End Area Volume Calculation

Introduction to end area volume calculation:

In math, Volume plays vital role in geometry. Volume is used to find how much space occupied by the shape. It is measured in cubic units. Understanding Area of Circle Equation is always challenging for me but thanks to all math help websites to help me out.

In this article, we shall see about calculation of cross section measurement by the average end area method.

We are going to calculate volume of cross section shape by the average end area method using formula.

Let us see about end area volume calculation in detail.

Average End Area Volume Calculation:

The volume formula for average end area is very exact method calculation only for end areas are equal. For end areas are not equal cases, this volume formula calculates slightly larger than its original values. For example, we are going to calculate volume of pyramid, the calculation of volume will be equal 50% of correct volume values.

The formula for calculation of volume by average end area:

Volume = L x `1/2` (A1 + A2) cubic meter.

L – Distance in meters

A1 and A2 – area in Square meters

This average end area calculation is used to calculate volume between two cross sections. That is, two cross sectional areas are averaged and multiplied by the length(distance) between two cross sections to get the volume. Is this topic how to solve linear equations and inequalities hard for you? Watch out for my coming posts.

Let us discuss example problems of average end area calculation.

Example Problem - Average End Area Volume Calculation:

Example Problem:

Find the volume of two cross sections using average end area method for following values.

Length is 20 m, two areas are 150 m2 , 180 m2.

Solution:

Given:

L = 20 m

A1 = 150 m2

A2 = 180 m2

Formula for calculation:

Volume = L x `1/2` (A1 + A2)

Substituting values for length and areas into above formula.

Volume = 20 x `1/2` (150 + 180)

Volume = 20 x `1/2` (330)

Volume = 20 x 165

Volume = 3300

Therefore, volume of two cross sections is 3300 m3.

Thus we solved volume calculation by using average end area method.

Percentage Variance

Introduction for percentage variance:

The percentage of the number is a method of showing the number with the denominator 100 as fraction. The percentage of the number is represented by the symbol “%” or “pct”. Percentage shows the relation of the two quantities, the first quantity is associated with the second quantity. So, first quantity should be larger than zero. But in this article we are going to see percentage variance. I like to share this Definition of Variance with you all through my article.

Equation for percentage variance is `("new value"-"previous value")/"previous"` `xx` 100

Examples for Percentage Variance:

Example 1:

The last year of the bag was soled at the price 57. In this year, the bag is selling with the price 28. Calculate percentage variance by using formula.

Solution:

Percentage variance = ` ("new value-previous value")/"previous value"` `xx` 100

= `(28-57)/57` ` xx` 100

= `-29/57` `xx` 100

= -0.5088 `xx` 100

= -50.88%

Therefore, 50.88% loss in selling of bag in this year when we compare with the last year.

Example 2:

The last year of the basket was soled at the price 39. In this year, the basket is selling with the price 58.

Solution:

Percentage variance = `("new value-previous value")/"previous value"` `xx` 100

= `(58-39)/39` `xx` 100

= `19/39` `xx` 100

= 0.4872 `xx` 100

= 48.72%

Therefore, 48.72% percentage is profit in selling of basket in this year when we compare with the last year.
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Practice Problems for Percentage Variance:

Problem 1:

The last year of the washing machine was soled at the price 48. In this year, the washing machine is selling with the price 52. Calculate percentage variance by using formula.

Solution:

8.33% profit in selling of washing maching in this year when we compare with last year

Problem 2:

The last year of the TV was soled at the 50 . In this year, the TV is selling with the price 72. Calculate percentage variance by using formula.

Solution:

44% profin in selling of TV in this year when we compare with the last year.

Problem 3:

The last year of the laptop was soled at the price 89. In this year, the laptop is selling with price 20.

Solution:

77.53% loss in selling of laptop in this year when we compare with the last year.

Problem 4:

The last year of the computer was soled at the price 89. In this year, the computer is selling with the price 83. Calculate percentage variance by using formula.

Solution:

6.74% loss in selling of computer in this year when we compared with the last year

Information on Math Functions

Introduction to information on math functions:

In this article information on math functions, we will refer definition of a function and some worked example problems. Function is defined as a equation where the terms present in one side will vary with the terms present on the other side. For example x=5b+6c here x is a dependent variable of b and c, since b and c are independent variables.

Information on Math Function

1.Function is defined as a collection of mathematical function which produce the result after performing the calculation.

For example y=(4x+5x)+`(5x)/2` -4(x+2)

After performing the right hand side calculation which produces the result.

2.The function is used to identify or compare the relationship between two numbers

y=f(x)=x+7

Here the value y is greater than x.

3.There are lot of functions like

logarithmic function Example: f(x)=log x

Trigonometric function Example: f(x)= sin x

Geometric function Example `y=x^2`

Exponential function `f(x)=e^x`. Understanding square root of 7 is always challenging for me but thanks to all math help websites to help me out.

Worked Example Problems - Information on Math Functions

Example problem 1- information on math functions:

The price of one lemon and watermelon is $80 then what is the price of 5 jack fruit and 5 watermelon?

Solution:

Consider lemon as x

Watermelon as y

So from the given statement x + y = 80-------1

The price of 5 lemon and 5 watermelon, 5x + 5y = ?

We can write the above equation as

The price of 5 lemon and 5 watermelon = 5(x+y)

From the first equation we can substitute the x+y value

The price of 5 lemon and 5 watermelon =5(x+y)

= 5(80) = 400

The price of 5 lemon and 5 watermelon = $400

Example problem 2 - information on math functions

If log8=0.903 then find the value of log16?

Solution:

The number 8 can also be written as` 2^3`

They have given that log` 2^3=0.903`

3 log2=0.903

So log2=0.3010

The value of log 16 is given by `log 2^4 =4 xx log2`

`=4 xx 0.3010`

log 16 =1.204

Example problem 3- information on math functions

Calculate the value of `e^2 xx e^3` =?

Solution:

From the above statement `e^x xx e^y=e^(x+y)`

So `e^2 xx e^3=e^(2+3)=e^5=148.4`

Example problem 4- information on math functions

Consider `cos x = (2/5)` .Find out the value for sin x=?

Solution:

From the statement `sin^2 x+cos^2 x=1`

`sin^2 x=1- cos^2 x`

`sin x=sqrt(1- cos^2 x)`

=`sqrt[1-(2/5)^2]`

=`sqrt[1-(4/25)]`

=`sqrt(21/25)`

Divide Imaginary Numbers

Introduction about imaginary numbers:

The term "numbers" are used to measure the quantity. The term "numbers" are a fixed value, if it is integers or constants. It is also refer to a complex numbers, real number, imaginary numbers, etc. The complex numbers are numbers involving, there is no number that when squared equals -1. The square root of minus one is denoted by mathematical symbol ' i '. A number contains, i  am the co-efficient of that numbers are called imaginary numbers. Having problem with What is Composite Number keep reading my upcoming posts, i will try to help you.


Evaluation of Imaginary Numbers:

In the definition (`sqrt(-1)` )2should equal -1, but it does not, therefore the symbol i =`sqrt(-1)` is introduced,

i.e.,         `sqrt(-x)`      = i `sqrt(x)`  Where x is a positive number and (i)2 = -1.

The square roots of other than -1 of the negative numbers format is, `sqrt(-n)` = i`sqrt(n)` .

The important rules for symbol 'i' :

i-1 = `1/ i` = `i / (i^2)` = `i /(-1)` =  -i      ( reciprocal of symbol ' i ' is ' -i ' )

The powers of symbol ' i ' :

i1 = i

i2 = -1

i3 = i2 i = (-1)(i) = -i

i4 = i3 i= (-i)(i) = -(i)2 = -(-1) = 1

i5 = i4 i = (1)( i) = i

..........

Complex numbers are numbers involving i and are generally in the form:

iy `hArr` (real number)(i)  `hArr` imaginary number

x + iy   (real number) + (i)(real number) `hArr` complex number

Dividing Imaginary numbers

In the division of complex number operations performed through learn complex numbers online, first find out the complex conjugate of the denominator. Second we multiply the conjugate of complex number to both numerator and denominator of complex number. The conjugate of complex number of (x + iy) is (x - iy) and vice versa.

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Example 1:dividing Imaginary Numbers.

