Thursday, January 24

Elementary Functions Math


The mathematical concept of a function expresses the intuitive idea that one quantity (the argument of the function, also known as the input) completely determines another quantity (the value, or the output). A function assigns a unique value to each input of a specified type. The argument and the value may be real numbers. (Source: Wikipedia)
Example Problems for Elementary Functions Math :

Elementary functions math - problem 1:

Subtract x3 – 5x2 – 7 from 7x3 + 8x2 – 2x – 9.

Solution:

7x3 + 8x2 – 2x – 9 – x3 + 5x2 +7

6x3 + 13x2 – 2x – 2

Elementary functions math - problem 2:

Find the sum of 6x4 – 7x2 + 9x + 11 and 5x + 7x3 – 6x2 – 9.

Solution:

6x4 + 7x3 – 7x3 – 6x2 + 9x + 11 – 9

6x4  – 13x2 + 9x + 2

Elementary functions math - problem 3:

Factorize: x2 + 10x + 16

Solution:

x2 + 10x + 16

x2 + 8x + 2x + 16

x(x+8) + 2(x + 8)

(x + 2) (x + 8)

Elementary functions math - problem 4:

Simplify 7 * 3 - 2(4)3 ÷ (-6)

Solution:

` (7 * 3 - 2(4)^3 )/ -6`

= =>`(21 - 2(64) )/ -6`

= =>`(21- 128 )/ -6`

= =>`-107/ -6 `

= => `107/6` .

Elementary functions math - problem 5:

Simplify (5y + x)(8y – x)

Solution:

(5y + x)(8y – x)

= = > 40y2 – 5xy + 8xy – x2

= =>40y2 + 3xy – x2

Elementary functions math - problem 6:

Solve the given algebraic equation

3(-5x - 6) - (x - 7) = -8(4x + 7) + 21

Solution:

Given equation is 3(-5x - 6) - (x - 7) = -8(4x + 7) + 21

Multiply the terms

-15 x-18-x+7 = -32x-56+21

Make them as a group

-14x -11= -32x - 35

-14x + 32x = 11 - 35

18x = -24

x = `-24/18`

X = `-12/9` .
Practice Problems for Elementary Functions Math :

1. Subtract x3 – 3x2 – 1 from 3x3 + 6x2 – 4x – 8.

Answer: 2x3 + 9x2 – 4x – 7

2. Find the sum of 5x4 – 8x2 + 7x + 8 and 7x + 6x3 – 3x2 – 1.

Answer: 5x4 – 2x3 – 3x2 + 7

3. Factorize: x2 + 6x + 8

Answer: (X + 2) (X + 4)

4. Simplify 8 * 5 - 5(4)3 ÷ (-8)

Answer: 35.

5. Simplify (4y + x)(4y – x)

Answer: (4y + x)(4y – x) = 16y2 – x2

6. Solve the given algebraic equation  6(-2x - 3) - (x - 2) = -5(2x + 3) + 19

Answer: X = 10.

Wednesday, January 23

Parabola Problems and Solutions


In mathematics, the parabola is a conic section, the intersection of a right circular conical surface and a plane to a generating straight line of that surface. Given a point (the focus) and a corresponding line (the directrix) on the plane, the locus of points in that plane that are equidistant from them is a parabola (Source: Wikipedia). In this topic we discuss about parabola problems and solutions, parabola example problems

Parabola Grapher
Parabola Example Problems with Solutions:

Problem: 1

What is the minimum value of the expression 2x2 – 20x + 17?

Solution:Parabola Example 1 Diagram

Consider the function y = 2x2 – 20x + 17. This function is defined by a second degree equation. This xo-efficient of its x2 term is positive. Hence the curve is a parabola opening up ward.

`(-coefficient fo X term)/(2.coefficient fo X^2 term)` = `-b/(2a)` =` (-(-20))/(2(2))` = `20/4` = 5.

For x = 5, y = 2(5)2 – 20(5) + 17 = - 33. Therefore the minimum value of the expression 2x2- 20x + 17 for any value of x is – 33. This minimum value is assumed only when x = 5.
Problem: 2

Find the coordinates of maximum point of the curve y = - 3x2 – 12x + 5, and locate the axis of symmetry.

