Adjacent and Opposite Angles

Adjacent angle:

Two angles having a general side and general vertex of that angle is adjacent angles. In adjacent angle, angle that forces never meets. The adj is the short form of the adjacent angle.

Opposite angle:

Intersect two line is framed by four angle.The angles are straightly opposite to other angle is known as opposite angle. The opposite angle is also known as vertically opposite angels.

In this article we see in detail about adjacent and opposite angles.

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Example 1: Adjacent and Opposite Angles:

Following diagram to find the unknown angle a, b , c

example problem for adjacent and opposites angles

Solution:

In this problem given angle is the EFG = 114 degree

Given angle is EOF

Unknown angle is FOG, GOH and HOE

In this diagram angle EOF and angle GOH is equal, since it is opposite angle, written as EOF = GOH

Therefore

Angle EOF = 114

Angle GOH = 114

Angle dimension is 180 degree

EOF+FOG=180

114+c=180

c=180-114

c=66

Angle c and angle a is opposite angle so a=66.

Answer is:

a=66

b=114

c=66
Example 2: Adjacent and Opposite Angles:

Following diagram to find the unknown angle a, b , c

example problem for adjacent and opposite angles

Solution:

In this problem given angle is the EFG = 101 degree

Given angle the angles EOF.

Unknown angle is FOG, GOH and HOE

In this diagram angle EOF and angle GOH is equal since they are opposite angles, written as EOF = GOH

Therefore

Angle EOF = 101

Angle GOH = 101

Angle dimension is 180 degree

EOF+FOG=180

101+c=180

c=180-101

c=79

Angle c and angle a is opposite angle so a=79.

Answer is:

a=79

b=101

c=79

My forthcoming post is on solve my math problem step by step and Algebra Calculator Online will give you more understanding about Algebra.

Example 3: Adjacent and Opposite Angles:

Given complementary angle, angle E is the 74 degree, find the unknown adjacent angle

example for adjacent and opposites angles

Solution:

In complementary angle measurement of the angle is 90 degree

Given the value of angle is 74°

Subtracting form angle measurement of the complementary angle and given angle

90° – 74° = 16°

Answer for this problem is 16 degree.
Example 4: Adjacent and Opposite Angles:

Given complementary angle, angle E is the 71 degree, find the unknown adjacent angle

example for adjacent and opposites angles

Solution:

In complementary angle measurement of the angle is 90 degree

Given the value of angle are 71°

Subtracting form angle measurement of the complementary angle and given angle

90° – 71° = 19°

Answer for this problem is 19 degree.

Running Distance Calculator

How to find a distance covered by a object if it's speed and time  is given,first we will see the formula for running distance calculator.

use the following formula

Distance = Speed x Time

if the speed of the object is running or moving with the speed of 100 km per hr and in 2 hrs that object is going to be cover ------------------------distance?

as we know speed = 100 km per hr and time = 2 hrs

so we know the formula for  distance = speed x time = 100 x 2  = 200 kms

so if the object is moving with 100 km per hr than in 2 hrs it is going to be cover 200 kms distance

when distance covered by the object is asked in miles than use below notation

1 mile =1.6 kilo meters

If the distance i have to find in miles than what i have to do?

As we know 1 mile =1.6 km

so  in 200 km there will be how many miles?

1mile--------------------->1.6km

xmile----------------------->200 km

so by using cross multiplication

x = 200 / 1.6 miles  = 125  miles

I like to share this Critical Value Calculator with you all through my article.

Examples on Running Distance Calculator:

Ex:2 If the speed of a object(plane) is 20 mile per sec than what is the distance traveled by the object in 1 hour.

here speed = 20 miles per sec

so object is traveling 20 mile in 1 sec

so in 1 hour there are 60 sec so it will travel 20 x 60 mile per hour

so it will travel 1200 mile in each hour

if you want to convert this 1200 miles into kilo meters just multiply this 1200 with 1.6

so plane covered the distance of 1200 x1.6 =120 x 16 = 1920 kilo meters

some more examples

example 1: if jhon is driving a car with 90 kilo meter per hour and this is the average speed of his one side journey , for one side he take 5 hours to reach his destination than what is the distance at one side?

answer: jhon is driving with 90 kmph  so speed =90 kmph

and he taken 5 hours of time so time =5 hrs

distance formula is :

distance = speed x time =90 x 5 = 450 km

so his one side journey is 450 kilo meters(answer)

example 2 : if a helicopter is moving with 400 kilo meter per hour  and it traveled about 6 hours than helicopter covered distance in 5 hrs in miles__________?

here helicopter traveled the distance we have to find that to be in miles

for that first we will calculate the distance covered by the helicopter in 5 hrs with the speed of 400 km

than we convert the distance which is in km into miles for that we will use following formula

distance in miles=distance in km/1.6

by this formula distance will change from kilo meters to miles

helicopter  speed = 400 kmph

it taken time about of 5 hrs to reach the destination

so speed =400 kmph and time =5 hrs

distance = speed x time

=400 x 5

=1200 km

so distance in miles =1200/1.6 =12000/16 =750 miles(answer)

so helicopter covered distance of 750 miles in 5 hrs

example 3. pal has to reach his office in exact 15 mins from his home and for that he is moving with his bicycle with the speed if 20 kmph than what is distance  between his house and office?

here bicycle speed is 20 kmph and pal takes 15 mins to reach his office

so speed =20kmph

time=15 mins

here time is in minutes so we have to convert this time in to hours before we substitute time value in formula distance = speed x time , because speed is in km per hours

so distance=20 x15/60 =15/3=5

so distance between pal house and his office =5 kilo meters

example4:  peter average car speed is 60 kmph and he traveled about 3 hours than , distance covered by peter in  this time period=__________in miles

answer: speed = 60 kmph

time = 3 hrs

distance = speed x time = 60 x 3 =180 kilo meters

so peter covered the distance of 180 km but we have to calculate this in miles

so distance =180/1.6 = 1800/16 =900 / 8= 450 / 4 =225 / 2 =112.5 miles

so peter covered distance in miles = 112.5 miles

example 5: robo car is moving with 10 cm/ sec than in 1 minute that robo car is going to cover distance of _________in meters

speed = 10 cm /sec

time = 60 secs

so distance = speed x time = 10 x 60 = 600 cms

but we have to calculate this distance in meters

we know that 100 cms(centi meter)- - - - - - - - -- - - -- - - >1 meter

so 600cms-------------------------------------------------------------->?(take unknown value as X)

so 600 x 1 = 100 x X

X=600/100 meters

distance in meters = 600/100 =6 meters(by cross multiplication)

so fill the blank with the 6 meters

I am planning to write more post on Interpretation of Confidence Interval and Interpreting Bar Graphs. Keep checking my blog.

example 6: distance between two metro stations is _____________in miles, if train is moving with 60kmph and it taken 2 and 1/2 hr to reach the destination

so we have to calculate the distance between two metro stations

here data is given in the question that train is moving with the average speed of 60kmph

and time taken by the train to reach it's destination is 2 and 1/2 hrs

so speed = 60 kmph and time =(2+1/2)= (2x2+1)/2=5/2 hrs

we know distance formula as distance = speed x time

so distance = 60 x 5/2 = 30 x 5 =150 kilo meters

but we have to calculate distance in miles

so convert the kilometers in to miles

we know that 1 mile------------------------------->1.6 km

?       <-------------------------------150 km="" p="">
by cross multiplication

? x 1.6 = 1 x 150

? = 150/1.6

?=150 x 10/16

?=150 x 5/8

?=75 x 5/4

after calculating above expression we will get the distance between two stations as 93.75 miles(this is the required answer)
Problems for Practice on Running Distance Calculator:

some more problems for practice......

