Answer:
The truck travel must to have a constant speed of 
Step-by-step explanation:
we have
where
d expresses a car's distance in feet
t is the number of seconds
<em>Find the distance d for t=8 sec</em>
<em>Find the distance d for t=8.2 sec</em>
The total distance in this interval of 0.2 sec is
<em>Find the speed of the car</em>
Divide the total distance by the time
therefore
The truck travel must to have a constant speed of
Answer:
28y^3 -35y^5
Step-by-step explanation:
= 7y^3(4) -7y^3(5y^2)
= 28y^3 -35y^5
_____
The applicable rule of exponents is ...
(y^a)(y^b) = y^(a+b)
Answer:
Morning's average rate = 50 mph, and Afternoon's average rate = 25 mph.
Step-by-step explanation:
Suppose he drove 150 miles for X hours, then his average rate in the morning was (150/X) mph.
Given that he spent 5 hours in driving.
And he drove 50 miles for (5-X) hours, then his average rate in the afternoon was 50/(5-X) mph.
Given that his average rate in the morning was twice his average rate in the afternoon.
(150/x) = 2 * 50/(5-x)
150/x = 100/(5-x)
Cross multiplying terms, we get:-
150*(5-x) = 100*x
750 - 150x = 100x
750 = 100x + 150x
750 = 250x
x = 750/250 = 3.
It means he spent 3 hours in the morning and 2 hours in the afternoon.
So morning's average rate = 150/3 = 50 mph.
and afternoon's average rate = 50/(5-3) = 25 mph.
