Dot Product of Two Matrices

Introduction about dot product of two matrices:

The dot product is an algebraic operation that takes two equal-length sequences of numbers (usually denoted as the coordinate vectors) and returns a single number obtained by multiplying corresponding entries and adding up those products. The name is derived from the dot "·" that is often used to designate this operation. Here the dot product of the two matrices are represent the product of two matrices that give finally one matrix as result. Understanding Derivative Product Rule is always challenging for me but thanks to all math help websites to help me out.

(Source from Wikipedia)

Explanation about "dot Product of Two Matrices "

Two matrices S having order x x y and D having order a x b are said to be conformable for the process ofmultiplication if the columns count of the first matrix S (y)  is equal to the rows count of the second matrix B(a).

Then the order of SD is x × b =rows count for matrix A × columns count for the matrix B

On considering  the matrices given  having same rows and columns, perform the multiplication operation
`[[-1,-2],[-1,-4]]`  and   `[[-1,1],[-1,1]]`

Solution:

We take the first matrix as G and G= `[[-1,-2],[-1,-4]]` has the order  of  2 x 2.

And take the second matrix as H and H= `[[-1,1],[-1,1]]` has the order  of  2 x2

For multiplication operation on G x H

Here the orders present are  2 x 2  in G and 2 x 2  in H so it is possible

G x H =   `[[-1,-2],[-1,-4]]` `[[-1,1],[-1,1]]`

=  `[[1+2,-1-2],[1+4,-1-4]]`

= `[[3,-3],[5,-5]]`

Is this topic math 3rd grade word problems hard for you? Watch out for my coming posts.

Problems to Explain "dot Product of Two Matrices "

On considering  the matrices given  having same rows and columns, perform the multiplication operation
`[[32,-11,-13,0],[-11,10,-11,-11],[-11,-11,10,-11]]` and `.``[[0,0,0,0,0],[0,0,0,0,0],[0,0,0,0,0],[0,0,0,0,0],[0,0,0,0,0]]`

Solution:

We take the first matrix as G and G= `[[32,-11,-13,0],[-11,10,-11,-11],[-11,-11,10,-11]]` has the order  of  3 x 4.

And take the second matrix as H and H= `[[0,0,0,0,0],[0,0,0,0,0],[0,0,0,0,0],[0,0,0,0,0],[0,0,0,0,0]]`    has the order  of  5 x5

For multiplication operation on G x H

We must  have equal value on the columns for the 1st matrix 'G' and the rows of the 2nd matrix 'R'.

We have the columns of the 1st matrix 'S' is 4  with  the rows of the 2nd matrix 'R' is 5.

So, the operation on the multiplication is not comfortable.

Types of Events

Introduction to types of events:

The word ‘event’ literally means ‘happening. This is an accepted fact that anything may happen at any time. But, some are preset and some are instant. Some are the markers of fortune and others are the markers of misfortune, According to types of the occurrence, events can be categorized as special, temporal and accidental. Please express your views of this topic Probability of Compound Events by commenting on blog.

Types of Events-special Events
Special events are marked as the renewable variety. This type of events is remembered by all. The examples of such events are birthday, marriage anniversary, the moment of meeting someone who has a great impact on one’s life, or the day of receiving some award. These days have a lasting influence on one’s life and these are the major sources of inspiration in future.

Types of Events-temporal Events

A category of events can be made on the basis of the time of an action. It can be called ‘temporal events’. The temporal events may be of three types: past, present and future. The past action points at the event happened before and may be working on the action that is happening now, as we call it a present action. The relation between the past and the present can be explained as the cause and the effect. For example, ‘one studied well and he passed a test very well.’ Here, ‘the activity of study’ is the past event and ‘passing the test’ is the present event. And, based on the present events the career options or future study options are chosen and that choice is a future event as it is to happen next to the present event. So, in a sense, the past, the present and the future events are related by cause and effect. I have recently faced lot of problem while learning live online math help, But thank to online resources of math which helped me to learn myself easily on net.

Types of Events-accidental Events

There are a few events that are not preset or programmed but fully based on chance. The events happening by chance has a positive or negative outcome. The events may be like, winning a lottery makes a happy end and it may be a mishap marking a tragedy or a sad episode. This two-pronged intersection is an important junction in the eventful lifestyle whose basis is experience or accident which is either special or disappointing.

Statistics Poisson Distribution

Introduction to statistics Poisson distribution

The Poisson distribution in statistics is named for the French mathematician S. D. Poisson. It is used to describe a number of processes like the distribution of telephone calls going through a switchboard system, the demand of customers for service at a restaurant, the arrival of customer at a shop and the number of accidents at a junction.

Characteristics of a Statistics Poisson Distribution

1. The experiment consists of counting x the number of  times a particular event occurs during a given unit of time.

2. The probability of the occurrence of  an event in a given unit of time is the same for all units.

3. The number of events that occur in one unit of time  is independent of the number that occur in other units.

4. The mean number of events in each unit will be denoted by the Greek letter `lambda` .

Statistics of Poisson Distribution-mean and Variance

The probability distribution :

P ( x )  = ( `lambda` x e -`lambda` ) / x !

where `lambda` = mean umber of events during the given time period,

e = 2.71828 ( the base of natural logarithm )

the mean `mu` = `lambda`

the variance `sigma` 2 = `lambda`

Having problem with Determining Sample Size for Research Activities keep reading my upcoming posts, i will try to help you.

Example of Statistics Poisson Distribution

Let us investigate the safety of a dangerous curve. Past police records indicate a mean of 5 accidents per month at this curve. Suppose the number of accidents is distributed according to a Poisson distribution. Calculate the probability in any month of exactly 0, 1, 2, 3 or 4 accidents.