1. convert 110 kilometers into miles_________(answer?)

2.car moving with speed 90 kmph and to cover particular distance car taken about 1 and half hours than what is the distance of car starting point to the destination?

3. train is moving with the average speed of 110 kmph and at each stop 5 minutes and total stops between two major stations are 3 and train taken about 3 and 1/4 hr to reach it's destination station , then find the distance between two major stations in miles?

4. peter is ride  his bicycle with average speed of 30 kmph in first 10 minutes of total time and next 5 minutes he ride his bycle with 15 kmph average speed . than he accelerated his speed about 15 kmph and he reached his destination with the same speed in 7 minutes. than find the total distance between his starting and ending point of that race?

5.convert 1200 miles to kilometers?(answer 1920)

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.

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


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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

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)

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

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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)

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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)

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

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.

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

Logarithmic Expression Calculator

In mathematics, the logarithmic expression of an integer with a provided base is the power or exponent to which the base must be increased in order to produce the integer. For example, the logarithmic function of 1000x to base 10 is 3x, because 3 is the power to which ten must be raised to produce 1000: 103 = 1000, so log101000x  = 3x. Only positive real integers will have the real integer values; negative and complex integers hold complex logarithms.Now we are going to discuss about logarithmic expression calculator in detail.



Logarithmic Expression Calculator:

Logarithmic expression calculator is used to calculate the expression given in terms of logarithmic form. Logarithmic calculator calculates the values for the logarithmic problems. For every logarithmic function there is a value present in logarithm form. sample lograthmic calculator is given below. Here the calculate button is used to find the value of the given question. Reset button is used to clear the values and the question present in the calculator, which is useful to enter the next problem. sample calculator is given below,

Logarithmic Expression Calculator

Rules of lograthmic expression solving.

If ax = N, then loga N = x

logx mn = logx m + logx n

logx (m/n) = logx m - logx n

logx mn = n logx m

logn m = logx m/ logx n

logn m * logo n * logp o = logp m

logn m * logm n = 1, logm m = 1

logn m = `1/ (log_m n)`

logx 1 = 0 . Log 1 to any base is always 0.

these are the formulas for expression using in logarithmic expression calculator.
Exampl for Logarithmic Expression Calculator:

Exampl for logarithmic expression calculator 1:

Solve logarithmic function log4(9x)5

Solution:

Step 1: The given function is log4(9x)5

Step 2: To solve this function using log function rules

Rule 3: logb (mn) = n· logb (m)

Step 3: log4 (9x)5= 5 log4 (9x)

So the solution is log4(9x)5 = 5 log4 (9x)

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Exampl for logarithmic expression calculator 2:

Solve logarithmic function log3 `((99/2)x)`

Solution:

Step 1: The given function is log3`((99/2)x)`

Step 2: To solve this function by help of using log function rules

Rule 1:  logb (mn) = logb (m) + logb (n)

Step 3: log3 `((99/2)x)` = log3 `(99/2)`+ log3(x)

Rule 2:  logb (m/n) = logb (m) - logb (n)

log `(99/2)` = log (99) - log (2)

So the solution is log3  `((99/2)x)` = log3 (99) - log3 (2) + log3(x)

Exampl for logarithmic expression calculator 3:

Solve logarithmic functionlog2 (99/x)

Solution:

Step 1: The given function is log2 (99/x)

Step 2: To solve this function by help of using log function rules

Rule 2: logb (m/n)) = logb (m) – logb (n)

Step 3:  log2 (99/x) = log2 (99) - log2(x)

So the solution is log2 (99/x) = log2 (99) - log2(x)

Solving Simple Differential Equation

Solving simple differential equations involve the process of differentiating the algebraic function with respect to the input function. The algebraic function which is differentiable is known as differential equations. The differential equation comes under calculus category whereas to find the rate of change of the given function with respect to the input function. The following are simple example differential equations for solving.


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Simple Differential Equations Examples for Solving:

The following are the example problems with simple differential equations for solving.

Example 1:

Solve the simple differential equation.

f(k) = k2 – 4k + 8

Solution:

The given equation is

f(k) = k 2 – 4k + 8

The first derivative f ' for the algebraic function is

f '(k) = 2 k  – 4

Example 2:

Solve the simple differential equation.

f(k) = k 3 – 5 k 2  + 11k

Solution:

The given function is

f(k) = k 3 – 5 k 2  + 11k

The first derivative f ' for the algebraic function is

f '(k) = 3k 2 – 5(2 k  ) + 11

f '(k) = 3k 2 – 10 k + 11

Example 3:

Solve the simple differential equation.

f(k) = k4 – 3k 3 – 4k 2  + k

Solution:

The given function is

f(k) = k4 – 3k 3 – 4 k 2  + k

The first derivative f ' for the algebraic function is

f '(k) = 4 k 3 – 3(3k 2 ) – 4( 2 k  ) + 1

f '(k) = 4 k 3 – 9k 2  – 8 k  + 1

My forthcoming post is on Example of Hypothesis Testing and Dividing Fraction will give you more understanding about Algebra.

Example 4:

Solve the simple differential equation.

f(k) = k 5 – 6 k 3  + 11

Solution:

The given function is

f(k) = k 5 – 6 k 3  + 10

The first derivative f ' for the algebraic function is

f '(k) = 5k 4 – 6(3 k 2 )

f '(k) = 5k 4 – 18 k 2
Simple Differential Practice Equations for Solving:

1) Solve the simple differential equation.

f(k) = k 3 – 6 k 2  + 11k

Answer: f '(k) = 3k 2 – 12 k

2) Solve the simple differential equation.

f(k) = k 2 – 6 k   + 11

Answer: f '(k) = 2k – 6

About Metric Volume Units

The quantities used to find area, length, width capacities and volume of things etc are called measures. Many countries have their own system of measures. But Metric System of measures is very simple and easy to calculate. The area is measured in square unit. In metric system the volume is measured in cubic units.
Example Problems - Metric Volume Units:

The triangular prism has width 6 cm, height 9 cm and length 11 cm. find the volume of triangular prism.

Solution:

Given:

Width (w) = 6 cm

Height (h) = 9 cm

Length (l) = 11 cm

Formula:

Volume of triangular prism (V) = `1/2` (l x w x h) cubic units

= `1/2` (11 x 6 x 9)

= `1/2` (594)

= 297

Volume of triangular prism (V) = 297 cm3

2. figure out the volume trapezoidal prism whose length 11 cm, height 8cm, length of parallel sides a=7 cm and b=4cm.

