# Math Help Online Tutor Free

Online math help is gaining in popularity as students and parents all over the globe realize the benefits that the internet offers. Majority of students look for extra help in math, whether it’s on a daily basis or study sessions every few weeks. Math tutor help puts students in touch with experienced math tutors who teach them over the internet. Free math tutor help will ensure that you understand the concepts and formulae and also get help out with your homework every day.

## Free Online Math Tutor Help

Math tutor help online is fast becoming one of the easiest ways to learn math. Shown to be as effective as class room learning, math help online tutors make complex math simple and interesting. Online tools like interactive whiteboards, math games and puzzles make the subject engaging and keep students coming back. Schedule sessions any time you want and get all your doubts cleared. Math help tutors assist students during exam season by providing exam prep and mock tests which assess your standard.

## Solved Examples

Question 1: Solve (x - 2)(x - 4)2
Solution:
Given, (x - 2)(x - 4)2

=> (x - 2)(x - 4)2  = (x - 2)(x2 + 16 - 8x)

[Using identity, (a - b)2 = a2 + b2 - 2ab]

=> (x - 2)(x2 + 16 - 8x) = x(x2 + 16 - 8x) - 2(x2 + 16 - 8x)

= x3 + 16x - 8x2 - 2x2 - 32 + 16x

= x- 10x2 + 32x - 32

=> (x - 2)(x - 4)2 = x- 10x2 + 32x - 32

Question 2: The perimeter of a rectangle is 50 inch. The length of the rectangle is 5 inch less than twice the width. Find the length and width of the rectangle.

Solution:
Given
Perimeter of rectangle = 50

Let width of the rectangle = x

Then length of rectangle = 2x - 5

[Perimeter of rectangle = 2(Length + Breadth)]

=> 50 = 2(2x - 5 + x)

=> 50 = 2(3x - 5)

=> 50 = 6x - 10

=> 50 + 10 = 6x

=> 60 = 6x

=> x = 10

So length of rectangle = 2x - 5 = 2 * 10 - 5

= 20 - 5 = 15

=> Width of the rectangle = 10 inch

and length of rectangle = 15 inch

Question 3: A pair of dice is rolled. What is the probability of getting a sum of 2 ?
Solution:
Sample Space = {(1, 1), (1, 2), .......(1, 6), (2, 1), (2, 2),......,(2, 6), (3, 1), (3, 2),.........., (3, 6), (4, 1), (4, 2), .............., (4, 6), (5, 1), (5, 2), ......, (5, 6), (6, 1), (6, 2), .........., (6, 6)}

Total number of possible outcomes = 36

Possible outcomes getting sum of 2 = {(1,1)}

Number of favorable outcomes = 1

P(getting sum of 2) = $\frac{Number of Favorable Outcomes}{Number of Possible Outcomes}$

= $\frac{1}{36}$

=> P(getting sum of 2) = $\frac{1}{36}$.