ALGEBRA 2 HOMEWORK QUESTION HELP!!!?

8.312^2 = x^2 + (10.32-x)^2

69.09 = x^2 + 106.5 - 20.64x + x^2

0 = 2x^2 - 20.64x + 37.41

0 = x^2 - 10.32x + 18.7

x = [10.32 +/- *sqrt(10.32^2 - 4(1)(18.7))] / [2x1]

....

After trying to solve problem after problem on yahooanswers, I honestly think most people dont learn a lot if I give them the answers. Heres a VERY similar problem. Just fill in the values from your own problem. (Hint: you will need to use the Quadratic Formula at the end!) Goodluck!!!

QUESTION: The hypotenuse of a right triangle has a length of 13 cm. The sum of the lengths of the two legs is 17 cm. Find the lengths of the legs.

Let a = length of one leg

Let b = length of other leg

We are told that: a + b = 10.23 --> b = 10.23 - b [1]

The hypotenuse is 8.312.

From Pythagorus: a² + b² = 8.312² [2]

Substitute [1] into [2]: a² + (8.312 - b)² = 169

This simplifies to: 2a² - 34a + 120 = 0

Divide by 2: a² - 17a + 60 = 0

which factors: (a - 5)(a - 12) = 0

and has roots: a = 5, 12

Substitute into [1] and get: b = 12, 5

The lengths of the legs are 5 and 12 cm.

Trey, give me a moment while I grab my abacus, he-he. Just kidding. This is a tricky one. You must've pulled a prank on your teacher to have something like this assigned. Well if you have it you have it, so ...

You know that the legs, a and b, sum to 10.23, so

a + b = 10.23.

You also know that a^2 + b^2 = 8.312^2 since it's a right triangle. Two equations, two unknowns, have at it son.

isn't that 10.23 divided by 2? sine its the sum of the two legs i'm thinking you need to divide the sum by 2 to get the two lengths of the legs. 10.23 divided by 2 is 5.115 you can round.

Pothygreum Therum is A2+B2=C2

(the 2's are exponents)

hope this helps.

x^2 + (10.23-x)^2 = 8.312^2

solve for x

[first, for the sake of clarity, switch f(x) with y] y=5x+4 [then, because this is how you find inverses, switch the (x)s and the (y)s] x=5y+4 [now solve for, or "isolate" y] x-4=5y (x-4)/5=y [so there's your answer, but I think inverse notation looks like this] f^-1(x)=(x-4)/5 [alternatively,] f^-1(x)=x/5+0.8 [or] f^-1(x)=x/5+4/5

the phythagoram theorm

5.115 cm