Solution:

Given:

Length (l) = 11cm

Height (h) = 8cm

Parallel sides a=7cm and b=4cm

Formula:

Volume of trapezoidal prism = l x area of the base cubic units

Area of the base:

Area of the base = `1/2` x (a + b) x h

= `1/2` x (7 + 4) x 8

=`1/2` x 11 x 8

= 44 cm2

Volume of trapezoidal prism = 11x 44

= 484

Volume of trapezoidal prism = 484 cm3

3. The cylinder has the radius r = 10 feet, h=23 feet. Find the volume of cylinder.

Solution:

Given:

r=10 cm

h=23 cm

Formula:

The volume of the cylinder = π x r2 x h cubic unit

=3.14 x (10)2 x 23

The volume of the cylinder = 7222 ft3.
Example Problems - Metric Volume Units:

Cone:

4. The cone has the radius = 10 feet and height = 23 feet. Find the volume of the cone.

Solution:

Given:

Radius (r) = 10 feet

Height (h) = 23 feet

Formula:

The volume of the cone =`1/3` x π x r2  x  h

= `1/3` x 3.14 x (10)2 x 23

The volume of the cone = 2407.33 ft3

5. What is the volume rectangular prism with length 8 cm width 5 cm and height 6 cm?

Solution:

Given:

Length =8 cm

Width = 5 cm

Height = 6 cm

Formula:

Volume of rectangular solid (v) = l x w x h

= 8 x 5 x 6

= 240

Volume of rectangular solid (v) = 240 cm3

radius calculation formula

Pi is also one of the important concepts in algebra. The value of pi is 22/7 or 3.14. Using these values we can easily find the answer for complex problems in algebra. We have different kinds of problems using pi in algebra. Commonly the pi formulas are used to find area or volume of the specific shape in algebra. Here we are going to learn some problems for calculate radius.
Radius Calculation Formulas:

Perimeter of a circle = 2 * pi * radius.
Area of circle = Pi * radius 2.
Area of ellipse = pi *radius1 *radius 2.
Surface area of sphere = 4*pi* radius 2.
Surface area of cylinder = 2 * pi *radius *height.
Surface area of cone: pi* radius* side.
Surface area of torus = pi2 * (radius2 2 –radius12).
Volume of sphere = `(4)/(3)` *pi*radius 3.
Volume of ellipsoid = `(4)/(3)` *pi*radius 1*radius 2*radius3.
Volume of cylinder =pi*radius*height.
Volume of cone = `(1)/(3)` * pi* radius 2 *height.
Volume of torus = `(1)/(4)` * pi2 *(radius1 + radius2) *(radius1-radius2)2
Volume of hemisphere = `(2)/(3)` pi r3.

These are the formulas to calculate radius.
Example Problems of Radius Calculation Formula:

Example 1:

Perimeter of a circle is 40 m .calculate the radius of the circle?

Solution:

Step 1: Perimeter of a circle = 2 * pi * radius.

Step 2: We know that perimeter = 40 m.

Step 3: Plug the perimeter value in to the formula.

Step 4: 40 = 2 * 3.14 * radius.

Step 5: 40 = 6.28 *radius. (Divide using 6.28 on both the sides)

Step 6: Therefore, radius = 6.4 m.

Example 2:

Surface area of cone is 120 cm^2 and side = 7cm.Calculate the radius of the cone?

Solution:

Step 1: Surface area of cone: pi* radius* side.

Step 2: We know that surface area and side.

Step 3: Plug the surface area and side value in to the formula.

Step 4: Therefore, surface area of cone is 120 = 3.14 * radius *7.

Step 5: So, 120 = 21.98 *radius. (Divide using 21.98 on both the sides)

Step 5: Therefore radius = 5.5 cm.

These are the example problems of calculate radius using formula.

Practice problems of calculate radius using formula.

1)   Volume of cylinder = 160 m and height = 9 cm .calculate the radius of the cylinder?

2)   Suppose the area of circle is 210 m. calculate the radius of the circle?

Answer key

1)   Radius = 5.7 m

2)   Radius = 8.17 m

Dividing Decimal Word Problems

Decimal is one of the number systems which have the base value 10. Division is one of the basic arithmetic operations which is the inverse of multiplication operation. Dividing decimal word problems occurs in such circumstances where we are trying to find out how many times a decimal number go into another. In this article we will see few examples of dividing decimal word problems.

Dividing Decimal Word Problems Examples:

Example 1:

If 2.5 kg of apples cost `$` 15.50 what is the price of one kilogram of apple?

Solution:

Here to find the cost of 1kg of apple we need to divide 15.50 by 2.5

Division of 15.50 by 2.5

By dividing `15.50/2.5` = 6.2

Hence the price of 1 kg apple = `$` 6.2

Example 2:

Tom wants to buy chocolates. The cost of one chocolate is `$` 0.5. The total amount tom has is `$` 20.25. How many chocolates he can buy?

Solution:

Total amount Tom has =` $` 20.25

Cost of one chocolate = `$` 0.5

Number of chocolate can buy =` 20.25 / 0.5`

Division of 20.25 by 0.5

= 40.5

Hence Tom can buy 40 chocolates for `$` 20.25

Example 3:

Each bag of cherries weighs 2.25 pounds. How many bags of cherries would be needed to make 92.25 pounds?

Solution:

Weight of 1 bag of cherries = 2.25 pounds

Total pounds = 92.25

To find number of bags of cherries,

Divide 92.25 by 2.25

Division of 92.25 by 2.5

By dividing` 92.25 /2.25` = 41

Hence 41 bags of cherries needed to make 92.25 pounds.


Dividing Decimal Word Problems Examples Continued:

Example 4:

There are 15 students in a class.  Together they weigh 1925.90 pounds.  Find the average weight?

Solution:

Total Number of students = 15

Total weight = 1925.90 pounds

Average weight = `1925.90/15`

Division of 1925.90 by15

=128.39pounds

Example 5:

There are 2.54 centimeters in one inch.  How many inches are there in 75.24 centimeters?

Solution:

1 inch = 2.54cm

75.24cm= `75.24/2.54`

Division of 75.24 by 2.54

= 29.62inches

Hence there are 29.62inches in 75.24cm

Example 6:

The scores of three persons in diving competition are 3.4, 5.75, and 6.61.  What is the average score?

Solution:

Average score = Total score/ number of persons

Total score =   3.4+5.75+6.61=15.76

Average score = `15.76/3`

Division of 15.76 by3

= 5.25

Two Equations in Standard Form

Linear equation is an algebraic equation which has constants and variables together. Linear equation has 1 or more variables. Linear equation has more forms. Standard form is one of the form of linear equation.

Standard form of linear equation is Ax + By = C.

Here A, B and C are constants. X and y are variables. A and B are not zero.

