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Vitek1552 [10]
3 years ago
7

A parabolic arch sculpture is on top of a city bank. A model of the arch is y = −0.005x2 + 0.3x where x and y are in feet.

Mathematics
1 answer:
o-na [289]3 years ago
7 0

Answer:

A. a. 34.5 feet

b. 60 feet

Step-by-step explanation:

Parabola equation

y=-0.005x^2+0.3x

Differentiating with respect to x we get

\dfrac{dy}{dx}=-0.01x+0.3

Equating with zero

-0.01x+0.3=0\\\Rightarrow -0.01x=-0.3\\\Rightarrow x=\dfrac{0.3}{0.01}\\\Rightarrow x=30

Double derivative of the parabolic equation

\dfrac{d^2y}{dx^2}=-0.01

So, x=30 is maximum.

y=-0.005\times 30^2+0.3\times 30\\\Rightarrow y=4.5

So, the maximum height of the arch will be 4.5 feet.

From the ground the highest point of the arch will be 30+4.5=34.5\ \text{ft}

We are taking the x axis as the width of the bank.

0=-0.005x^2+0.3x\\\Rightarrow 0.005x^2=0.3x\\\Rightarrow x=\dfrac{0.3}{0.005}\\\Rightarrow x=60

So, the width of the bank will be 60 feet.

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Step-by-step explanation:

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Since, S is being dilated with a scale factor to form T,

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Answer:

\boxed{\sf x=\cfrac{1\pm \sqrt{-7} }{2}}

Step-by-step explanation:

\sf x^2-x+2=0

<u>Use quadratic formula:</u>

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___________________________

\sf x_1,_2=\cfrac{-\left(-1\right)\pm \sqrt{\left(-1\right)^2-4\times \:1\times \:2}}{2\times \:1}

  • \sf \sqrt{\left(-1\right)^2-4\times \:1\times \:2}
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_________________

\sf x_1,_2\cfrac{-\left(-1\right)\pm \sqrt{-7}}{2\times \:1}

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\sf x_1,_2=\cfrac{1\pm \sqrt{-7} }{2}

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<u>Learn more: brainly.com/question/22286698</u>

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