This is a distance problem so we can use distance = rate x time.
Given: the distance to the telephone is the same as the distance back from the telephone. We also know the two speeds and the total time for the trip.
But first let's convert 15 minutes into seconds: 15*60 = 900 seconds.
So: d = 5(t1) (t1 is the time he took to run to the telephone)
d = 4(t2) (t2 is the time he took to come back from the telephone
t1 + t2 = 900 or t1 = 900 - t2
Now use substitution:
5t1 = 4t2
5*(900 - t2) = 4t2
4500 - 5t2 = 4t2
4500 = 9t2
500 = t2
So therefore t1 = 400
So the distance to the telephone is 5*400 = 2000 meters
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<u>step </u><u>by </u><u>step:</u>
A and C are linear equations. A linear equestrian is an equation where the exponent has a term of one and the equation results in a straight line. The function changes at a common rate.
B and D are not linear because the power of their exponents are 3 and 2 respectively making them exponential.
Answer:
6 Weeks
Step-by-step explanation:
