help
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- anonymous

help
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- chestercat

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- anonymous

how to find the second one?

- anonymous

The diagonals are in the middle of the triangle so its basically it is breaking it up in 4 equal peices.

- anonymous

pok

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## More answers

- anonymous

ok

- anonymous

then

- anonymous

So if you are trying to find ac it would what half of it twice.

- anonymous

so (x+4)(x+4)

- anonymous

simplify?

- anonymous

Since you have the sum of the length of the diaganols you have to make an equation equal to 20.

- anonymous

(2x-3)(2x-3)+(x+4)(x+4)=20

- anonymous

You have to solve for x before moving on

- anonymous

is the answer 7?

- anonymous

Uh one sec still solving!

- anonymous

Plug it in to check!

- anonymous

I dont think thats it?

- anonymous

idk

- anonymous

i solve this for x (2x-3)(2x-3)+(x+4)(x+4)=20

- anonymous

You have to do the quadratic eqaution...

- anonymous

idk how to do quad

- anonymous

did you get 5x^2-4x+25=20 when you simplified?

- anonymous

yep

- anonymous

Okay you need quad to find x give me a minute and ill do it...

- anonymous

ok

- anonymous

Its an imaginary number...

- anonymous

Assuming the rhombus is ABCD and E is the intersection of the diagonals, then we know the diagonals perpendicularly bisect each other, so AE is half of one diagonal and BE is half the other diagonal. That allows us to solve for x:
(x+4) + (2x-3) = 20/2
3x + 1 = 10
3x = 9
x = 3
and since
BE = 2x - 3 = 3 is half of BD,
BD = 2*3 = 6 (answer C).

- anonymous

Figured out the better way to solve it lol

- anonymous

thank you

- anonymous

DO you understand how i got there?

- anonymous

so BE is half and BD is the other half?

- anonymous

each is 3?

- anonymous

Wait!!! i messed up lol you plug the three in x+4 and multiply it by 2 because you are finding ac

- anonymous

For some reason i thought it was the other one. But it will be 14 in that case.

- anonymous

thanks

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