#1

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...

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...

#2

Is that the chemical compound of meth?

#3

Is that the chemical compound of meth?

Did somebody say meth?

#4

Is that the chemical compound of meth?

Hilarious.

#5

I have to solve

Well why did you ask us? Solve it yourself, it's easy enough.

#6

baking a cake? here, watch this: http://www.youtube.com/watch?v=-azqXygCzO8

#7

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.

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.

*Last edited by Calgone at Feb 1, 2009,*

#8

> 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.

> 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.

#9

Throw cancer at the teacher. He sounds like a dick.

#10

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?

How do i find the inverse of this problem:

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

is it 3 5th Roots of x?

#11

The Pit isn't the internet equivalent of Einstein.

#12

The Pit isn't the internet equivalent of Einstein.

apparently you aren't, but there are actually intelligent people who get on this forum.

albeit rare, but they exist nonetheless

#13

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

#14

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