Solution : Since the number of accidents is distributed according to a Poisson distribution and the mean number of accidents per month is 5, we have the probability of accidents happening in any  month

p ( x )  = (` 5 ^ x e ^ ( -5 )) / x !`

`By this formula, we can calculate`

`p ( 0 ) = 0.00674 p ( 7 ) = 0.104445`

`p ( 1 ) = 0.3370 p( 8 ) = 0.065278`

`p ( 2) = 0.08425 p ( 9 ) = 0.036266`

`p ( 3 ) = 0.14042 p ( 10 ) = 0.018133`

`p( 4 ) = 0.17552 p ( 11 ) = 0.008242`

`p ( 5) = 0.175467 p ( 12 ) = 0.003434`

p ( 6 ) = 0.146223

Solve and Check Inequalities

Introduction to solve and check inequalities:

Inequality is a one the concept of algebra.Inequality in the form of equation with greathearted or lesser than symbol. . Inequalities are described with the symbol of <, >, <=, and >=It is algebra expression. Inequalities is a combination of variables, numbers ,constants , conditions( < or >) with operations( + or -)

Four forms of checking inequalities:

1. Ax + By  <  C

2. Ax + By  >  C

3. Ax + By  <  C

4. Ax + By  >  C

Basic Concepts of Checking Inequalities:

Types of inequalities with examples

Less Than (<)

Example of solve and check inequalities:

x<6 br="br">
Here variable x is less than the value of six ,It mean x may contains the value of like 5,4,3,2,1,0,-1,-2,-3 … etc

Less Than (≤)

Example of solve and check inequalities:

x ≤ 8,

Here  variable x is less than or equal to 8 mean , that is x may contain the value of 8,7,6,5,4,3,2,……….. so on.

Greater than (>)

Example of solve and check inequalities:

x > 10,

Here the variable x  is great than value of  10, that is x may be any one of the following value 11,12,13……so on.

Greater Than or equal to (≥)

Example of solve and check inequalities:

x ≥ 3,

Here  variable x  is great than or equal to 3 means that, that is x may be contain any one of the following value 3,4,5,6……so on.

Example Problems in Inequalities:

Solving and check the inequalities

1. solve and check: 4x+5 < -2x+14

Solution:

4x+5 < -2x+14(given problem)

4x+5-5 < -2x+14-5( subtract both sid e by 5)

4x < -2x+9

4x+2x < -2x+2x+9( add both side by 2x)

6x < 9

6x /9< 9/9( divide both side by 9)

x < 3/2  in otherwords x<1 .5=".5" br="br">
So x values are may 1.4.1.3,1.2,1.1,1…..and so on

Check the ineuqlities:

Substitute x value to the given problem

Substitute x=1 in  4x+5 < -2x+14

4(1)+5 < -2(1)+14

5< -12

Condition are not satisfied here 5 is graeterthan the -12

So given problem is not inequalities.

Exampl 2:

Solve  and check x + 3 < 2x +7.

Sol:

x + 3 < 2x + 7

x +3 – 3 < 2x + 7 – 3 (Rule 1)

x < 2x + 4

x-2x < 2x-2x+4

- x<4 br="br">
x > – 1

The solution set is {0, 1, 2, 3…}

Verification:

Substitute x =0,1,2

Substitute x = 0 in given eqaution

X+3<2x br="br">
0+3<2 br="br">
3<7 br="br">
Here condition is satisfied so given equation is inequality

Discrete Math Counting

Introduction for discrete math counting:

Counting means the process of a counting, the number of samples or number of outcomes or number of probable ways to do a thing. The counting techniques are permutation, combination and factorial. Permutation means the process of rearranging the given number of objects or elements. Combination means the process of selection of elements or objects from a collection. In combination order of elements is irrelevant. Factorial means the value of a number is the product of a given number and all smaller positive numbers. Let us prepare for a discrete math counting techniques formulas and example problems.I like to share this Permutations and Combinations Word Problems with you all through my article.

Formulas for Discrete Math Counting:

1) Formula for Permutation:

P(n,r) = `(n!) / ((n-r)!)`

2) Formula for Combination:

C(n,r) = `(n!) / (r!(n-r)!)`

3) Formula for Factorial:

n! = `n*(n-1)*(n-2)...3*2 *1`. Is this topic binomial probability distribution hard for you? Watch out for my coming posts.

Example Problems for Discrete Math Counting:

Example problem 1:

How many ways 3 mouses can be taken from among 8 mouses?

Solution:

Permutation: P(n,r) = `(n!) / ((n-r)!)`

Here, n = 8, r = 3

P(10, 6) = `(8!)/((8 - 3)!)`

= `(8xx7xx6xx5!) / (5!)`

After simplify this, we get

= 336

In 336 ways 3 mouses can be taken from among 8 mouses.

Example problem 2:

In how many ways 5-chairs can be arrange?

Solution:

Factorial: n! = `n*(n-1)*(n-2)...3*2 *1`

5! =  5 x 4 x 3 x2 x 1

After simplify this, we get

5! = 120

Example problem 3:

How many ways 2 books can be chosen from among 10 books?

Sol:

Combination: C(n,r) = `(n!) / (r!(n-r)!)`,

Here n = 10, r =2

C(10, 2) = `(10!) / (2!(10 - 2)!)`

= `(10xx9xx8!) / (8!xx2xx1)`

After simplify this, we get

= 45

In 45 ways 2 guides can be chosen from among 10 guides.

Example problem 4:

In how many ways 6 lights can be arrange?

Solution:

By using the Factorial formula

: n! = `n*(n-1)*(n-2)...3*2 *1`

6! = 6 x 5 x 4 x 3 x 2 x 1

After simplify this, we get

6! = 720

The above examples are helpful to study of discrete math counting techniques.

Prime Example Definition

Introduction to prime example definition:

In this section we have prime example definition. Prime numbers are divisible by itself and one. The non prime numbers are known as composite numbers. We are going to solve some based on the prime numbers. Let us study about prime example definition with some solved problems along with step by step answer and exercise problems.

Example Problems for Prime Example Definition:

Example problem 1: What is the prime factorization of 10?

Solution:

Given number is 10

Divide by prime factors until the quotient is 1.

10 ÷ 2 = 5

5 ÷ 5 = 1

The prime factorization of 10 is: 10 = 2 × 5

Answer: The prime factorization of 10 is: 10 = 2 × 5

Example problem 2: Is 37 a prime number or composite number?

Solution:

Given number is 37

Prime numbers are divisible by itself and one.

37 is divisible by 37 and 1 only.

Therefore, 37 is prime number.

Answer: 37 is prime number.Please express your views of this topic Roman Number by commenting on blog.

Practice Problems for Prime Example Definition:

Practice problem 1: Can you tell the prime factorization of 2?

Practice problem 2: Write the prime factorization of 54. Use exponents when appropriate and order the factors from least to greatest (for example, 22 × 3 × 5).

Practice problem 3: Which of the following numbers are not prime numbers? 12, 37, 88, 90, and 101

Practice problem 4: Is 727 a prime number or composite number?

Practice problem 5: Write the prime factorization of 6. Use exponents when appropriate and order the factors from least to greatest.

