I have this math problem that I have to solve by tomorrow. Here's a picture of what I'm trying to solve.

The whole thing is a cake I have to bake. But then I have to cut off the four corners, throw away 2 of them, and put the remaining 2 triangles on the sides of the cake with the lengths 9. X is the length of the hypotenuses and also the length of the side that remains after I cut off the triangles. X is the main length I'm trying to find, because if I have x I can figure out the other 2 sides. Any and all help is appreciated.

Edit: sorry for the massive picture, but at least you get the idea...
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Is that the chemical compound of meth?

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Quote by forsaknazrael
You should probably mug John Frusciante or Ritchie Blackmore. They're small guys, we could take 'em.

Just look out for that other guy in the Red Hot Chili Peppers, Will Farrel. He's a tall mofo, got a long reach.

I give up.

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I have to solve

Well why did you ask us? Solve it yourself, it's easy enough.
X=(9√2)/(2+√2) or approximately 3.7279.

Edit: A more elegant exact answer is 9(√2-1), in case you're as picky as I am.

This assumes, of course, that the triangles are supposed to be isosceles.
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Last edited by Calgone at Feb 1, 2009,
> eq1:=a^2+b^2=x^2;
> eq2:=2*b+x=9;
> eq3:=2*a+y=13;
> eq4:=(2*b+x)*(2*a+y)=117;

{a = 4.089992981, b = 2.182145341, x = 4.635709317, y = 4.820014038}

This is the answer without supossing that the triangles are isosceles.
I too need help.

How do i find the inverse of this problem:

f(x) = x^(3/5)

is it 3 5th Roots of x?
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i've always found pop to be harder to play than metal... especially shred metal... it's just really fast tremolo picking and the occasional palm mute... and the only chords you have to worry about are power chords...
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I too need help.

How do i find the inverse of this problem:

f(x) = x^(3/5)

is it 3 5th Roots of x?

im not too sure about this one but i got the cube root of x to the fifth
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im not too sure about this one but i got the cube root of x to the fifth

makes sense, i didnt even think about leaving the denominator part of the exponent on there.

thanks, im writing both answers down lol
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