hectictar

width = 2 inches

length = 3*width = 3*2 = 6 inches

area = length * width

area = 6 inches * 2 inches = 12 square inches

The fastest method looks like "elimination," where you line the equations up and add straight down.

5x + 7y = -11

5x - 3y = 19

If we added straight down right now, that would give us 10x + 4y = 8, which doesn't help any. We want one of the variables to go away.

Multiply the top equation by negative one all the way through.

(-1)(5x + 7y) = (-1)(-11)

-5x - 7y = 11

Now line up both equations and add straight down:

-5x - 7y = 11

5x - 3y = 19    

0 - 10y = 30

-10y = 30

y = -3

Plug in y into either one of the equations to find x:

5x - 3(-3) = 19

5x + 9 = 19

5x = 10

x = 2

The additive inverse of -4 is 4 . (I think this is the one you want.)

The mutliplicative inverse, or reciprocal, of -4 is -¼ .

The negative reciprocal of -4 is ¼ .

That's all I can think of :)

Hey!! Glad you could join!! Nice username, haha 

6x⁴y² - 4x²y + 2xy² + 8

Ascending means from least to greatest.

8 - 4x²y + 2xy² + 6x⁴y²

The last two terms have the same power of y, so I guess it doesn't matter which comes first.

Also the normal way to arrange an expression is to put the powers in descending order, which just means from greatest to least.

What FirstLast said is true, x > 9.

Also, since it is the absolute value of 10x+15,

the other option is:

10x + 15 < -105

10x < -120

x < -12

x > 9     OR    x < -12

Hmm, I had some issues logging on with the main computer at my house. It logs on but as soon as you click anything it logs out. I read that some other people had this exact same issue somewhere. It works fine on my phone and laptop though.

I actually created my account using the main computer and when I tried to log in it kept logging me right back out so I thought that something had gone wrong in the creation of the account, but actually it did that part just fine. So if you still remember the username and password I would try logging in using some other device and see if it works then.

Thank you!!!

I wish you had an account so I could give you points. Ah well I will give points anyway! :)

I also would like to thank some really cool peeps:

ElectricPavlov

CPhill

heureka

Melody

MaxWong

asinus

\(7\frac{7}{9} + 2\frac{1}{9} \\~\\ = 7 + \frac{7}{9} + 2 + \frac{1}{9} \\~\\ = 7 + 2 + \frac{7}{9} + \frac{1}{9} \\~\\ = 9 + \frac{8}{9} \\~\\ = 9 \frac{8}{9}\)

I hope this made sense :)

\(\frac{\sin A}{a} = \frac{\sin B}{b} = \frac{\sin C}{c} \\~\\ \frac{\sin 56}{17} = \frac{\sin C}{14} \\~\\ \frac{14\sin 56}{17} = \sin C \\~\\ \arcsin{(\frac{14\sin 56}{17})} = C \\~\\ C \approx 43.058^{\circ}\)

Since there are 180 degrees in every triangle,

180 = A + B + C

A = 180 - B - C

A ≈ 180 - 56 - 43.058

A ≈ 80.942º

All that's left to find is side a.

\(\frac{\sin 56}{17} \approx \frac{\sin 80.942}{a} \\~\\ (a)\frac{\sin 56}{17} \approx \sin 80.942 \\~\\ a \approx \frac{17\sin80.942}{sin56} \\~\\ a \approx 20.250\)

Recap:

A ≈ 80.942º          a ≈ 20.25

B = 56º                   b = 17

C ≈ 43.058º            c = 14

*edit* I put c = 4 instead of c = 14