(a). `(2 + i3) /(4-i2)`

Solution:

Given:

` (2 + i3) /(4-i2)`

Dividing imaginary numbers:

`(2 + i3) /(4-i2)` = `(2 + i3) /(4-i2)` . `(4 + i2) /(4 +i2)`

= `((2 + i3)(4 + i2))/((4-i2)(4 + i2))`

= `((2 + i3)(4 + i2)) / (4^2 + 2^2)`

= `(8 + i4 + i12 -6)/(16 + 4)`

= `(2 + i16) / (20)`

= `(2)/(20)` + `i(16)/(20)`

= `1/(10)` + `i (4/5)`

Example 2:dividing Imaginary Numbers.
(a). `(6 + i2)/(-2+i3)`

Solution:

Given:

`(6 + i2)/(-2+i3)`

Dividing imaginary numbers:

`(6 + i2)/(-2+i3)`   = `(6 + i2)/(-2 + i3)` . `(-2-i3)/(-2-i3)`

= `((6 + i2)(-2-i3))/((-2 + i3)(-2-i3))`

= `(-12 +i18-4i -6)/(-2^2 + 3^2)`

= `(-20 + i14)/(4+9)`

= `(-20 + i14)/(13)`

=`(-20)/(13)` +`i(14)/(13)`    [Answer.]

Introduction to variance of discrete random variable:

The discrete random variables have the countable values for their probability. The variance value for the discrete random variable is calculated with the help of the mean. The random variables are said to discrete random variable when the sum of their probability is one. The online produces the link between the tutor and the students. With the help of the online the students clarify their doubts. This article contains the information about the discrete random variables in the probability theory and statistics.

Formula Used for Solve Online Discrete Random Variables:

The formulas used to determine variance for the discrete random variable are

Variance = `sum p(x) (x^2) - (sum x p(x)) ^2`

In this above mentioned formula the `(sum x p(x)) ^2` is the squared value of the mean in the statistics.

In the above formula x denotes the given set of discrete random values and p(x) denotes the probability value for the discrete random variables in the statistics.Having problem with what are exponents keep reading my upcoming posts, i will try to help you.

Examples for Solve Online Discrete Random Variables:

Example 1 to solve online discrete random variables:

Predict the mean, variance and standard deviation for the discrete random variables.

x	2	4	6	8	9
P(x)	0.23	0.21	0.30	0.11	0.15

Solution:

Mean = `sum x p(x)`

Mean = 2 (0.23) +4 (0.21) +6 (0.30) +8 (0.11) + 9(0.05)

Mean = 0.46+ 0.84 + 1.8 + 0.88 + 0.45

Mean = 4. 43

Variance = `sum p(x) (x^2) - (sum x p(x)) ^2`

Variance = ((0.23) (2) (2) + (0.21) (4) (4) + (0.30) (6) (6) + (0.11) (8) (8) + (0.15) (9) (9)) - (4.43)2

Variance = (0.92+ 3.36+ 10.8 + 7.04+ 12.15) - 19.6249

Variance = 34.27 -21.5296

Variance = 46.42

The variance for the discrete random variable is 46.42.

Example 2 to solve online discrete random variables:

Predict the mean, variance and standard deviation of discrete random variables.

x	10	20	-30	40	50
P(x)	0.22	0.12	0.25	0.05	0.36

Solution:

Mean = `sum x p(x)`

Mean = 10(0.22) +20(0.12) - 30(0.25) +40(0.05) + 50(0.36)

Mean = 2.2 +2.4 -7.5+ 2+ 18

Mean = 17.1

Variance = `sum p(x) (x^2) - (sum x p(x)) ^2`

Variance = ((0.22) (10) (10) + (0.12) (20) (20) + (0.25) (-30) (-30) + (0.05) (40) (40) + (0.36) (50) (50)) - (17.1)2

Variance = (22+48+ 225 + 80 +900) -292.41

Variance = 1275 - 292.41

Variance = 982.59

The variance for the discrete random variable is 982.59.

Open Number Line

Introduction to open number line
A definition of number line is placed in the correct position of numbers. The number line is generally represented the positive and negative numbers. Number line is commonly performing by the addition and subtraction operation. The general format of the line number is {-5, -4, -3, -2, -1, 0, 1, 2, 3, 4, 5}. Next we discuss about this articles open number line.

Open Number Line

The open number line is defined the empty line used in the operations are addition and subtraction. Open number line is generally used in the three parts. Those parts are organ of open line number, positive value of open line number and negative open line number.

The positive value means larges values of the open line number it is represented the 25, 35 and etc.

The negative line number means smallest values of the open line number it is represented the -50, -75 and etc.Understanding how to solve matrices is always challenging for me but thanks to all math help websites to help me out.

Example Problem of Open Number Line

Problem 1:

Calculate the following numbers using open line numbers.

(77 + 2 + 8 + 3)

Solution

The number is 77 + 2 + 8 + 3

The above all numbers are positive number. So that numbers are move from the open line number is right side only.

Step 1: Move 2 units from right side of the open line number. The answer is 79.

Step 2: Move 8 units from right side, the answer is 87

Step 3: Move 3 units from right side, the answer is 90.

The answer of the 77 + 2 + 8 + 3 is 90.

Problem 2:

Calculate the following numbers using open line numbers.

(63 + 12 - 3)

Solution

The number is 63 + 12 - 3

The above numbers are positive and negative number. So positive numbers are move from the right side and negative number are move from left side of the open line number.

Step 1: Move 12 units from right side of the open line number. The answer is 75.

Step 2: Move 3 units from left side, the answer is 72

The answer of the 63 + 12 - 3 is 72.

Addition of Whole Numbers

Introduction to addition of whole numbers:

One of the arithmetic operations is addition. In mathematics, addition represents to combining group of numbers together into a bigger collection. Add symbol denoted by the plus sign (+). For example, assume there are 2 + 2 objects meaning two objects and two other objects, add the form of object is four objects. Therefore, 2 + 2 = 2.

Whole numbers includes all integers. Addition of whole numbers is nothing but it is addition operation between whole numbers. For example, 3 + 2 + 5 = 10. The following are the arithmetic operation like addition, subtraction, multiplication and division performed on whole numbers.

Examples Problem for Addition of Whole Numbers:

1. Find the sum of value for given whole numbers, 2552, 5542, 7624, and 6751 using addition operation.

Solution:

Given whole numbers are 2552, 5542, 7624, and 6751,

Steps for addition of whole number,

First, arrange the numbers with placing values like ones under ones, tens under tens.
Addition operation starts with right side of the column.
Then add the each column in which sum is 9 or less than 9 means write the sum.
Else, sum is 10 or more than 10 means re-group the carry tens on the next column and write ones of that sum.Please express your views of this topic cbse class 12 question papers by commenting on blog.
2 1  --> carry

2,552

5,542

7,624

`+` 6,751

22,469

2  --> carry     1 --> carry

2 thousands +  5 hundreds +  5 tens +  2 ones

5 thousands +  5 hundreds +  4 tens +  2 ones

7 thousands +  6 hundreds +  2 tens +  4 ones

6 thousands +  7 hundreds +  5 tens +  1 ones

22 thousands +  4 hundreds +  6 tens +  9 ones

Answer for addition of whole numbers is 22469 or twenty two thousand four hundred and sixty nine.

2. Find the sum of value for given whole numbers, 6552, 9252, and 2751 using addition operation.

Solution:

Given whole numbers are 6552, 9252, and 2751,

Steps for addition of whole number,

First, arrange the numbers with placing values like ones under ones, tens under tens.
Addition operation starts with right side of the column.
Then add the each column in which sum is 9 or less than 9 means write the sum.
Else, sum is 10 or more than 10 means re-group the carry tens on the next column and write ones of that sum.
1 1 --> Carry

6,552

9,252

`+` 2,751

18,555

1 -> carry        1  --> carry

6 thousands +  5 hundreds +  5 tens +  2 ones

9 thousands +  2 hundreds +  5 tens +  2 ones

2 thousands +  7 hundreds +  5 tens +  1 ones

18 thousands + 5 hundreds +  5 tens +  5 ones

Answer for addition of whole numbers is 18555 or eighteen thousand five hundred and fifty-five.

Practice Problems for Addition of Whole Numbers:

1. Find the sum of value for given whole numbers, 52, 542, 624, and 651 using addition operation.

Answer for addition of whole numbers is 1869.

2. Find the sum of value for given whole numbers, 25, 54, 72, and 751 using addition operation.

Answer for addition of whole numbers is 902.

Frequency Tables and Histograms

Introduction of frequency tables and histogram:

Frequency distribution represent  a histogram is by means of the rectangles whose width represents class intervals and areas are directly proportional to the corresponding frequencies.” A frequency tables distribution applied for continuous quantitative data. The tabular collection of data is frequency distribution viewing how many interpretation lie above, or below the records. Histograms are helpful for interpreting tasks. The collections of groups of data called classes, and in the context of a histogram, they called as bins.

