Tuesday, December 1, 2009

Example for complex conjugates

In grade 10 math , complex conjugates are a pair of complex numbers, both having the same real part, but with imaginary parts of opposite signs. For example, 3 + 4i and 3 - 4i are complex conjugates.
The conjugate of the complex number z
 z=a+ib, \,
where a and b are real numbers, is
\overline{z} = a - ib.\,
An alternate notation for the complex conjugate is z * . However, the \bar z notation avoids confusion with the notation for the conjugate transpose of a matrix, which can be thought of as a generalization of complex conjugation. If a complex number is represented as a 2×2 matrix, the notations are identical.We also can use scientific notation converter to find the number.

Let's an example problem from numeric and algebraic operations


Question:-

How do you use complex conjugates to find (3+7i)/(2-i) ?


Answer:-


In the Given problem the denominator is (2-i)

So ,it's conjugate is (2+i)

Multiplying both numerator and denominator with (2+i)

= (3+7i)(2+i)/(2-i)(2+i)

= (5+17i)/(4-i2)

We know that i2 = -1

= (5+17i)/4+1

= (5+17i)/5 Answer

Wednesday, September 16, 2009

Problem on Order of Operations

In mathematics and computer programming, an expression or string of symbols is intended to represent a numerical value; a properly-formed expression may be evaluated in an unambiguous way. But in practice, an expression with multiple terms and operators may be written without parentheses, in which case the intended value of the expression is determined by convention. When a term in the expression is both preceded and followed by an operator such as minus or times, a convention is needed to clarify which operator should be applied first; this rule is known as a precedence rule, or more informally order of operation. From the earliest use of mathematical notation[citation needed], multiplication took precedence over addition, whichever side of a number it appeared on. Thus 3 + 4 × 5 = 5 × 4 + 3 = 23. When numeric and algebraic operations were first introduced, in the 16th and 17th centuries, exponents took precedence over both addition and multiplication, and could be placed only as a superscript to the right of their base. Thus 3 + 5 2 = 28 and 3 × 5 2 = 75. To change the order of operations calculator , a vinculum (an over line or underline) was originally used. Today one uses parentheses (). Thus, if one wants to force addition to precede multiplication, one writes (3 + 4) × 5 = 35.

Let's see an example from algebra answers

Question:-

solve 24-(24+4+2)+2 x (4 x 2)

by using PEMDAS rule

Answer:-

In PEMDAS

P - Parentheses

E - Exponents

M - Multiplication

D - Division

A - Addition

S - Subtraction

let's solve the given equation step by step
in the same order ...

24-(30)+2 x (16)

24-32+30

combine the same terms

24+32-30

56 - 30

26 is the answer.

Tuesday, August 25, 2009

Equation of the line which is passing through two points

Question :-

Find the equation of the line which is passing through two points (-3,7)(5,-1)

Answer:-

We have to use the point formula to find the equation of the line which is similar to midpoint formula



y-y1     x-x1
------ = ------
y2-y1     x2-x1

We have 2 points

( -3 , 7 )  and ( 5 , -1 )
  x1  y1          x2  y2

So the equation is

y-7      x-(-3)
------ = ------
-1-7      5-(-3)

y-7      x+3
------ = ------
-8        8

We can further simplify it by cross multiplication.that comes under indices maths


similarly we can find all points having an x-coordinate of 2 whose distance from the point 2 1 is 5

Monday, August 17, 2009

Simple quadratic equation

Topic: Quadratic Equation


An equation of the form ax2+bx+c=0 where a, b, c are real numbers and a =/ 0, is called a quadratic equation.

Question:

Solve: x2 - y = 3, x - y = -3

Answer:

x2 - y = 3, x - y = -3

x2 - y = 3

x - y = -3

+ y + y
__________
x = y - 3

Putting this in the first equation

(y -3)x2 - y = 3

y2 - 6x + 9 - = 3

y2 - 6x + 9 - y = 3

y2 - 7y + 9 = 3

-3 -3
____________

Factoring (y - 6) (y -1) = 0

y = 6 on 1

x = y - 3 = 6 - 3 = 3

x = 1 - 3 = - 2

The solution set is : {(-2,1) , (3,6)}

Wednesday, August 12, 2009

Graphing trigonometric functions

Graphs are useful for analyzing properties of various trigonometric functions and are valuable in many applications.The most common use of these functions is in analyzing waves ,sound and electric current and voltage.Although any trig function can be graphed,the emphasis here is on graphing trigonometric functions.

