Friday, May 17, 2013

Help for 3rd Grade Math

Introduction to help for 3rd grade math:
In this article  we are going to help for 3rd grade  math student ,those 3rd grade students need help in basic operation of math like place value, compare numbers, addition, subtraction, multiplication, division, conversion of units like kilometer, meter, centimeter, millimeter, how to draw the graph  and also they need help in geometry.

I like to share this Tangent Line Equation with you all through my article.

Example - help for 3rd grade math


Which symbols are used for compare the numbers?
Solution:
Generally in comparison of number we use the symbol as < (less than),> (greater than), <=(less than or equal to),>= (greater than or equal to)
suppose take the  number as 24 and 12 ,here the number 24 is less than the 12 ,so it will be used in the form of 24>12

Example - help for 3rd grade math
8 hundreds+4 tens. How those expression will be written in number format
Solution:
Here 8 hundreds will be written as 800 then 4 tens will be written as 40 after that we have to note one thing in-between the plus sign will be there so after converging in the number format we have to add 800+40=840
So the answer is 840.

Example - help for 3rd grade math
What is the value of the underlined digit 789?
Solution:
Here place value of 7 will be written as 70, place value of 8 will be written as 80 and place value of 9 will be written as 9, so the value of the underlined digit is 80.

Understanding hard math word problems is always challenging for me but thanks to all math help websites to help me out.

Example - help for 3rd grade math


Solve 32-11
Solution:
Here we are doing the operation as subtraction it means removing some of the objects from the group here 11 will be removed from 22
First we subtract ones digit place 2-1=1 then do the operation is tens digit place 3-1=2
Combine these two digits we get the answer 21, it will be shown below.

Math Notes for Grade 8

Math notes for grade 8:

In this article we are going to discuss about find the math notes for grade 8.Math notes for grade 8 are easy to understand and solve. Math notes for grade 8 problems involve basic addition, subtraction, multiplication and division problems. Those arithmetic problems involve biggest numbers. The following topics are studied in the 8 grade,

Numbers
Measures
Algebra
Geometry
Handling data.
The math notes for grade 8 solving problems are given below.

Is this topic How to Find the Surface Area of a Cube hard for you? Watch out for my coming posts.

Example problems of math notes for grade 8:


Example 1:

Sum of 3 consecutive odd numbers is 51. Find the numbers.

Solution:

Framing the equation:

Let the first odd numbers be x.

Then the second and third odd numbers are (x+2) and (x+4)

There sum is 51.

x+(x+2)+(x+4) = 51

x+x+2+x+4 = 51

3x + 6 = 51(adding the like terms)

3x+6 = 51 is the required equation.

Solving the equation:

3x+6 = 51

3x = 51 – 6 (by rule 1)

= 45

x = 45 * `1/3 ` (by rule 3)

x = 15

The consecutive odd numbers are x, (x=2), (x=4)

15, (15+2) and (15+4)

The required numbers are 15, 17 and 19.

Example 2:

If office managers fixed in the pay scale 3200 – 85 – 4900, when will he reach his maximum?

Solution:

Pay scale: 3200 – 85 – 4900

Starting salary = $ 3200= a, Annual increment = $85 = d,

Maximum salary = $4900 = tn

tn = a + (n – 1)d = 4900 = 3200+(n-1) 85

n – 1 = 1700/85 = 20

n =20+1 = 21

The manager will reach his maximum in his 21st year of service.Having problem with math problems for 3rd grade keep reading my upcoming posts, i will try to help you.


Example 3:


The base and height of a right triangular ground are 60 m and 45 m. Find the cost of leveling the ground at Rs 150 per are ( 1 are = 100 sq.m)

Solution:

Given base b = 60m; height, h = 45m

Cost of leveling 1 are = Rs.150

Area of the triangular, A = `1/2` * 60 * 45

= 1350m2

100m2 = 1are

1350m2 = `1350/100` = 13.5are

Cost of leveling 1 are = Rs.150

Cost of leveling 13.5 Ares =Rs. 13.5 * 150 = Rs. 2025

Cost of leveling the right triangular ground = Rs. 2025

Example 4:

Find the circumference of a circle whose radius is 7cm.

Solution:

Radius r = 7cm

Circumference, C = 2`pi` r

= 2 * 22/7 * 7

= 44

Circumference of the circle = 44cm.

Example 5:

Solve 6x + 12 = 4x – 2

Solution:

Given expression 6x + 12 = 4x – 2

Subtract 12 on both sides of the equation

6x +12 – 12 = 4x – 2 -12

6x = 4x -14

Subtract 4x on both sides of the equation

6x – 4x = 4x – 4x -14

2x = -14

Divide 2 on both sides of the equation

`(2x)/2` =` -14/2`

x = -7

Solution is x = -7