Solution:

The curve is defined by a second degree equation. The coefficient of x2 term is negative.

`(-coefficient fo X term)/(2.coefficient fo X^2 term)` = `-b/(2a)` = `(-(-12))/(2(-3))` = `12/-6` = -2.

For x = -2, y = -3 (-2)2 – 12(- 2) + 5 = 17. Hence the coordinates of the vertex are (- 2, 17). The curve is symmetric with respect to the vertical line through its vertex, through the point (-2, 17), i.e., the line x = -2


I am planning to write more post on math homework help free online and Find Arc Length. Keep checking my blog.

Problem: 3

If a parabolic reflector is 16 cm in diameter and 4cm deep, find the focus.

Solution:

let POQ be the vertical section of the reflector. Mid - point of PQ is M. Let OX be along OM and OY parallel to MP.

Let the equation of the parabola be y2 = 4ax.

The coordinates of P are (4, 8)

(8)2= 4a (4) or a = 4

Focus = (a, 0) = (4, 0).

Focus coincides with M, the mid-point of PQ

Monday, January 21

Solve Equations to Vertex Form


Solve equations to vertex form article deals with how to write the standard equation to vertex form and the model problems that helps to understand vertex form.

General form of equation and vertex form:

For solving equation to vertex form, we should have idea in general form

1. The general form of the quadratic function is f(x) =ax2+bx+c

2. The general form of the vertex equation is

f(x) = a(x-h)2 +k

Here (h, k) is the vertex.
Steps in Solving Equations to Vertex Form:

In solving the equation tovertex form, the following steps are very important

Step1: First, we need to factor out the leading coefficient.

Step2:  Add and subtract by common number `(b/2) ^2`

Step 3: simplify the equation to the vertex form.
Model Problems to Solve Equation to Vertex Form:

1. Solve the equation to vertex form f(x) = 2x2 +8x +35.

Solution:

Let f(x) be y

y= 2x2 +8x +35.

Step1: First, we need to factor out the leading coefficient

y = [2x2 +8x]+35
y = 2[x2 +4x]+35

Step2:  add and subtract by common number `(b/2) ^2`

y = 2[x2 +4x + (2)2 -(2)2]+35

Step 3: simplify the equation to the vertex form.

y = 2[(x2 +4x + 4] +35-8
y= 2[(x +2)2] +35-8

y = 2[(x +3)2] +27

Here the vertex form is y =2[(x +3)2] +27 , vertex is (-3,27)

2. Solve the equation to vertex form f (x) = x2+4x+21

Solution:

Let f(x) be y

y= x2+4x+21

Step1: First, we need to factor out the leading coefficient

y = (x2+4x) +21

y = (x2+4x) +21

Step2:  add and subtract by common number `(b/2) ^2`

y = (x2+4x+4) +21-4

Step 3: simplify the equation to the vertex form.

y = (x2+4x+4) +17

y = (x+2)2+17

Here the vertex form is f(x) = (x+2)2+17, vertes is(-2,17)

3. Solve the equation to vertex form y= x2+16x+30

Solution:

y= x2+16x+30

Step1: First, we need to factor out the leading coefficient

y= (x2+16x) +30

y= (x2+16x) +30

Step2:  add and subtract by common number `(b/2) ^2`

y= (x2+16x+64) +30-64

Step 3: simplify the equation to the vertex form.

y= (x2+16x+64)-34

y= (x+8)2-34

Here the vertex form is f(x) = (x+8)2-34, vertex is (-8,-34)

Friday, January 18

Simple Algebra Calculator


Algebra is one of the most important topics in mathematics. Algebra covers quadratic equation, arithmetic progression, cubic equation, quadratic equation, arithmetic progression, sum of consecutive cubes, equation solver and etc., In this document we shall discus about “simple algebra calculator” with suitable example problems.

Simple Algebraic Equation:

A algebraic equation is the set of the variables, constant and arithmetic operators like addition, subtraction, multiplication, division and equal operators.