We can solve standard form of equation using substitution or elimination method. But here we use two standard form equations. Let us see how to solve.

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Standard Form of Two Equations – Substitution Method:

Problem:

Solve these two standard forms of equation using substitution method.

2x + 4y – 12 = 0

X – 2y – 4 = 0

Solution:

The given standard form of two equations are,

2x + 4y = 12  (1)

X – 2y = 4  (2)

From (2), we rewrite the equation as

X = 4 + 2y  (3)

Substitute equation (3) into (1) to find the variable y.

2(4+2y) + 4y = 12

Apply the distributive property, we get

8 + 4y + 4y = 12

Combine like terms,

8+ 8y = 12

Subtract 8 from each side.

8 – 8 + 8y = 12 – 8

8y = 4

Divide by 8 each side.

`(8y)/8` = `4/8`

Y = `1/2`

Substitute y = `1/2` into equation (2)

X – 2y = 4

X – 2(`1/2` ) = 4

X – 1 = 4

Add 1 to each side.

X – 1 + 1 = 4 + 1

X = 5.

Therefore, the solutions are 5 and 1/2.

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Standard Form of Two Equations – Elimination Method:

Problem :

Use elimination method to determine the solutions of the following the systems of equations.

x + y – 16 = 0 and 4x – 2y – 4 = 0

Solution:

The given standard forms of two equations are

x + y = 16  (1)

4x – 2y = 4  (2)

Step 1:

Multiply the equation (1) by 2 and Equation (2) by 1 to get the coefficients of variable y same. So the equations are,

2x + 2y = 32

4x – 2y = 4

Step 2:

Add the two equations for eliminating y variable.

2x + 2y + 4x – 2y = 32 + 4

2x + 4x + 2y – 2y = 36

6x = 36

Divide by 6 both sides.

x = 6

Step 3:

Substitute the x value into the equation (1) to get value of y variable.

x + y = 16

6 + y = 16

Subtract 6 from each side.

y = 10.

The solutions are x = 6 and y = 10.

Rational Zeros of a Function

Normally zeros of a function mean when we plug the values for the variables the functions values tends to be zero.  Let us consider if we are having a function with variable x and we have a set of solution p(x) if we plug the solution for the given variables present in the function we will get the function f(x) = 0. To find the rational zeros we have to use the rational zeros theorem.

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Rational Zeros of a Function – Examples:

The rational theorem states that if we are having the polynomial p(x) wit the integer coefficients and we are having the zeros of the polynomial `p / q` then we can say `p(p / q) = 0` . Here p is nothing but the constant term of the polynomial and the q I nothing but the leading coefficient of the polynomial p(x). We will see some examples for finding the rational zeros of a function.

Example 1 for rational zeros of a function:

Find the rational zeros of the following function.

2x2 + 12x + 10

Solution:

The given function is

2x2 + 12x + 10

First we have to find the rots of the constant term. ±1, ±2, ±5, ±10

Now the leading co – efficient of the constant term is 2. So we have to divide by 2.

±`p / q` = ±`1/ 2` , ±`2/ 2` , ±`5/ 2` , ±`10/ 2`

= ±`1/2` , ±1, ±`5/2` , ±5

Now we have to use the synthetic division method to find the rational zeros.

1 / 2 | 2           12        10

|               1         13/2

|_____________________

2           13        33/2      = not a zero

-1 / 2 | 2           12        10

|               -1         -11/2

|_____________________

2           11        9/2      = not a zero



1 | 2           12        10

|               2         24

|_____________________

2           24        34      = not a zero

-1 | 2           12        10

|               -2        -10

|_____________________

2           10        0      = is a zero

5/2 | 2           12        10

|               5        85/2

|_____________________

2           17        105/2      = not a zero

- 5/2 | 2           12        10

|               -5        -35/2

|_____________________

2           7        15/2      = not a zero

5 | 2           12        10

|              10        110

|_____________________

2           22        120      = not a zero

-5 | 2           12        10

|              -10      -10

|_____________________

2           2        0      = is a zero

So from the above the rational zeros of the functions are p(x) is -1 and -5

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Rational Zeros of a Function – more Examples:

Example 2 for rational zeros of a function:

Find the rational zeros of the following function.

x2 + 4x + 3

Solution:

The given function is

x2 + 4x + 3

First we have to find the rots of the constant term. ±1, ±3

Now the leading co – efficient of the constant term is 1. So we have to divide by 1.

±`p / q` = ±`1/ 1` , ±`3 / 1`

= ±1, ±3

Now we have to use the synthetic division method to find the rational zeros.

1 | 1           4        3

|               1       5

|_____________________

1           5       8      = not a zero

-1| 1           4        3

|              -1      -3

|_____________________

1            3        0      = is a zero

3 | 1           4        3

|              3        21

|_____________________

1          7       24      = not a zero

-3| 1           4        3

|             -3      -3

|_____________________

1            1        0      = is a zero

So from the above the rational zeros of the functions are p(x) is -1 and -3

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Interval of Convergence for Taylor Series

The interval of convergence for the given series is the set of all values such that the series converges if the values are within the interval and diverges if the value exceeds the interval.
The interval of convergence series must have the interval a - R < x < a + R since at this interval power series will converge.

In this article, we are going to see few example and practice problems of Taylor series to find interval of convergence which help you to learn interval of convergence.
Example Problems to Find the Interval of Convergence for Taylor Series:

Example problem 1:

Solve and determine the interval of convergence for Taylor seriessum_(n = 0)^oo(x^n)/(n!) .

Solution:

Step 1: Given series

sum_(n = 0)^oo (x^n)/(n!)  .

Step 2: Find L using the ratio test

L =  lim_(n->oo) | ((x^(n+1))/((n+1)!))/((x^n)/(n!)) |

= lim_(n->oo) |   (xn!)/((n + 1)(n!))  |

= lim_(n->oo) |   x/(n + 1)  |

= 0

So, this series converge for all value of x. Therefore, the interval of convergence is (-∞, ∞).

Step 3: Solution

Hence, the interval of convergence for the given Taylor series is (-∞, ∞).

Example problem 2:

Solve and determine the interval of convergence for Taylor series sum_(n = 0)^oo(-1)n (x^(2n + 1))/((2n + 1)!) .

Solution:

Step 1: Given series

sum_(n = 0)^oo(-1)n (x^(2n + 1))/((2n + 1)!)  .

Step 2: Find L using the ratio test

L =  lim_(n->oo) | (-1)^(n + 1)(x^(2(n + 1)+1))/((2n + 1)!) |

= lim_(n->oo) |  (-1)^(n + 1)(x^(2n + 3))/((2n + 1)!)   |

= 0

So, this series converge for all value of x. Therefore, the interval of convergence is (-∞, ∞).


Step 3: Solution

Hence, the interval of convergence for the given Taylor series is (-∞, ∞).

Practice Problems to Find the Interval of Convergence for Taylor Series:

1) Determine the interval of convergence for Taylor series sum_(n = 0)^oo(-1)n (x^(2n))/(2n!)  .

