Online Calculus Courses

Introduction :

The online calculus courses is common and open to everyone around the world. Students from anywhere can enroll in this online calculus courses. There is no need of class meetings in online calculus courses. Online calculus courses include limits, derivatives, applications of derivatives, integrals and application of integrals. In online calculus courses, students can attend final exam at their home or anywhere around the world. Derivatives and integrals are explained below with its applications to show how online courses helpful to you.

Derivatives and its applications:


In derivatives, we will study the changes in the values of y corresponding to small changes in the values of x where y and x are related to each other by the equation y = f(x).

Application of derivatives:

• Rate of change of quantities
• Errors and approximation
• Rolle's and Lagrange's theorem
• Maxima and minima functions
• Increasing function and decreasing functions
• Tangents and normal
• Optimization
• Graph shape

Example problem:

Find the derivative of the function y = 2x^3 + x^2 + 3.

Solution:

Step 1: Given function

y = 2x^3 + x^2 + 3

Step 2: Differentiate the given function y = 2x^3 + x^2 + 3   with respect to ' x ', to get `(dy)/dx`

`(dy)/dx` = 6x^2 + 2x

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Integrals and its applications:


Integration is the reverse process of that of differentiation. The process of finding derivative of the given function is called differentiation whereas finding the function whose derivative is known is called integration. This function is called as integral of the given function.

Application of integrals:

• To find area under simple curves
• To find areas of circles
• To find areas of parabolas
• To find areas of ellipses
• To find area between the curves
• Average function value
• Volumes of solids of revolution

Example problem:

Find the integration of the function,  f(x) = 3x + 8.

Solution:

Step 1: Given function

f(x) = 3x + 8

`int` f(x) dx = `int` 3x + 8 dx

Step 2: Separate the integral function

`int` 3x + 8 dx= `int` 3x dx + `int` 8 dx

Step 3: Integrate each function with respect to ' x',

`int` 3x + 8 = `(3x^2)/2` + 8x + C

Online Classroom Learning

Introduction :

Math solving is a mechanism of solving math problems. It consists of an in-numerous number of problems for calculation. It consists of many fields related to problem solving methodology. It is related to vector, difference equation, integrals, sets, numerals, and diagrams etc. the online classroom learning article offers you some solved examples and some techniques in mathematics.online is the best tool for learning.Through online we can get instant help at the second.

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Basic operations - online classroom learning:


The following are the basic operations of online classroom learning,

• Addition,
• Subtraction,
• Multiplication and
• Division.

Online Classroom learning - Solved examples:


Example 1:

Solve the equation y +6 = 5.

Solution:

it is easy to solve within one step by grouping like terms

y = 5 - 6

Solution :y = -1

Example 2:

Solve the equation 5y + 5= 25

Solution:

it is easy to solve within one step by grouping like terms

5y + 5 = 25

5y = 25 - 5

5y = 20

Divide by 5

Solution :y = 4

Example 3:

Find the value: 25 y - 10 = 5 y - 5

Solution:

It is easy to solve within one step by grouping like terms

25y - 10 = 5y - 5

25y - 5y = 10 - 5

20 y = 5

Divide by 20

y = 1/4.

The answer is y = 1/4.

Example 4:

Find the sum of 1000 + 5000?

Solution:

1000 + 5000 = 6000.

Answer = 6000.

Example 5:

Solve the equation y +16 = 15.

Solution:

It is easy to solve within one step by grouping like terms

y = 15 - 16

solution :y = -1

Example 6:

Solve the equation 5y + 15= 45

Solution:

it is easy to solve within one step by grouping like terms

5y + 15 = 45

5y = 45 - 15

5y = 30

Divide by 5

solution :y = 6

Example 7:

Find the value: 25 y - 15 = 15 y - 5

Solution:

it is easy to solve within one step by grouping like terms

25y - 15 = 15y - 5

25y - 15y = 15 - 5

5y = 10

Divide by 5

y = 1/2.

The answer is y = 1/2.


