Find the missing length in the right triangle. Round to the nearest tenth. Legs: 15, 20 cm
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- anonymous

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- anonymous

So you know the Pythagorean theorem right?

- anonymous

NO!!

- anonymous

Well let's say that the two legs are a and b and the hypotenuse is c. In the Pythagorean theorem, we can figure out the hypotenuse with the two legs. It looks like this \[\large a^{2}+ \large b^{2}= \large c^{2}\]

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## More answers

- anonymous

So then the answer would be?????

- anonymous

We'll get to that. So let's plug in our leg values, a and b. We'll get \[\large 15^{2}+20^{2}=c^{2}\]

- anonymous

So thats the answer!!!!

- anonymous

Our answer is the c. We're solving for that currently.

- anonymous

kk thanks!!!

- anonymous

Thanks for what? Did you figure out the answer?

- anonymous

NO!!!!!!

- anonymous

So do you still want my help or?

- anonymous

DUDE can u just tell me answer please!!!!! i will fan u!!!!!

- anonymous

wait so the answer is 4,900???

- anonymous

Giving the answer directly is against the code of conduct so no, i can't just give you the answer. http://openstudy.com/code-of-conduct

- anonymous

GOtcha but the answer then is 4,900???

- anonymous

Well what is \[\large 15^{2}?\]

- anonymous

225

- anonymous

And what is \[\large 20^{2}?\]

- anonymous

400

- anonymous

so do i add them toghether to get the answer??

- anonymous

Close. One more step after that. 225+400=?

- anonymous

kk and i get 625!!

- anonymous

and that the answer!!

- anonymous

Ok so right now we have \[\large 625=c^{2}\] So to find c, we need to get the square root of both sides \[\large \sqrt{625}=c\] THIS will give you the answer

- anonymous

312.5

- anonymous

??????

- anonymous

We aren't dividing by 2, we're getting the square root of it.

- anonymous

ooh so then.....

- anonymous

Go to http://web2.0calc.com/ and put sqrt(625) into the bar

- anonymous

this is 25!!!

- anonymous

Yep. So 25 is your answer :)

- anonymous

man thanks bro u teached me something new man thanks ill will fan u!!

- anonymous

No problem :)

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