Constructing a Frequency Tables and Histogram:

Step 1 – Calculate the quantity of data
Step 2 - Review on a tally sheet
Step 3 – Calculate the range
Step 4 – Find the number of intervals
Step 5 - Calculate interval width
Step 6 - Identify interval of initial points
Step 7 - Calculate number of data in each interval
Step 8 - Plot the data
Step 9 - Add designation and tale


Explanation about Frequency Tables and Histogram:
A range of the Histogram equally separates all possible values in a data set of groups or classes. For each set of  a 4-sided figure constructed with a base length, which is identical to the collection of values in that definite group, and an area directly proportional to the number of observations decreasing into that group.

The height of a four-sided figure, which is the same to the frequency density of the interval, i.e. the frequency, should split by the breadth of the interval. The total region of the histogram that exactly is identical to the number of data.I like to share this 6th grade math homework with you all through my article.

Frequency Tables:

Marks           Frequency       Cumulative frequency

0-5                     3                 3

5-10                    9                12

10-15                  18                30

15-20                  26                56

20-25                  30                86

25-30                  38               124

i) Plot a points of coordinates such as (0,5),(5,10),(10,15),(15,20),(20,25),(25,30) and the related frequency can be identified and make the points.

ii) Take x -axis 1 cm = 5 unit, y -axis 1cm = 1 c.f.

iii) The number line represent by the horizontal axis,  which displays the data in equal interval. The vertical axis represents the frequency of each bar.

Trigonometric Functions Calculus

Introduction of Trigonometric Functions Calculus:

Trigonometry came from the Greek words ‘trigon-triangle’ and ‘metron-measure’. Trigonometric functions generally define the functions of angle. An equality that satisfies for any value of the variable is generally defined as an identity. An equation is an equality, which is true only for several values of the variable. An identity gives more clear and convenient form than the expression. Trigonometric function calculus are generally applied in modeling periodic phenomena and study of triangles.

Types of Identities in Trigonometric Functions Calculus:

There are several identities in trigonometry function calculus. They are as follows,

Reciprocal identity,
Tangent and Cotangent identity,
Pythagorean identity,
Co-function identity,
Even-Odd identity.

Example Problems for Trigonometric Function Calculus:

Ex 1:

Prove that, sec^2 a + csc^2 a = sec^2 a × csc^2 a, Using Trigonometric Functions calculus.

Proof:         L.H.S. = sec^2 a + csc^2 a

=> (1 / cos^2 a) + (1 / sin^2 a)                                    [Using Reciprocal Identity ]

=> (sin^2 a + cos^2 a) / sin^2 a × cos^2 a                   [Taking L.C.M. ]

=> 1 / sin^2 a × cos^2 a                                               [By Pythagorean Identity]

=> 1 / sin^2 a × 1 / cos^2 a                                          [Separating it Into Two Terms ]

=> csc^2 a × sec^2 a                                                   [Using Reciprocal Identity ]

=> sec^2 a × csc^2 a

=> R.H.S.

Hence proved that, sec^2 a + csc^2 a = sec^2 a × csc^2 a

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Ex 2:

Prove that, sin4 a - 2sin^2 a cos^2 a + cos4 a = cos^2 (2 a), Using Trigonometric Functions calculus.

Proof:

L.H.S. = sin4 a - 2 sin^2 a cos^2 a + cos^4 a

=> (sin^2 a)2 - 2sin^2 a cos^2 a + (cos^2 a)2

=> (sin^2 a - cos^2 a)2                                                [ Using ( a - b )2 = a2 - 2ab + b2 ]

=> ((1 - cos^2 a) - cos^2 a)2                                       [By Pythagorean identity ]

=> (1 - 2cos^2 a)2

=>( -(2cos^2 a - 1))2

=>( -cos 2a)2                                                          [ Using Formula: cos 2T = 2 cos^2T - 1 ]

=>cos^2 (2 a)

=> R.H.S.

Hence proved that, sin4 a - 2sin^2 a cos^2 a + cos^4 a = cos^2 (2 a).

Standard Deviation Dispersion

Introduction to standard deviation dispersion:

Standard Deviation can be defined as the measurement of describing the variability of given data set. Standard deviation is also used for measuring the exact value of the number in the given data set. Dispersion is defined as the measuring that’s also including the average deviation, variance, and then standard deviation. The standard deviation and the variance are most widely used for measuring the dispersion.Please express your views of this topic What is Dispersion by commenting on blog.

Formula for Standard Deviation Dispersion:

The formula for the standard deviation dispersion are given below,

`Variance = (sigma)^2 = (sum(x_r - mu)^2)/N`

The standard deviation, `(sigma)` , is the square root of the variance.

What the formula means:
(1)` x_r -mu` – take the mean value and subtract the mean value with each value given.
(2) `(x_r - mu)^2` – square the each of the results obtained from the step -1.this is to get or remove of any minus signs from the value.
(3) `(sum(x_r - mu)^2)` – this is for adding all the values that is obtained from step-2.
(4) Divide step-3 by N, which is the total number of the given values
(5) For obtaining the standard deviation square root the answer to the step-4.

Example Problems for Standard Deviation Dispersion:

Example problem 1:

Calculate the variance and also standard deviation for the following values: 1, 3, 5, 6, 6, 8, 9, and 10.

Solution:

The mean = `(1+3+5+6+6+8+9+10)/ 8 `
= `48/8`
=6

Therefore (`mu` =6)

Step 1: Subtract the mean value from given values

(1-6)=-5
(3-6)=-3
(5-6)=-1
(6-6)= 0
(6-6)= 0
(8-6)= 2
(9-6)= 3
(10-6)=4

Step 2: In this step we have to calculate the square for the mean values 25, 9, 1, 0,  0, 4, 9, 16.

Step 3: After finding square mean ,sum the square mean 25+9+1+0+0+4+9+16=64.

Step 4: Here , N = 8, therefore variance should be =` 64/ 8` = 8

Step 5: Therefore standard deviation = 2.8284.

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Example problem 2:

Calculate the variance and also standard deviation for the following values: 1, 2, 3, 4, and 5.

Solution:

The mean = `(1+2+3+4+5)/ 5`
= `15/5`
=3

Therefore (`mu` =3)

Step 1: Subtract the mean value from given values

(1-3)=-2
(2-3)=-1
(3-3)=0
(4-3)= 1
(5-3)= 2

Step 2: In this step we have to calculate the square for the mean values 4, 1, 0, 1, 2

Step 3: After finding square mean ,sum the square mean 4+1+0+1+2=8.

Step 4: Here , N = 5, therefore variance should be = `8/5` = 1.6

Step 5: Therefore standard deviation = 1.264.

Eccentricity of Conic Sections

Introduction to eccentricity of conic sections:
A conic is the locus of a point in a plane such that the ratio of its distance from a fixed point in the plane to its distance from a fixed line in the plane is a constant. Here, the fixed point is called the focus, the fixed line is called the directrix and the constant ratio is called the eccentricity of the conic. It is denoted by ‘e’.

(i)     If e = 1, the conic is called the parabola.

(ii)    If e < 1, the conic is called the ellipse

(iii)   If e > 1, the conic is called the hyperbola

(iv)   If e = 0, the conic is called the circle.

Now, let us see few problems on this topic eccentricity of conic sections.

Example Problems on Eccentricity of Conic Sections

Ex 1. Find the eccentricity of conic  x2/ 36 + y 2 / 64 = 1

Soln: Given: `x^ 2 / 64` +` y ^2 / 36 ` = 1 ………… (1)

Here (1) represents an ellipse. To get the eccentricity, are follow this formula: e 2 = 1 – `b ^2 /a^ 2` .

Here a = `sqrt 64` , b = `sqrt 36`     [This we get from the standard form of an ellipse

`x^ 2 / a^ 2` +` y ^2 / b ^2` = 1        Here a >b]

a = 8, b = 6

Therefore, e 2 = 1 – `6 ^2 / 8^ 2`

= 1 – `36 / 64`    = 1 – `9 / 16`    = 16 – `9 / 16`    = `7 / 16`

Therefore e = `sqrt7 / 16`    [Eccentricity is always positive]

Therefore e = `sqrt7 / 4`

Ex 2: Find the eccentricity of the conic:` x ^2 / 49` +` y ^2 / 225` = 1

Sol: Given: ` x^2 / 49` +` y ^2 / 225` = 1

Here a = 7, b = 15,   = a < b

Therefore e 2 = 1 – `a^ 2 / 15^ 2` = 1 – ` 49 / 225`

= `(225 ** 49) / 225`   = `176 / 225`

Therefore  e = `sqrt (176 / 225)` = `sqrt ((16 * 11) / 225)`    =  `(4sqrt 11) / 15`

Ex 3: Find the eccentricity of the conic: ` x^ 2 / 625 **y ^2 / 144`   = 1

Soln: Given:` x^ 2 / 625 ** y ^2 / 144` = 1 ………… (1)

Here (1) represents a hyperbola

The eccentricity is given by e 2 = 1 + `(b^2/ a^2)`

= 1 + `144 / 625`      =` 769 / 625`

Therefore e = `sqrt 769 / 25` .

Ex 4: Find the eccentricity of the conic` x ^2/ 25 ** y ^2 / 36` = 1

Soln: Given:` x ^2 / 25** y ^2 / 36` = 1 ………….. (1)

This is a hyperbola.  a = 5, b = 6

Therefore e 2 = 1 + `b ^2 / a^ 2` = 1 + `36 / 25` = `61 / 25`

Therefore e = `(sqrt 61) / 5`

Note: Here the fraction  b 2 /  a 2 will change as whether a < b or a >b.

Practice Problems on Eccentricity of Conic Sections

Find the eccentricity of the following conics:
A) `x^2/16 + y^2/9` = 1

[Ans: e = `sqrt 7 / 4` ]

B)    ` x ^2 / 25 + y ^2 / 64` = 1

[Ans: e = `sqrt 39 / 8` ]

2. Find the eccentricity of the following conics:

A)    `x ^2 / 64 ** y ^2 / 36` = 1

[Ans: e = `5 / 2` ]

B)    ` x ^2 / 81** y ^2 / 144` = 1

[Ans: `5/3` ]

Sample Algebra Functions

Introduction of sample algebra functions

Algebra is the branch of mathematics deals with finding unknown variable from the given expression with the help of known values. The variable of algebraic are represents alphabetic letters.In algebra the numbers consider as constants, algebraic expression may include  real number, complex number, matrices and polynomials. In algebra several identities need to find the x values by using this we can simply find  the algebraic expression of the particular function. The sample algebra functions may include in the function of f(x), p(x),… to find the x value of the algebra functions.

For example

Sample algebra functions f(x) = 2x2+3x + 4. In this sample algebra function we have  to find the function of f(2).

Sample Algebra Functions Problem

Problems using the square of the sample algebra functions

Problem1:Using  square  problem in sample algebra functions

f(x) = x2 +2x +4 find the f(4).

Solution :

Given the function of f(x) there is x value is given find the sample algebra functions of

f(x) = x2 +2x +4 find the f(4).

The value of x is4 is given

f(4) = 42 + 2*4 + 4

f(4) = 16 + 8 + 4 In this step 4 square is 16 it is calculate and 2*4 is 8 be added

f(4) = 28

Sample algebra functions of f(4) = x2 +2x +4 find the f(4) is 28.

Problem2; Using  square  problem in sample algebra problem

f(x) = x2 + 5x + 5 find the f(5).

Solution :

Given the function of f(x) there is x value is given find the sample algebra functions of

f(x) = x2 + 5x + 5 find the f(5).

The value of x is 5 is given

f(5) = 52 + 5*5 + 5

f(5) = 25 + 25 + 5 In this step 5 square is 25 it is calculate and 5*5 is 25 be added

f(5) = 55

The algebra 2 function of the f(5) = x2 + 5x + 5 find the f(5) is 55.

Having problem with cbse board exam question papers keep reading my upcoming posts, i will try to help you.

Sample Algebra Functions Using the Cubic Equation

Problems 1: using the cubes equation in sample algebra functions:

f(x) = x3 + 2x2 + 2x + 4 find the sample algebra function f(2).

Solution

Given the function of  f(x) there is x value is given find the sample algebra functions

f(x) = x3 +2x2 + 2x + 4 find the f(2).

Here the value of x is given as 2

f(2) = 23 + 2*22 + 2*2 +4

f(2) = 8 + 8 + 4 + 4 In this step 23 is calculated  as 8 and 2 square is 4

f(2) = 24.

The sample algebra functions of the f(x) = x3 +2x2 + 2x + 4 find the f(2) = 24.

Problems 2: using the cubes of the sample algebra functions

f(x) = 2x3 +4x2 + 3x + 4 find the sample algebra functions of f(4).

Solution

Given the function of  f(x) there is x value is given find the sample algebra functions

f(x) = 2x3 + 4x2 + 3x + 4 find the f(4).

Here the value of x is given as 4

f(4) = 2*43 +  4*42 + 3*4 + 4

f(4) = 128 + 64 + 12 + 4 In this step 4 cubes  is calculated  as 64 and multiplied by 2 we get as 128.Then the 4 square is 16 and multiplied by 4 we get 64.

f(4) = 208

Sample algebra functions of the f(x) = 2x3 +4x2 + 3x + 4 find the f(4) = 208.

Multiplication Rule for Independent Events

Multiplication Rule for Independent Events

Probability is the likelihood of the occurrence of an event. An event is a one or more possible outcomes of a certain experiment. An event is called independent event if one event does not affect the other event. For example, choosing a 7 and 8 in the deck of card with replacement is two independent events. An event consisting of more than one simple event is called compound event.

Multiplication rule for two independent events:

If A and B are two independent event then; P(A and B) = P(A) · P(B)

Multiplication rule for three independent events:

If A, B, and B are three independent events then; P(A and B and C) = P(A) · P(B) · P(C)

Multiplication Rule for Independent Events - Example Problems

Example 1: Two coins are drawn at one by one with replacement of previous coin in a bag of 12 nickels and  9 quarters. Find the probability of first coin is nickels and second coin is quarters.

Solution:

Lest S be the sample space, n(S) = 12 + 9 = 21

A be the event of drawing a nickels, n(A) = 12

B be the event of drawing a quarters, n(B) = 9

P(A) = `(n(A))/(n(S))` = `12/21` = `4/7`

P(B) = `(n(B))/(n(S))` = `9/21 ` = `3/7`

P(A and B) = P(A) · P(B) = `4/7` · `3/7` = `12/49`

P(A and B) = `12/49`

Example 2: A jar contains 6 dark, 8 milk, and 10 white chocolates. Three chocolates are drawn one by one with replacement. What is the probability of drawing each kind is one time?

Solution:

Lest S be the sample space, n(S) = 6 + 8 + 10 = 24

A be the event of drawing a dark chocolate, n(A) = 6

B be the event of drawing a milk chocolate, n(B) = 8

C be the event of drawing a white chocolate, n(B) = 10

P(A) = `(n(A))/(n(S))` = `6/24` = `1/4`

P(B) = `(n(B))/(n(S))` = `8/24` = `1/3`

P(C) = `(n(C))/(n(S))` = `10/24` = `5/12`

P(A and B and C) = P(A) · P(B) · P(C) = `1/4` · `1/3` · `1/12` = `1/144`

P(A and B and C) = `1/144`
I am planning to write more post on solving complex rational expressions, geometric probability formula. Keep checking my blog.
Multiplication Rule for Independent Events - Practice Problems

Problem 1:   Two chocolates are drawn one by one with replacement of previous chocolate in a jar of 10 bitter and  5 dark chocolates. Find the probability of first one is bitter and dark chocolate.

Problem 2: A jar contains 11 dark, 12 milk, and 13 white chocolates. Three chocolates are drawn one by one with replacement.What is the probability of drawing each kind exactly one time.

Answer: 1) `2/9` 2) `143/3888`

Example of Conjunction

Introduction to conjunction example:

The words which combine simple statements to form compound statements are called connectives. We use the connectives ‘and’, ‘or’, etc., to form new statements by combining two or more statements. But the use of these connectives in English language is not always precise and unambiguous. Hence it is necessary to define a set of connectives with definite meanings in the language of logic, called object language. Three basic connectives a conjunction which related to the English word ‘and’, ‘disjunction’ which related to the word ‘or’ ‘negation’ which corresponds to the word ‘not’.

Conjunction Example
Conjunction:

Use the symbol “?” to denote conjunction, “?” to denote disjunction and “~” to denote negation. If two simple statements p and q are connected by the word ‘and’, then the resulting compound statement ‘p and q’ is called the conjunction of p and q and is written in the symbolic form as ‘p ? q’.

Rule:

(A1) the statement p ? q has the truth value T whenever both p and q have the truth value T.

(A2) The statement p ? q has the truth value F whenever either p or q or both have the truth value F.

Ex 1:  Form the conjunction of the following simple statements

P: jasmine is beautiful.

q: Joseph is intelligent.

P ? q: Jasmine beautiful and Joseph is intelligent.