Solutions for prime example definition:

Solution 1: The prime factorization of 2 is: 2

Solution 2: The prime factorization of 54 is: 2 × 3 × 3 × 3

Solution 3: 12, 88, and 90 are composite numbers.

Solution 4: 727 is a prime number.

Solution 5: The prime factorization of 6 is: 2 × 3

Uses of a Quadratic Equation

Introduction to Uses of a quadratic equation:

In this article we are going to discuss the uses of quadratic equation solving. The general form of quadratic equation is ax2 +bx +c, the value of is not equal to zero. The uses graph of this function is a parabola have vertical axis. For example f(x) =2x2 + x – 30 = 0 is a quadratic equation.

Uses of a Quadratic Equation –quadratic Equation Formula:

Formula for finding the roots of the quadratic equation is,

` -b +- sqrt(b^2-4ac)/(2a)`

Mathematic quadratic equation example problems are given below.

Uses of a Quadratic Equation -example Problems:

Example 1:

f(x)=x2+8x+16 = 0 solve by uses of factorize method for quadratic equation.

Solution:

The quadratic equation solving by uses of factoring method, to split the middle part (8x) into two parts so that the product of their co-efficient is equivalent to the constant part (16).

Like 8x = (4x) and (4x)

So, 4x + 4x = 8x and

4 * 4 (coefficients of 4x and 4x) = 16 (constant term)

So, now the equation becomes

x2 + 4x + 4x + 16 = 0

Here ‘x’ in 1’st part and 4 in last two parts are equal, by taking both part of equal out, we get

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

Now we have (x+4) in common

(x+4) (x+4)  = 0

Now x+4 = 0 or x+4 = 0

x= -4 (or) x= -4

The solution of quadratic equation x is -4 (or) -4

Example 2:

Solve 3x2 - 6x = -2 for x, uses of quadratic equation formula,

Solution:

Using standard form of ax2+bx+c=0

3x2 - 6x + 2 = 0

a =3

b = -6

c = 2 Plugging the values you found for a, b, c in the quadratic equation formula.

x =` -b +- sqrt(b^2-4ac)/(2a)`

x = `6 +- sqrt(36-24)/ 6`

x = `6 +- sqrt(12)/6`

x = `6 +- 2 sqrt(3)/6`

The solutions are as follows:

The solution of quadratic equation is x = 6 + 2 `sqrt(3)/6` and 6 - 2 `sqrt(3)/6`

Length of Right Triangle Sides

Introduction to length of right triangle sides:

In a triangle the one angle having 90 degree and the sum of other two angles is 90 degree means, that triangle is known as right angle triangle. And right triangle satisfies the Pythagoras theorem. The sides of the right triangle is in the manner of one largest side known as hypotenuse  and two legs called adjacent side and opposite side .



Explanation of Length of Right Triangle Sides:

Types of Right triangle:

General Right triangle
Isosceles Right triangle
30-60-90 Right triangle.
General Right triangle:
General Right triangle having one angle 90 degree and other two angles can be any measure that total yields 90 degree.

Isosceles Right triangle:

In Isosceles Right triangle the other two angles are in the measure of 45 degree each. In isosceles right triangle the sides are in the ratio of 1:1: `sqrt(2)` (adjacent side: opposite side: hypotenuse)

30-60-90 Right triangle:

As the name itself noted that this kind of right triangle having one angle is 90 degree mandatory and in other angles one is 30 degree and the other one is 60 degree. The ratios of the sides are 1:`sqrt(3)` :2 (adjacent side: opposite side: hypotenuse).





Pythagoras Theorem for the Length of Right Triangle Sides:

The Pythagoras theorem states that,

In a right angle triangle the total sum of the squares of the two sides (adjacent and opposite) are equal to the square of the longest side (hypotenuse) understanding hard math problems for 9th graders is always challenging for me but thanks to all math help websites to help me out.

Let a, b, c are the three sides of a right triangle where a, b are two legs and c is the longest side . Then the formula of Pythagoras theorem is,

c2= a2+b2

c= `sqrt(a^2+b^2)`

Examples on Length of Right Triangle Sides

Ex:1 In a right triangle the length of the adjacent side is 12cm and opposite side is 14cm. Find the length of the hypotenuse.

Sol:   Let a= 12cm, b=14cm and c= hypotenuse

By Pythagoras theorem,

c= `sqrt(a^2+b^2)`

= `sqrt(12^2+14^2)`

= `sqrt(144+196)`

= `sqrt(340)`

= 18.43

Hence the length of the longest side is 18.43 .

Ex:2 The length of the adjacent side and opposite side is 15cm in a right triangle. Find the length of the longest side.

Sol:  Let a= b=15 cm and c= hypotenuse

By Pythagoras theorem,

c= `sqrt(a^2+b^2)`

=  `sqrt(15^2+15^2)`

= `sqrt(225+225)`

=`sqrt(450)`

=21.21

Hence the length of the hypotenuse is 21.21

Summary on Quadratic Functions

Introduction on summary on quadratic functions

Quadratic functions are polynomial functions containing a quadratic expression. A quadratic expression is an algebraic expression of the degree 2.
A quadratic function can be written in different forms,

Standard form

A quadratic function of the form of `f(x) = ax^2 + bx + c` is in the standard form. In this form, a, b, and c are real numbers and `a != 0` . Furthermore, a, b and c are called the Coefficient of x square, Coefficient of x and Constant term respectively. The vertex of the parabola formed by graphing a quadratic function can be obtained by the formula `((-b)/(2a), (-1(b^2 - 4ac))/(4a))` .

Summary on Quadratic Functions-2 Forms

Intercept form

A quadratic function of the form of `f(x) = k(x - a)(x - b)` , that is, of the form of the product of two linear expressions, is in the intercept form. To graph a quadratic function, this form is converted into the standard form by expanding (multiplying the two linear expressions and number 'k').

Vertex form

A quadratic function of the form of `f(x) = a(x - h)^2 + k` is in the vertex form. The vertex of the parabola formed by graphing a quadratic function is given by `(h, k)` .

Graphing a Quadratic Function -summary on Quadratic Functions

The graph of a quadratic function gets the shape of a parabola, which is a conic section. A conic section is the surface obtained by intersecting a cone or a conical figure.
Let us learn the method of graphing a quadratic function by graphing the function `f(x) = x^2 - 5x + 6`

On comparing the given function with `f(x) = ax^2 + bx + c` , we get `a = 1` , `b = -5` and `c = 6` . Vertex of the parabola to be formed is given by `((-b)/(2a), (-1(b^2 - 4ac))/(4a))` .
Thus, vertex = `((5/2), (-1)/(4))`
First plot the vertex on the graph.

