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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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Let the number of orders Eric served be x

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\dfrac{dy}{dx} = \dfrac{dy}{dt} \cdot \dfrac{dt}{dx} = \dfrac{\frac{dy}{dt}}{\frac{dx}{dt}}

We have

x=2t \implies \dfrac{dx}{dt}=2

y=t^4+1 \implies \dfrac{dy}{dt}=4t^3

The slope of the tangent line to the curve at t=1 is then

\dfrac{dy}{dx} \bigg|_{t=1} = \dfrac{4t^3}{2} \bigg|_{t=1} = 2t^3\bigg|_{t=1} = 2

so the slope of the normal line is -\frac12. When t=1, we have

x\bigg|_{t=1} = 2t\bigg|_{t=1} = 2

y\bigg|_{t=1} = (t^4+1)\bigg|_{t=1} = 2

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y - 2 = -\dfrac12 (x - 2) \implies y = -\dfrac12 x + 3

Q3. Differentiating with the product, power, and chain rules, we have

y = x(x+1)^{1/2} \implies \dfrac{dy}{dx} = \dfrac{3x+2}{2\sqrt{x+1}} \implies \dfrac{dy}{dx}\bigg|_{x=3} = \dfrac{11}4

The derivative vanishes when

\dfrac{3x+2}{2\sqrt{x+1}} = 0 \implies 3x+2=0 \implies x = -\dfrac23

Q4. Differentiating  with the product and chain rules, we have

y = (2x+1)e^{-2x} \implies \dfrac{dy}{dx} = -4xe^{-2x}

The stationary points occur where the derivative is zero.

-4xe^{-2x} = 0 \implies x = 0

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y = (2x+1)e^{-2x} \bigg|_{x=0} = 1

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