Ex 2 : Form the conjunction of the following simple statements

P: Joseph goes to college.

q: Jake is intelligent.

P ? q: Joseph goes to college and Jake is intelligent.

Ex 2 : Convert the following statement into symbolic form:

‘Kim and John are going to hotel’.

The given statement can be rewritten as:

‘Kim is going to hotel’, and

‘John is going to hotel’.

Let p: Kim is going to hotel.

q: John is going to hotel.

The given statement in symbolic form is p ? q.

Conjunction Practice Problems

Form the conjunction of

Pro 1:  p: Jessica reads newspaper,  q : Jessica plays cricket.

Pro 2:  p: I like coffee. q: I like ice-cream.

Probability Formula Wiki

Introduction to Probability Formula Wiki

The probability is the sketch of match up information or scheme that an occurrence will happen. The probability of an event A should be represented as P(A). The probability value is always the range from 0 and 1. The wiki answers are used for checking our resultant of problems. In this article, we are going to learn about formula and examples in probability wiki.

Formula for Probability Wiki

The formula for probability of an event A is measured as the following formula,

This is the formula that is used for calculating probability of any event.

Example Problems - Probability Formula Wiki

Example 1:

Jockey tossing three coins at concurrently. Calculate the probability of receiving one head or more than one tail?

Solution:

The sample space, when three fair coins tossed is,

S = {HHH, HHT, HTH, THH, HTT, THT, TTH, TTT}

n(S) = 8

Let us consider A will be the event of getting one head,

n(A) = {HTT, THT, TTH} = 3

Let us consider B will be the event of getting more than one tail,

n(B) = {HTT, THT, TTH, TTT} = 4

P(A) = `(n(A))/(n(S))` = `3 / 8`

P(B) = `(n(B))/(n(S))` = `4 / 8`

P(A or B) = P(A) + P(B)

= `3/8` + `4/8`

= `7/8`

= 0.875

Example 2:

Calculate the probability of selecting a letter ‘S’ from the word ‘RESPONSIBILITY’?

Solution:

Given word is, “RESPONSIBILITY”.

Here, total letters, n(S) = 14

Let consider G be the event of selecting a letter ‘S’ from the given word.

Total number of letter ‘S’, n(G) = 2

So, probability of selecting letter ‘S’, P(G) = `(n(G))/(n(S))`

= `2/14`

= `1/7`
Between, if you have problem on these topics how many prime numbers from 1 to 100, please browse expert math related websites for more help on solving systems of linear differential equations by elimination.
Example 3:

There are totally 34 handbags are in a shop. In those handbags, 8 are Black color, 12 are Maroon color, 5 are Grey color and remaining are Purple color. Calculate the probability for selecting i) Black color handbags ii) Grey color handbags?

Solution:

Total number of handbags, n(S) = 34

Number of Black color handbags = 8

Number of Maroon color handbags = 12

Number of Grey color handbags = 5

Number of Purple color handbags = 34 – (8+12+5)

= 34 – 25

= 9

Let us consider G be the event of selecting Black color handbags.

So, n(G) = 8

P(G) = `(n(G))/(n(S))`

= `8/34`

= `4/17`

Let us consider H be the event of selecting Grey color handbags.

So, n(H) = 5

P(H) = `(n(H))/(n(S))`

= `5/34`

That’s all about probability formula wiki.

Various Forms of the Equation of a Line

Introduction to various forms of the equation of a line:

Definitions

The standard form of the equation line is defines as

Ax+By =C

Where A,B and C are represented as real numbers.
The point slope form of the equation of a line is represented given below
The equation of a line with slope represents m and passing through the point (x1,y1) is given by
y-y1 = m (x=x1)

Where m is the slope and x1, y1 is the point given
The slope intercept form of the equation of a line:
The equation of a line with slope m and y intercept (0,b) is given by
Y= mx+b

Where m is the slope and (0,b)is the y intercept.
If an equation represents a straight line is referred as the equation of the line. The following conditions are used to solve the lne equation.
when the slope and  y intercept is given
when the slope and a point is given.
when two points given for the line

Practice Problems for Various Forms of the Equation of a Line:

Solving the Equation of a Line when slope and y intercept is given:

The slope-intercept form of a line is y= mx + b,
where m is the slope

b is the y-intercept.

Example

Find the equation of the line with slope -4 and y-intercept 6.

Solution:                m = -4  and  b = 6

The general equation is y = mx + b
Substitute the values into equation, we get
y = -4x + 6

Hence, the equation of the line is y = -4x + 6

Solving  the Equation of a Line if slope and a point is given:

various forms of the equation of a line - Example: 1

Find the equation of a line in slope intercept form, if the slope is  -3 and passes through (5, 8)

Step 1: Use the slope-intercept form of a line: y = mx + b

Given m = -3     Hence y = - 3x + b

Step 2 : Substitute values into equation:

The y-intercept was not given. However, we are given the point, (5, 8). Thus x = 5 and y = 8
Substitute the values to find the y-intercept b
y  =  mx + b
8  = -3(5) + b
8  = -15 + b
23 = b
Step 3: Solution

apply  the value of b, we get  y = -3x + 23

Hence, the equation of the line is y = -3x + 23

Solving the Equation of a Line passing through two given points:

Example:

Find out the line that passes through (3, 5) and (7, 25)

Steps:

1)      Use the two points to find the slope using slope formula

2)      Use the slope and either one of the points to find the value the y-intercept.

Step 1:                       Slope          =   (y2– y1) / (x2 – x1)

=   (25– 5) / (7 – 3)

= 20/4 = 5

Step 2 :  Let’s choose the point (3,5)
y   =  mx + b
5  =  5(3) + b
5  = 15 + b
5 – 15 = b
-10 = b

Substituting the values in the general equation, we get

y = 5x -10

Hence the equation of the line is y = 5x -10.


Example for Various Forms of the Equation of a Line:

Example Problem 1

solve the standard equation of the line that passes through the points
( -2, 5 ) and ( 3, 8 ).

Solution:

The first step is to determine the slope of the line. According to determine the slope of the line, we must use the formula
m = ( y2 - y1 ) / ( x2 - x1 ). This gives us,

m = ( y2 - y1 ) / ( x2 - x1 )
m = ( 8 - 5 ) / ( 3 - ( -2 ) )
m = ( 8 - 5 ) / ( 3 + 2 )
m = 3/5

Therefore, the slope of the line is equal to 3/5.

by applying the point-slope formula. To do so, we must choose one of the points, ( x1, y1 ), and insert it and the slope into the formula which will give us,

y - y1 = m ( x - x1 )
y - 5 = ( 3/5 )( x - ( -2 ) )
y - 5 = ( 3/5 )( x + 2 )

Finally, now that we have the equation of the line in
point-slope form, we will want to convert the equation into
slope-intercept form. ( This will allow us to determine the
y-intercept directly from the formula. )

y - 5 = ( 3/5 )( x + 2 )
y - 5 = ( 3/5 )x +6/5
y = ( 3/5 )x6/5 + 5
y = ( 3/5 )x6/5 + 25/5
y = ( 3/5 )x +31/5

`(3/5)` x-y+`31/5` =0 is the equation of the line is standard form.

Basic Math Division

Introduction to basic math division:

In math, algebra is a fundamental arithmetic operation (addition, subtraction, multiplication, and division) generally used in day-to-day life. Division is capable of considered as repeated subtraction or equal distribution. In these articles, we are going to discuss about basic math division.

Simple division can teach to find the how frequently one whole number called the divisor, is contain in an additional whole number, called the dividend, or to divide a whole number into some proposed number of equivalent parts, and is a small method of performing frequent subtraction.

Basic Math Division – Steps with Examples:

Division process is inverse of multiplication operation. The division is classifying in 2 methods, dividend, and divisor. 

Basic math division – Steps:

Dividend:

The number that is to be divided is known as dividend.

Example:  `15 / 3`

Here, 15 is the dividend.

Divisor:

The number that divides the dividend is known as divisor.

Example: `15 / 3`

Here, 3 is the divisor.

Quotient:

Quotient defined as the number of times dividend is divided by the divisor.

Example: `21 / 3 = 7`

Here 7 is the quotient.

Remainder:

Remainder defined as, the number which is left behind when the dividend is not completely divided by the divisor.

Example: `29 / 4`

`7 xx 4 = 28`

`28 + 1 = 29`

Here, the number 1 is the remainder.

