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3 years ago

5

A Parachutes speed during a freefall reaches 108 mph. What is the speed in feet per second? At this speed, how many feet with a

parachutist fall during 15 seconds of free fall?

1 answer:

Slav-nsk [51]3 years ago

7 0

The speed would be 158.4fps and the parachutist would fall 2,376 feet during 15 seconds.

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Convert 3 pints per second to gallons per minute

exis [7]

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3 years ago

What is the lowest and highest grade average in the class respectively?

sineoko [7]

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3 years ago

a contractor needs to purchase 500 bricks. the dimensions of each brick are 5.1 cm by 10.2 by 20.3 cm, and the density of each b

AveGali [126]

No, the trailer cannot hold the weight of the bricks. It is beyond the 900kg capacity of the trailer. The total weight of the bricks is 1,013.77 kilograms. The total weight was derived from getting the volume of the brick (0.051m x 0.102m x 0.203m), then multiplying the volume to the density of each brick (1.056 x 10^3m^3 x 1920kg/m^3). The weight of each brick is 2.03kg. Lastly, multiply the total number of bricks to the weight of each brick to get the total weight.

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3 years ago

A rectangular swimming pool is twice as long as it is wide. A small concrete walkway surrounds the pool. The walkway is a 2 feet

Ksivusya [100]

Answer:

The width and the length of the pool are 12 ft and 24 ft respectively.

Step-by-step explanation:

The length (L) of the rectangular swimming pool is twice its wide (W):

$L_{1} = 2W_{1}$

Also, the area of the walkway of 2 feet wide is 448:

$W_{2} = 2 ft$

$A_{T} = W_{2}*L_{2} = 448 ft^{2}$

Where 1 is for the swimming pool (lower rectangle) and 2 is for the walkway more the pool (bigger rectangle).

The total area is related to the pool area and the walkway area as follows:

$A_{T} = A_{1} + A_{w}$ (1)

The area of the pool is given by:

$A_{1} = L_{1}*W_{1}$

$A_{1} = (2W_{1})*W_{1} = 2W_{1}^{2}$ (2)

And the area of the walkway is:

$A_{w} = 2(L_{2}*2 + W_{1}*2) = 4L_{2} + 4W_{1}$ (3)

Where the length of the bigger rectangle is related to the lower rectangle as follows:

$L_{2} = 4 + L_{1} = 4 + 2W_{1}$ (4)

By entering equations (4), (3), and (2) into equation (1) we have:

$A_{T} = A_{1} + A_{w}$

$A_{T} = 2W_{1}^{2} + 4L_{2} + 4W_{1}$

$448 = 2W_{1}^{2} + 4(4 + 2W_{1}) + 4W_{1}$

$224 = W_{1}^{2} + 8 + 4W_{1} + 2W_{1}$

$224 = W_{1}^{2} + 8 + 6W_{1}$

By solving the above quadratic equation we have:

W₁ = 12 ft

Hence, the width of the pool is 12 feet, and the length is:

$L_{1} = 2W_{1} = 2*12 ft = 24 ft$

Therefore, the width and the length of the pool are 12 ft and 24 ft respectively.

I hope it helps you!

8 0

3 years ago

If 30,000 cm2 of material is available to make a box with a square base and an open top, what is the largest possible volume (in

Cloud [144]

Answer:

The largest possible volume of the box is 2000000 cubic meters.

Step-by-step explanation:

The volume ( $V$ ), in cubic centimeters, and surface area ( $A_{s}$ ), in square centimeters, of the box with a square base are described below:

$A_{s} = l^{2}+h\cdot l$ (1)

$V = l^{2}\cdot h$ (2)

Where:

$l$ - Side length of the base, in centimeters.

$h$ - Height of the box, in centimeters.

By (2), we clear $h$ within the formula:

$h = \frac{V}{l^{2}}$

And we apply in (1) and simplify the resulting expression:

$A_{s} = l^{2}+ \frac{V}{l}$

$A_{s}\cdot l = l^{3}+V$

$V = A_{s}\cdot l -l^{3}$ (3)

Then, we find the first and second derivatives of this expression:

$V' = A_{s}-3\cdot l^{2}$ (4)

$V'' = -6\cdot l$ (5)

If $V' = 0$ and $A_{s} = 30000\,cm^{2}$ , then we find the critical value of the side length of the base is:

$30000-3\cdot l^{2} = 0$

$3\cdot l^{2} = 30000$

$l = 100\,cm$

Then, we evaluate this result in the expression of the second derivative:

$V'' = -600$

By Second Derivative Test, we conclude that critical value leads to an absolute maximum. The maximum possible volume of the box is:

$V = 30000\cdot l - l^{3}$

$V = 2000000\,cm^{3}$

The largest possible volume of the box is 2000000 cubic meters.

4 0

2 years ago