Nine Grade Math Slope

Introduction to nine-grade math slope:

Nine- grade math slope article deals with how to find the slope from the slope intercept from of the equation and from the two given points of the line. It includes the slope of the parallel and perpendicular line. We have the different formula to find the slope of the given line

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Formula related to nine grade math slope.

When two coordinate points on the line is given

Formula to find slope, m = `((y2-y1)/(x2-x1))`

When slope intercept form is given

The slope intercept form is y = mx+b

m = `((y-b)/x)`

Where m is the slope of the line

b is the y intercept.

When two lines are parallel to each other, the slope of the both lines is equal.

When two lines are perpendicular to each other, the product of slope is -1[ (m1*m2)= -1]

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Model problems for nine grade math slope


Example problems for nine grade math

Find the slope of the line whose coordinate points are (2, 3) and (4, 5).


Solution:

(x1,y1) = (2, 3)

(x2,y2) = (4, 5)

Formula:

Formula to find slope, m = `((y2-y1)/(x2-x1))`

= `((5-3)/ (4-2))`

= `(2/2)`

= 1

The slope of the line is 1

2.find the slope of the line whose slope intercept form is 3y = 6x +9

Solution:

The slope intercept form is y= mx+b

The given equation is 3y= 6x +9
rewrite the equation of the line

y= `(6/3)x+(9/3)`

y= 2x+3

Now it is in the standard form.

Therefore, the slope of the line is 2

3. Line AB and CD are perpendicular to each other, find the slope of the line CD when the two points on the lineAB is (2, 7) and (6, 4)

Solution:

(x1,y1) = (2, 7)

(x2,y2) = (6, 4)

Formula:

Formula to find slope, m = `((y2-y1)/(x2-x1))`

= `((4-7)/(6-2))`

= `((-3)/4)`

= `(-3/4)`

The slope of the line is `(-3/4)`

Since AB and CD are perpendicular to each other.

so, the slope of the CD =` (- 1/"slope of AB")`

= `(-1/ ((-3/4)))`

= `(4/3)`

Slope of the line CD is `(4/3)`

4th Grade Learning Math

INTRODUCTION TO FOURTH GRADE LEARNING MATH

Fourth grade learning math is the learning basic of some more brief of mathematics that was learned in third grade. Fourth grade learning math is the learning based on topic such as addition, subtraction, multiplication, division, equations and variables. Fourth grade learning math consists of chapters like algebra, charts, graph, fractions, decimals and geometry. Fourth grade learning math work consists of more subtopic under each chapter. Here we can learn some of the fourth grade math on some chapters. Fourth grade learning math is advance of third grade learning math.

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FOURTH GRADE LEARNING MATH


1.    Add 7 2 1 0 + 2 4 2 5

Solution

7 2 1 0

2 4 2 5 +

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

9 6 3 5

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

2.  Subtract 9 7 5 3 – 2 6 3 0

Solution

9 7 5 3

2 6 3 0 –

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

7 1 2 3

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

3.  Tony had 15 biscuit packets. Each packet consists of 7 biscuits. How many biscuits totally he had?

Solution

Number of biscuits in one packet = 7 biscuits

So 15 biscuit packets consists = 15 x 7= 105 biscuits.

4. Add the decimal value 75. 36 + 23.21

Solution

75. 3 6

23. 2 1 +

-----------

98. 5 7

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

5. Subtract decimal value given 54.34 – 24.51

Solution

5 4.3 4

2 4.5 1   -

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

2 9.8 3

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

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MORE PROBLEM OF FOURTH GRADE LEARNING MATH


6.   Solve the equation 5x + x = 24

Solution

5x + x = 24

6x = 24

Divide 6 on both sides

x = 4



7. Add `3/11` +`6/11`

Solution

Here the denominators are same. So just add the numerator alone

` (3+6)/11`

`9/11` is the answer.

8. Subtract `8/9` – `4/9`

Solution

Here the denominators are same. So just subtract the numerator alone.

`(8 - 4)/9`

`4/9`

9. Multiply` 3/5` * `3/ 4 `

Solution

`3 / 5 ` * `3 / 4`

3* 3 / 5* 4

= `9 / 20 `

10. Convert 100 centimeter =? Meter

Solution

100 centimeter = 1 meter.

Commission Math Problems

Introduction to commission math problems:

Commission math problems article deals with the definition of the commission and the math problems related to the commission.

Definition of commission math problems:

This math problems deals with the rate of money, which we can earn by selling some product or doing the work behind the target fixed.This is very important things in selling and buying the products and it is used to find the commission amount.

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Formula to commission math problem


In these problems the commission rate r % is fixed, by multiply the commission rate with the original cost of the product, we can get the commission amount on that particular product,

Commission amount = r% *(original cost of the product).

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Model problems to the commission math problems:


Find the monthly income of the john if his monthly salary is $2000 and he gets 4% of commission if he sells a car of cost about $4500.
Solution:

The monthly salary of the john is $2000

The commission rate is r= 4%

Commission amount = r% *(original cost of the product).

= 4 %*( 4500)

= (4/100)*(4500)

=(0.04)*(4500)

= $180

Commission amount is 180 $

The monthly income of the john = monthly income +commission amount

= $ 2000 +$180

= $2180

The monthly income of the john is $2180.

2.Find the income of the sam if his daily salary is $200 and he gets 3% of commission if he sells the product amount morethan $500.

Solution:

The daily salary of the john is $200

The commission rate is r= 3%

Commission amount = r% *(500).

= 3 %*( 500)

= (3/100)*(500)

=(0.03)*(500)

= $15

Commission amount is $15

The daily income of the john = daily income +commission amount

= $ 200 +$15

= $215

The monthly income of the john is $215.

3.Find the commission price of the product of cost $700 and the tax commission is 6%.

Solution:

The tax commission is 6%

The cost of the product is $700

Commission amount = r% *(original cost of the product).

= (6/100)*700

= (0.06)*700

= 42

commission amount is $42.

Learn Gaussian Elimination

Introduction to learn Gaussian elimination:

In linear algebra, Learning Gaussian elimination is an algorithm for solving systems of linear equations, finding the rank of a matrix, and calculating the inverse of an invertible square matrix. Gaussian elimination is named after German mathematician and scientist Carl Friedrich Gauss.

Elementary row operations are used to reduce a matrix to row echelon form. Gauss–Jordan elimination, an extension of this algorithm, reduces the matrix further to reduced row echelon form. Gaussian elimination alone is sufficient for many applications.

I like to share this Gaussian Standard Deviation with you all through my article.

Solving by Learning Gaussian Elimination:


A technique of solve a method of n linear equations in a n unknowns, in which there are first n - 1 steps, the math step of which consists of subtracting a lots of the math equation from both of the pursue ones so as to remove one variable, ensuing in a triangular set of equations which can be solved by turn around substitution, compute the nth variable from the nth equation, the (n-1)st variable from the (n-1)st equation.

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Example problems for learning Gaussian Elimination:


Example problems for learning Gaussian Elimination are as follows:

1) Solve the following system equation using Gaussian Elimination method.

3x + y = 9

3x – y = 15

Solution:

If add down, the y determination cancel out. So sketch an "equals" bar below the system, and add down:

3x + y = 9

3x – y = 15

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

6x = 24

x = 24 / 6

x = 4

At the present divide from side to side to solve for x = 4, and then back-solve, using either of the original equations, to find the value of y. The first equation have lesser facts, so back - explain in that one:

2(4) + y = 10

8 + y = 10

y = 2

Then the solution is (x, y) = (4, 2)

2) Solve the following system using Elimination method.

2x + 2y = 4 --- (1)

4x – 3y = 8 ---- (2)

Solution:

Multiply equation 1 with (3) and multiply equation, 2 with (2)

6x + 6y = 12

8x – 6y = 16

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

14x = 28

x = 28/14

x = 2

Apply x=2 in equation (1)

2(2) +2 y = 4

4 + 2y = 4

2y = 4 -4

y=0

Then the solution is (x, y) = (2, 0).

The World of Math

Introduction to world math :

In this article we are going to discuss about world math  Now a days math is one of the widely used part in the world. Sometimes math will challenge to solve the problems but every math has a solution to prove it. Let we see some math challenge problems to world math.

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Example Problems for math problems:


world math problem 1

Find the value for x in the expression `x/8` +x-`4/8` =0

Solution:

The given expression is `x/8` +x-`4/8` =0

We need to find the value of x

`x/8` +`(8x-4)/8` =0

The above expression can be written has

`x/8` +`(8x)/8` -`4/8` =0

Now add `4/8` on both sides of above equation

`x/8` +`(8x)/8` - `4/8` +`4/8` =`(0+ 4)/8`

`x/8` +`(8x)/8` =`4/8`

`(9x)/8` =`4/8`

Multiply by 8 on both sides of above equation

`(9x)/(8)xx(8)=` `4/8 xx8`

9x=4

Now divide by 9 on both side of the above

x=`4/9`

world math problem 2

Multiply the fraction `120/65` and `148/96`

Solution:-

Given we need to multiply fractions`120/65` and `148/96`

Fractions must multiplied using the formula `(axxc)/(bxxd).`

Here a = 120, b =65, c= 148, d=96

= `(axxc)/(bxxd).` =`(120xx148)/(65xx96) ` by solving it we get

A =`17760/6240`

= `37/13`

world math problem 3

Find the area of the circle with radius 54 feet. Important use symbol pie value approximately as 3.14

Solution:

We know that area of circle is equal to `pi` r2

Here the given radius is 54

Substitute the r value in the formula

=3.14x 542

=3.14x 54x 54

=9156.24

Therefore the area of circle is 9156.24 feet square

world math problem 4

Fine the surface area of cube in kilometers for the side is 30 m

Solution:

Area of cube A=6 * 302

A=6 *900

A=5400sq meters

1 square meter = 0.000001 square kilometer

5400 square meter = 5400 * 0.000001  = 0.0054

now surface area in kilometer is 0.005400 sq kilometers


Few More Example Problems for math problems


world math problems 5

Subtract the two fractions `32/15` and `94/15`

Solution:

The two given fractions are`32/15` and `94/15`

Step1: The given two fractions

`32/15` - `94/15`

Step2: Now we need to find the differences of `32/15` and `94/15`

`(32-94)/15`

Step3: The difference of 32 and 94 is -62

=-`62/15`

world math problem 6

Find equivalent fraction `7333/336`

Solution:

Equivalent fraction has to multiply numerator and denominator by same number

Denominator is 336 and the numerator is 7333

Multiply the denominator and numerator by 160

= `(7333xx160)/(336xx160)`

= `1173280/53760`

So the equivalent fraction is `7333/336` is` 1173280/53760`

Learn to Read a Ruler

Introduction of learn to read a ruler:

Math can understand through various equipment. The ruler is one of the main basic equipment for math. In every math problems the ruler plays a role to compute the answers. The ruler is measurable equipment for math. Ruler learns has two measures with cm in one side and mm in another end. Using the two measures we can have the two types of measuring terms.

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Learn to read a ruler:


The ruler is a measuring tool. The origin of the measuring tool is made from the man’s foot. To find the area’s measurement the man is asked to cross through the area for measuring the area’s measurement. The measure read in the yard can be done through the average man’s normal foot steps. Make the both measures to view, they won’t be exact. Till now the horses are made to measures in hands. This idea shows the origin of the ruler. We can have measuring idea with the ruler through the project on measuring the classroom. The classroom can be measured with the ruler that makes a measure which can be noted through the read that given over the scale. The measures about the book can be made along the measures with the scale.


More Learning to read a ruler:


Hence the learn of ruler can made to have the measures in the two ways. The first way is in the centimeters and the second way is in millimeters. The ruler is the simple measuring tool which is meant for measuring. Ruler is made to have in the way which gives the various changes that makes measures. The measure in the yard can be done through the average man’s normal foot steps. Make the both measures to view, they won’t be exact. Till now the horses are made to measures in hands. This idea shows the origin of the ruler.

Learning Support Math

Introduction for math learning support:

Math support learning  is a most interesting subject comparing to other subjects.It is the lesson for learning support math patterns, numbers, and shapes. Basic concepts of math are  Addition (a + b), subtraction (a-b), multiplication (a*b) and division (a/b) a and b are the numbers or integers. In this article, we are going to see some solved math support learning. I like to share this list of the prime numbers with you all through my article.


Example Problems for math problems:


Learning for support math problem 1:

Perform the Plus operation for 300 and 584

Solution:

Given we need to find the sum of 300 and 584

300

584  (+)

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

884

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

So the answer for 300 and 584 is 884

Learning for support math problem 2:

Multiplying two numbers 168 and 43

Solution:

The given two numbers 168 and 43

We need to find the product of two numbers

By Multiplying  168 and 43

168 × 43

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

5    0  4

6  7   2

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

7  2   2  4

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

We get  7  2   2  4

Learning for support math problem 3:

Find equivalent fraction `166/65`

Solution:

Equivalent fraction has to multiply numerator and denominator by same number

Denominator is 65 and the numerator is 166

Multiply the denominator and numerator by 24

= ` (166xx24)/(65xx24)`

=`3984/1560`

So the equivalent fraction is `166/65` is  `3984/1560`

Learning for support math problem 4.

Subtract the two fractions `174/112` and `224/112`

Solution:

The two given fractions are `174/112` and `224/112`

Step1: The given two fractions

`174/112` - `224/112`

Step2: Now we need to find the differences of   `174/112` and `224/112`

`(174-224)/112`

Step3: The difference of 174 and 224 is 50

= - `50/112`

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Few More Example Problems for math problems


Learning for support math problem 5:

Round the number 136 to the nearest 10

Solution:

The given number is 136

The number in 10 places is 3 and the number 1s places is 6

Since the number ten places is 6 so we make the number 6 as zero and we add 1 to 3

So the number 136 grounded to nearest 10 becomes  140

Learning for support math problem 6:

Identify 135 is a prime number or not

Solution:

The given number is 135

We need to identify prime number or not

To identify 135  is a prime number we need to determine in factors

Factors of 135  are 1  3  5  9  15  27  45  135

The number to be a prime number it must have only two factors 1 and itself

So here 135  has eight factors so it is not a prime number