Algebra Rational Exponents

Algebra is one of the most basic element of mathematics in which, we switch from basic arithmetic to variables.  Algebra covers a large number of subdivisions like polynomials, rational, exponents, logarithms, expressions etc under it. Exponents are in the form of 'ab ' where a is the base and b is the power (exponent). Exponents in rational form are called as rational exponents. For example: a 1/2 , 'a' has a rational exponent of 1/2. The rules , representation and examples on algebra ational exponents is given in the following sections.

Rational Exponents:

As said earlier rational exponents are in the form  ab/c, where b/c is the rational exponent. Algebra rational exponents can be represented in the following ways.

root(n)(x)  = x^(1/n)

root(3)(x)  = x^(1/3)

root(3)(x^2)  = x^(2/3)

an = b  ===>  a = b^(1/n)

Examples on Rational exponents:

ALgebra Example 1:

Simplify the expression(root(3)(2^3))^4

Solution:

The given expression is (root(3)(2^3))^4

It can be represented as  ((2^3)^(1/3) )4

Therefore, (2^(3/3) )4

= 24

= 2*2*2*2 = 16

Therefore, The simplified answer for the expression is 16.

Algebra Example 2:

Simplify the expression(root(2)(2^3))^4

Solution:

The given expression is (root(2)(2^3))^4

It can be represented as  ((2^3)^(1/2) )4

Therefore, (2^(3/2) )4

= 2^(12/2)

= 26 = 2*2*2*2*2*2 = 64

Therefore, The simplified answer for the expression is 64.

Algebra Example 3:

Simplify the expression(root(2)(2^3))^2

Solution:

The given expression is (root(2)(2^3))^2

It can be represented as  ((2^3)^(1/2) )2

Therefore, (2^(3/2) )2

= 2^(6/2)

= 23 = 2*2*2 = 8

Therefore, The simplified answer for the expression is 8.

Practice problems on rational exponents:

Here are few practice problems given to make sure that the students have learned  the  above mentioned rational exponents concept,

1. Simplify the expression root(5)(x) = 2 , and find the value of 'x'

2. SImplify the expression root(3)(27)

Solution:

1. x =32

2. 3

Solving Quadratic Equations

An equation of the form ax2+bx+c=y where ‘a’ is not equal to zero; a, b are the coefficients and c is a constant is called a Quadratic Equation.

There are different methods used in solving these equations. They are, factorization method, grouping factors method, completing square method, using the quadratic formula if the equation does not factor and graphing method.

Solving Quadratic Equations Examples are as follows: Solve 2x2+5x-3=0
Let us solve this using the factorization method. Here first find the product of the coefficient of x2 term and the constant term, (2)(-3)=-6. Now find the factors of -6 such that the sum of the factors is equal to the coefficient of the middle term(x). The list of factors of -6 can be given as, -2x3, -3x2, -1x6, 6x-1.


Now select the appropriate set of factors whose sum would be equal to 5.
That would be 6, -1 as 6-1=5. Let us now re-write the given equation by splitting the middle term into the factors, which gives
2x2+6x –x +3=0. Now re-grouping the terms we get, 2x(x+3)-1(x+3). Taking (x+3) results in, (x+3)(2x-1). So the factors of the given equation are (2x-1) and (x+3). Each of these factors is equated to zero and solved to arrive to the solution set of the equation. 2x-1=0; x=1/2 and x+3=0; x=-3. So, the solution set is x=1/2, x=-3.

The above method can be used only when the given equation can be factored. There are some cases where the equation cannot be factored. In such cases the Quadratic Formula Help solving Quadratic Equations. The standard Quadratic Equation being ax2+bx+c=0, the quadratic formula is given by,
x= -bsqrt[b2-4ac] divided by 2a.

There would be two solutions, one with positive square root and other with negative square root. Let us consider an example problem, solve x2+3x-5=0. Here first the given equation is compared with the standard form to get the values of a,b and c respectively.