Tips - online classroom learning:

The below are the tips for online classroom learning.

at first, solve all the operations with in the parenthesis and braces.
then  the exponents are calculated
next perform the multiplication and division operations
at last do the addition and subtraction.

Online Geometry Calculator

Introduction to Geometry:-

Geometry "Earth-measuring" is a part of mathematics concerned with questions of size, shape, relative position of figures, and the properties of space. Geometry is one of the oldest sciences. Initially a body of practical knowledge concerning lengths, areas, and volumes, in the 3rd century BC geometry was put into an axiomatic form by Euclid, whose treatment—Euclidean geometry—set a standard for many centuries to follow.                                                                                                                                                                            Source: - Wikipedia.

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Online geometry calculator - Example problems:


Online geometry calculator - Problem 1

Calculate the area of rectangle whose side’s measure 12 centimeters and 8 centimeters respectively.

Solution:-

Given

Length – 12 centimeter.

Breadth – 8 centimeter.

The formula used to find the area of rectangle is length * breadth.

Area = length * breadth.

By plugging the given values in to the formula.

We get,

Area   = 12 * 8

The product of 12 and 8 yields 96 so the area of the given rectangle is 96 square centimeter.

Online geometry calculator - Problem 2

Calculate the volume of the cube  whose side measure 12 centimeter.

Solution:-

Given:-

Measure of cube length = 12.

The formula used to find the volume of the cube is `a^3` .

By plugging the value in to formula we get the result as

Volume =  `12 ^3` .

=  1728.

So the volume of given cube is 1728 cubic centimeters.

Online geometry calculator - Problem 3

Calculate the volume of the cylinder which has the height is 12 centimeter and radius of 6 centimeter.

Solution:-

Given:-

Height = 12 centimeter.

Radius = 6 centimeter.

The formula used to calculate the volume of the cylinder is  `pi r^2 h`



`Volume = pi r^2 h`

r– radius of the cylinder

H – height of the cylinder.

By plugging in the given values in to the formula we get

` Volume = pi (6)^2 * 12`

`6^2` can be written as 6 * 6 = 36

Volume = `pi` 36 * 12

The product of 36 and 12 is 432

Volume =` pi` 432.

So the volume of the given cylinder is `432pi`

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Online geometry calculator - Practice Problems:


Problem 1

Calculate the volume of the cylinder which has the height is 2 centimeter and radius of 1 centimeter.

Answer:-

2`pi` .

Problem 2

Calculate the area of rectangle whose side’s measure 14 centimeters and 2 centimeters respectively.

Answer:-

28.

Online Distance Measuring

Introduction to online distance measuring:

Distance is a numerical description of how far apart objects are. In mathematics, a distance function is a generalization of the concept of physical distance.

The distance d between any two points (x1, y1) and (x 2, y2) can be calculated using the following formula,

Distance, d = `sqrt ((x2 - x1)^2 + (y2 - y1)^2)`

If the speed and time is known, then distance d is given by

Distance, d = Speed * Time

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Example problems for online distance measuring:


Measuring distance through online is very simple. In online, if  we enters the data for which the distance is to be measured, it performs the specified operations automatically and generates the required solution for the problem. The example problems are given below which shows online distance measuring with steps in detail.

Example 1:

Using distance formula, calculate the distance between the points (2, 5) and (5, 6).

Solution:

Step 1: Assign variables

x1 = 2     x2 =5

y1= 5      y2 = 6

Step 2: Plug all values in the distance formula

d = `sqrt((x2 - x1)^2 + (y2 - y1)^2)` ........... Distance formula

=  `sqrt(((5 - 2)^2+(6 - 5)^2))`

Step 3: Solve the above equation and find distance

= `sqrt((3)^2+ (1)^2)`

= `sqrt (9 + 1)`

= `sqrt10`

= 3.162

Step 5: Solution

The distance between the points (2, 5) and (5, 6) is 3.162

Example 2:

If a bike runs at the speed of 50kmph and it covers the certain distance in 6 hours, then find out the distance covered by it?