Since a parabola is a curve, we need the coordinates of many points lying on it. To obain these points, choose different (at least 4) values for the variable `x` , and plug in those values in the function to get the corresponding values of the function. For the function `f(x) = x^2 - 5x + 6` , we choose the values `x = 1` , `0` , `-1` , and `2` and obtain the following pairs of coordinates:-
Input value (x)    Output value (f(x))
0    6
1    2
-1    12
2
0

All the above pairs of coordinates are of the points lying on the parabola. Graph them, and then join them to form a parabola. This parabola is the graph of the function `f(x) = x^2 - 5x + 6` . It will look as follows:-

PreCalculus Problem Solver

Introduction to Precalculus Problem Solver:

The Precalculus is one of the foundation classes of mathematics. The topics involved in Precalculus test are real  complex numbers, binomial theorem, composite & polynomial functions, vectors, parametric equations, polar coordinates, matrices, rational functions, solving inequalities & equations and trigonometry. In this article we shall discuss about the examples involved in Precalculus Problem Solver. The following are the examples involved in Precalculus Problem Solver.

Precalculus Problem Solver:

Example 1:  Evaluate the distance between the points (2, 5) and (-1, 2) using distance formula.

Solution :   Given (x1, y1) = (2, 5)

(x2, y2) = (-1, 2)

The distance formula D = `sqrt((x2-x1)^2+(y2-y1)^2)`

= `sqrt((-1-2)^2+(2-5)^2)`

=  `sqrt((-3)^2+(-3)^2)`

= `sqrt(9 + 9)`

= `sqrt(18)`

Example 2:  Express the real and imaginary parts for 19 - i `sqrt(5) `

Solution:        Let z= 19 - i `sqrt(5) `      

Re(z) = 19    Im(z) = `sqrt(5)`

Example 3:   Find the vertex of the parabola y = 2x2 – 16x + 6

Solution:  Given:  y = 2x2 – 16x + 6

We know that x = -`(b)/(2a)` ,

Here a = 2, b = -16

So that,   X =` -b/(2a)` = `-((-16))/(2*2)` = 4

And then y = 2(22) – 16(2) + 6 = 8 – 32 + 6 = -18

So, x = 4 and y = -18.

Example 4:   If f(x) = 7x-1, find f-1(y)?

Solution:Let y = 7x - 1

The given equation can be rewritten as

y + 1 =7x

=> x = y + 1
7

Therefore the f-1(y) = y + 1
7

Example 5: Write the given number in complex form `sqrt(-44)`

Solution: `sqrt(-44)` = `sqrt((-1)(44))`

= `sqrt(-1)` * `sqrt(44)`

= i `sqrt(44)`

Having problem with solve math problem for me keep reading my upcoming posts, i will try to help you.

Precalculus Problem Solver:

Problem 1:Write the given number in complex form `sqrt(-175)`

Answer: = i `sqrt(175)`

Problem 2:Write the real and imaginary parts for 22 - i `sqrt(7) `

Answer:   Re(z) = 22    Im(z) =`sqrt(7)`

Problem 3:   If f(x) = x-9, find f-1(y)?

Answer: y + 9

Problem 4: Evaluate distance between the points (4, 5) and (-1, 2) using distance formula.

Answer: D= `sqrt(34)`

Steps for Rational Expressions

Introduction to steps for rational expressions:

Rational expressions are defined as the fractional number but instead of number the polynomials are present. By using the rational expressions we can able to do all the arithmetic operations that we performing in mathematics. For example, the rational expressions are in the form of `A/B` , where the numerator and denominator are called the polynomials terms.

Steps for Rational Expressions

Steps of rational expressions are follows,

By using the rational expression, we can able to do all the arithmetic operations.
Certain rules are followed for each of the operations performed for rational expressions.
The operation performed on the rational expressions are,
Addition
Subtraction
Division
Multiplication
Cancellation
Reciprocals

Example Problem for Steps of Rational Expressions

Problem 1: Add the following rational expressions, `(2x)/(3x^2)`  +  `(3x)/(3x^2)`.

Solution:

Step 1: Write the given rational expressions, we get,

`(2x)/(3x^2)`  +  `(3x)/(3x^2)`

Step 2: Check the denominator same or not,

Step 3: Add the numerator terms, we get,

`(2x + 3x)/(3x^2)` 

Step 4: Solve the obtained result, we get,

`(5x)/(3x^2)`

This is the required answer for the rational expressions.

Problem 2: Add the following rational expressions, `(4x)/(2y^2)`  +  `(4x)/(2y^2)`.

Solution:

Step 1: Write the given rational expressions, we get,

`(4x)/(2y^2)`  +  `(4x)/(2y^2)`

Step 2: Check the denominator same or not,

Step 3: Add the numerator terms, we get,

`(4x + 4x)/(2y^2)` 

Step 4: Solve the obtained result, we get,

`(8x)/(2x^2)`

This is the required answer for the rational expressions.

Problem 3: Add the following rational expressions, `(y^2)/(8x^2)`  +  `(4y^2)/(8x^2)`.

Solution:

Step 1: Write the given rational expressions, we get,

`(y^2)/(8x^2)`  +  `(4y^2)/(8x^2)`

Step 2: Check the denominator same or not,

Step 3: Add the numerator terms, we get,

`(y^2 + 4y^2)/(8x^2)` 

Step 4: Solve the obtained result, we get,

`(5y^2)/(8x^2)`

This is the required answer for the rational expressions.

Practice Problem for Steps of Rational Expressions

Problem 1: Add the following rational expressions, `(x)/(5x^2)`  +  `(x)/(5x^2)`.

Answer: The answer for the above problem is `(2x)/(5x^2)`

Problem 2: Add the following rational expressions, `(6x)/(2y^2)`  +  `(8x)/(2y^2)`.

Answer: The answer for the above problem is  `(14x)/(2y^2)`

Integral Change of Variable

Introduction to integral change of variable:

In this article, we study about integral change of variable and their example problems. Integral means finding the anti derivative of the function. Integral can be classified as two types. They are indefinte integral and definite integral. Improper integral also used in change of variable. Change of variable means change the variable of the given integral function. If the variable of the changed means, we also change the limit values of the given integral function. Change of variable is mainly used in definite integral problems.