To solve the simple algebraic equation using calculator, we have to enter the coefficient of x and y and the constant term. After clicking the calculate button the value of x and y will be displayed on answer box. The algebraic equation calculator is shown below.

Simple Algebraic Equation

Example problem on algebraic equation:

Evaluate the unknown variable for the following algebraic equation.

4x - 5y = 12 ----- (1)

2x+5y = 24 ------ (2)
Solution:

Step 1: Add equation (1) and (2) ,we get
4x - 5y = 12

2x+5y = 24

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

6x        = 36

Divide by 6 on both sides,

`(6x)/6 = 36/6`

X = 6
Step 2: Plug x value in second equation;
2x+5y = 24

2 (6) + 5y = 24
Step 3: Expand and simplify the equation:
12 + 5y = 24
12 +5y – 12 = 24 - 12
5y = 12

Divide by 5 on both sides, we get

` (5y)/5 =12/5`

y = 2.4

Algebraic equation of two variable x =6 and y =2.4
Simple Quadratic Equation Calculator:

An equation is collection of  more than one terms are squared but no higher power in terms, having the syntax, ax2+bx+c where a represents the numerical coefficient of x2, b represents the numerical coefficient of x, and c represents the constant term

Example: 3x2- 5x + 15

To evaluate the quadratic equation using calculator, we have to enter the value of coefficient of x2 and x and the constant term. After clicking the calculate button the value of x will be displayed on answer box. The quadratic equation calculator

Please express your views of this topic Algebra Equation Solver by commenting on blog.

Example problem of quadratic equation:

Calculate the unknown value in quadratic equation

x2 + 11x - 42

Solution:

Step 1: Multiply the coefficient of x2 and the constant term,

1*-42 = -42 (product term)

Step 2: Find the factors for the product term

-42 --- > 14*-3 = -42 (factors are 14 and -3)

-42 --- > 14-3 = 11 (11 is equal to the coefficient of x)

I am planning to write more post on Examples of Mean Median and Mode. Keep checking my blog.

Step 3: Separate the coefficient of x

x2 + 11x - 42

x2-3x+14x-42

Step 4: The common term x for the first two terms and 14 for the next two terms are taking outside

x(x - 3) + 14 (x - 3)

(x - 3) (x + 14)

Roots of the quadratic equations (x - 3) (x + 14)

Thursday, January 17

Pi Fraction


Fractions :

A certain part of the whole is called as fractions. The fractions can be denoted as a/b , Where a, b are integers. We can multiply two or more fractions. There are three types of fractions in math,

1) Proper fractions

2) Improper fractions

3) Mixed fractions

Pi( pi  ):

We can express pi as an improper fraction.

pi  = 22 / 7

Here we are going to see some problems using pi fraction .


Example Problems for Pi Fraction

Problem 1:

Find the area of the circle with the radius of 5 cm. using the area formula πr2. (Note: Here substitute the fraction value for π =(22/7)

Solution:

Given radius of the circle is 5 cm.

Formula:

Area of the circle  = πr2

Substitute the given radius value and π in the given formula, we get

Area of the circle = (22/7) * (52) cm2

= 3.14 * 25 cm2

Multiplying the both the values, we get

=  78.5 cm2

Answer: The area of the circle is 78.5 cm2

Problem 2:

Find the surface area of the sphere with the radius of 12 cm. using the area formula 4πr2. (Note: Here substitute the fraction value for π =(22/7) )

Solution:

Given radius of the sphere is 12 cm.

Formula:

Surface area of the sphere = 4πr2

Substitute the given radius value and π in the given formula, we get

Surface area of the sphere = 4 * (22/7) * (122) cm2

= 4 *  3.14 * 144 cm2

Multiplying the both the values, we get

= 1808.64 cm2

Answer:

The surface area of the sphere is 1808.64 cm2

My forthcoming post is on Geometry Calculator will give you more understanding about Algebra.

Practice Problems for Pi Fraction

Practice problem 1:

Find the area of the given circle has the radius length is 7m. (Note: Here substitute the fraction value for π =(22/7) ).

Answer:

The area of the circle is 154 m2

Practice problem 2:

Find the surface area of the given sphere has the radius length is 21m. (Note: Here substitute the fraction value for π = (22/7) ).