2) Determine the interval of convergence for Taylor series sum_(n = 0)^ooxn.

Solutions:

1) The interval of convergence for the given Taylor series is (-∞, ∞).

2) The interval of convergence for the given Taylor series is |x| < 1.

Number Properties

Number properties and number operations is nothing but we are going to learn about the basic number properties and number operation. We are having the three basic properties and we are having the four basic operations. In this we are going to learn about the number properties and number operations using some example. It is very useful to performing the operations on the numbers easily.

Number Properties:

Basically we are having three basic number properties. We will see all the properties using some examples.

Distributive property:

Distributive property is nothing but giving the values. We can easily remember the distributive property using the following words multiplication distributes over addition.

2 X (3 + 5) = 2 X 3 + 2 X 5

2 X (8)        = 6 + 10 = 16

We are getting the equal value.

Associative property:

Associative property is nothing but grouping the numbers. We can use this for addition and multiplication.

2 + (3 + 5) = (2 + 3) + 5

2 + 8          = 5 + 5 = 10

This is the associative property of numbers.

Commutative property:

Commutative property of numbers is nothing but the value of numbers won’t change when the place of the numbers change.

2 + 3 = 3 + 2 = 5
Number Operations:

In numbers we can perform four types of operations.  Those operations are addition operation, subtraction operation, multiplication operation and division operation.



Addition operations

Addition operation is nothing but we are adding the value of any two numbers. If we take 5 and 6 if we want to perform addition operation we have to add the quantities of the numbers. The addition operation is denoted by the symbol`+`

So 5 + 6 = 11

Subtraction operation

Subtraction operation is nothing but we are going to subtract the value of numbers. The subtraction operation is denoted by `-`

Example:

9 – 6 = 3

Multiplication operation

Multiplication operation is nothing the repeated addition. The multiplication operation is denoted by `xx`

Example: 5 X 3

5 X 3 = 5 + 5 + 5 (We have to add the value of 5 three times)

= 15

Division operation

Division operation is nothing but the repeated subtraction. The division operation is denoted by `-:`

For example take 15 `-:` 3

We have to subtract the value of 3 from 15 again and again.

So 15 – 3 = 12

12 – 3 = 9

9 – 3 = 6

6 – 3 = 3

3 – 3 = 0

So totally this operation performed in 5 steps we got the remainder as 0.

So the quotient is 5 and the remainder is 0.

Decimal Numbers

The decimal number system (also called base ten or occasionally denary) has ten as its base. It is the numerical base most widely used by modern civilizations.

Decimal notation often refers to the base-10 positional notation such as the Hindu-Arabic numeral system; however it can also be used more generally to refer to non-positional systems such as Roman or Chinese numerals which are also based on powers of ten.

(Source – Wikipedia.)
Example Problems for Dividing Decimal Numbers:

Example problems 1 for dividing decimal numbers:

Divide 3.5 by 0.7

Solution:

Step 1: First we begin with 3.5 /0.7

Step 2: and Then we divide without decimal points, we attain

= 35 / 7

= 5

Step 3: and then we put decimal places

3.5 have one decimal place.

0.7 has one decimal place.

Step 4: Final answer has two decimal places 5

Answer 5.

Example problems 2 for dividing decimal numbers:

Divide 4.4 by 4

Solution:

Step 1: First we begin with 4.4 / 4

Step 2: and Then we divide without decimal points, we attain

= 44 / 4

= 11


Step 3: and then we put decimal places

4.4 have one decimal place.

4 have no decimal place.

Step 4: Final answer has one decimal places 1.1

Answer 1.1.

Example problems 3 for dividing decimal numbers:

Divide 1.2 by 0.2

Solution:

Step 1: First we begin with 1.2 / 0.2

Step 2: and Then we divide without decimal points, we attain

= 12 / 2

= 6

Step 3: and then we put decimal places

1.2 have one decimal place.

0.2 have one decimal place.

Step 4: Final answer has one decimal places 6.

Answer 2.

Example problems 4 for dividing decimal numbers:

Divide 6.4 by 8

Solution:

Step 1: First we begin with 6.4 / 8

Step 2: and Then we divide without decimal points, we attain

= 64 / 8

= 8

Step 3: and then we put decimal places

6.4 have one decimal place.

8 have no decimal place.

Step 4: Final answer has one decimal places 8

Answer 0.8.
Practice Problems for Dividing Decimal Numbers:

Practice problem 1 for dividing decimal numbers:

1) Divide 3.6 by 4

Ans: 0.9

Practice problem 2 for dividing decimal numbers:

2) Divide 7.2 by 0.8

Ans: 9

Practice problem 3 for dividing decimal numbers:

3) Divide 6.2 by 2

Ans: 3.1

Practice problem 4 for dividing decimal numbers:

4) Divide 42 by 0.7

Ans: 60

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Practice problem 5 for dividing decimal numbers:

5) Divide 72 by 0.8

Ans: 90.

Factoring Polynomials

Polynomials:

A polynomial is called an algebraic expression having the power of each variable is a only positive integral numbers.

For example,  8x2 -  16x3y2 + 24 y4 + 10, 9y2 - 16, y2 + 20x + 100.

These examples are a polynomial expression in two variables x and y and polynomial expression in one variable y.

Let us discuss about factoring polynomial .
Factoring Polynomials:

Factoring Polynomials:

Factoring a polynomial is the reverse process of multiplying polynomials.

When we factor a real number, First we are getting for prime factors for real number which multiply together to give the same real number.

For example,   28 = 7 x 2 x 2

When we are going to factor a polynomial,  First we need  to look for simpler polynomial expressions which are multiplied each other and give the same polynomial expression what we started with.

For example,  y5x2 + 25xy = 5xy(x + 5)

Let us solve sample problems on factoring polynomials.
Sample Problems: Factoring Polynomials

Steps to factoring Polynomials:

Step 1: First we factor the polynomial

Step 2: Expand the factorized polynomial using algebraic identities formula.

Step 3: Solve it if possible.

Problem 1:

Factor the polynomial expression: 25m2 – 64

Solution:

The given polynomial expression is 25m2 – 64.

Step 1: Factoring it,

= 52m2 – 82

= (5m)2 - 82

Step 2: Expand it

This is in the form of a2 – b2 = (a + b) (a – b)

So,

= (5x + 8)(5x – 8)

Therefore, the factors are (5x + 8)(5x – 8).

Step 3: Solving it,

5x + 8 = 0

5x = - 8

x = - `8/5`

5x - 8 = 0

5x = 8

x = `8/5`

Therefore, Solutions are x = `8/5` , `- 8/5` .

Problem 2:

Factor the polynomial expression: 9p2 – 81

Solution:

The given polynomial expression is 9p2 – 81.

Step 1: Factoring it,

= 32p2 – 92

= (3p)2 - 92

Step 2: Expand it

This is in the form of a2 – b2 = (a + b) (a – b)

So,

= (3p + 9)(3p – 9)

Therefore, the factors are (3p + 9)(3p – 9).