Solution:

Step 1: Given

Speed = 50kmph

Time = 6 hours

Step 2: Formula to find distance

Distance, d = Speed * Time

Step 3: Substitute all values in the formula

d = 50 *6

= 300

Step 4: Solution

Therefore, the total distance covered by the car is 300km

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Practice problems for online distance measuring:


The practice problems are given below which helps you for learning online distance measuring.

1) Using distance formula, calculate the distance between the points (-4, 5) and (- 3, 8).

2) Find out the distance covered by the train if it runs at the speed of 60kmph and it covers the certain distance in 5 hours?

3) Using distance formula, calculate the distance between the points (6, 8) and (1, - 7).

Solutions:

1) d = 3.162

2) d = 300km

3) d = 15.811

What is a Function in Algebra

What is a Function in Algebra

To find the area of a square we need the length of the side. So, we can say that the area of the square depends on the length of the side of the square. It means that if the length of the side changes, the area also changes then we can say that the first is the function of the other. So, we can conclude that the area of the square is the function of and it depends on the length of its side.  So, in mathematics algebraic functions can be defined as, a relationship between two variables something like ‘x’ and ‘y’ , function if there is a rule which assigns to each value of the variable ‘x’ there exists one and only one value of ‘y’. Then we can say that y is a function of x.

It is clear now that a function must be single valued, for instance, y=3x+2. Here to each value of x there is a unique value of y.  All the values which x can assume is called the domain.  These are the values for which the function can be defined. In the function y=3x+2, the domain include all the real numbers which means ‘x’ can be any real number.

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Once the domain is defined, the set of corresponding ‘y’ values for each value of ‘x’ is called the range of the function. For instance, if x=2 in the domain then y=(3.3)+2=11 is the corresponding value in the range.  So, ‘x’ can be called the independent variable and ‘y’ a dependent variable as its value depends on the value of x.
Consider an example of a algebra functions given by, x={-1,0,1,2} where each value is the value of the domain,  the rule being y=x2-1. Let us write the ordered pair of the given function.  Here we need to plug in each of the values of ‘x’ in the given rule to arrive to the corresponding ‘y’ value of the range.  When x=-1 then y=(-1)2-1 = 1-1=0; when x=0 then y=(0)2-1=-1; when x=1 then y=(1)2-1=0; when x=2 then y=(2)2-1=4-1=3. So, when the domain= {-1,0,1,2} the corresponding values of the range={0,-1,0,3} and the ordered pair of the function is (x,y)={(-1,0),(0,-1),(1,0),(2,3)}. So, functions algebra is a well behaved relation which means given an ‘x’ there is exactly one and only one ‘y’.

In the above functions algebra each ‘x’ to each ‘y’ we can see that there is only one arrow coming from each of ‘x’ of the domain and hence it is a function.

Trinomials Solving Online

A trinomial expression is any polynomial expression which has accurately three terms.
The equation or function or expression is in the structure of  ax^2+bx+c =0 where a?0, b, c are constants called as trinomials.
We can say it also as a quadratic function or quadratic equation.
The trinomials are having two roots. When we multiply the two binomials we can get the trinomial.

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STEPS FOR SOLVING TRINOMIALS


Factor out any factors common to all terms. The equation 8x^2 + 12x + 4 has 4 as a common factor, since every term can be divided by 4. Therefore, it can be factored as 4(2x^2 + 3x +1).
The equation x3 +2x^2 + x has x as a common factor. It can be factored as x(x^2 +2x +1).

Look for any other common factors you may have missed. In sometimes, an equation has both a integer and a variable that can be factored out. For example, 4x3 +12x^2 + 16x has both 4 and x as a factor. Factored out, it becomes 4x(x^2 + 3x + 4)

Locate out what type of trinomial equation you have left. If the maximum power of the unsatisfactory part is a squared variable like y2 or 7a2, you can factor it like a quadratic equation.

