Question

What is the value of x?

Round to the nearest tenth.

Answer 1

\(\large\color{black}{ \displaystyle a^2+b^2=c^2 }\)
where c is the hypotenuse of a right triangle, and a & b are the legs.

Answer 2

Would you then agree with the following set up (to find x)?

\(\large\color{black}{ \displaystyle 12^2+x^2=22^2 }\)

Answer 3

Yes

Answer 4

ok, then tell me what x is going to equal
(and disregard the negative x solution)

Answer 5

10

Answer 6

no

Answer 7

it is not 12+x=22
then it would have been correct, BUT///

Answer 8

it is 12²+x²=22²

Answer 9

oh ok

Answer 10

first, you have to simplify the powers.

Answer 11

12²=?
22²=?

Answer 12

144

Answer 13

484

Answer 14

yes

Answer 15

\(\large\color{black}{ \displaystyle 12^2+x^2=22^2 }\)

\(\large\color{black}{ \displaystyle 144+x^2=484 }\)

when you subtract 144 from both sides, you get:

\(\large\color{black}{ \displaystyle x^2=340 }\)

Answer 16

So, x² is 340

Answer 17

okay

Answer 18

then take the square root of both sides.

Answer 19

okay

Answer 20

then?

Answer 21

so what's the final answer? @SolomonZelman

Answer 22

@SolomonZelman i running out of time please

Answer 23

i'm*

Answer 24

\(\large\color{black}{ \displaystyle x^2=340 }\)
\(\large\color{black}{ \displaystyle \sqrt{x^2}=\sqrt{340} }\)
\(\large\color{black}{ \displaystyle x=\pm \sqrt{340} }\)
exclude the negative result, as a side can't be negagive.
\(\large\color{black}{ \displaystyle x=\sqrt{340} }\)
\(\large\color{black}{ \displaystyle x=\sqrt{4\times 85} }\)
\(\large\color{black}{ \displaystyle x=2\sqrt{85} }\)