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Step 3: Solving it,

3p + 9 = 0

3p = - 9

p = -` 9/3` = - 3

3p - 9 = 0

3p = 9

p = `9/3` = 3

Therefore, Solutions are p = 3 , - 3.
Problem 3:

Factor the polynomial expression: A2 – `1/49`

Solution:

The given polynomial expression is A2 – `1/49` .

Step 1: Factoring it,

= A2 – `1/7^2`

= A2 - `(1/7)^2`

Step 2 : Expand it,

This is in the form of a2 – b2 = (a + b) (a – b)

So,

= (A + `1/7` )(A – `1/7` )

Therefore, the factors are (A + `1/7` )(A – `1/7` ).

Step 3 : Solving it,

(A + `1/7` )(A - `1/7` ) = 0

A + `1/7` = 0

A = - `1/7`

X - `1/7` = 0

X = `1/7`

Therefore, Solutions are A = `1/7` , - `1/7` .

Problem 4:

Factor the polynomial expression: y2 + 2y - 15

Solution:

The given polynomial expression is y2 + 2y  - 15.

Step 1 : Factoring it,

y2 + 2y - 15

Step 2 : Expand the polynomial expression

= y2 - 3y + 5y - 15

= y(y - 3) + 5(y - 3)

Taking out common factor,

= (y - 3)(y + 5)

Therefore, the factors are (y + 5)(y – 3).

Step 3 : Solving it,

(y + 5)(y - 3) = 0

y + 5 = 0

y = - 5

y - 3 = 0

y = 3

Therefore, Solutions are y = 3 , - 5.

Problem 5:

Solve for x in the perfect square trinomial: 9x2 + 24x + 16 = 0

Solution:

The given perfect square trinomial is 9x2 + 24x + 16 =0

Step 1: First factoring this perfect square trinomial

(3X)2 + 2·3.4x + 42 = 0

(3x + 4)2 = 0

Step 2: Expand it

This is in the form (ax + b)2 = (ax + b) (ax + b)

(3x + 4)(3x + 4) = 0

Step 3: Solving x

3X + 4 = 0

3X = - 4

Divide by 3 each side.

X = `- 4/3` .

Prime Factorization Of Composite Numbers

Converting the composite numbers in to multiplication of numbers is known as factorization, simplifying till the prime numbers is known as prime factorization.

Prime Number

    A whole number greater than 1 that cannot be divided by any other number except the number and the same number is known as the prime number.
    Some of the prime numbers are : 2, 3, 5, 7, 11, 13, and 17 ...

Composite numbers:

      It is an integer that has more than one prime factor. They can be expressed as unique set of prime numbers. The first composite number is 4. Besides 1 each other natural number is either prime or composite number.



Prime Factorization of Composite Numbers
Prime Factorization is the method of splitting the composite number into prime numbers

Factors:

The factor is the numbers that are multiplied to get another number.

  5 x 3 = 15

 (5, 3 factor)

while factoring if we get the factored numbers as the prime number, then the numbers are called as prime factors.

Example: The prime factors of 15 will be 5 and 3 ( 5 x 3 =15, and 5 and 3 are prime numbers).

Examples of Prime Factorization of Composite Numbers:

Example1.

What are the prime factorization of 225?

Solution:

Here 225 is a composite number, we are going to convert into a equivalent prime number.

It is best to start with the smallest prime number that can divide the number, which is 3, so let's check:

225 ÷ 3 = 75

since 75 is not a prime number we have to factor 75

75 ÷ 3 = 25

since 25 is not a prime number we have to factor 25.

25 ÷ 5 = 5

here, 5 is a prime number, so we can stop with this

225 = 3 x 3 x 5 x 5

The prime factorization of 225 is 3 x 3 x 5 x 5



Example 2.

What are the prime factorization of 490?

Solution:

Here 490 is a composite number, we are going to convert into a equivalent prime number.

It is best to start with the smallest prime number that can divide the number, which is 5, so let's check:

490 ÷ 5 = 98

since 98 is not a prime number we have to factor 98

98 ÷ 2 = 49

since 49 is not a prime number we have to factor 49.

49 ÷ 7 = 7

here, 5 is a prime number, so we can stop with this

490 = 5 x 2 x 7 x 7

The prime factorization of 490 is 5 x 2 x 7 x 7

Antiderivatives: An Introduction to Indefinite Integrals

The rate of change of a function at a particular value x is known as the Derivative of that function. Anti-derivatives as the name suggests is the opposite of derivatives. An Anti derivative is commonly referred to as an Indefinite Integral. We can define an indefinite integral of F as follows: Any given function G is an indefinite integral of F or an indefinite integral of a function g if the derivative of that function G’ equals g. The notation of an indefinite integral of F or the indefinite integral of a function is, G(x) = Integral g(x) dx. From this notation, we can conclude that G(x) equals integral[g(x)]dx  if and only if G’(x) = g(x)

From the above we can understand that an Antiderivative is basically an Integral of a given function, which is set into a formula which helps us to take the indefinite integral of F. So, when we say the Integral it means indefinite integral of F. Here, we have to remember to add a constant “c” as every integral has an unknown constant which is added to the equation. Let us consider an example for a better understanding, given function y = x^2 + 3x + 5. The derivative y’ would be 2x +3. Let us now find the antiderivative of y’, that gives us integral[2x +3] dx = 2. X^(1+1)/(1+1) + 3. x^(0+1)/(0+1) + c = 2x^/2 + 3x/1 + c = x^2+3x +c, this function is same as the original function except that the constant 5 is missing, this is the reason why we need to add the constant “c” to the Integral of a function.  From this we can conclude that Anti Derivative is the reverse derivative or the indefinite integral.

When we solve an Integral, we eliminate the integral sign and dx to arrive to a function G(x), this function is the antiderivative.  For instance, indefinite integral of F of the function x^3 is given by x raised to the power (3+1) whole divided by (3+1), that is, integral(x^3)dx = x^(3+1)/(3+1)= x^4/4. In general we can write the indefinite integral of F of x^n  as x^(n+1)/(n+1)
Antiderivative of Sec X
indefinite integral of F is the Integral of Sec X
Multiplying sec(x) with 1 which is [sec(x)+tan(x)]/[tan(x)+sec(x)]
Integral[sec(x)] dx = Integral[Sec x][sec(x) +tan(x)]/[tan(x)+sec(x)]      

Let u= sec(x) + tan(x)
Differentiating on both sides,
du = [sec(x)tan(x) + sec^2(x)]dx
Substituting u= [sec(x) + tan(x)]du = [sec(x)tan(x) + sec^2(x)]dx ,
Integral[sec(x)][sec(x)+tan(x)]/[tan(x)+sec(x)] dx
 = Integral[sec^2(x)+sec(x)tan(x)]dx/[sec(x)+tan(x)]
= Integral[du/u]
Solving the integral we get,
 = ln|u|+ c
Again substituting u= sec(x) + tan(x)
= ln|sec(x)+tan(x)| + c
So, the Antiderivative of Sec X is the Integral[sec x]dx = ln|sec(x)+tan(x)| + c

A short note on derivative of Cot function

Cot is the short for the trigonometric function cotangent. It is the complementary function of the tan also called the tangent function. Cot is defined as the ratio of the adjacent side to the opposite side of the acute angle in a right triangle. It is also the reciprocal of the tan function.