Basic Math Division – Example Problems:

Problem 1:    

Dividing the given values,

` 980 -: 10`

Solution:

Dividing the given values,

`980 -: 10`

98       

--------

10   | 980

| 90

------

80

80

--------

0

---------

So, the final answer is 98 quotients and 0 is remainder



Problem 2:

Evaluate the given values,

`1216 -: 19`

Solution:

Dividing the given values,

`1216 -: 19`

64     

--------

19   |  1216

|  114

---------

76

76

--------

0

---------

So, the final answer is 64 quotients and 0 is remainder

Problem 3:

Dividing the given values,

`1817 -: 23`

Solution:

Dividing the given values,

`1817 -: 23`

79         

--------

23   |1817

|161

------

207

207

--------

0

---------

Therefore, the final answer is 79 quotients and 0 is remainder.

Problem 4:

Dividing the given values,

`1898 -: 13`

Solution:

Dividing the given values,

`1898 -: 13`

146       

--------

13 | 1898

| 13

------

59

52

--------

78

78                                                                    

---------

0

---------

Therefore, the answer is 146 quotients and 0 remainder

Example 5:

Dividing the given values

`3872 -: 22`

Solution:

`3872 -: 22`

176       

----------

22 |   3872

|   22

167

154

---------

132

132

----------

0

---------

Therefore, the answer is 176 quotients and 0 remainder

Quantitative Reasoning Math

Introduction to quantitative reasoning math

Quantitative reasoning is one of  the most useful application in mathematics concepts and skills.It is used to solve the  real-world problems. The students must solve this quantitative reasoning for improve their skills. It is very useful to solve the real life problems also.Here some of the quantitative reasoning problems has been solved. The practice problems are very useful to the students to improve their knowledge in the quantitative reasoning mathematics.

Number Theory Problems in Quantitative Reasoning Math

Find the largest number which perfectly divide 10110-1

Solution:

Given number is 10110-1

The simple way to solve the math problem is given below

1012 = 10201.
1012 - 1 = 10200. This is divisible by 100.

Similarly   for 1013 - 1 = 1030301 - 1 = 1030300.

So you can conclude that (1011 - 1) to (1019 - 1) will be divisible by 100.

(10110 - 1) to (10199 - 1) will be divisible by 1000.

Therefore, (10110 - 1) will be divisible by the large number 10^10

2.Find the minimum value of tiles required in a top floor of breadth 3 meters 78 cm and height 1 meters 74 cm?

Solution for math:

The tiles used in top floor are square. Therefore, the breadth of the tile = height of the tile. As we have to usewhole number of tiles, the side of the tiles should a factor of both 3 m 78 cm and 1m 74. And it should be the highest factor of 3m 78 cm and 1m 74.

3 m 78 cm = 378 cm and 1 m 74 cm = 174 cm.

The HCF of 378 and 174 = 6

Hence, the side of the square is 6.

The number of such tiles required =` (378xx174)/(6xx6)`

The total tiles required = 1827 marbles.

Percentage Problems in Quantitative Reasoning Math

If the price of book increases by 10% and Raj intends to spend only an additional 5% on book, by how much % will he reduce the quantity of book purchased?

Solution for math:

Let the price of 1 book be Rs.x and let Raj initially buy 'y' books.

Therefore, he would have spent Rs. xy on book

When the price of book increases by 10%, the new price per l book is 1.10x.


Raj intends to increase the amount he spends on book by 5%.

i.e., he is willing to spend xy + 5% of xy = 1.05xy


Let the new quantity of book that he can get be 'q'.

Then, `1.10x * q = 1.05xy`
Or `q = (1.05xy)/(1.10x)=(1.05)/(1.10)` y= 0.95y.


As the new quantity that he can buy is 0.95y, he gets 0.05y lesser than what he used to get earlier.

Or a reduction of 5%.

2. A student  who gets 10% marks fails by 5 marks but another student who gets 21% marks gets 6% more than the passing marks. Find the maximum marks.

Solution for math

From the given data pass percentage is 21% - 6% = 15%

By hypothesis, 15% of x – 10% of x = 5 (marks)

i.e., 5% of x = 5

Therefore, x = 100 marks.

Fraction Strips for Math

Introduction to fraction strips for math:

Fractions:

A fraction is a number that can represent part of a whole. The earliest fractions were reciprocals of integers: ancient symbols representing one part of two, one part of three, one part of four, and so on. A much later development was the common or "vulgar" fractions which are still used today (`1/2` , `2/3` , `3/4` , etc.) and which consist of a numerator and a denominator. (Source: Wikipedia)

Objective of fraction strips for math :

The main objective of this article is to learn the fraction strips for math. This article gives you types of fractions , some solved math problems on fraction strips .

Types of Fraction Strips in Math:

The following are three types of fractions in math,

Proper fraction

Improper fraction

Mixed fraction

Proper fraction:

Fractions are in the form a/b  ( where a, b are the integers), b is always greater than a(b > a) called as proper fractions.

Example: `7/8` , `16/26`

Improper fractions:

The fraction a/b ( where a, b are the integers) a is always greater than b( a > b)  called as improper fractions.

Example: `5/3` ,`12/5` , `156/46`

Mixed fraction:

The fractions are in the form of a b/c ( where a, b ,c are the integers) is called as mixed fraction. Here a is called as quotient, b is known as remainder and c is known as divisor.

Example: 2 `3/4` , 6 `7/5`

Problems on Fraction Strips for Math:

Problem 1:

Add the following negative fractions ` - 3/5 ` and `-9/5`

Solution:

Given, Add` - 3/5` and `-9/5`

That is `-3/5` + `( -9/5)` =  `-3/5 ` - `9/5`

= - ( `3/5 ` + `9/5` )

Here both fractions are have common denominator. So we can add the numerator juat like integers and keep the denominator as it is.

- ( `3/5` + `9/5` ) = - `( 3+9 )/ 5`

= `- 12/5`

Answer: `-3/5` - `9/5` =` - 12/5`

Problem 2:

Add the following negative fractions` -17/15` and `-21/ 26`

Solution:

Given, Add `-17/15` and` -21/ 26`

That is,` -17/15` + (` -21/ 26 ` ) = `-17/15`   `-21/ 26`

= - (`17/15`   +` 21/ 26` )

Here we are going to add `17/15` ,  `21/ 26` . But both fractions have different denominators. We cannot add directly. We need to make common denominator. For that we need to find the lcd of 15 and 26.

Multiples of 15 = 15, 30, 45, 60, 75, 90, 105, 120, 135, 150, 165, 180, 195, 210, 225, 240, 255,…..390............

Multiples of 26 = 26, 52, 78, 104, 130, 156, 182, 208, 234, 260, 286, 312, 338, 364, 390, 416 ...

Here 390 is the least common factor of 15 and 26.

Multiply `17/15` by 26 on both numerator and denominator,

`17/15` = (17 * 26) / ( 15 * 26)

= `442 / 390`

Multiply `21/ 26` by 15 on both numerator and denominator,

`21/ 26` = ( 21 * 15) / ( 26 * 15)

=   `315 / 390`

Now we can add the given fractions,

- (`17/15`   + `21/ 26` ) = - ( `442 / 390 ` + `315 / 390` )

= - `( 442 + 315) / 390`

= -` 757 / 390`

Answer:  -`17/15` + ( `-21/ 26` ) = - `757 / 390`

Problem 3:

Multiply `(x+2)/( y-3)` * `(x)/(5y)`

Solution:

Given, `(x+2)/( y-3)` * `(x)/(5y)`

We can multiply just like integers, But multiply the Both numerator and multiply the both denominator separately.

`(x+2)/( y-3)` * `(x)/(5y)` = `(x(x+2))/(5y(y-3)) ` 

= `(x^2+2x)/(5y^2-15y)`

Answer: `(x+2)/( y-3)` * `(x)/(5y)` =`(x^2+2x)/(5y^2-15y)`

Vertical Line Slope

Introduction to vertical line slope:

Vertical line slope is nothing but the line which is parallel to y – axis. If the given line is parallel to the y – axis and it is perpendicular to x – axis we will say that is a vertical line. In this x value is constant it won’t get any variation the changes will be in y value. The equation is like x = a constant value. We will see some example problems for vertical line slope.

Example Problems for Vertical Line Slope:

We are having the general formula to find the slope of the line = `(y2 - y1) / (x2 - x1)`

Here (x1, x2) and (y1, y2) are the points which the line is passing.

Example 1 for vertical line slope:

Find the slope of the line which is passing through the following points (5, 8) ad (5, 9)

Solution:

We know if any line is having the constant value on x then it is known as vertical line. So the given line is a vertical line. We have to find the slope value of the given line.

Slope of the given vertical line =` (y2 - y1) / (x2 - x1)`

(x1, x2) = (5, 8) and (y1, y2) = (5, 9)

Slope of the given vertical line = `(9 - 8) / (5 - 5) = 1 / 0 ` = undefined.