Graphing a Sine function

The graph of y= sinx(where x equals the angle) can be sketched by simply constructing a table ,like the one shown below,selecting values for the angle,x and solve for y,then plotting the points on a graph .

x(2)    : -90   45    0    45      90   180   270   360 

y(sin2) : -1   -0.8   0   0.707     1     0    -1     0
The result graph below,which continues indefinitely in both directions


Note:- The x-axis is set in increments of standard angles in degrees,although radians can also be used .
The y- axis is set in decimal increments ,equal to function values of the corresponding angles.

For more help on this, you can reply me.

Monday, August 3, 2009

Maximum and Minimum value of a function

Maximum and minimum value of a function can be determined by simplifying the function with given intervals of time.

Function is a concept which expresses the idea that one quantity completely determines another quantity. Here is one such problem for your practice and understand how the value of function varies with respect to given interval of time.

Question : Find the maximum and minimum value of the function y = x3 - x on the interval [-3,3]

On substituting the value of x as -3 and 3, maxima and minima of function are found.

Solution :

When x = -3

y = x3 - x

= (-3)3 - (-3)

= -27 + 3

= - 24

And when x = 3

y = (3)3 - 3

= 27 - 3

= 24

Hence maximum value of function is 24 and minimum is - 24

Wednesday, July 15, 2009

A simple problem on ratio's

Topic:- Ratio's

If we wish to divide a amount or a number based on specific parts we use ratios.This will help us to share the values to different people in different quantities.

Here is an example which explain this better.

Question:-


How to share 35 between two people with the ration of 1:6

Answer:-

The given ratio is 1:6

The sum of the ratios is 1+6=7

Let us assume it is divided between A and B

1
A= ----- * 35
7

35
= -----
7

= 5

6
B= ----- * 35
7

210
= -------
7

So A gets 5 and b gets 30

For more help on this ,Please reply me.

Wednesday, July 1, 2009

Problem on Mixed fractions

Topic:-Mixed fractions

A fraction is a number that can represent part of a whole.
The earliest fractions were reciprocals of integers, symbols representing one half, one third, one quarter, and so on.A much later development were the common or vulgar fractions which are still used today, and which consist of a numerator and a denominator.

In this algebra help ,we are giving an example of mixed fraction problem.
Topic:- Mixed Fractions
Question:-
 
 
  3      10
8--- - 4---- = ? 
  5      15


Answer:-

  3      10
8--- - 4---- = ? 
  5      15

(5*8)+3      (15*4)+10
--------  -  ----------
  5              15

 43          70
------  -  ------- 
  5          15

Multiply the numarator and the 
denominator of the first fraction 
with 3 to make equal denomitors 


   43*3         70
 -------  -   ------
   5*3          15

 
    129       70
   -----  -  -----
    15        15

      129-70
    -----------
        15

          59        14
         ----  or 3------
          15        15


For more help on this ,you can reply me.

Wednesday, June 24, 2009

Examples of Locus

Topic:- Locus

Locus refers to a collection of points.

It's generally used to define a figure

Example 1:


A circle is a locus of all points ,Which are equidistant from a fixed point ,called the center of the circle.


Example 2:

A line is the locus of all points equidistant from two fixed points or from two parallel lines.

Let us consider a line

C_____________________________D >

-   -   -   -   -   -  -
A_____________________________B >
 k  l  m  o  n  p  q

Here,AB is a collection of points k,l,m,n,o,p,q
So,AB is called as Locus of all these points.

Commom point:
All the points are equidistant from line CD.


For more math help please reply me

Thursday, May 14, 2009

Using Shell or Disc Method to Find Volume of the Solid

Disc method and Shell(cylinder) method of integration are the two different methods of finding volume of solid of a revolution, using rectangular coordination system the functions are defined in terms of x in the below problem.

Topic : Disc or Cylinder Method of Finding Volume of the Sphere.

Problem : Use the disc or shell method to find the volume of the solid generated by revolving the regions bounded by the graphs of the equations about the x axis. y=x3 y=0, x=2

Solution :

y = x3 => 3√y = x
or (y)1/3 = x
or x = y1/3

Volume of a Solid by rotating about x-axis is given by:

V = 2πabp(y)h(y) dy
here p(y)=y1/3, h(y)=y
when x = 2 and y = 33 = 8
So a = 0 and b = 8
Plugging in all the values in the formula, we get

V = 2π08(y)1/3.y dy

= 2π08(y)4/3 dy (as 1/3 + 1/1 = (1+3)/3 = 4/3)

= 2π[y7/3/(7/3)0]8 (as 4/3 + 1/1 = (4+3)/3 = 7/3)

= 2π[(8)7/3/(7/3)- (0)7/3/(7/3)]

= 2π[((2)3)7/3/(7/3)]

= 2π(27/(7/3))

= 2π(128/(7/3))

= 2 * 3 * 128 * π / 7

= 768π /7


So this how the volume of Solid of revolution is determined when the equations about the x axis.
For more help write to our calculus help.

