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If you need sum of it use gauss theorem 8,5(1+2...x)=8,5(x.x+1/2)

5*5=25  then add the 5 zeros

2,500,000

AS x goes up by 2s, y goes dfdown by 1s

So I'm thinking I will start with    y=-(1/2)x

now what do I have

x      -(1/2)x

2         -1           1

4         -2           0

6        -3            -1

8        -4            -2

looks like I need to add 2

y=-(1/2)x+2

or  rearranged we get

y=2-(x/2)

The ordered pair (3, -1) is a solution of 5y + 2x = 1. Which of the following is true

Try (2,1)      left hand side = 5*2+2*1=10+2=12   That does not equal 1      So that is not right.

you have to try each ot the others in the same way and see which left hand side will equal 1     

You can finish it yourself.  

4/5 +3 does not equal 8 and that is the honest truth.

I already answered this questions for you!

Please do not repost so quickly!

Because

sin 10=sin (360+10)=sin(2*360+10)=....... sin(360n+10)

n is any integer

and this is in degrees.

This is just a demonstration not a proof.  Is it good enough?

The perimeter of a rectangle is 26 meters. The difference between the length and the width is 5 meters. Find the width ?

$$\small{\text{
l=length w=width:
$
\begin{array}{rrcrr}
&2l+2w &=&26\ m & \quad | \quad :2\\
(1)&l+w&=&13\ m \\
(2)&l-w&=&5\ m
\end{array}
$
}}
\\\\
\text{
$\frac{(1)+(2)}{2}=\frac{l+\not{w}+l-\not{w}}{2}=l=\frac{13+5}{2}=9 \ m$
}
\\\\
\text{
$
\frac{(1)-(2)}{2} =\frac{l+w-(l-w)}{2}=\frac{\not{l}+w-\not{l}+w}{2} =w=\frac{13-5}{2}=4\ m
$
}$$

Use the web2 calc on the home page. 

 (60x-48,000)*.7

I am not sure what the problem is.

0.7*(60x-48000)

=(7/10)*60x - (7/10)*48000

=7*6x - 7*4800

=42x - 33600

Is that what you intended?

The compound interest formula is:  A  = P(1 + r/n)^(n·t)

where          A = final amount          P = amount invested        r = rate (as a decimal)

                   n = number of times compounded per year          t = number of years

You can choose any number that you want for P; I will choose 1.  (This procedure works for any number that you choose for P.)                 The value of A is twice that of P, so A = 2             r = .028              n = 4

Substituting:  2  =  1(1 + .028/4)^(4t)     --->     2  =  (1.007)^(4t)

Since your unknown is in an exponent, use logs:    log(2)  =  log[(1.007)^4t]

An exponent in a log comes out as a multiplier:      log(2)  =  4t·log(1.007)

                                                                          log(2) / log(1.007)  =  4t

                                                                         t  =  [ log(2) / log(1.007) ] ÷ 4

t  ≈ 24.8 years

$$g(-1/2)=4*(-1/2)-9 = -2-9=-11$$

To go from the x-intercept (-6, 0) to the y-intercept (0, 4), you have a rise of 4 and a run of 6 (both positive because you are moving to the right and moving upward).

slope  =  rise / run  =  4/6  =  2/3  

You forgot the negative sign in your answer anon!      So I can't give you the points.   sorry. 

This means that the point (2, y) is on the line; it says that x has the value of 2; so substitute 2 in for x:

y  =  3x - 2     --->     y  = 3(2) - 2     --->   y  =  4

There is a point-slope form for a line  y - y1 = m(x - x1)

point = (x1, y1) = (4, 2)

slope = m = 3

Substituting:  y - 2  =  3(x - 4)

                     y - 2  =  3x - 12

                      y       =  3x - 10

The point-slope form is probably the most useful of all the forms; it would save you a lot of time if you could learn it.

$$\\a^2/a^2\;\;-4 /3*a\;\;-a^2/a^2\;\;-5a+6\\\\
=1\;\;-4/3*a\;\;-1\;\;-5a+6\\\\
=-4/3*a\;\;-5a+6\\\\
=-(4a)/3\;\;-5a+6\\\\
=-(4a)/3\;\;-(15a)/3\;\;+6\\\\
=-(19a)/3\;\;+6\\\\
=6-\frac{19a}{3}$$

I am going to give you 3 points for your good effort unknown.  

2x - 3y  =  11          and         x  =  4y + 9

Substituting the second equation into the first:  2(4y + 9) - 3y  =  11

                                                                        8y + 18 - 3y  =  11

                                                                                 5y + 18  =  11

                                                                                        5y  =  -7

                                                                                          y  =  -7/5

Substituting this into the second equation:  x  =  4(-7/5) + 9  =  -28/5 + 45/5  =  17/5

it is just 5.1

The slope formula is:  m  =  (y2 - y1) / (x2 - x1).

Choose either point to be the first point; I am choosing (x1, y1) = (5, -4) and (x2, y2) = (-2, -5).

This means that x1 = 5, y1 = -4, x2 = -2, and y2 = -5.

Putting these values into the formula:  m  =  (-5 - -4) /(-2 - 5)  =  (-5 + 4) / (-2 - 5)  =  -1 / -7  =  1/7.

If you are in algebra class, you should memorize this formula.

@ ALAN, maybe this one is for you.    

I don't even understand what it means.  

Your question is incomplete.  

i searched google yesterday.

I think I put

download of web2.0calc

i just did it again and look what I found.

x = 2 - 3y

3x + 9y  =  6  --->  3(2 - 3y) + 9y  =  6

                                 6 - 9y + 9y  =  6

                                        6 - 0 = 6

                                              6  =  6   

There is no single answer; there are an infinite number of values that work.

If you rearranged the first equation to get  x + 3y  =  2  and then multiply this equation by 3, you would get     3x + 9y  =  6  which is exactly the same equation as the second equation. So any combination of numbers for x and y that solve the first equation also solves the second, and there are an infinite number of possible answers.     

Consideremos y = x + 2

cada valor de x tiene un valor y correspondiente.

Hay un número infinito de x valores así que debe haber un número infinito desoluciones.

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Help please: true or false?, why? and if you can please an example, thank you. Is there avariable equations that have infinite solutions?.

All have an infinite number of solutions if there is one equation and two or more vairables.

consider y=x+2

every value of x has a corresponding y value.

There are an infinite number of x values so there must be an infinite number of solutions.  

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