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tester [92]
3 years ago
14

In rectangle $ABCD$, $AD=1$, $P$ is on $\overline{AB}$, and $\overline{DB}$ and $\overline{DP}$ trisect $\angle ADC$. Write the

perimeter of $\triangle BDP$ in simplest form as: $w + \frac{x \cdot \sqrt{y}}{z}$, where $w, x, y, z$ are nonnegative integers. What is $w + x + y + z$?
Mathematics
1 answer:
Studentka2010 [4]3 years ago
8 0
<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

... = 2 + (4√3)/3 . . . . . . . . w=2, x=4, y=3, z=3

w + x + y + z = 2+4+3+3 = 12

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zepelin [54]

Answer:

<em>x = 2</em>

Step-by-step explanation:

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Solve:

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Operating:

\displaystyle \frac{5^{2x}}{5}+5^{x}5=250

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Rearranging:

5^{2x}+25\cdot5^x-1250=0

Recall that:

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kozerog [31]
Q1. The answer is  \frac{8x^{3}y^{6}  }{27}

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\frac{(16)^{ \frac{3}{4} }}{(18)^{ \frac{3}{4} }}*(x^{4})^{ \frac{3}{4} }*(y^{8})^{ \frac{3}{4} }=\frac{( 2^{4} )^{ \frac{3}{4} }}{( 3^{4} )^{ \frac{3}{4} }}*(x^{4})^{ \frac{3}{4} }*(y^{8})^{ \frac{3}{4} } \\  \\ &#10; (x^{a} )^{b} = x^{a*b}  \\  \\ &#10;\frac{( 2^{4} )^{ \frac{3}{4} }}{( 3^{4} )^{ \frac{3}{4} }}*(x^{4})^{ \frac{3}{4} }*(y^{8})^{ \frac{3}{4} } =  \frac{ 2^{4* \frac{3}{4} } }{ 3^{4* \frac{3}{4} } } * x^{4* \frac{3}{4} } * y^{8*\frac{3}{4}} = \frac{ 2^{3} }{ 3^{3} } * x^{3} *y^{6} = 
= \frac{8x^{3}y^{6}  }{27}

Q2. The answer is 1/16

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Q3. The answer is a^{ \frac{7}{6} }

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7 0
2 years ago
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