Therefore symbolically it can be written like this:
Cot(x) = adjacent side/opposite = 1/tan(x)
We can graph the cot function using a table of values as follows:
X -pi -3pi/4 -pi/2 -pi/4 0 Pi/4 Pi/2 3pi/4 pi
Cot(x) Inf 1 0 -1 Inf 1 0 -1 inf

The graph would look as follows:

 From the graph we note that the cot function is not defined at the points –pi, 0 and pi. That is because at these points, the tan function has the values of 0,0 and 0 respectively. Since the cot function is the reciprocal of tan function, the cot is not defined at these points.

Derivative Cot X  function:

The above graph is of the function y = Cot(x). Now the derivative of Cot X would be the slope of tangent to the above curve at any point x. Therefore for example if we were to find the derivative of Cotx at the point x = pi/4, then the graph of the tangent line to the curve at that point would like this:

The blue line in the above picture is the tangent to the cot function at the point x= pi/4. The slope of this tangent is the derivative of cot x. In other words we can also say that the derivative of the cot function is the rate of change of cot function at a given point. This is also called the instantaneous rate of change of cot function at a point x=a. (In this case it is x = pi/4). Another way of stating the same thing is like this: the gradient of the curve of the function y = cot x at a point x=a, is called the derivative of the function at that point.

Understanding Trigonometry Problems is always challenging for me but thanks to all math help websites to help me out.

The derivative of the function y = cot x at the point x = 0 can be found by making a tangent to the curve at the point x=0. Again from the above graph we see that, at x=0, the tangent would be vertical. So slope of tangent = not defined. Therefore derivative at this point is not defined.

Homogeneous and non homogeneous differential equation

Non Homogeneous Differential Equation: A differential equation is a linear differential equation if it is expressible in the form P0 d^n y/dx^n + P1 d^n-1 y/dx^n-1 + P2 d^n-2 y/dx^n-2 + … + Pn-1 dy/dx + Pn y = Q, Where P0, P1, P2, …., Pn-1, Pn and Q are either constants or functions of independent variable x.Thus, if a differential equation when expressed in the form of a polynomial involves the derivatives and dependent variable in the first power and there are no product of these, and also the coefficient  of the various terms are either constants or functions of the independent variable, then it is said to be linear differential equation. Otherwise, it is a non-differential equation.

Homogeneous differential equation: A function f(x, y) is called a homogeneous function of degree n if f(lambda x, lambda y) = lambda^n f(x, y).

For example, f(x, y) = x^2 – y^2 + 3xy is a homogeneous function degree 2, because f(lambda x, lambda y) = lambda^2 x^2 – lambda^2 y^2 + 3. Lambda x. lambda y = lambda^2 f(x, y).Let us understand Homogeneous Differential Equation Examples. Suppose we have to Solve x^2 y dx – (x^3 + y^3) dy = 0 . The given differential equation is x^2 y dx – (x^3 + y^3) dy = 0, dy/dx = (x^2 y)/(x^3 + y^3)……(i), Since each of the functions x^2 y and x^3 + y^3 is a homogeneous function of degree 3, so the given differential equation is homogeneous. Putting y = vx and dy/dx = v + x dv/dx in (i), we get  v + x dv/dx = vx^3 /(x^3 + v^3 x^3), v + x dv/dx = v/1 + v^3, x dv/dx = [v/(1 + v^3)] – v, x dv/dx = (v – v – v^4)/(1 + v^3), x dv/dx = -v^4/(1 + v^3), x (1 + v^3) dv = -v^4 dx, (1 + v^3)/v^4 dv = -dx/x, (1/v^4 + 1/v) dv = -dx/x, Integrating both sides we get v^-3/-3 + log v = -log x + C=> -1/3v^3 + log v + log x = C=>-1/3 x^3/y^3 + log(y/x . x) = C=-1/3 . x^3/y^3 + log y = C, which is the required Homogeneous Solution Differential Equation.

Linear Homogeneous Differential Equation:  A differential equation is linear if the dependent variable (y) and its derivative appear only in first degree. The general form of a linear differential equation of first order is  dy/dx + Py = Q, where P and Q are functions of x (or constants).For example, dy/dx + xy = x^3, x dy/dx + 2y = x^3, dy/dx + 2y = sin x etc. are linear differential equations. Let us Find the General Solution of the Homogeneous Differential Equation  (1 + y^2) dx + (1 + x^2) dy = 0. Now   (1 + y^2) dx + (1 + x^2) dy = 0 => dx/(1 + x^2) + dy/(1 + y^2) = 0.On integration, we get  tan^-1 x + tan^-1 y = tan^-1 C, (x + y)/(1 – xy) = C, X + y = C (1 – xy).

P values

P values
P values are obtained from either the tables or programmable calculators. We can use standard excel sheet as well to calculate the P values. Rather it is tedious to evaluate manually and requires a great mathematical skills.


From Z score
Step 1: Under Normal Distribution, we try to identify the z score for the respective mean and standard deviation value of the null hypothesis using the formula
z= (Test value – mean)/ standard deviation
Step 2: Refer to the Z table and find the P value directly.
To read the P values from the normal distribution table take the first two digits of the z score and locate it on the left most column, move along that row to locate the intersection of third digit of the z score along the column.


Table of P values:
We have two z tables to find the p values. As the normal distribution is symmetric about its mean, and assumes a bell shaped curve, the area right of the mean is equal to the area left of the mean. We have table for the z values varying between 0 and positive infinity, and for the values varying between negative infinity to positive infinity.

P Values Table

P values Significance
The importance of P values in statistical analysis is to test the authenticity of the hypothesis. We need to identify the hypothesis related to the study. In case of null hypothesis, the P value signifies whether the null hypothesis is true or not subject the allowed deviation.
It is imperative to note that the P values calculated should not be less than the allowed significance level to accept the null hypothesis else the alternate hypothesis is accepted.


Finding P values
We adopt the following algorithm to evaluate the P value
Step 1: Propose a null hypothesis
Step 2: Indentify the means and standard deviation
Step 3: Find the Z score
Step 4: Find the P value from the table.

P values in Statistics:
We need a parameter to take an impartial decision. In statistical significance testing, P value gives us the probability of obtaining maximum value of the test statistic as the one that is actually either desired or observed with an assumption that the null hypothesis made is true. We reject the hypothesis if the p- value obtained is less than the significance level, which is usually, 0.05 or 0.01.When the null hypothesis is rejected, and the result proposed is statistically significant.

What is absolute value?