By comparing the terms to the standard form ax2+bx+c=0 we get, a=1, b=3 and c=-5. Now these values are substituted in the quadratic formula, x= [-bsqrt(b2-4ac)]/2a which gives x= [-3sqrt(9 –(4)(1)(-5))]/2(1)
= -3sqrt[9+20]/2 = [-3]/2. The solutions for the given equation are x=[3+29]/2 and x=[3-29]/2

Thus using the quadratic formula it becomes easy to solve equations which cannot be factored. This formula can also be used to an equation that can be factored too. Some of the Quadratic Equation Practice Problems are:
Solve 2x2-4x-3=0
Solve x2-14x+45=0
Solve -3x2+x+5=0
Solve 4x-5-3x2=0
Solve x2-x-6=0

Algebra is widely used in day to day activities watch out for my forthcoming posts on Finding Volume of a Cube and cbse 11th sample paper. I am sure they will be helpful.

Composite Function Solver

In mathematics, composite function is the application of one function to the results of another. For instance, the functions f: X → Y and g: Y → Z can be composed by computing the output of g when it has an argument of f(x) instead of x. intuitively, if z is a function g of y and y is a function f of x, and then z is a function of x                                  .( Source: Wikipedia )

In the following section we are going to see properties of composite function and some problems on composite function by using composite function solver.


Composite function solver

Fig(i) Composite function

Properties of composite function solver:

The composite function solver has the following properties.

The notation for indicate the composite function is  `@` .That is (f ∘ g).Where f and g are two functions. We can also say that f `@` g as f circle g or g composed with f, g after f , g of f.

f ∘ (g ∘ h) = (f ∘ g) ∘ h

(f ∘ g)(x) =f(g(x))

If the function f and g are commute each other ,we can define (f ∘ g) = (g ∘ f)

(f ∘ f)(x) =f 2 ( x )

(f ∘ f ∘ f )(x) = f 3 (x)


Problems on composite function solver :

Problem 1:

f(x) = 2x+5 and g(x) = 4x. Find f(g(x)).

Solution:

Given, f(x) = 2x+5

g(x) = 4x

We need to find f(g(x)) .

That is in the function f(x) ,substitute x =g(x) = 4x

f(x) = 2x+5

f(g(x)) =  2(4x) + 5

= 8x + 5

f(g(x)) = 8x + 5

Answer: 8x + 5

Problem 2:

f(x) = 3x2 + 5x -6 and g(x) = x -5 .Find (f ∘ g)(x)

Solution:

Given,  f(x) = 3x2 + 5x -6

g(x) = x -5

We need to find (f ∘ g)(x),

According to composite function solver property , (f ∘ g)(x) = f(g(x))

To obtain  f(g(x)) ,Substitute x =g(x) in f(x),

f(x) = 3x2 + 5x -6

f(g(x)) = 3( x -5 )2 + 5 ( x -5 ) -6

= 3 ( x2 - 10x +25) + 5x -25 - 6

=3x2 - 30x +75 + 5x -31

= 3x2 - 25 x +44

Answer: f(g(x)) = 3x2 - 25 x +44


Problem 3:

h(x) = x+2 and g(x) = -5x .Find h(g(6))

Solution:

Given , h(x) = x+2

g(x) = -5x

g( 6 ) = -5(6)

= -30

f(g(6)) = x+2

= -30 + 2

= -28

Answer: f(g(6)) = -28

Exponential Scientific Notation

This article is showing the exponential scientific notation. Scientific notation is nothing but transform very big numbers into simplest form of a number. That is called as scientific notation. Scientific notation is nothing but exponential of the base integer 10 have expressed in the number scientific notation. Example for scientific notation is 1860000000 this can be note down as 1.86 x 109. And the one more method in decimal is 0.0000081 = 8.1 x 10-6

Steps to find exponential scientific notation:

The steps to find exponential scientific notation are given below that:

The scientific notation has one number to the left of the decimal place.

We have to set the decimal point following the initial digit.

How many digits the decimal point was moved point out during the exponential of ten (10).


Example problems based on exponential scientific notation:

The example problems based on exponential scientific notation is given below that,

Example 1:

Change 82000 in exponential scientific notation.

Solution:

Step 1:

Specified that 82000.

Step 2:

Specified number is not a big number, but it is simple to convert scientific notation.