Sunday, May 3, 2009

Problem on Trigonometric Identity

Trigonometric Identities are the standard equations used to solve trigonometric problems.
Below is as one such problem.

Topic : Trigonometric Identity

A proof for Trigonometric Identity

Problem : Prove the given identity 2 Cos x - 2 Cos³ x = Sin x Sin 2x

Solution :

LHS : 2 Cos x - 2 Cos³ x

= 2 Cos x (1 - Cos²x)
= 2 Cos x . Sin² x
= 2 Cos x . Sin x . Sinx
= Sin x . 2 Sin x Cos x
= Sin x . Sin 2x (using Sin 2x = 2 Sinx Cox x)

So we proved 2 Cos x - 2 Cos³ x = Sin x Sin 2x

If you have any queries related to the proof please leave a comment and we will get back to you soon.

Sunday, April 5, 2009

A Solved Problem on Simplification

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

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 +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

Using Factorization


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 !

Thursday, March 19, 2009

Solved Problem on Partial Differentiation

Topic : Partial Differentiation
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 />

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

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.

Thursday, March 5, 2009

Question on Box Whisker Plot

Topic : Box Whisker Plot

Question :
Explain Box Whisker Plot for the given sequence of numbers.

4,6,7,9,15,18,20,23,27,31,33,35

Solution :


4,6,7,9,15,18,20,23,27,31,33,35

Minimum value = 4
Maximum value = 35
The middle numbers are 18 and 20

Let Q2 is the middle term of whole sequence is 18 and 20
=(18+20)/2 = 38/2 = 19

Q2 = 19

And Q1 = (7+9)/2 = 16/2

Q1 = 8

Also Q3 = (27+31)/2 = 58/2

Q3 = 29

Friday, February 27, 2009

Problem on Calculating Percentage

Topic : Percentile Problem

Question : What is the % of 64 is 24?

Answer :

Let x%of 64 =24
So x/100*64=24
64x/100=24
0.64x=24 (divide both sides by 0.64)
x =24/0.64
x =37.5%

Hope the Answer is helpful.

Friday, February 6, 2009

simple Question on Combinations

Topic : Chess Board Problem

Question : Determine the number of rectangles that can be formed on a chess board.

Answer :

There are nine horizontal lines and there are nine vertical lines.To form a rectangle we need 2 vertical lines and 2 horizontal lines from each set.This can be done in 9C2 *9C2 ways= 36*36= 1296.

Hope the Answer is Helpful.

Word Problem about Distance travelled by a Camel with some Constraints

Topic : Word Problem on Distance

Question : A camel must travel 1000 miles across a desert to the nearest city. She has 3000 bananas but can only carry 1000 at a time. For every mile she walks, she needs to eat a banana. What is the maximum number of bananas she can transport to the city?

Answer :
Suppose the camel does this:
1. Pick up 1000 bananas.
2. Carry them one mile into the desert, eating 1 on the way.
3. coming back 1 mile and eating 1 banana;
4. pick up 1000 bananas and carry one mile into the desert , eating 1 banana;
5. coming back 1 mile and eating 1 banana;
6. pick up 1000 bananas and carry one mile into the desert , eating 1 banana;
At the end of this, there will be 2995 bananas, which are 999 miles from the city.
The camel will also be 999 miles from the city.Camel will need 5 bananas to transfer all the bananas 1 mile away…
5 as for 1st 1000 he'll go n come back then for another 1000 he'll go n come back but for last 1000 he'll only go as nothing will be left behind…
For second mile too it'll take 5 bananas…
Now it'll require 5 bananas per mile for total banana count > 2000and for banana count <= 2000 it'll eat 3 bananas…
and for <= 1000 it'll eat 1 banana….
so 1st 1000 bananas will be finished in 1000/5 = 200 Miles next 1000 in 1000/3 = 334 Miles…finally he'll have 1000 bananas and 466 miles to go…
At B camel will have 534 bananas


Hope the explanation helps you to solve the problem, if you have any queries leave your comments.