What is absolute value?
Absolute value of number is defined asthe difference between it and zero and is denoted by symbol of |x| or sometime we can use abs(x). Absolute valueismost commonly used by mathematicians and scientists as a tool which is used to separate magnitude and direction when only magnitude matters. Absolute value is positive. The absolute value of a real number is the difference between that real number and zero

If x is the real number then the absolute value for real number |x| = x, when it is positive. |x| = -X, when x is not positive. The absolute value is the distance of x from 0.
The equation for absolute value is,
K = |x-b| ------------- (1)
If x-b is positive
Then k = x - b
X = b + k   ------------- (2)
If x-b is negative
-k = x - b
X = b – k ---------------- (3)
Finding the absolute values.
Example: | x-5 | = 7
From equations 2 and 3 the absolute values are
| x-5 | = 7 | x-5| = -7
X = 7 + 5 x = -7 + 5
X = 12 x = -2
So the absolute values are {12, -2}

Definition for absolute value

The absolute value of a number measures its distance to the origin/zero on the real number line. From above the figure x is 5 units away from the zero. The absolute value is always positive, so it can change the sign of the negative number in to positive number. But the absolute value of zero has no sign, since it is equal to its absolute value.

Absolute value equations:
Isolate the absolute value expression to solve absolute value equations. The expression is called as k. Then write both the equations in which the expression provided within the absolute value symbol is one of the two expressions.  Write the first equation k should be equal to the expression inside the absolute value symbol, and then write the other equation should be equal to –k. Then we get the set of solution obtaining the solution for the equations.

Example: | 2x+12 | = 4x
| x – b | = k
From above equation 2 and 3
| x – b | = k | x – b | = - k
2x + 12 = 4x 2x + 12 = -4x
X = 6 x = -2
So the solution of equation is {-2, 6}

Absolute value in algebra
Absolute value is defined as; it is functions which measure the size of the element as a field or integral domine (D).
|x| = 0
|x| = 0 if and only if x =0
|xy| = |x||y|
|x+y| = |x| + |y|

Example: |1|  = 1
  |-1| = 1

Statistical Mean

In Mathematics in the branch of Statistics, the expression for the mathematical mean of a statistical distribution is the mathematical average of all the terms in the data. To calculate this, we add up the values of the terms given and divide the sum by the number of terms in the data. This expression is also called the
Arithmetic Mean.


While solving word problems, we come across problems like, Scott drove from Bloomington to Chicago driving at 6 different speeds on the same freeway; 63mph, 72mph,42mph, 53mph, 67mph, 59mph. how do we calculate the Mean driving speed? It is very simple, we know that Arithmetic Mean is the regular average, so we just add up the different speeds and divide the sum by the number of different speeds given. Let us see the steps involved,
 The Mean driving speed = total sum of driving speeds /total number of speeds
Total sum of driving speeds = 63+72+56+68+59+48 =366
Total number of different speeds = 6
So, Mean driving speed = 366/6 = 61mph


Definition of Mean Or Definition Mean
To define mean, let us first learn about average in mathematics. Given a data of values, the average would be the total sum of the values divided by the total number of values given in the data. This in statistics is called the Mean or Arithmetic Mean. So, Definition of Mean is the average of the values given in a data.
Mean Formula
The Arithmetic Mean is the regular average of the values given in a data.
So, Mean Formula = Total sum of the values/Total number of values

Example: A student took 8 tests in a subject in one marking period. What is the mean test score?
                88, 75, 82, 90, 86, 78, 94, 87
Solution:  Number of tests = 8
  Sum of the marks =88+75+82+90+86+78+94+87= 680
   Mean = sum of the marks/number of tests
= 680/8 = 85

Sample Mean Formula
The sample mean in statistics, branch of Mathematics is the sum of all observed outcomes from the sample divided by the total number of events (Arithmetic Average). It is denoted by the symbol x with a (bar) above it. The Sample Mean formula is as follows:
X (bar) = (1/n) (x1+x2+x3………xn)

Example: At a car rental shop data was collected on the number of rentals on each of 10 consecutive Saturdays.  50, 44, 96, 38, 39, 40,50,46,47,42. Find the Sample Mean of the rentals.
Solution:   n = 10
Sigma(rentals) = 50+44+96+38+39+40+50+46+47+42=492
X(bar) =sigma(rentals)/n
            = 492/10
            = 49.2 is the Sample mean

Decimal Numbers

Decimals:
The Number system that we follow is called base 10 number system. In this system, we have numbers between 0 and 9, which defines every other number.  Fractional part of a number can be expressed as decimals.
What is a decimal?
Decimal is way of representing a fraction without numerator and denominator. The period or decimal point signifies that the number following it is fractional part. Decimal number can be either greater than or lesser than one. In case of decimal numbers, which are less than one, the leading number is zero.
Example:
39/10 = 3.9 [Decimal number greater than one]
79/100 = 0.79 [Decimal number less than one]
Decimal Place value:
In numbers, position of a number decides its value.
For example in counting number 345, the 3 represents three hundreds, 4 represents four tens and 5 represents 5 units. Similarly, in decimal numbers also, we have place value
Naming decimals:
One tenths = 1/10
One hundredths = 1/100
In 0.7, there are seven tenths.
In 0.79, there are seventy nine hundredths
In 0.798, there are seven hundred and ninety eight thousandths.
Decimal to fraction:
Let us learn how to convert a decimal into fraction.
Step 1: Find the number digits after the decimal point
Step 2: Write the number without the decimal point in the numerator
Step 3: in denominator write 1, followed by as many zeros as the number of digits after the decimal point in the given number
Example: Convert 0.347 into a fraction
Step 1: Number of digits after the decimal point is 3
Step 2: Numerator = 347
Step 3: Denominator = 1000
Hence, the fraction is 347/1000
Decimal to percent:
Let us learn how to convert a decimal number into a percentage.
To convert decimal to percent, multiply the decimal number by 100 and add % sign.
In other words, shift the decimal point to the right and add % sign.
Example: Convert 0.95 into percentage.
0.95 x 100 = 95%
Convert 0.895 into percentage.
0.895 x 100 = 89.5%
Rounding decimals:
Reducing the number of digits without any appreciable change in its value is called Rounding off.  You need to know how many digits you would like to have after the decimal point to round off.
It involves two steps.
Step 1: Decide the number of digits that want to keep in a decimal number.
Step 2: Increase by 1 if the next digit is greater than or equal to 5.
Step 3: Leave it, as it is if the next digit is less than 5.
Example:
Round off 0.789 to nearest hundredths.
0.789 is rounded off to 0.79, as the next digit 9 is greater than 5.
Round off 0.5674 to nearest thousandths.
0.5674 is rounded off to 0.567, as the next digit 4 is less than 5.

Measures of Central Tendency

Central Tendency is one of the main topics of statistics. There are three measures of statistics central tendency. The three measures of central tendency are mentioned bellow:

• Mean
• Median
• Mode

There are again three different types of mean: arithmetic mean, geometric mean and harmonic mean.

Next time i will help you with inference statistics. Till then understand this concept better. For more help you can connect with an online tutor and get your required help. Not just statistics, but you can get help from a trig tutor for trigonometry concepts.

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