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i would use pythagorean theorem for a most of these \[a^2+b^2=c^2\] where a and b are two sides, c is the hypotenuse
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Crazy
First one: Setting up a proportion: \( \frac{77}{87.5} = \frac{22}{y} \) Solving for y, you found \( y = 25 \). Then, to find x, you added y to the given 22: \( x = 22 + 25 = 47 \) feet.
@umm wrote:
First one:
Setting up a proportion:
\( \frac{77}{87.5} = \frac{22}{y} \)
Solving for y, you found \( y = 25 \). Then, to find x, you added y to the given 22:
\( x = 22 + 25 = 47 \) feet.
for second triangles MAB and MNP are similar. \[\frac{ MB }{ MP }=\frac{ MA }{ MN }\] \[\frac{ x }{ 72.6 }=\frac{ 46.2-14 }{ 46.2 }\] \[\frac{ x }{ 72.6 }=\frac{ 32.2 }{ 46.2 }\] find x
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