Ask your own question, for FREE!
Mathematics 8 Online
OpenStudy (anonymous):

If x = 17 cm and y = 8 cm, what is the length of z?

OpenStudy (anonymous):

OpenStudy (anonymous):

A 13 cm B 15 cm C 14 cm

OpenStudy (whpalmer4):

Okay, we showed you how to do it, time for you to do one yourself :-)

OpenStudy (whpalmer4):

This is a little bit different, as this time you know the length of the hypotenuse, and don't know the length of one of the sides. Same formula, though.

OpenStudy (anonymous):

^

OpenStudy (anonymous):

@whpalmer4

OpenStudy (anonymous):

yea hes right try it for yourself !

OpenStudy (anonymous):

im sure you can do it

OpenStudy (whpalmer4):

Come on, @katieeeeeeeee !

OpenStudy (anonymous):

I don't know what to do

OpenStudy (whpalmer4):

As I told you, the sum of the squares of the lengths of the two legs of the triangle will equal the square of the length of the hypotenuse. Write out the equation that states that.

OpenStudy (anonymous):

z2 +x2=y2

OpenStudy (whpalmer4):

Close, but no cigar. First, we use ^ to indicate an exponent. z^2 = z*z Second, which are the two legs?

OpenStudy (anonymous):

Your getting it...

OpenStudy (anonymous):

Okie doke, square root 17 and 8

OpenStudy (whpalmer4):

y and z, right? So you need to square y, square z, and add them together. The sum is the same as x squared. Can you write the equation for that?

OpenStudy (anonymous):

y2 and z2

OpenStudy (anonymous):

Correct as well whpalmer4 :)

OpenStudy (whpalmer4):

Remember, katie, you need the ^ (found on the 6 key, usually) between the letter and the number, if you mean an exponent.

OpenStudy (anonymous):

\[z^{2}+y^{2} = x^{2}\]

OpenStudy (the_fizicx99):

\[a^2 + b^2 = c^2\] |dw:1395623470012:dw| It's just backwards, \[8^2 + b^2 = 17^2\] Solve for B by squaring both 8 and 17 8^2 = 64 17^2 = 289 64 + b = 289 Basically just subtract 64 from both sides and that's B

OpenStudy (anonymous):

Y^2 and Z^w

OpenStudy (anonymous):

Yep simple algebra at the end. :)

OpenStudy (whpalmer4):

And an equation needs an = sign. You have three quantities: x^2, y^2, and z^2 You have two equation symbols: +, = Write an equation that uses all 5 of those items and equates the sum of the squares of the two legs (y^2 and z^2) with the square of the hypotenuse (x^2)

OpenStudy (anonymous):

z^2 + (8)^2 = (17)^2

OpenStudy (anonymous):

z^2 + 64 = 289

OpenStudy (anonymous):

z^2 = 225

OpenStudy (whpalmer4):

I'm really impressed that everyone else knows how to solve this problem, but it would be most helpful for @katieeeeeeeee if she had the opportunity to do it...

OpenStudy (anonymous):

z = 15

OpenStudy (the_fizicx99):

You're given y, where y is it's where A is -- the original formula is A^2 + B^2 = C^2 Technically it would be y^2 + x^2 = z^2

OpenStudy (anonymous):

i square rooted z^2 and 225

OpenStudy (anonymous):

so I got z = 15

OpenStudy (anonymous):

Teamwork here. :)

OpenStudy (whpalmer4):

@tHe_FiZiCx99 No, the equation is \[y^x + z^2 = x^2\]because \(x\) is the hypotenuse of this triangle, not \(z\).

OpenStudy (whpalmer4):

make that \[y^2 + z^2 = x^2\]

OpenStudy (anonymous):

We, did that already

OpenStudy (the_fizicx99):

I kept mixing them up, :/ the z shoudl have copied over to the right side

OpenStudy (anonymous):

the problem is solved

OpenStudy (anonymous):

if you read through the answers, i left a step by step solution

OpenStudy (whpalmer4):

if by "the problem is solved" you mean "katie got an answer, but probably couldn't do the problem on her own", then I agree.

OpenStudy (anonymous):

as well as other people

OpenStudy (the_fizicx99):

^^

OpenStudy (anonymous):

no, everyone left a step by step solution

OpenStudy (whpalmer4):

The point of OpenStudy is to get it so that katie can find the solution herself.

OpenStudy (anonymous):

lol

OpenStudy (anonymous):

if katie doesn't get it now, she will need to ask her teacher

Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!
Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!