Answer:

The area of the circle is 5544 m2

Friday, January 11

Linear Regression Line


A regression line is the straight line which gives the best fit by using the least square cocept to the given sets of data.

To get the line of regression we need to chose the sum of the squares of derivations parallel to the axis of y and minimize the same. It is called that the line of regression of y on x and it gives the best estimate of y for any given value of x.

The equation of y on x is given byy = a + bx.

To get this equation, we use the following normal equation.

sum y = na + bsumx

sum xy = asum x + bsum x 2

By solving them, we get ‘a’ and ‘b’, hence we get the regression line y on x as y = a + bx. Same principle will be applied to get the line x on y.

Now let us see few problems on linear regression line.

Example Problem on Linear Regression Line.

Ex 1: Find the regression line y on x for the following data

Table_1

Soln:

Regression Table1

Here n = 7.

The normal equations are:

sum y = an + bsum x = 56 => 7a + 56b ……….. (1)

sum xy = asum x + bsum x 2 => 469 = 56 a + 476 b ………. (2)

Therefore 8 xx (1) => 448 = 56a + 448 b

(2)=> 469 = 56a + 476 b

Simplifying them, we get: 21 = 28 b => b = 21 / 28  = 3 / 4 .

Therefore (1) => 56 = 7a + 56 (3 / 4)

=> 7a = 56 – 42 = 14 => a = 2

Therefore the equation y on x is y = 2 + 3 / 4 x.

I am planning to write more post on Finding Limits at Infinity . Keep checking my blog.

Example Problem on Linear Regression Line.

Ex 2: Find the regression line x on y for the following data

Table_2

soln:

Regression Table2

Here n = 9.

The normal equations are:

Σx = an + bΣy = 45 = 9a + 96b ……….. (1)

Σxy = aΣy + bΣy 2 = 536 = 96 a + 1224 b ………. (2)

Therefore 96 xx (1) = 4320 = 864a + 9216 b

9 xx (2) = 4824 = 864a + 11016b

Therefore on simplification, we get: 504 = 1800 b => b = 7 / 25 .

Therefore (1) => 45 = 9a + 96 (7 / 25)

=> 9a = 45 – 26.88 = 18.12 => a = 2.01

Therefore the equation x on y is x = 2.01 + 7 / 25 y.

Wednesday, January 9

Exponential Equations with Different Bases


Exponential equations with different bases mean we are going to solve the exponential functions with different bases. Normally exponential functions have the base as e. Here we are going to solve the equations other than e. We will see some example problems for exponential functions with different bases. It is better to understand how to work on the different base. If we want to solve the exponential equations we will use the logarithmic values.


Examples for Exponential Equations with Different Bases:

Solve the following equation 3x = 275x + 3

Solution:

The given equation is 3x = 275x + 3

So we get 3x = ((3)3)5x + 3

From this 3x = (3)15x + 9

From the above we can get x = 15x + 9

15x – x = -9

14x = -9

x = `-9 / 14`

Example 2 for exponential equations with different bases:

Solve the following equation 2x = 42x + 4

Solution:

The given equation is 2x = 42x + 4

So we get 2x = ((2)2)2x + 4

From this 2x = (2)4x + 8

From the above we can get x = 4x + 8

4x – x = -8

3x = -8

So x = `-8 / 3`
More Examples for Exponential Equations with Different Bases:

Example 3 for exponential equations with different bases:

Solve the following equation 5x = 125x + 3

Solution:

The given equation is 5x = 125x + 3

So we get 5x = ((5)3) x + 3

From this 5x = (5)3x + 9

From the above we can get x = 3x + 9

3x – x = -9

2x = -9

So x = `-9 / 2`

Between, if you have problem on these topics Right Triangle Formulas, please browse expert math related websites for more help on algebra 2 problems and solutions.

Example 4 for exponential equations with different bases:

Solve the following equation 4x = 16x + 4

Solution:

The given equation is 4x = 16x + 4

So we get 4x = ((4)2) x + 4

From this 4x = (4)2x + 4

From the above we can get x = 2x + 4

x – 2x = 4

-x = 4

So x = 4