If you’re maximum power term is a cubed number or higher, you have a higher order equation. By this point, you will most likely not have anything greater than a cubed variable to deal with.
Many trinomial quadratics are easy sums of squares. Using an example from step one: 8x^2 + 12x + 4 = 4(2x^2 + 3x + 1).


EXAMPLES OF TRINOMIALS

1) Factors of trinomials of the form (Ax^2 +Bx +c)

We find two numbers a and b such that ab = C, and a+b = B

3 x^ 2 + 11x + 10

Coefficient of first term is 3, last term is 10

3 x 10 = 30

6x5 = 30, 6+5 = 11

We split middle term 11z = 6x + 5x

3x^2 + 6x +5x +10

3x ( x +2) + 5( x +2)

(3x +5) (x +2)

2)  Factors of trinomials of the form (A x^2 +Bx - C)

We find two numbers a and b such that a (-b) = C, a-b = B

8x^2 +2x - 3

8x^2 +6x - 4x - 3

2x ( 4x +3) -1(4x +3)

(2x - 1) ( 4x +3)

3) Factors of trinomial of the form (Ax^2 - Bx +C)

We find two numbers a and b such that (-a) (-b) = C, -a-b =-B

14 x^ 2 -23x +8

14 x^ 2 -16x -7x +8

2x (7x -8) -1(7x -8)

(2x -1) (7x - 8)

4) Factors of the trinomial of the form (Ax^2 -Bx -C)

We find two numbers a and b such that (-a) (b) = C -a+b = -B

12 x^ 2 -x -35

12x^2 -21x +20x -35

3x (4x - 7) +5(4x -7)

(3x +5) (4x - 7)

Study Online Factors

Introduction to factors:

In online study on factors, it is essential to know the definition of factors.The set of numbers which produce the remainder zero by dividing a particular number. These sets of numbers are called as factors. We can express a number by the multiples of another two numbers called as factors.

For Example,

10= 2 x 5

32=4 x 8

Here 2, 5 are the factors of 10 and 4, 8 are the factors of 32.


Types of Factors and Sample problems:


For the online study on factors, knowing the types of factors is an essential one. The types of factors for online study are as follows:

There are two types of factors,

• Prime Factors
• Composite factors

Online study on Prime Factors:

The numbers that can be expressed as the multiple of one and that number itself are called as prime numbers. The factors of prime numbers are called as prime factors.

Consider the following Examples,

3= 1 x 3 (1, 3 are the prime factors of 3)

5= 1 x 5 (1, 5 are the prime factors of 5)

7= 1 x 7 (1, 7 are the prime factors of 7)

Online study on Composite Factors:

The factors that are not a prime numbers are called as composite factors.

Consider the following example,

24=4 x 6

Here the numbers 4, 6 are not a prime number. Therefore, 4, 6 are called as composite factors of 24.

Example:

Find the factors of 130.
Solution:

130= 1 x  130

130= 2 x  65

130= 5 x  26

130= 10 x 13

So the factors of 130 =1, 2,5,10,13,26,65,130.

Important note: We can express all the numbers except “1” as a product of prime numbers are called as prime factorization.

Consider the following Example,

24 = 4 x 6

= (2 x 2) x (2 x 3)

24 = 23 x 3

Here 2, 3 are the prime numbers. Hence, we conclude that all the numbers can be expressed as a product of prime numbers.

Prime factorization tree:

By using the prime factorization tree, we can get all the prime factors of the given number.

Example:

Find the prime Factors of 124
Solution:

124

/  \

2     62

/  \

2   31

124 = 2 x 2 x 31

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Practice Problems on factors:


Following are the practice problems given for online study on factors.

1. Find the factors of 524.

2. Find the Prime factors of 900

3. Find the composite factors of 1056.

Answer Key:

1. (1, 2, 4, 131, 262, 524)

2.  (2 x 2 x 3 x 3 x 5 x 5)

3. (2  3  4  6  8  11  12  16  22  24  32  33  44  48  66  88  96  132  176  264  352  528  1056)