Example Problems for Integral Change of Variable

Integral change of variable example problem 1:

Find the integral of the given function

` int_0^4 sin(4x)dx`

Solution:

Given function is `int_0^4 sin(4x)dx`

Using the change of variable method,

Take u = 4x

Therefore, for finding dx differentiate the u value, we get

du = 4 dx

Rearrange the above value, we get

dx = `(du)/(4)`

The limits are also changed,

When x = 0, u = 0

When x = 4, u = 16

The given integral function can be written as,

`int_0^4 sin(4x) dx` = `int_0^16 sinu (du)/(4)`

After rearranging the above function, we get

= `(1/4) int_0^16 sinu du`

Integrating the above function, we get

= `(1/4) [- cosu]^16_0`

Substituting the limit values, we get

= `(1/4) ((- cos 16) - (- cos0))`

= `(1/4) ((- 0.96) - (- 1))`

After simplifying, we get

= 0.01

Answer:

The final answer of the given function is 0.01

Integral Change of Variable Example Problem 2:

Find the integral of the given function

`int_0^3 ((1)/(2x + 3))dx`

Solution:

Given function is `int_0^3 ((1)/(2x + 3))dx`

Using the change of variable method,

Take u = (2x + 3)

Therefore, for finding dx differentiate the u value, we get

du = 2 dx

Rearrange the above value, we get

dx = `((du)/(2))`

The limits are also changed,

When x = 0, u = 3

When x = 3, u = 9

The given integral function can be written as,

`int_0^3 ((1)/(2x + 3))dx` = `int_3^9 ((1)/(u))(du)/(2)`

After rearranging the above function, we get

= `(1/2) int_3^9 (1/u) du`

Integrating the above function, we get

= `(1/2) [logu]^9_3`

Substituting the limit values, we get

= `(1/2) ((log 9) - (log3))`

= `(1/2) ((0.95) - (0.47))`

After simplifying, we get

= 0.24

Answer:

The final answer of the given function is 0.24

My Previous Blog :- http://advancemath.blogspot.in/2012/08/complex-fraction-calculator.html

Area of a Triangle by Heron's Formula

Introduction to area of a triangle by heron's formula:    
Heron formula defined by heron of alexandria and in METRICA book in 60 A.D.It is found by chinese.They published in 1247 A.D.Area of a triangle is calculated by using this formula.The length of sides are b,c,d and it is used in area of triangle A.It is most widely used in geometry.The formula for finding area of triangle is

A = vS(S-b)(S-c)(S-d) , where  S is the semiperimeter of the triangle.

S= (b+c+d) / 2.

Area of a Triangle by Heron's Formula-example

Problem 1: Find the area of below triangle:



Answer: Using heron's formula,

p= 34.5+16.4+19.6 / 2 = 35.26.

Area = `sqrt(35.26*0.76*18.86*15.66)`

= 89.75.

Therefore, the area of triangle is 89.75.

I am planning to write more post on examples of substitution method, prime numbers under 1000. Keep checking my blog.

Problem 2: Find the area of triangle from given value.

A=34.5, B= 28.8, C= 12.0

Answer: Using heron's theorem,

p = 34.5+28.8+12.0 / 2 = 37.64.

Area = `sqrt(37.64*3.14*8.84*25.64)`

= 164.

Therefore the area of triangle is 164.

Excercise for heron's theorem:

Problem: Find the area of triangle from A= 15.7, B= 34.5 and C =19.5 using heron's theorem.

Answer: Area of triangle =62.35.

Adding Several Numbers

Introduction to Adding Several Numbers:

An integer is a set of whole numbers. Whole numbers above zero is said to be positive numbers denoted as ‘+’ sign and whole numbers below zero is said to be negative numbers denoted as ‘-‘. A number with zero is said to be neither negative nor positive and it does not have any sign in math. In addition there may be different digits numbers to add. The positive numbers can be written with or without the sign. Let us see about adding numbers in this article.

Rules for Adding Several Numbers

For adding several numbers, write the numbers one by one in column-wise.
Write the numbers of one’s digit in the right side of the column.
Write the numbers of tens place; hundred’s place one before the unit’s place and ten’s place in a row.
First add the one’s place and the sum of one’s place is more than one digit add the last digit to the columns top of the ten’s place, and vice versa.

Example Problems to Adding Several Numbers

Example 1:

Adding several numbers 6 + 7 + 3 + 5 + 9

Solution:

Adding all these numbers we get the carry of 3 and write before the number zero.

6

7

3

5

9

----------

30

----------



Example 2:

Adding several numbers 45 + 23 + 78 + 55+ 68+ 19

Solution:

Adding all these numbers we get the carry of 3 to the ten’s place column and add it to the ten’ place digit.

45

23

78

55

68

19

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

258

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

Example 3:

Adding several numbers 3564 + 4767 + 2433 + 7548

Solution:

Add all these numbers together to get 2 carry by adding one’s place and then carry of 2 by adding ten’s place and then the carry of 2 by adding hundred’s place and adding thousands place we get the carry of 1 it can be placed before the one’s digit by adding thousand place sum because there is no place digits before the thousands place.

3564

4767

2433

7548

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

18312

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



Problems to Practice for Adding Several Numbers

1. Adding 64 + 47 + 9 + 32 + 81

Key: 233

2. Adding 476 + 568 + 324 + 978

Key: 2346

3. Adding 7586 +9798 + 2434 + 9791

Key: 29609

4. Adding 94079 + 23247 + 86485 + 7607

Key: 211418

Probability Histograms

Introduction to probability histograms:

Histograms are used to plot density of an data, and often for density estimation: estimating the probability density function of the underlying variable. The total area of in a histogram used for probability density is always normalized to 1. If the length of the intervals on the x-axis are all 1, then a histogram is identical to a relative frequency plot. Now let us  see about the probability histogram.    

Solving Problem for Probability Histograms:

Example

When we toss a coin for three times. What is the probability of getting a heads? And also find the mean, standard deviation, expected value, standard error using probability histogram.

Solution:

Let us consider a coin to toss. If we toss a coin. We will get head or tail. Assume we calculate the number of heads. If we got head means probability is 1. Otherwise probability is 0.

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Calculate mean:

Mean = 0 + 1 / 2

= 1 / 2

probability histograms value of   Mean = 0.5.

Calculate Standard deviation:

Standard deviation = (1 – 0) v (1 / 2) (1 /2)

probability histograms value of standard deviation = 0.5.

