Can you tell me how to set up the equation so I can determine what % 1365 is of 6962?

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Posted on Jan 02, 2017

5x-y=-5 answer x=-1, y= 5
http://www.wolframalpha.com/input/?i=Determine+the+x+and+y+intercept+of+the+graph+for+equation%3A+5x-y%3D-5

-3x+2y=7 answer x = -7/3, y = 7/2 http://www.wolframalpha.com/input/?i=Determine+the+x+and+y+intercept+of+the+graph+for+equation%3A+-3x%2B2y%3D7

-3x+2y=7 answer x = -7/3, y = 7/2 http://www.wolframalpha.com/input/?i=Determine+the+x+and+y+intercept+of+the+graph+for+equation%3A+-3x%2B2y%3D7

Feb 23, 2015 | Office Equipment & Supplies

First, we will find y in terms of x. We will use the first equation to determine this.

4x+2y=2

We can subtract 4x from both sides:

2y=2-4x

And then divide both sides of the equation by two:

y=1-2x

Since we now have y in terms of x, we can substitute this into our second equation.

-3x-y=-3

-3x-(1-2x)=-3

Then, we can distribute the minus sign

-3x-1+2x=-3

-x-1=-3

Next, we can add 1 to both sides of the equation.

-x=-2

Finally, we divide both sides by negative one to isolate x.

x=2

Now that we have x's value, we can find y's value.

The first thing that we determined is:

y=1-2x

We can substitute in the value of x to this equation.

y=1-2x

y=1-4

y=-3

Therefore, we now have the values of both variables.

x=2

y=-3

4x+2y=2

We can subtract 4x from both sides:

2y=2-4x

And then divide both sides of the equation by two:

y=1-2x

Since we now have y in terms of x, we can substitute this into our second equation.

-3x-y=-3

-3x-(1-2x)=-3

Then, we can distribute the minus sign

-3x-1+2x=-3

-x-1=-3

Next, we can add 1 to both sides of the equation.

-x=-2

Finally, we divide both sides by negative one to isolate x.

x=2

Now that we have x's value, we can find y's value.

The first thing that we determined is:

y=1-2x

We can substitute in the value of x to this equation.

y=1-2x

y=1-4

y=-3

Therefore, we now have the values of both variables.

x=2

y=-3

Jan 13, 2015 | SoftMath Algebrator - Algebra Homework...

For two linear equations, one can use

comparison, substitution, or addition/combination.

For more linear equations one uses the Cramer Rule that involves matrices and their determinants.

comparison, substitution, or addition/combination.

For more linear equations one uses the Cramer Rule that involves matrices and their determinants.

Sep 16, 2014 | Office Equipment & Supplies

You need to find out what number x is 100%.

Start out with the facts known, 50 is 20% of this x, and write it down in mathematical symbols. This is just a translation of the words in your problem to the shortcut equation language used in math, the meaning is exactly the same, and with a little more experience you will easily move between these two.

50 = x * ( 20 / 100 )

Remember, 'percent' is from latin "per one hundred".

Multplying left and right side of the equation by 100 yields:

50 * 100 = x * ( 20 / 100 ) * 100, consolidation results in:

5000 = x * 20. Divide both sides by 20:

5000 / 20 = x * 20 / 20, consolidate:

250 = x.

This is your result, 250.

Start out with the facts known, 50 is 20% of this x, and write it down in mathematical symbols. This is just a translation of the words in your problem to the shortcut equation language used in math, the meaning is exactly the same, and with a little more experience you will easily move between these two.

50 = x * ( 20 / 100 )

Remember, 'percent' is from latin "per one hundred".

Multplying left and right side of the equation by 100 yields:

50 * 100 = x * ( 20 / 100 ) * 100, consolidation results in:

5000 = x * 20. Divide both sides by 20:

5000 / 20 = x * 20 / 20, consolidate:

250 = x.

This is your result, 250.

Mar 19, 2014 | Office Equipment & Supplies

x will be my paycheck and y will be coworker's paycheck:

1.8y=890

II hope I helped you :)

- my paycheck is 20 percent less than coworker's so x is 20% smaller then y:

- These two paychecks total 890:

- We substitute x in previous equation:

1.8y=890

- y=890/1.8=494.4

- x=890-y=395.6

II hope I helped you :)

Sep 13, 2011 | Office Equipment & Supplies

Hi,

Here's a simple equation to get the answer.

Just divide 80.99 by 109. You will get an answer in decimal point. Move 2 places to the right and you will get the answer in percentage.

The answer is 74.30%.

Hope this helps.

Here's a simple equation to get the answer.

Just divide 80.99 by 109. You will get an answer in decimal point. Move 2 places to the right and you will get the answer in percentage.

The answer is 74.30%.

Hope this helps.

Jun 01, 2011 | Office Equipment & Supplies

equation for series R1 + R2 = 6
equation for parallel 1/R1 + 1/R2 = 1/25
substitute first equation for R2 in second equation:
1/R1 + 1/(6 - R1) = 1/25
I will leave the algebra solution to you.

Oct 01, 2010 | Cycling

Hello,

Let us assume you have two simultaneous linear equations :

**a_1*x+ b_1*y+c_1=0**

a_2*x +b_2*y+c_2=0

where a_1, a_2, b_1, b_2, c_1,c_2 are coefficients (numerical or algebraic).

The problem is to obtain the particular values of the unknowns x and y for which the two equations are both satisfied: If you substitute the particular values of x and y you find in any of the two equations you discover that both equalities are true.

A small system of equations like the one above can be solved by some very simple algorithms (elimination, substitution, combination) which can be carried out by hand.

The solution of large systems of linear equations can be sought by making use of the concepts of matrices (plural of matrix), determinants, and certain rules called Cramer's rules.

Due to its repetitive nature, the algorithm ( a well defined, limited sequence of steps) is suitable for a calculating machine (computer or calculator).

Certain calculators have, embedded in their ROM, a program that solves linear systems of simultaneous equations. Usually you are asked to enter the values of the coefficients a_1, etc. in a set order, then you press ENTER or EXE (Casio) . If a solution exits (not all linear systems have solutions) the calculator displays it.

Hope that satisfies your curiosity.

Let us assume you have two simultaneous linear equations :

a_2*x +b_2*y+c_2=0

where a_1, a_2, b_1, b_2, c_1,c_2 are coefficients (numerical or algebraic).

The problem is to obtain the particular values of the unknowns x and y for which the two equations are both satisfied: If you substitute the particular values of x and y you find in any of the two equations you discover that both equalities are true.

A small system of equations like the one above can be solved by some very simple algorithms (elimination, substitution, combination) which can be carried out by hand.

The solution of large systems of linear equations can be sought by making use of the concepts of matrices (plural of matrix), determinants, and certain rules called Cramer's rules.

Due to its repetitive nature, the algorithm ( a well defined, limited sequence of steps) is suitable for a calculating machine (computer or calculator).

Certain calculators have, embedded in their ROM, a program that solves linear systems of simultaneous equations. Usually you are asked to enter the values of the coefficients a_1, etc. in a set order, then you press ENTER or EXE (Casio) . If a solution exits (not all linear systems have solutions) the calculator displays it.

Hope that satisfies your curiosity.

Aug 12, 2009 | Sharp EL-531VB Calculator

Check the manual @ http://www.mykameleon.com/downloads/user_manuals/RS6in1_manual.pdf

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