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xxMikexx [17]
3 years ago
7

Enter your answer and show all the steps that you use to solve this problem in the space provided.

Mathematics
1 answer:
dexar [7]3 years ago
8 0

Answer:14!

Step-by-step explanation:22-8=...?

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\sqrt{q^3}

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How many centimeters(cm) of rain fell in this region during the course of the year​
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January through

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3 years ago
Demand for Tablet Computers The quantity demanded per month, x, of a certain make of tablet computer is related to the average u
soldier1979 [14.2K]

x = f ( p ) = \frac { 100 } { 9 } \sqrt { 810,000 - p ^ { 2 } } } \\\\ \qquad { p ( t ) = \dfrac { 400 } { 1 + \dfrac { 1 } { 8 } \sqrt { t } } + 200 \quad ( 0 \leq t \leq 60 ) }

Answer:

12.0 tablet computers/month

Step-by-step explanation:

The average price of the tablet 25 months from now will be:

p ( 25) = \dfrac { 400 } { 1 + \dfrac { 1 } { 8 } \sqrt { 25 } } + 200 \\= \dfrac { 400 } { 1 + \dfrac { 1 } { 8 } \times 5 } + 200\\\\=\dfrac { 400 } { 1 + \dfrac { 5 } { 8 } } + 200\\p(25)=\dfrac { 5800 } {13}

Next, we determine the rate at which the quantity demanded changes with respect to time.

Using Chain Rule (and a calculator)

\dfrac{dx}{dt}= \dfrac{dx}{dp}\dfrac{dp}{dt}

\dfrac{dx}{dp}= \dfrac{d}{dp}\left[{ \dfrac { 100 } { 9 } \sqrt { 810,000 - p ^ { 2 } } }\right] =-\dfrac{100}{9}p(810,000-p^2)^{-1/2}

\dfrac{dp}{dt}=\dfrac{d}{dt}\left[\dfrac { 400 } { 1 + \dfrac { 1 } { 8 } \sqrt { t } } + 200 \right]=-25\left[1 + \dfrac { 1 } { 8 } \sqrt { t } \right]^{-2}t^{-1/2}

Therefore:

\dfrac{dx}{dt}= \left[-\dfrac{100}{9}p(810,000-p^2)^{-1/2}\right]\left[-25\left[1 + \dfrac { 1 } { 8 } \sqrt { t } \right]^{-2}t^{-1/2}\right]

Recall that at t=25, p(25)=\dfrac { 5800 } {13} \approx 446.15

Therefore:

\dfrac{dx}{dt}(25)= \left[-\dfrac{100}{9}\times 446.15(810,000-446.15^2)^{-1/2}\right]\left[-25\left[1 + \dfrac { 1 } { 8 } \sqrt {25} \right]^{-2}25^{-1/2}\right]\\=12.009

The quantity demanded per month of the tablet computers will be changing at a rate of 12 tablet computers/month correct to 1 decimal place.

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