Calculate Expected value:

Expected value = number of tosses × 0.5

If we toss a coin for three times means expected value is

= 3 × 0.5

probability histograms value of expected vaue  = 1

Calculate standard error:

Calculate standard error = vnumber of toss ×0.5

probability histograms value of  standard error  = v3 × 0.5

Practice Problems for Solving Probability Histograms:

Problem 1:

Compare the probability histogram and expect the sum with the actual results. What do conclude about that 4 dice result?

Answer :

Result of the 4 dice is not fair with the actual size.

Problem 2:

When we toss a coin for 13 times. What is the probability to get a tails? And also find the mean, standard deviation, expected value, standard error using probability histogram.

Answer:

Mean = 0.76 Standard deviation = 0.76 Expected value = 0.76 Standard error = 0.76

Trinomial Factoring Program

Introduction to trinomial factoring program:-
An algebraic expression containing three terms is called a trinomial.for example 2x+3y + 4z is a trinomial.
When we multiply two binomials, we get a trinomial. For ex (2x+5)(x+4) = 2x2 + 13x + 20 which is a trinomial.
So it follows that  we can find the factors of a trinomial.
Let us see some formulas that we use for trinomial factoring program.I like to share this What is a Trinomial with you all through my article.

Formulas Used in Trinomial Factoring Program:-

There are certain formulas that assist us when we need to factorise the trinomials.v
1. a2 + 2ab + b2  = (a + b)2
Let us do a problem bases on this formula.
# Factorize 25x2 + 30xy + 9y2
Using the above formula we find a2 = 25x2 => a = 5x;  2ab = 30xy = 2*5*3*x*y ; b2 = 9y2 = > b = 3y
Hence the solution is (a+b)2  =  (5x+3y)2= (5x+3y)(5x+3y)
Solution: The factors of 25x2+ 30xy + 9y2 = (5x+3y)(5x+3y)
2. Here is another formula to assist us to do the trinomial factoring program.
It is a2 - 2ab + b2 = (a - b)2
# Factorize 16x2 - 24xy + 9y2
This problem confirms to the second formula given here
Hence if a2 = 16x2 then a = 4x; b2 = 9y2 => b= 3y and  -2ab = -24xy => 2*4*3*x*y
Hence the factors of 16x2 - 24xy + 9y2  = (4x - 3y)2  = (4x-3y)(4x -3y)
Solution: The factors of 16x2 - 24xy + 9y2 = (4x - 3y)(4x - 3y)
Here is another formula to help trinomial factoring program
3. a2 + (a+b)x + ab =  (x+a)(x+b)
The second term is addition of two factors and the third term is the multiplication of two terms.
Hence our steps would be (1) to find the factors of a and b (2) select the factors that satisfy a+b
# Factorise x2 + 12x + 35
Solution:-
Here a+b= 12 and ab = 35
Step 1 : Let us find the factors of ab that is 35
Factors of 35 are (1,35), (5,7)
Step 2 : Let us add the factors and see which satisfies (a+b)
If we add 1+35 = 36 which is not what we want.
If we add 5+7= 12 which is what we want.
So the factors are (x+5)(x+7)
Hence  x2 + 12x + 35  can be factored as (x+5)(x+7)
Solution of the problem is (x+5)(x+7).

Probles Based on Trinomial Factoring Program:

Let us do one more problem based on the third formula .
# Factorize x2 +6x - 27
Solution:-
In this problem (a+b) = 6 and ab = -27
Step 1 find the factors of 27
(1,27), (3,9)
Let us add them 1+27=28 which is not the 2nd term 6 So we discard it.
Let us add 3+9 = 12 which is again not +6
But the third term has a negative sign.
So let us do 9-3 = +6
Now our formula changes slightly.  we need to put a negative number also
Hence we write it as (x+9)(x-3) and note that 9-3=6 which is the middle term and 9*-3 = -27 which is the last term             Hence the solution is x2 + 6x - 27 = (x+9)(x-3)
Answer : (x+9)(x-3)
Thus trinomial factoring program can be made easy if we learn the formulas that assit the factoring.

How to Solve Matrix Equality

Introduction for how to solve matrix equality:

Let us see how to solve the matrix equality in this article. The matrix equality normally represents the problems that involving the matrices for equating the right hand side with the left hand side using some identities in matrix like  `[[1,0,0,0],[0,1,0,0],[0,0,1,0],[0,0,0,1]]` which is generally represented as I having the diagonal elements equal to one. Some of the problems using with some properties that exist with the matrices.

Consider the two matrices P and Q are said to be equal if

(i) both the matrices P and Q are of the same order or size.

(ii) the corresponding entries in both the matrices P and Q are equal.

i.e. the matrices P = [pij]m × n and Q = [qij]a × b are equal if m = a, n = b and pij = qij  for every i and j.

Examples to Explain "how to Solve Matrix Equality"

Let us see some of the example problems about how to solve the matrix equality.

If  `[[p,o],[i,u]]`  =`[[-115,215],[115,-415]]`  then find the values of p, o, i, u .
Solution:

Here we know that the two matrices are equal, their corresponding entries are also equal.

`[[p,o],[i,u]]`  =`[[-115,215],[115,-415]]` 

? The answer is p = -115, o = 215, i = 115, u = -415

Find the value of  a by solving if `[[p,3p - q],[2p + r,3q - r]]` = `[[0,-7],[3,2a]]`
Solution:

Given `[[p,3p - q],[2p + r,3q - r]]` = `[[0,-7],[3,2a]]`

On equating we have

p = 0              ---------------(1)

3p - q = -7     ---------------(2)

2p + r = 3      ---------------(3)

3q - r = 2a    ---------------(4)

Put (1) i.e p = 0  in (2)

3(0) - q = -7

q = 7   ---------------(4)

Put (1) i.e p = 0  in (3)

2p + r = 3

2(0) + r = 3

r = 3        ---------------(6)

Put r=3 and q=7 in (4)

3q - r = 2a

3(7) - (3) = 2a

21 - 3 = 2a

18 = 2a

2a = 18

a = `18/2`

a=9

Hence we find the a value using the matrix equality after solving the equations.

My forthcoming post is on Linear Systems of Equations, Linear Programming Examples will give you more understanding about Algebra.

Practice Problems to Explain "how to Solve Matrix Equality"

If  `[[p,o],[u,u+a]]`  =`[[-105,25],[15,-45]]`  then find the value of a .


The answer is that a = -60

Find the value of  a by solving if `[[p,3p - q],[2p + r,3q + r]]` = `[[0,-7],[3,2a]]`


The answer is that  a = 12.

