MomoKiki1
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I need to solve for x using the pythagorean theorem some how (I think) please only answer if you can give a hint or the answer with an explanation.

I need to solve for x using the pythagorean theorem some how I think please only answer if you can give a hint or the answer with an explanation class=

Respuesta :

[tex]\bf \textit{using the pythagorean theorem} \\\\ c^2=a^2+b^2\implies \sqrt{c^2-a^2}=b \qquad \begin{cases} c=\stackrel{hypotenuse}{\stackrel{3959+5.9}{3964.9}}\\ a=\stackrel{adjacent}{3959}\\ b=\stackrel{opposite}{x}\\ \end{cases} \\\\\\ \sqrt{3964.9^2-3959^2}=x\implies \sqrt{46751.01}=x\implies 216.22\approx x[/tex]

jcherry99

Answer:


Step-by-step explanation:

Formula

a^2 + b^2 = c^2

Remark

The horizon point is where x and 3959 meet.

Givens

a = 3959

b = x

c =3959 + 5.9 = 3964.9

Solution

3959^2 + x^2 = 3964.9 ^2

15673681 + x^2 = 15720432.01       Subtract 15673681 from both sides

x^2 = 15720432.01 - 15673681        Combine

x^2 =46751.01                                  Take the square root of both sides

sqrt(x^2) = sqrt(46751.01)

x = 216.2

If you round anywhere but at the end, an error will creep in that will make your answer incorrect.


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