First note down 8.2 x 104 = 82000 [104 = 10000]

Step 3:

After that it can be write in scientific notation for 8.2 x 104.

Example 2:

Change 456000 in exponential scientific notation.

Solution:

Step 1:

Specified that 456000.

Step 2:

Specified number is a big number, but it is simple to convert scientific notation.

First note down 4.56 this is not equivalent to specified number.

Step 3:

Identify how many numbers available following the decimal point in previous step.

Present are 5 digits subsequent the decimal point.

4.56 × (100000)

Therefore the notation will be 105.

Step 4:

The scientific notation can be note down as 4.56 x 105

Example 3:

Change 987000 in exponential scientific notation.

Solution:

Step 1:

Specified that 987000.

Step 2:

Specified number is not a big number, but it is simple to convert scientific notation.

First note down the number as 9.87

Step 3:

There are 5 digit following the number is there so we can write 105.

Step 4:

The scientific notation can be note down as 9.87 x 105.
Practice problems based on exponential scientific notation:

The practice problems based on exponential scientific notation is given below that,

Problem 1:

Change 8960000 in exponential scientific notation.

Answer: The concluding answer for this problem is 8.96 x 106.

Problem 2:

Change 963000 in exponential scientific notation.

Answer: The concluding answer for this problem is 9.63 x 105.

Degree of Product Differentiation

The degree of the product differentiation represents the operations in the differentiation which deals with the product of the terms. For example if we have the function like xy then the degree of the function is 2 [x^1y^1 = ^2  ]. The homogeneous equation represents the degreee of the functions in the numerator and the denominator are same. In this article we are going to discuss about the degree of product differentiation for the students in detail with the homogeneous equations in different form with clear view.


Examples for the degree of product differentiation

Review on the degree of product differentiation for the differentiable function  dy/dx = [y^2-x^2]/[2xy]

Solution:

The function given for differentiation  is   dy/dx = [y^2-x^2]/[2xy]

Here it is present in the homogeneous form.

Since the numerator degree[ [y^2-x^2]] is two and the denominator [ [2x^1y^1]] degree is also two

Put y = vx and  dy/dx = v + xdv/dx in the equation given we get,

=>  v + x[dv]/dx   =    [(vx)^2-x^2]/[2x(vx)]

=>  v + x[dv]/dx   =    [v^2x^2-x^2]/[2vx^2]

=>  v + x[dv]/dx   =    x^2[v^2-1]/[2vx^2]

=>  v + x[dv]/dx   =    [v^2-1]/[2v]

=>   x[dv]/dx   =    [v^2-1]/[2v] - v

=>   x[dv]/dx   =    [v^2-1-2v^2]/[2v]

=>   x[dv]/dx   =    [-1-v^2]/[2v]

=>  [2v]/[-[1+v^2]] dv  =  dx/x

=>  [2v]/[1+v^2] dv  =  -1/x dx

Apply integration on both sides, we get

=>  int   [2v]/[1+v^2] dv  = int  -1/x dx

=>  int   [2v]/[1+v^2] dv  = -int  1/x dx

=>   log|1 + v^2|  =  - log |x| + log c

=>   log|1 + v^2|   + log |x| =  log c

=>   log|x[1 + v^2]|  =  log c

=>   x[1 + v^2]  =  +-   c

Put y = vx and v = y/x

=>   x[1 + [y/x]^2]  =  C_1

=>   x[1 + y^2/x^2]  =  C_1

=>   x[x^2 + y^2]/x^2  =  C_1

=>   [x^2 + y^2]/x  =  C_1

=>   x^2 + y^2  = xC_1           is the answer for the differentiation for the degree of product.

Problems for the degree of product differentiation

Review on the degree of product differentiation for the differentiable function dy/dx = [x^2y]/[x^3 + y^3 ]

Solution:

The answer is      [-x^3]/[3y^3] + log|y| =  C

Sectional Area of a Cylinder

A cylinder is one of the most basic curvilinear geometric shapes, the surface formed by the points at a fixed distance from a given straight line, the axis of the cylinder. The solid enclosed by this surface and by two planes perpendicular to the axis is also called a cylinder. The surface area and the volume of a cylinder have been known since deep antiquity. let us disuse about how to find the cross sectional area of cylinder. source-wikipedia


Cross section area of cylinder is nothing but the base area of cylinder. The base of cylinder is circle. So to find cross sectional area of cylinder, we use the area of circle formula.

