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Karo-lina-s [1.5K]
3 years ago
11

Helpppppp it’s due in 30 minutes.

Mathematics
1 answer:
dezoksy [38]3 years ago
7 0

Answer:

i think it is 27

Step-by-step explanation:

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A dog chases a squirrel. The dog is originally 200 feet away from the squirrel. The dog's speed is 150 feet per minute. The squi
Dafna11 [192]

Answer:

15 seconds

Step-by-step explanation:

If you make a table of values for the dog and the squirrel using d = rt, then the rates are easy:  the dog's rate is 150 and the squirrel's is 100.  The t is what we are looking for, so that's our unknown, and the distance is a bit tricky, but let's look at what we know:  the dog is 200 feet behind the squirrel, so when the dog catches up to the squirrel, he has run some distance d plus the 200 feet to catch up.  Since we don't know what d is, we will just call it d!  Now it seems as though we have 2 unknowns which is a problem.  However, if we solve both equations (the one for the dog and the one for the squirrel) for t, we can set them equal to each other.  Here's the dog's equation:

d = rt

d+200 = 150t

And the squirrel's:

d = 100t

If we solve both for t and set them equal to each other we have:

\frac{150}{d+200} =\frac{100}{d}

Now we can cross multiply to solve for d:

150d = 100d + 20,000 and

50d = 20,000

d = 400

But we're not looking for the distance the squirrel traveled before the dog caught it, we are looking for how long it took.  So sub that d value back into one of the equations we have solved for t and do the math:

t=\frac{100}{d}=\frac{100}{400} =\frac{1}{4}

That's 1/4 of a minute which is 15 seconds.

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Answer:4 is your answer

Step-by-step explanation:

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