Respuesta :

gmany

Answer:

[tex]\large\boxed{x=8\sqrt3}[/tex]

Step-by-step explanation:

The Pythagorean theorem:

[tex]leg^2+leg^2=hypotenuse^2[/tex]

We have:

[tex]leg=8, leg=x,\ hypotenuse=16[/tex]

Substitute:

[tex]8^2+x^2=16^2[/tex]

[tex]64+x^2=256[/tex]        subtract 64 from both sides

[tex]x^2=192\to x=\sqrt{192}\\\\x=\sqrt{(64)(3)}\\\\x=\sqrt{64}\cdot\sqrt3\\\\x=8\sqrt3[/tex]

Plip

We are missing one of the legs. Thus, we rearrange the original formula of the pythagorean theorem...

[tex]\boxed{A^2+B^2=C^2}[/tex] → [tex]\boxed{A^2=C^2-B^2}[/tex]

Let's solve it...

[tex]Plug\;in\;the\;given\;numbers\;to\;the\;formula...\\\\A^2=16^2-8^2\\\\Find\;what\;the\;squared\;numbers\;equal\\\\\left[\begin{array}{ccc}16^2(16*16)=256\\8^2(8*8)=64\end{array}\right]\\\\\\Now,\;since\;we\;are\;trying\;to\;find;\a\;missing\;leg\;we\;subtract\;256\;from\;64...\\\\\boxed{256-64=192}\\\\\\Then,\;we\;find\;the\;square\;root\;of\;192\;(\sqrt{x} )\\\\\boxed{\sqrt{192}=13.856 }\\\\\\Since\;we\;have\;a\;decimal\;we\;need\;to\;round\;it\;to\;the\;nearest\;tenths...\\\\\boxed{13.856=13.9}[/tex]

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