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anzhelika [568]
3 years ago
9

A toll-free consumer information line answered 105 of

Mathematics
1 answer:
Stolb23 [73]3 years ago
7 0

(105/120)*100=87.5

………………..

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In rectangle $ABCD$, $AD=1$, $P$ is on $\overline{AB}$, and $\overline{DB}$ and $\overline{DP}$ trisect $\angle ADC$. Write the
Studentka2010 [4]
<h3>Answer:</h3>

w+x+y+z = 12

<h3>Explanation:</h3>

When right angle ADC is divided into three equal parts (trisected), each of those parts is 90°/3 = 30°. Thus, ∠CDB = ∠PBD = ∠PDB = 30°.

The sides of a 30°-60°-90° triangle are in the proportion 1 : √3 : 2. Hence BD = 2, and PD = PB = 2/√3 = (2/3)√3.

Then the perimeter of ∆BDP is ...

... perimeter ∆BDP = BD + DP + PB

... = 2 + (2/3)√3 + (2/3)√3

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3 years ago
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