Prime Number Algorithm

Introduction to prime number algorithm:

A prime number is a positive natural number with no other divisors except one and itself but the number 1 is excluded by definition. That is to say one is not a prime number. That leaves us with the first prime number as 2  and it is quite obvious that all other prime numbers after 2 have to be odd numbers only. The concept of prime numbers has fascinated mathematicians from the earliest times. Euclid proved infinitude of prime numbers as early as 300 B.C. There have been many failed attempts to find a formula that will generate prime numbers. A search for efficient algorithms to find prime numbers gained momentum with the advent of computers and today many efficient programs exist to calculate prime numbers limited only by the processor memory and speed.

Finding Prime Numbers by Division:

In order to check whether a given number n is a prime or not, it has to be divided one by one by all prime numbers less than `sqrt(n)`. The reasoning is simple; if the number is not a prime then there will be at least one prime factor which is less than  `sqrt(n)`. This can be used to create a simple algorithm for checkng numbers one by one and discarding compound numbers will leave a list of prime numbers.

In simple language we can write the algorithm as:

Take numbers 2 to N (one by one). Let the number by x

Divide x by all prime numbers less than or equal to `sqrt(x)`

If it is divisible by any of the prime numbers, discard it as non-prime and go to next number.

Simple! Isn't it? But it is not a very useful algorithm for use with computers, where too much processor is lost in division operation at every stage of this algorithm. Processors are more comfortable with read and write commands and simple addition and substraction operations. It is interesting to note that one of the oldest algorithms Sieve of Eratosthenes is one of the efficient algorithms for use with modern computers.

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Sieve of Eratosthenes of Prime Number Algorithm:

Concept behind this prime number algorithm is quite simple and uses tool of skip counting learnt by all of us in elementary school. It has only two steps. Write all the numbers up to a specified number. 2 is a prime, so after 2 strike out every second number. Next is 3 again a prime, so after 3 strike out every third number; and so on. Every time you complete the process for a prime p, you would be left with only primes up to the number p2.

Step 1:

2,  3,  4,  5,  6,  7,  8,  9,  10,  11,  12,  13,  14,  15,  16,  17,  18,  19,  20,  21,  22,  23,  24,  25

Step 2: Striking out every second, then every third and then every fifth number, we are left with prime numbers less than 52 = 25 (shown in bold).

2,  3,  4,  5,  6,  7,  8,  9,  10,  11,  12,  13,  14,  15,  16,  17,  18,  19,  20,  21,  22,  23,  24,  25

We are left with prime numbers up to 25 as: 2, 3, 5, 7, 11, 13, 17, 19, 23.

It is interesting to note that this algorithm is found quite suitable for listing prime numbers using present day computing devices.

Perimeter Semi Circle

Introduction to Semi circle perimeter:

A perimeter is a path that surrounds an area. The word comes from the Greek peri (around) and meter (measure). The term may be used either for the path or its length - it can be thought of as the length of the outline of a shape. The perimeter of a circular area is called circumference.

(Source: wikipedia)

Semi circle formula:

Fomula for finding semi circle is p =  1/2 p × d + d.

Semi Circle Perimeter Definition - Examples:

Semi circle perimeter definition - Example 1:

Find the perimeter of a circle known that its diameter is 6

Solution:

The perimeter is always multiplied by ½.`pi` into diameter then adds with diameter

Perimeter = 1.57142857 x 6 + 6 = 15.4285714

Semi circle perimeter definition - Example 2:

Find the perimeter of a circle known that its diameter is 25

Solution:

The perimeter is always multiplied by ½.`pi` into diameter then adds with diameter

Perimeter = 1.57142857 x 25 + 25 = 64.2857142

Semi circle perimeter definition - Example 3:

Find the perimeter of a circle known that its diameter is 54

Solution:

The perimeter is always multiplied by ½.`pi` into diameter then adds with diameter

Perimeter = 1.57142857 x 54 + 54 = 138.857143

Semi circle perimeter definition - Example 4:

Find the perimeter of a circle known that its diameter is 45

Solution:

The perimeter is always multiplied by ½.`pi` into diameter then adds with diameter

Perimeter = 1.57142857 x 45 + 45 = 115.714286

Semi circle perimeter definition - Example 5:

Find the perimeter of a circle known that its diameter is 63

Solution:

The perimeter is always multiplied by ½.`pi` into diameter then adds with diameter

Perimeter = 1.57142857 x 63 + 63 = 162

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Semi Circle Perimeter Definition - Practice Problems:

Practice Problem 1:

Find the perimeter of a circle known that its diameter is 5

Answer:

12.8571428

Practice Problem 2:

Find the perimeter of a circle known that its diameter is 8

Answer:

20.5714286

Line Segments with Numbers

Introduction for line segments with numbers:
Line segments are the important one in geometry chapters of mathematics subject. Line segments have more definitions; Line is an endless straight mark. Line segment is a part of one line, which are both directions having an end points with name. And line segments have so many names or types. Here, we use numbers for line segments names.

Line Segments:

A line segment is one part of line, which is having two end points. And it is defined as a straight line, which is joining with two points with coordinates and with out extending the line after that the point. It means all the given points between the given two numbers.

The above figure is having a line segment like 3,6                     

Generally we know a normal line is extending in both two directions, so this word “segment” is very important in this “line segment”.

In the above figure 3, 6 are the two end points, the length of the line segment has used with numbers for name as two end points 3, 6.

Line Segments distance formula:

In line segments we have to find the distance of line segments with end points name or numbers,

Line segment distance formula,

XY = √(x2-x1)2+ (y2-y1)2

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Examples of Line Segments with Numbers:

Example 1:

Using the distance formula Find line segment between two end point numbers, end points are

M = (3, 2), N = (5, 2)

Solution:

x1=3, y1= 2, x2=5, y2=2

Line segment distance formula  

XY = √(x2-x1)2+ (y2-y1)2

= √ (5-3)2 + (2-2)2

= √ (2)2 + (0)2

= √4+ 0

= √4

XY distance = 2 units

Example 2:

Using the distance formula Find line segment between two end point numbers, end points are

S = (3, 2), T = (6, 4)

Solution:

x1=3, y1= 2, x2=6, y2=4

Line segment distance formula  

ST = √(x2-x1)2+ (y2-y1)2

= √ (6-3)2 + (4-2)2

= √ (3)2 + (2)2

= √9+ 4

Answer is ST distance = √13 units.

