Multiply Rational Expressions

How to Multiply Rational Expressions

Multiplication of rational expressions is similar to the multiplication of fractions of rational numbers. Let us consider an example of multiplying fractions, 3/4 x 7/2. Here the method to be followed is, first multiply the terms in the numerator and then multiply the terms in the denominator which gives 3x7/4x2 = 21/8.

Then any further simplification possible is carried out. In this example there was no possibility of cancelations as there were no common factors and hence the fraction got by multiplying the numbers in the numerator and the numbers in the denominators would be the final answer. Similar set of steps are used while multiplying rational expressions.

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A rational expression  consist of coefficients which are constants for a given term and hence first the coefficient parts are simplified then the product of the variable parts in the numerator is found, then the product of the variable part in the denominator is found and finally the products are written in the simplified form. In this process we use exponential laws and common factors for simplification.

Let us consider an example for a better understanding. Multiply Rational Expressions 4x2y2/3x and 3x/4y.  The product of the given expressions would be,  4x2y2/3x . 3x/4y Here first the coefficients are simplified and any cancelations possible are carried out. 4 and 3 get canceled and we get x2y2/x . x/y, this can be written as x.x.y.y/x. x/y; x and y are the common factors which can be canceled and we get x2y which is the final answer. Here we can use the exponential rule in simplification.

To multiply rational expressions involving polynomials the steps to be followed are:
First factor the terms in the numerator and the denominator
Reduce all the common factors
Further simplify the terms by either multiplying the numerators and denominators or leave the product in the fraction form

Example: Multiply (x2-9)/(x2+6x+9) and (3x+9)/(3x-9)
Solution: First factor the terms in the numerator and denominator
(x2- 9)= (x+3)(x-3) [special products]
(x2+6x+9)= (x+3)(x+3) [(a+b)2=a2+2ab+b2]
(3x+9)=3(x+3)                               [taking out common factor]
(3x-9) = 3(x-3)                               [taking out common factor]

The product of the expressions would be in the form, (x2-9)/(x2+6x+9) . (3x+9)/(3x-9)
Re-writing the given expressions as factors gives,
(x+3)(x-3)/(x+3)(x+3)  .  3(x+3)/(x-3)
Reducing the common factors gives,
(x+3)(x-3)/(x+3)(x+3)  .  3(x+3)/3(x-3)
Canceling the common terms gives 3/3 = 1, the final answer!

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Multiplying Rational Expressions Solver performs the multiplication and division of rational numbers when the expressions are entered in the given fields. This helps to check the answers.

Exam for Total Differentiation

The total differentiation of a function, f of several variables. For example, t, x, y etc., with respect to t is different from the partial derivative. The total differentiation of f with respect to t does not assume and other variables are constant while t varies. The  total differentiation adds in these indirect dependencies to find the overall dependency of f on t. For example, the total differentiation of f(t,x,y) with respect to t is

(df)/(dt) = (df)/(dt) +  ((df)/(dx)) ((dx)/(dt)) + ((df)/(dy))  ((dy)/(dt)) .

Exam for total differentiation problems:

Let us see some example problems for total differentiation and its helps to exam preparation.

Exam for total differentiation problem 1:

Find the total differentiation of trigonometric term (cot x2) with respect to x.

Solution:

Given trigonometric term is  (cot x2)

Let      y = (cot x2)

Putting x2 = u and  cot x2 = cot u = y

y = cot u,        and       u = x2

dy/(du) = - cosec2 u, and  (du)/(dx) = 2x

The total differentiation is  dy/dx

dy / dx = [  dy/(du) ×  (du)/(dx)]

= [( - cosec2 u) × 2x]

=  2x ( - cosec2 u)                                                         (u = x2)

= - 2x cosec2 (x2).

dy/dx = - 2x cosec2 (x2).

Answer:    dy/dx =  - 2x cosec2 (x2)

Exam for total differentiation problem 2:

Find the total differentiation of  function  P = xe^(y + z) ,  x = uv,  y = u - v , z = u + v

Find     (dP)/(dv)      when u = 1  and v = -1

Solution:

Given function is    P =  xe^(y+ z),

x = uv,        y = u - v        and     z = u + v

((dx)/(dv))= u = 1          ((dy)/(dv))  = -1     and    ((dz)/(du)) = 1

We know the chain rule of partial differentiation. it is the Total differentiation

 (dP)/(du) =  ((dP)/(dx)) ((dx)/(du)) +  ((dP)/(dy))   ((dy)/(du))  + ((dP)/(dz))   ((dz)/(du)) .

