Question

@jim_thompson5910 need help number 10 please

Answer 1

to find x, solve


\[\Large x^2 + x^2 = 12^2\]


for x

Answer 2

x2+x2= 144

Answer 3

good, what's next

Answer 4

idk

Answer 5

\[\Large 2x^2 = 144\]

Answer 6

now divide both sides by 2

Answer 7

\[\Large x^2 = \frac{144}{2}\]


\[\Large x^2 = 72\]

Answer 8

and take the square root of both sides


\[\Large x = \sqrt{72}\]


\[\Large x = 6\sqrt{2}\]

Answer 9

yay we did it!

Answer 10

now we hav to find z and y

Answer 11

to find y, solve the equation


\[\Large y\sqrt{3} = 12\]


for y


\[\Large y\sqrt{3} = 12\]


\[\Large y = \frac{12}{\sqrt{3}}\]


\[\Large y = \frac{12\sqrt{3}}{3}\]


\[\Large y = 4\sqrt{3}\]

Answer 12

I'll let you do the last part. To find z, solve


\[\Large a^2 + b^2 = c^2\]


\[\Large 12^2 + (4\sqrt{3})^2 = z^2\]


for z

Answer 13

wat about z??

Answer 14

\[\Large 12^2 + (4\sqrt{3})^2 = z^2\]

\[\Large 144 + 16(3) = z^2\]

\[\Large 144 + 48 = z^2\]

Answer 15

@jim_thompson5910 i got 192 thts the final answer for z?

Answer 16

or is it 36864??

Answer 17

\[\Large 144 + 48 = z^2\]

\[\Large 192 = z^2\]

\[\Large z^2 = 192\]

\[\Large z = \sqrt{192}\]

\[\Large z = 8\sqrt{3}\]

Answer 18

you take the square root of 192, not square it