So the slope of the given vertical line is undefined.

Algebra is widely used in day to day activities watch out for my forthcoming posts on free algebra equation solver and algebra math problem solver. I am sure they will be helpful.

More Problems for Vertical Line Slope:

Find the slope of the line which is passing through the following points (3, 16) ad (3, 24)

Solution:

Here x value of the given points are constant. So thhe given line is a vertical line. We have to find the slope of the given vertical line.

Slope of the given vertical line = `(y2 - y1) / (x2 - x1)`

(x1, x2) = (3, 16) and (y1, y2) = (3, 24)

Slope of the given vertical line = `(24 - 16) / (3 - 3) = 8 / 0 ` = undefined.

So the slope of the given vertical line is undefined.

From the above we can conclude slope of any vertical line is undefined.

Graphing Absolute Value Inequalities

Introduction to graphing absolute value inequalities:

Graphing absolute value inequalities is nothing but we are going to graph the solutions of the absolute value inequalities. To graph the absolute value inequalities first we have to find the solutions and using that we have to graph. Here we are going to learn about the graphing of absolute value inequalities. We will see some example problems for graphing the absolute value inequalities.

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Example Problems for Graphing Absolute Value Inequalities:

The main difference between the inequalities and equalities is we can say the range of the solutions in the inequalities. But in equalities we can determine the exact solution.

Example 1 for graphing absolute value inequalities:

Graph the absolute values of the inequalities. |x +2| `lt=` 8

Solution:

Given inequality is |x + 2| `lt=` 8

To find the absolute value for the given inequality

(x + 2) `lt=` 8 ………… (1) And –(x + 2) `lt=` 8 …………………. (2)

Equation 1:

(x + 2) `lt=` 8 ………… (1)

Add - 2 on both sides

x + 2 - 2 `lt=` 8 - 2

x `lt=` 6

Equation 2:

-(x + 2) `lt=` 8

-x - 2 `lt=` 8

Add +2 on both sides.

-x - 2 + 2 `lt=` 8 + 2

-x `lt=` 10

Divide by -1.

x `gt=` -10

So x lies between -10 `lt=` x `lt=` 6

We will see some more example problems bfor graphing the absolute value inequalities.

Example 2 for Absolute Value Inequalities:

Find the absolute values of the inequalities. |x - 6| `gt=` 5

Solution:

Given inequality is |x - 6| `gt=` 5

To find the absolute value for the given inequality

(x – 6) `gt= ` 5 ………… (1) And –(x – 6) `gt=` 5 …………………. (2)

Equation 1:

(x – 6) `gt=` 5 ………… (1)

Add +6 on both sides

x - 6 + 6 `gt=` 5 + 6

x `gt=` 11

Equation 2:

-(x – 6) `gt=` 5

-x + 6 `gt=` 5

Add -6 on both sides.

-x + 6 - 6 `gt=` 5 - 6

-x `gt=` -1

Divide by -1.

x `lt=` 1

So x lies between 1 `gt=` x `gt=` 11     

Quadratic Equations and Graphs

Definition of quadratic equation : An equation in which one or more of the terms is squared but raised to no higher power, having the general form ax2 + bx + c = 0, where a, b, and c are constants.

Quadratic Equations and Graphs:

The general form of a quadratic equation is

ax2 + bx + c =0

Where a, b and c are real numbers and a`!=` 0

It is called so because it has a second degree term.

The values of the variable involves in a quadratic equation are called its solution.

The following method are usually used to solve a quadratic equation.

completing square method
using the quadratic formula
Factorization method


In general, the shape of the graph of a quadratic equations a parabola.

The steps we follow to sketch the graph of the quadratic equation are:

1. Check if a > 0 or a < 0

If a > 0, then the parabola is u-shaped ie it opens upwards

If a < 0, then the parabola is  n-shaped ie it opens downwards

2. Find Vertex

The x-coordinate of the vertex is –b/2a

Substitute the value of x in the function to have the y coordinate.

Thus we have the (x, y) coordinates of the vertex.

3. Find y-intercept

The coordinates of the y-intercept can be found by substituting x = 0.

4. Find x-intercept

The coordinates of the x-intercept can be found by substituting y = 0 and solving the quadratic equation

5. Finding other coordinates

Substitute any possible value for x to get the corresponding value of y or vice versa

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Quadratic Equations and Graphs-example

Graph of the function y =  2x2 - 8x + 6

Here a= 2 b=-8 c=6

1. Since a > 0, then the parabola is u-shaped i.e. it opens upwards

2. The x-coordinate of the vertex is –b/2a = 8/(2*2) = 2

Substituting the value of x in the function y = 2x2 - 8x + 6 to have the y coordinate.

We get y = 2*22 - 8*2 + 6 = -2

So the vertex is (2,-2)

3. The coordinates of the y-intercept can be found by substituting x = 0.

We get y = 2*02 - 8*0+ 6 =6

So the coordinates of the y-intercept is (0,6)

4. The coordinates of the x-intercept can be found by substituting y = 0 and solving   the quadratic equation

We get

2x2 - 8x + 6=0

2(x2 - 4x + 3)=0

2(x - 1)(x - 3) = 0

So x = 1, or x = 3.

So the coordinates of the x-intercept is (1,0)  and (3,0)

5. Finding some other coordinates by substituting any possible value for x

We can now plot the graph.

Quadratic Equations and Graphs-graphs

Exponents to Write Prime Factors

Introduction of exponents to write prime factors:
In mathematics a number or an expression can be factored. A number can be factored only by prime numbers or by a non – prime number or by a combination of prime numbers and a non prime number. A prime number is a number which should divide by itself or by 1. When factorization of a number takes place it is written as exponents finally. Prime factoring is done from the smallest prime and goes upwards.


Exponents to Write Prime Factors Example 1

Write the number 10000 in prime numbers with exponents.

Solution: Given 10000,

Let’s start from the lowest prime 2.

10000 can be written as 2 * 5000.

Now take 5000 and write it in prime factors

5000 is same as 2 * 2500

Take 2500 and write it in prime factors.

2500 is equal to 2 *1250.

Take 1250 and write it in prime factors.

2* 625.

Now take 625 which cannot be factor with 2.

So, take the next prime 3, which is also not possible.

Now take next prime 5 and write 625 in factor form with 5

625 = 5 * 125.

Take 125 and prime factor it now and start from the lowest prime

125 = 5 * 25

Take 25 which is 5 * 5

Take 5 which is 5 * 1.

So the prime factors of 10000 is

=10000

=2 * 5000

=2 * 2 * 2500

=2 * 2 * 2 *1250

=2 * 2 * 2 * 625

=2 * 2 * 2 * 5 * 125

=2 * 2 * 2 * 5 * 5 * 25

=2 * 2 * 2 *5 * 5 * 5 * 5

=23*54, which is the exponents to write prime factors.

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Exponents to Write Prime Factors Example 2 and 3

Write the number 75 in prime numbers with exponents.

Solution: Given 75,

To factor using prime numbers take the least prime 2 first which is not possible.

So take the next prime 3 and write the 75 in terms of 3

75 = 3 * 25.

Now factor the 25 with the prime numbers.

25 cannot be factor in terms of 3. So take the next prime number 5 and write 25 in terms of 5

25 = 5 * 5

=75

=3 * 25

=3* 5 * 5

=31 * 52, which is the exponents to write prime factors.

Write the number 175 in prime number with exponents.

Solution: Given 175,

175 can be written as 5 * 35

Now factor 35 with prime numbers.

35 can be written as 5 * 7

7 can be factor as 7* 1

=35

=5 * 35

=5 * 5 * 7

=5 * 5 * 7 *1

=52 * 71

Which is the exponents to write prime factors.

Volume of Sphere Calculator

Introduction about volume sphere calculator:

The space occupied by an object is known as volume of that object. The volume of the sphere can be calculated easily by using online calculator. If the radius of sphere is given the volume of the sphere can find using online calculator. By using that calculator the operation is carried out robotically once the value is entered. The sphere calculator makes the calculation very easy.  In this article we are going to disuse about how to calculate the volume of sphere formula.

Examples Problems in Volume of Sphere Calculator:
Volume of the sphere = 4/3 p r 3

r = radius.

