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Pepsi [2]
3 years ago
10

A satellite camera takes a rectangular-shaped picture. The smallest region that can be photographed is a 4-km by 4-km rectangle.

As the camera zooms out, the length l and width w of the . rectangle increase at a rate of 3 km/sec. How long does it take for the area A to be at least 4 times its original size?
A.) 4.94 sec
B.) 3.28 sec
C.) 9.7 sec
D.) 1.33 sec
Mathematics
1 answer:
Sliva [168]3 years ago
6 0
Given:

L1 = original length = 4 km
W1 = original width = 4 km
A1 = original area = 4*4 = 16 km^2
A2 = 4(16) = 64 km^2

Given the second area, we can conclude that the lengths and widths of the zoomed out photograph should both be 8 km.

Given that the zooming occurs at 3 km/sec, the amount of time needed for the lengths and widths to zoom out from 4 km to 8 km is shown below:

8km - 4km / 3km/s = 4 / 3 = 1.33 seconds

Therefore, the correct answer is D.

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vichka [17]

Answer:

14878.04878miles/hours^2

Step-by-step explanation:

Let's find a solution by understanding the following:

The acceleration rate is defined as the change of velocity within a time interval, which can be written as:

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Using the problem's data we have:

Vf=65miles/hour

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Using the acceleration rate equation we have:

A=(65miles/hour - 6miles/hour)/14.8seconds, but look that velocities use 'hours' unit while 'T' uses 'seconds'.

So we need to transform 14.8seconds into Xhours, as follows:

X=(14.8seconds)*(1hours/60minutes)*(1minute/60seconds)

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