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klio [65]
3 years ago
13

A cheetah can run at top speed for only about 20 seconds. If an antelope is too far away for a cheetah to catch it in 20 seconds

, the antelope is probably safe. Your friend claims an antelope running 60 feet per second will not be safe if the cheetah running at 90 feet per second is 650 feet behind it. Is your friend correct?
Mathematics
1 answer:
Lisa [10]3 years ago
6 0
The friend would be incorrect. The antelope  running 60 feet per second will be safe if the cheetah running at 90 feet per second is 650 feet behind it. To prove this, we can calculate the time it takes for the cheetah to reach the initial position of the antelope which is 650 ft from its initial position. We do as follows:

time to reach 650 ft = 650 ft / 90 ft/s = 7.22 s

After 7.22 s, the antelope would be,

distance of the antelope = 60 ft /s (7.22 s) = 433.33 ft

It would be 433.33 ft away from its initial position. So, the cheetah would not be able to reach the antelope when it travels 650 ft.
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Jennifer had 7/8 foot board she cut off 1/4 foot peice that was for a project.in feet how much board was left?
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\frac{5}{8} \text{ feet board is left }

<em><u>Solution:</u></em>

Given that, Jennifer had 7/8 foot board she cut off 1/4 foot peice that was for a project

To find: Board that was left

From given,

\text{Total length of board } =\frac{7}{8} \text{ feet }

\text{Foot that was cut off } = \frac{1}{4} \text{ feet }

Remaining feet of board is found by finding the difference between total length of board and foot that was cut off

\text{Reamining board } = \text{Total length of board - feet of board that was cut off}\\

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\text{Remaining board } = \frac{7}{8} - \frac{1}{4}\\\\\text{Remaining board } =\frac{7}{8} - \frac{1 \times 2}{4 \times 2}\\\\\text{Remaining board } =\frac{7}{8} -\frac{2}{8} = \frac{7-2}{8}\\\\\text{Remaining board } =\frac{5}{8}

Thus \frac{5}{8} feet board is left

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