Problems:

1. The sphere has the radius 7 cm. find the volume of the sphere.

Solution:

Given:

Radius (r) = 7cm

Formula:

Volume of the sphere= 4/3 p r 3

= 4/3 * 3.14 * (7) 3

=4/3 * 3.14 * 343

= 1436.0266

Volume of the sphere=1436.0266 cm2

2. The sphere has the radius 4 cm. find the volume of the sphere.

Solution:

Given:

Radius (r) = 4cm

Formula:

Volume of the sphere = 4/3 p r 3

= 4/3 * 3.14 * (4) 3

=4/3 * 3.14 * 64

Volume of the sphere = 268.08cm3

3. The sphere has the radius 11 m. find the volume of the sphere.

Solution:

Given:

Radius (r) = 7m

Formula:

Volume of the sphere = 4/3 p r 3

= 4/3 * 3.14 * (11) 3

=4/3 * 3.14 * 1331

Volume of the sphere = 5575.27976m3

4. The sphere has the radius 2.5 cm. find the volume of the sphere.

Solution:

Given:

Radius (r) = 2.5cm

Formula:

Volume of the sphere = 4/3 p r 3

= 4/3 * 3.14 * (2.5) 3

=4/3 * 3.14 * 15.62

Volume of the sphere = 65.44985cm3

5. The sphere has the radius 3.7 cm. find the volume of the sphere.

Solution:

Given:

Radius (r) = 3.7 cm

Formula:

Volume of the sphere = 4/3 p r 3

= 4/3 * 3.14 * (3.7) 3

=4/3 * 3.14 * 50.653

Volume of the sphere  = 212.174cm3

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Practice Problems in Volume of Sphere Calculator:

1. The sphere has the radius 5m. Find the volume of the sphere.

Answer:  Volume of the sphere =523.59878 m3

2. The sphere has the radius 6m. Find the volume of the sphere.

Answer: Volume of the sphere = 904.77868 m3

3. The sphere has the radius 7m. Find the volume of the sphere.

Answer: Volume of the sphere = 1436.75504 m3

4. The sphere has the radius 8m. Find the volume of the sphere.

Answer: Volume of the sphere = 2144.66058 m3

The Product of Two Consecutive Integers

Introduction to Product of two consecutive integers:

Consecutive integer is the number after the same number. Let take the a number a  and its consecutive number can be said as      a + 1. 

Example: (4,5), (6,7) (5,6), (21, 22)

Product of two consecutive integers is nothing but multiplication of the number and its consecutive number. Thus the result of the multiplication of the number and its consecutive is called as product of two consecutive integers. In this article, we see about the product of two consecutive integers.

Product of Two Consecutive Integers:

Formula: multiplication of integer and its consecutive

Let take the number as ‘’a’’ and its consecutive is ‘’a + 1’’.

Step 1: Multiplication of the number ‘’a’’ and its consecutive

a * (a + 1) = a + a2

Thus the product of the integer and its consecutive is equal to the sum of the integer and its square of the same integer.

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Example Problems – Product of Two Consecutive Integers:

Example problem 1:

What is the product of the integer 10 and its consecutive?

Solution:

Formula: a * (a + 1) = a + a2

Here a = 10

10 * (10 + 1) = 10 + 102

Square of 10 =100

10 + 100 = 110

Answer: 110

Check the answer:

Step 1: The consecutive integer of 10 is 11

10 * (11) = 110

Therefore the answer is correct.

Example 2:

The product of two consecutive positive integers is 420. Find the integer and its consecutive integer.

Solution: Let take the integer as ‘’a’’ and its consecutive is ‘’a + 1’’

Formula: a * (a + 1) = a + a2

a * (a + 1) = a + a2 =  420

a2 + a - 420 = 0 --- > equation 1

Step 1: Factoring the equation1, we get the integer.

a2 + a - 420 = 0

a2 -20a + 21a - 420 = 0

a (a - 20) + 21 (a - 20) = 0

(a -20) (a + 21) = 0

Possibilities of ‘’a’’ value is

a = 20

a = -21

Note: in given problem the integer is positive. Hence the value of the integer is 20 and its consecutive is 20 +1 = 21

Answer: 20 and 21

Practice Problem – Product of Two Consecutive Integers:

1. What is the product of the integer -5 and its consecutive?

Answer: 20

2. What is the product of the integer 5 and its consecutive?

Answer: 30

Converting Quadratic Equations

INTRODUCTION

Any equation which has been written in the form  ax 2+bx + c = 0 is called as quadratic equations. In quadratic equations the numbers a, b, and c are real numbers and a ? zero. If p(x) is an equation and which is the polynomial having the degree of 2 and in the form of p(x)= 0 , then the equation is  quadratic.

Steps for Converting Quadratic Equations:

The General form of quadratic equation is: y=ax2+bx+c

The Vertex form   is: y= a(x-h) 2 +k.

The following steps are used for converting quadratic equations,

Step 1: The initial step is to factor out the leading co-efficient from the first two terms of the given equation.

Step 2:  Complete the square to which we have factored out (First two terms).

Step 3: Balance the constant term we get after completing the square, by multiplying it with the co-efficient.

Step 4: Add the opposite to the constant term of given equation.

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Example Problems on Converting Quadratic Equations :
Converting quadratic equations Example 1:

Find the vertex form of the given quadratic equation y = 4x² - 16x + 6

Solution:

Step 1: Factor out the leading co-efficient from the first two terms of the given equation.

Y= 4 (x2-4x) + 6

Step 2:  Complete the square to which we have factored out (First two terms).

Y= 4 (x2-4x) + 6

(x2-4x) = (x-2)2   { (a-b)2 = a2-2ab+b2  , here a= x, b=2 }

= x2-(2)(x)(2) +22

y = x2-4x+4

Step 3: Balance the constant term 4, by multiplying it with the co-efficient 4.

y  = 4(x2-4x+4)  +6 +4(4)

Step 4: Add the opposite to the constant term of given equation.

y= 4(x-2)2   +6-16

Vertex form is   :  y= 4(x-2)2   -10

Vertex            :   (h=2, k= -10).

Converting quadratic equationsExample 2:

Convert the quadratic equation y = -5x2+30 into vertex form.

Solution:

Step 1: Factor out the leading co-efficient from the first two terms of the given equation.

y = -5x2+30

y= -5(x2-6x)

Step 2:  Complete the square to which we have factored out

y= -5(x2-6x)

(x2-6x) = (x-3)2   { (a-b)2 = a2-2ab+b2   , Here a= x, b= 3}

= x2-(2)(x)(3)+ 32

y= x2-6x+9

Step 3: Balance the constant term 9, by multiplying it with the co-efficient -5

y= -5(x2-6x+9) + (-5)(9)

y  = -5(x2-6x+9)-45

Step 4: Add the opposite to the constant term of given equation.

Here there is no constant term .So just change the sign of -45.                                 

y= -5(x-3)2 +45

Vertex form is   :  y= -5(x-3)2 +45

Vertex      :  ( h=3, k=45)

Kinds of Polygons According to Sides

Introduction to Kinds of polygons according to sides

One of the all encompassing shapes in geometry is polygon. it has at least three sides that consists of a line segment in a closed shape. It does not intersect other than its vertices. The sum of the interior angle is 180 degree for n sides. And the sum of the exterior angle is 360 degree’s polygon which has all its sides and angles are equal than it is called regular polygon.

Different Kinds of Polygons According to Sides:

These are classified according to their convexity, angle s, and number of sides so on… A few types of polygons are

· Regular polygon

· Irregular polygon

· Congruent Polygon

· Concave polygon

· Crossed polygon

· Equiangular polygon.

Regular kinds of  polygons according to sides:

All its sides and interior angles are equal. It has the following types

· Triangle- it has 3 sides and 3 angles

·Quadrilateral- it has 4 sides and 4 angles.

· Pentagon- it has 5 sides and 5 angles

· Hexagon- it has 6 sides and 6 angles.

· Heptagon- it has 7 sides and 7 angles.

· Octagon- it has 8 sides and 8 angles.

· Nonagon- it has 9 sides and 9 angles.

· Decagon- it has 10 sides and 10 angles.

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Irregular Kinds of Polygons According to Sides:

These are classified according to their convexity, angle s, and number of sides so on… A few types of polygons are

· Regular polygon

· Irregular polygon

· Convex Polygon

· Concave polygon

· Crossed polygon

· Equiangular polygon.

Regular kinds of  polygon sides with their angles:

All its sides and interior angles are equal. It has the following types

· Triangle- it has 3 sides and 3 angles

·Quadrilateral- it has 4 sides and 4 angles.

· Pentagon- it has 5 sides and 5 angles

· Hexagon- it has 6 sides and 6 angles.

· Heptagon- it has 7 sides and 7 angles.

· Octagon- it has 8 sides and 8 angles.

· Nonagon- it has 9 sides and 9 angles.

· Decagon- it has 10 sides and 10 angles.