{\partial e^u}{\partial x} and u=x^2y\\\frac{\partial e^u}{\partial u} . \frac{\partial u}{\partial x}\\ as u=x^2y\\\frac{\partial e^u}{\partial u}=e^u=e^{x^2y}----------(a)\\\frac{\partial u}{\partial x}=2xy-------(b)\\ By Chain Rule\\\frac{\partial e^{x^2y}}{\partial x}=\frac{\partial e^u}{\partial u} . \frac{\partial u}{\partial x}\\=e^{x^2y} . 2xy where a = e^{x^2y} and b=2xy\\\frac{\partial f}{\partial x}=x \frac{\partial e^{x^2y}}{\partial x}+e^{x^2y}\frac{\partial x}{\partial x}\\=xe^{x^2y}(2xy)+e^{x^2y}(1)\\=e^{x^2y}[2x^2y+1]\\\frac{\partial f}{\partial x}=xe^{x^2y}(x^2)\\=x^3e^{x^2y}\\Therefore\frac{\partial f}{\partial x}_{(1,ln2)}=e^{1ln2}(i^2.ln2)=e^{ln2}=2ln2\\\frac{\partial f}{\partial y}_{(1,ln2)}=1^3 e^{1^2ln(2)}=e^{ln(2)}=2<br />](https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEiDA9G1EQoYK18sd1GibXDdHa8cpFAP50o9rwzNzbmdv5Xa2-Uz0wNPFBXtUc9AUkcbzkYKYksefcitwhs8g46nGNw6BAgNsijf7mJuiJwxi4syAbe7q8miLe71y4VTOX0_OET_ktTX4cM/s320/finaldydx.JPG)
Problem : Solve the function by differentiation f(x,y) = xe^(x^2 y) at the intervals (1, ln 2)
Solution:
Following are the steps to differentiate a given function.
![f(x,y) = xe^{x^2y}\\\frac{\partial f}{\partial x}=? And \frac{\partial f}{\partial y}=?\\\frac{\partial f}{\partial x}_{(1,2)}=? And \frac{\partial f}{\partial y}_{(1,ln 2)}=?\\ f(x,y) = xe^{x^2y}<br />\\Consider \frac{\partial e^{x^2y}}{\partial x}\\\frac{\partial e^u}{\partial x} and u=x^2y\\\frac{\partial e^u}{\partial u} . \frac{\partial u}{\partial x}\\ as u=x^2y\\\frac{\partial e^u}{\partial u}=e^u=e^{x^2y}----------(a)\\\frac{\partial u}{\partial x}=2xy-------(b)\\ By Chain Rule\\\frac{\partial e^{x^2y}}{\partial x}=\frac{\partial e^u}{\partial u} . \frac{\partial u}{\partial x}\\=e^{x^2y} . 2xy where a = e^{x^2y} and b=2xy\\\frac{\partial f}{\partial x}=x \frac{\partial e^{x^2y}}{\partial x}+e^{x^2y}\frac{\partial x}{\partial x}\\=xe^{x^2y}(2xy)+e^{x^2y}(1)\\=e^{x^2y}[2x^2y+1]\\\frac{\partial f}{\partial x}=xe^{x^2y}(x^2)\\=x^3e^{x^2y}\\Therefore\frac{\partial f}{\partial x}_{(1,ln2)}=e^{1ln2}(i^2.ln2)=e^{ln2}=2ln2\\\frac{\partial f}{\partial y}_{(1,ln2)}=1^3 e^{1^2ln(2)}=e^{ln(2)}=2<br />](https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEiDA9G1EQoYK18sd1GibXDdHa8cpFAP50o9rwzNzbmdv5Xa2-Uz0wNPFBXtUc9AUkcbzkYKYksefcitwhs8g46nGNw6BAgNsijf7mJuiJwxi4syAbe7q8miLe71y4VTOX0_OET_ktTX4cM/s1600-h/finaldydx.JPG)
Monday, March 16, 2009
Question on Probability of Students in a Class Room Under Different Contraints
Topic : Probability
Question : In a class of 72 students, 44 are girls and,of these, 12 are going to college. Of the 28 boys in the class, 9 are going to college. If a student is selected at random from the class, what is the possibility that the person chosen is going to college.
Solution :
12 girls college
9 girls college
No of students in a class = 72
Total number of students going to college = 12+9=21
probability that the student go to college = 21/72 = 7/24
Question : In a class of 72 students, 44 are girls and,of these, 12 are going to college. Of the 28 boys in the class, 9 are going to college. If a student is selected at random from the class, what is the possibility that the person chosen is going to college.
Solution :
12 girls college
9 girls college
No of students in a class = 72
Total number of students going to college = 12+9=21
probability that the student go to college = 21/72 = 7/24
Wednesday, March 11, 2009
Question on Simplifying an Exponent
Topic : Simplification
Question : Simplifying exponent (4xy)º
Answer :
(4xy)º = 1
According to law of Exponent (aº = 1), Any number or a variable to the indices zero will be equal to one.
Question : Simplifying exponent (4xy)º
Answer :
(4xy)º = 1
According to law of Exponent (aº = 1), Any number or a variable to the indices zero will be equal to one.
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