Cross sectional area of cylinder = π r2 square unit.

r – Radius of circle

Example problems:

Ex:1 Find the cross sectional area of cylinder with radius 5cm.

Sol:

Given:

Radius (r) = 5 cm

Formula:

Cross sectional area of cylinder = π r2 square unit.

= π x 52

= 3.14 x 25

= 78.5

Cross sectional area of cylinder=78.5 cm2

Ex:2 Find the cross sectional area of cylinder with radius 3cm.

Sol:

Given:

Radius (r) = 3 cm

Formula:

Cross sectional area of cylinder = π r2 square unit.

= π x 32

= 3.14 x 9

= 28.26

Cross sectional area of cylinder= 28.26 cm2


Ex:3 Find the cross sectional area of cylinder with radius 5.5 m.

Sol:

Given:

Radius (r) = 5.5 m

Formula:

Cross sectional area of cylinder = π r2 square unit.

= π x 5.52

= 3.14 x 30.25

= 94.985

Cross sectional area of cylinder= 94.985 m2

I am planning to write more post on solve a system of linear equations and cbse books for class 9th. Keep checking my blog.


Ex:4 Find the cross sectional area of cylinder with radius 8cm.

Sol:

Given:

Radius (r) = 8 cm

Formula:

Cross sectional area of cylinder = π r2 square unit.

= π x 82

= 3.14 x 64

= 200.96

Cross sectional area of cylinder= 200.96 cm2

Sectional area of a cylinder- practice problem:

Find the cross sectional area of cylinder with radius 2.5cm.

Ans: cross section area=19.625 cm2

2.   Find the cross sectional area of cylinder with radius 3.8cm

Ans: cross section area = 45.3416 cm2

3.   Find the cross sectional area of cylinder with radius 7.7m

Ans: cross section area = 186.1706 m2

5 Properties of Parallelogram

Let us study about the 5 properties of parallelogram. Parallelogram is also a type of quadrilateral. Parallelogram is said to have two pairs of parallel sides.
Each pair has equal length. Also their diagonals are seemed to bisect each other in the center point.
Parallelograms have 5 important properties used in their construction point and evaluation point. All those 5 properties of parallelogram are discussed in detail with their examples below.




5 properties of parallelogram:

The 5 important properties of parallelogram are as follows:

a)      Parallelogram opposite sides are said to be parallel.

b)      Opposite sides are seemed to be congruent in parallelogram.

c)      Also parallelogram is said to have congruent opposite angles.

d)     Parallelograms two consecutive angles are supplementary angles.

e)      Parallelogram diagonals seemed to be bisecting each other right at their center point.

Parallelogram opposite sides are said to be parallel:

two pairs of parallel sides

This is the 1st property of parallelogram.
The above figure ABCD is seemed to be having two pairs of parallel sides as (AD, BC) and (DC, AB). Therefore it is a parallelogram.



Opposite sides are seemed to be congruent in parallelogram:

two pairs of congruent sides

This is the 2nd property of parallelogram.
The above figure MNOP is seemed to have the two pairs of congruent opposite sides as (MP, NO) and (PO, MN).



Parallelogram is said to have congruent opposite angles:

two pairs of congruent opposite angles

This is the 3rd property of parallelogram.
The above figure KLMN is seemed to have the two opposite pairs of congruent angles as (angle K, angle M) and (angle L, angle N).


Parallelograms two consecutive angles are supplementary angles:

two consecutive angles are supplementary

This is the 4th property of parallelogram.
The above figure ABCD is showing that all their two consecutive angles are supplementary angles as (A, B) or (A, C) or (C, D) or (B, D).




Parallelogram diagonals bisect each other right at their center point:

diagonals bisecting each other

This is the 5th property of parallelogram.
From the above figure ABCD we prove that the two diagonals of the parallelogram one which extends form A and other which extends from B will meet C and D respectively.
Also they bisect each other exactly at the center position of that parallelogram.