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Lelu [443]
2 years ago
10

An open rectangular box with a square bottom must have a surface area of A square inches. Write a formula for the volume of the

box as a function of x. Your answer will also include the constant, A.
Image of problem is attached.

Mathematics
2 answers:
sveticcg [70]2 years ago
6 0
A= X x X
Volume= A x H
skelet666 [1.2K]2 years ago
5 0
The surface area of the box is being referred to as A
The surface area will be the sum of the four rectangular sides' areas and the base area.
Base area = x²
Rectangular side area = xh
A = x² + 4xh
Volume is, as you said,
V = x²h
Lets substitute h from the formula of A to obtain A in our final answer. Substituting x is possible, but more difficult
(A - x²)/4x = h
V = (x²) * (A - x²)/4x
V = (Ax - x³)/4

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10 to power of -8 so 0.00000001 I think

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Answer:

Step-by-step explanation:

5p + 2(p + 4) .......because pencil cases (p) sell for 5 bucks a piece....and mechanical pencils (p + 4), sell for 2 bucks a piece. She is basically selling 4 more mechanical pencils then she is pencil cases.

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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
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.

8 0
2 years ago
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