Substitute the u and v value we get   x = -1,    y = 2   and z = 0

P = xe^(y + z),

Differentiate P with respect to x .       (dP)/(dx)   = e^(y + z).

Differentiate P with respect to y .        (dP)/(dy) =  xe^(y + z)

Differentiate P with respect to z.       (dP)/(dz)   = xe^(y + z) .

Total differentiation is       (dP)/(dv) =  ((dP)/(dx)) ((dx)/(dv)) +  ((dP)/(dy))   ((dy)/(dv))  + ((dP)/(dz))   ((dz)/(dv)).

= e^(y + z).(1) +  xe^(y + z) .(-1) +  xe^(y + z).1

= e^(2 + 0).(1) +  (-1)e^(2 + 0) .(-1) +  (-1)e^(2 + 0).1

= e2 + e2 - e2

= e2

Answer:   e2

Addition of Decimals Methods

In this article we are going to learn about adding decimals concept.Decimal is base 10 number system. A number consists of decimal part and whole number part is known as decimal number. The point which splits the whole number part and decimal part is known as decimal point. The digit after a decimal point is decimal part and the left to the point is whole number part.

Addition of Decimals Methods

Th following are the steps for addition with decimals:

Step 1: Checking the decimal part of given numbers and making them equal in number of digits by adding zeros at the end.

Step 2: Write down the numbers vertically one under the other with the aligned decimal point

Step 3: Add the numbers as normal addition and place the decimal point in the result

Examples

Below are the examples based on Adding decimals:

Example 1:

1)      Add 12.43 with 5

Sol:

Write 5 as 5.00 (as 12.43 has 2 digits after decimal point)

Add 12.43+5.00

12.43

5.00

---------

17.43

Hence the sum is, 12.43 + 5 = 17.43

Example 2:

Add 1.34 + 6 + 12.7.

Sol:

Making decimal part equal ,

6= 6.00, 12.7 = 12.70

1.34

6.00

12.70

--------

20.04

Hence the sum is, 1.34 + 6 + 12.7 = 20.04

Example 3:

Add 0.803+ 3.1+ 12

Sol:

Making decimal part equal,

3.1 = 3.100, 12 = 12.000

0.803

3.100

12.000

------------

15.903

Hence the sum is 0.803+ 3.1+ 12 = 15.903

Example 4:

Add   7.09 +9.20 + 0.36

Sol:

7.09

9.20

0.36

---------

16.65

Hence the result is 7.09 +9.20 + 0.36 = 16.65

Example 5:

Add 45. 56 + 6.7+ 2

Sol:

Making decimal part equal,

6.7= 6.70, 2=2.00

45.56

6.70

2.00

-------------

54.26

Hence the sum is 45. 56 + 6.7+ 2 = 54.26

Addition with decimals Ex 5:

Add   0.5795 + 2.5301

Sol:

0.5795

2.5301

-----------

3.1096

Hence the sum is 0.5795 + 2.5301= 3.1096

Addition with decimals Ex 6:

Add 82.543+322.916

Sol:

82.543

322.916

----------------

405.459

Hence the sum is 82.543+322.916 = 405.459

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Addition with decimals Ex 7:

Add 369.2165 with 100

Making decimal part equal,

100 = 100 .000

369.2165

100 .0000

-------------

469.2165

Hence the sum is 369.2165+100 = 469.2165.

Addition with decimals Ex 8 :

Add 0.00013+3.902+56.7

Sol:

Making decimal part equal ,

3.902 = 3.90200 , 56.7 = 56.70000

0.00013

3.90200

56.70000

------------------------

60.60213

Hence the sum is 0.00013+3.902+56.7 =   60.60213.

Practice problems

Below are the practice problems on adding decimals:

Add 32+4.5+7.03

Answer: 43.53

Add 2.34+5.6+0.007

Answer: 7.947

Add 7.985 with 9.71485

Answer: 17.69985

Add 0.009+7.89+6.0

Answer: 13.899

Add 1.111+4.67+17

Answer: 31.899

Transformations of Functions

In this page we are going to discuss about transformations of functions concept . Function is defined as one quantity (input of transformations functions) associated with another quantity (output of transformations functions). The quantity can be a Real numbers or Elements from any given sets or the domain and the co domain of the function.

For Example: The function is defined as f : C -> D is a relation that assigns to each x belongs to C to y belongs to D. C is Domain of f and D is Range of f.


Types of transformations of functions

There are four classes of transformations,

1. Horizontal translation: The function is transformed along X axis.

g (x) = f (x + c)

It means that the graph is translated c values to the left side for c > 0 or to the right side for c < 0.

