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Dmitry_Shevchenko [17]
2 years ago
15

You are buying boxes of cookies at a bakery. Each box of cookies costs \$4$4dollar sign, 4. In the equation below, ccc represent

s the number of boxes of cookies you buy, and ddd represents the amount the cookies will cost you (in dollars). The relationship between these two variables can be expressed by the following equation: d
Mathematics
1 answer:
Basile [38]2 years ago
8 0

Answer:

The required equation is

d = 4c

where d represents the cost of cookies in dollar and c represents the number of boxes of cookies.

Step-by-step explanation:

Given that, the cost of each box of cookies is $4.

It means,

The cost of 1 box of cookies is $4.

The cost of 2 boxes of cookies is $(4+4)  =$(2×4)

The cost of 3 boxes of cookies is $(4+4+4)  =$(3×4)

The cost of cookies

=(Number of boxes × 4)

The required equation is

d = 4c

where d represents the cost of cookies in dollar and c represents the number of boxes of cookies.                                                    

   

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Negative dividing by negative gives positive,
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Demand for Tablet Computers The quantity demanded per month, x, of a certain make of tablet computer is related to the average u
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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

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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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