Jill traveled to the ferry office and back. the trip there took 5 hours and the trip back took 4 hours. she averaged 65 mph on the return trip. find the average speed of the trip there

We know from physics class that the formula for distance of a linear motion is given as:

d = v t

Where,

d = distance travelled

v = average velocity

t = time it took to reach the destination

Since the distance going to the office and back is just similar, therefore we can simply equate the two:

v1 t1 = v2 t2

Where 1 signifies going to the office and 2 signifies going back from the office. Therefore this yields to:

v1 * 5 hours = 65 mph * 4 hours

v1 = 52 mph

The average speed going to the office is 52 mph.

step-by-step explanation:

i think something is missing in your question because y-1=(x-12) would not pass through the point (-4,-3). because if you replace x by -4 your are supposed to find y=-3 which is not the case her. if you replace x by -4 you find =-15.

i assume the right equation is y-1=1/4*(x-12) because this one is passing through both your points (if you are interested to know how i found that out i can tell you).

in this equation if i replace x by -4 i find y=-3 and if i replace x by 12, i find y=1 so the line representing this equation is passing through those 2 points.

then the standard form of the equation is finding an expression of y in function of x

y-1=1/4*(x-12)

y=x/4-3+1

y=x/4-2

hope that .

do you confirm there was a mistake in the question?