2. Vertical translation: The transformation of function is along Y axis.

g (x) = f (x) + k

It means that the graph is translated k values upwards for k >0 or downwards for k < 0.

3. Change of amplitude:

g (x) = b f (x)

It means that the amplitude of the graph is increased by a factor of b if b > 1 and decreased by a factor of b if b < 1, if b < 0, then we get inverted graph.

4. Change of scale:

g (x) = f (cx)

It means that the graph is compressed if c > 1 and stretched out if c < 1. If c < 0 then we get the reflected graph about y axis.

Examples

Below are the examples on transformations of functions -

Examples:

Examples for functions include parabolas, trigonometric curves and polynomial functions.

* f (x) = 2x2 , for all x values are real

* f(x) = y + sin x, x ,y are real.

* f (x) = 1 / (x+1), for all real numbers except -1

*f (x) = x3-4x2+9x , Polynomial function


Even or Odd Functions

A function f:C-->D is said to be even if and only if f (-x) = f (x) for all x belongs to C.

A function is said to be odd if and only if f (-x) = -f(x) for all x belongs to C.

Even function is symmetric about the y-axis; an odd function is symmetric about the origin in the graph.

Example :  * f (x) = 2x2 is an even function.

* f (x) = x + sin x is odd.

Algebra Coefficient Variable

In mathematics, a coefficient is defined as the number in front of the variable. For example 2x is the given expression here the coefficient is 2. The coefficient is usually in numeral with the variable. The expression contains variable and coefficient of the variable. Here in this topic we are going to see about coefficient of variables.

Example problem for the coefficient:

Example 1:

Find the coefficient of the variables in the given algebraic expression:

x + 2y

Solution:

Given that x + 2y

Here x and y is the expression with the coefficient

The coefficient of x is 1

The coefficient of y is 2.

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Example 2:

Find the coefficient of variables in the given algebraic expression:

y2 + 2y + 3xy +1

Solution:

Given that y2 + 2y + 3xy +1

Here y2 , 2y,  3xy, is the expression with the coefficient and 1 is without variable

The 3xy having two variables that two variable consider as a one variable

1 is the constant term of the given algebraic expression

The coefficient of y2 is 1

The coefficient of y is 2

The coefficient of xy is 3

Example 3:

Find the coefficient of variables in the given algebraic expression:

z5 + z + y +6y

Solution:

Given that z5 + z + y

Here z5, z, y, 6y is the expression with the coefficient

The coefficient of z5 is 1

The coefficient of z   is 1

The coefficient of   y is 1

Example 4:

Find the coefficient of  variables in the given algebraic expression:

x3 + y 5 +2xy + 5yx

Solution:

Given that x3 + y 5 +2xy + 5yx

Here x3, y 5, 2xy, 5yx the expression with the coefficient

The coefficient of x3 is 1

The coefficient of y5 is 1

The coefficient of xy is 2

The coefficient of yx is 5.

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Example 5:

Find the coefficient of variables in the given algebraic expression:

100x2 + 2z3 + 102z

Solution:

Given that 100x2 + 2z3 + 102z

Here 100x2, 2z3, 102z the expression with the coefficient

The coefficient of x2 is 100

The coefficient of z3 is 2

The coefficient of z is 102.

Law of Cosines Calculator

Law of cosines calculator is a tool to calculate calculation applying law of cosines easily. First let's understand the concept of law of cosines. In trigonometry, the law of cosines (also known as the cosine formula or cosine rule) is a statement about a general triangle that relates the lengths of its sides to the cosine of one of its angles. The law of cosines states that,

c2 = a2 + b2 – 2ab cos γ,

The law of cosines generalizes the Pythagorean theorem, which holds only for right triangles: if the angle γ is a right angle (of measure 90° or `pi/2` radians), then cos(γ) = 0, and thus the law of cosines reduces to,

c2 = a2 + b2

Formula for Law of Cosine:

The law of cosines is used to solve the third side of a triangle, when other two sides and the angle are known.

By changing the sides of the triangle, one can find the following two formulas also to solve for the law of cosines,

a2 = b2 + c2 – 2bc cos α,

b2 = a2 + c2 – 2ac cos β.

Law of Cosines Using Distance Formula:

To solve for the law of cosines, consider a triangle with a side length of a, b, c, and θ is the measurement of the angle opposite to the side length c.

A = (bcosθ, bsinθ), B = (a, 0), and C = (0, 0).

By using distance formula, we have,

c = `sqrt((a - bcostheta)^2 + (0 - bsin theta)^2)` ,

Now, squaring on both sides, we get,

c2 = (a – bcos θ)2 + (–bsin θ)2

c2 = a2 – 2ab cos θ + b2cos2θ + b2sin2θ,

c2 = a2 – 2ab cos θ + b2(cos2θ + sin2θ)

c2 = a2 + b2 – 2ab cos θ.               [where, cos2θ + sin2θ = 1].

Law of Cosines Using Trigonometry:

To solve for the law of cosines, draw the perpendicular to the side c as shown in the figure,

c = a cosβ + b cosα.

Multiply by c, we get,

c2 = ac cosβ + bc cosα                                                 (1)

By considering the other two perpendiculars, we get,

a2 = ac cosβ + ab cosγ                                                (2)

b2 = bc cosα + ab cosγ                                                (3)

Adding equations (2), and (3), we get,

a2 + b2 = ac cosβ + ab cosγ + bc cosα + ab cosγ

a2 + b2 = ac cosβ + 2ab cosγ + bc cosα                       (4)

Subtracting equation (1) from the equation (4), we get,

a2 + b2 – c2 = (ac cosβ + 2ab cosγ + bc cosα) – (ac cosβ + bc cosα)

a2 + b2 – c2 = 2ab cosγ,

a2 + b2 – 2ab cosγ = c2,

c2 = a2 + b2 – 2ab cosγ.

Examples based on Law of Cosine:

Ex 1: Evaluate the length of side A using law of cosine formula.

Sol:

law of cosine example 1

Step 1:   A² = B² + C² −2(B)(C)cos (`-<`1) ( law of cosine formula)

Step 2:  Plug in the values of B and C

A² = 20² + 13² −2(20)(13)cos(66)

A² = 400 + 169 −520 cos(66)

Step 3: Add and subtract

A² = 569 −211

A² = 358

Step 4:  Take square root

A= √358 = 18.9

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Ex 2:  Evaluate x using law of cosine formula.

Sol:

law of cosine example 2

Step 1:   A² = B² + C² −2(B)(C)cos ( 1)

25² = 32² + 37² −2(32)(37)cos(x)

Step 2:  Solve the equation

625 = 2393 − 2,368 cos(x)

- 1760 = -2, 368 cos (x)

. 7432 = cos x

Cos-1 (.7432) = 42.0 º

Least Common Multiples

In arithmetic and number theory, the least common multiple or lowest common multiple (LCM) or smallest common multiple of two integers a and b is the smallest positive integer that is a multiple of both of a and of b. Since it is a multiple, it can be divided by a and b without a remainder. A few word problems for least common multiples is given below.

(Source: Wikipedia)

Example of word problems for least common multiples:

Word problem 1:

Find the largest number of four digits which when divided by 5, 10, 15, it leaves a remainder 4 in each case.

Solution:

Step 1: Given numbers

5, 10, 15

Step 2: Find least common multiple of 5, 10, 15

Multiples of 5 = 5, 10, 15, 20, 25, 30, 35, 40....

Multiples of 10 = 10, 20, 30, 40, 50, 60, 70, 80...

Multiples of 15 = 15, 30, 45, 60, 75, 90, 105....

From the list of multiples of 5, 10 and 15 the smallest common number in each list 30.

Therefore, the least common multiples of 5, 10 and 15 are 30.

Step 3: Find multiple of 30 which should be slightly less than five digit.

30 * 333 = 9990

So, 9990 is the largest number of four digit which is divisible by 5, 10, 15 and leaves a remainder 0.

Step 4: Find number which leaves a remainder 4.

To get remainder 4, we should add 4 to the obtained number.

Therefore, the required number is 9990 + 4 = 9994

Word problem 2:

Three children John, Nick and Shane run on a round  track. John takes 50 seconds, Nick takes 55 seconds and Shane takes 60 seconds to run a round. If all three of them start together at a point, when do they meet again?

Solution:

Step 1: Find least common multiple of 50, 55, 60

10 |       50       55       60

-----------------------------------------

5  |      5          55       6

-----------------------------------------

1           11       6

Least common multiple = 10 * 5 * 1 * 11 * 6 = 3300

Step 2: Solution

Therefore, they meet after 3300 seconds = 55 minutes.

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Homework of word problems for least common multiples:

1) Find the largest number of three digits which when divided by 7, 14, 28 it leaves a remainder 2 in each case.

2) Two children Joseph and Fleming run on a round track. Joseph takes 75 seconds and Fleming takes 80 seconds to run a round. If both of them start together at a point, when do they meet again?

Solutions:

1) 982

2) 1200 seconds