Kinds of Angles in Trigonometry

Introduction to kinds of angles in trigonometry:

In mathematics, angles are nothing but the combining of two lines in trigonometry. In trigonometry, there are different kinds of angles. In trigonometry, the different kinds of angles are named as right angles, obtuse angle, acute angles, straight angles etc. The classification of different kinds of angles with diagrams is given below.

Kinds of Angles in Trigonometry – Types:

The diagrams and explanation for the different kinds of angles are given below:

Right angles:                                                              

The Diagram this kind of angles is given below, 

           
These are right angles where the value of degree of angle is 90 degree.

Acute angles:

The Diagram this kind of angles is given below, 

             
These are acute angles where the value of degree of angle is smaller than 90 degree.

Obtuse angles:

The Diagram this kind of angles is given below,   

             
These are obtuse angles where the value of degree of angle is larger than 90 degree and less than 180 degrees.

Reflex angles:

The Diagram this kind of angles is given below, 

                
These are reflex angles where the value of degree of angle is larger then 180 degree.

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Kinds of Angles in Trigonometry – Examples:

Example 1: Find what kind of angle it is given that the value of degree is 120?

Solution:

Let as assume x is value of degree.

Then, x = 120 degree.

The degree of angle is larger than 90 degree and also less than 180 degree.

Therefore, it is obtuse angles.

Example 2: Find what kind of angle it is given that the value of degree is 220?

Solution:

Let as assume x is value of degree.

Then, x = 220 degree.

The degree of angle is larger than 180 degree.

Therefore, it is reflex angles.

Kinds of angles in trigonometry – Practice problems:

Problem 1: Find what kind of angle it is given that the value of degree is 90?

Answer is given below:

These kinds of angles are right angles.

Problem 2: Find what kind of angle it is given that the value of degree is 30?

Answer is given below:

These kind of angles are acute angles.

Elementary Math Methods

Introduction for elementary math methods:

Elementary math methods cover all the basic operations and function in algebra topic. The elementary math is the main part covered arithmetic. All the basic operation is presented in this area. This is covering all basic operation of addition multiplication, subtraction, and division. The elementary math methods cover kindergarten level to middle school level. This methods also helps to solve real life problems.

Example: 3x+ 2 = 10

Elementary Math Methods Cover

Elementary algebra lessons are contain this basic methods

Arithmetic Operations:

The real numbers have the following properties:

a + b= b +  a    ab  = ba                            (Commutative Law)

(a+ b)+ c= a+ (b + c)      (ab)c = a(bc)        (Associative Law)

a+(b +c)= ab +ac                                       (Distributive law)

Fractions:

To add two fractions numbers with the same denominator, we use the Distributive Law property:

` a/b+c/b` = `1/(b*a)` +`1/(b*c)`   =`1/(b(a+c))`   =`(a+c)/b`

To add two fraction with different denominators, we use a frequent denominator:

`a/b+c/d` =  `(ad+bc)/(bd)`

Factoring

Here we can make use of Distributive Law to expand certain algebraic conditions. In rare case we need to repeal this method (again using the Distributive Law) by factoring an expression as a product of simpler ones. The easiest condition occurs when the given expression has a common factor as given below,

3x(x-2)=3x2 – 6

Example Problems i Elementary Math Methods

Which is the larger number -13 or -16?

Solution:

The large number in the negative signed numbers we considered, which number is zero to negative side have been going that numbers small number. And near to zero numbers are called large numbers

-16,-15,,-14,-13,-12,-111,-10,-9,-8,-7,-6,-5,-4,-3,-2,-1,0

so,-13 is larger than -16.

2. List all the integers between -2 and 4.

Solution:

-2,-1, 0, 1,2,3,4 these number are present in the -2 ,4

The -2, 4 between numbers are -1,0,1,2,3

3. x-4=8?

We add +4 on both sides

x=8+4

x=12

4. Simplify 8 `-:` ` 2/3` ?

Solution:

We divide the given equation this is simple method the division inverse of multiplication

8

4*3   = 12

5. Reduce `14/35` .?

Solution:

`14/35`

We divide 7 on both sides

`2/5`

6.12x=4?

Solution:

x=`4/12`

x=`1/3`

7. Factorize the given expression x2-9   ?\

Solution;

The general form of

(a2-b2)=(a-b)(a+b)

So, x2-9=  (x-3)(x+3)

8. Simplify `1/2` + `2/3` ?

Solution   :

`1/2` + `2/3`

We take L.C. M   on 2, 3   

`(1*3)/(2*3)` +`(2*2)/(3*2) `

=`3/6` +`4/6`  =`(3+4)/6`

=`7/6`

9. 3X+4y=5x-2y ?

Solution:

4Y+2y=5x-3x

6y=2x

y=`(2x)/6`

y=`x/3`  

Triangular Prism Image

Introduction to triangular prism image

In geometry, a triangular prism is a three-sided prism; it is a polyhedron made of a triangular base, a translated copy, and 3 faces joining corresponding sides. Equivalently, it is a pentahedron of which two faces are parallel, while the surface normals of the other three are in the same plane. These three faces are parallelograms. All cross-sections parallel to the base faces are the same triangle. (Source: Wikipedia)



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Triangular Prism Image

Triangular prism

Volume of the prism

The method utilized for computing the volume of triangular prism, is identified as,

Volume of prism = Base area x Length of prism

Base area of prism = `(bh)/2`

Wherever, b = base of the triangle

h = height of the prism.

First we can find the bottom area of the triangular prism is to be computed then we have found the volume of the prism.

Examples for Triangular Prism Image

Example 1 for triangular prism image

Compute the volume of the triangular prism, whose base is 10 cm, height is 12 cm and length is 16 cm.

Solution

The method used for computing the volume of prism, is known as,

Volume of triangular prism = Base area x Length of prism

base = 10 cm, height = 12 cm and length = 16 cm

Base area = `(bh)/2`

= `(10xx12)/2`

= `(120)/2`

= 60 cm2

Volume     = 60 x 16

= 960 cm3.

Therefore the volume is 960 cm3

Example 2 for triangular prism image

Compute the volume of the triangular prism, whose base is 15 cm, height is 16 cm and length is 20 cm.

Solution

The method used for computing the volume of prism, is known as,

Volume of triangular prism = Base area x Length of prism

base = 15 cm, height = 16 cm and length = 20 cm

Base area = `(bh)/2`

= `(15xx16)/2`

= `(240)/2`

= 120 cm2

Volume     = 120 x 20

= 2400 cm3.

Therefore the volume is 2400 cm3