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2 answers:
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Here is the correct format for the question
At 2:00 PM a car's speedometer reads 30 mi/h. At 2:15 PM it reads 50 mi/h. Show that at some time between 2:00 and 2:15 the acceleration is exactly 80 mi/h².Let v(f) be the velocity of the car t hours after 2:00 PM.Then . By the Mean Value Theorem, there is a number c such that 0 < c <
with v'(c) =
. Since v'(t) is the acceleration at time t, the acceleration c hours after 2:00 PM is exactly 80 mi/h^2.
Answer:
Step-by-step explanation:
From the information given :
At 2:00 PM ;
a car's speedometer v(0) = 30 mi/h
At 2:15 PM;
a car's speedometer v(1/4) = 50 mi/h
Given that:
v(f) should be the velocity of the car t hours after 2:00 PM
Then will be:
= 20 × 4/1
= 80 mi/h²
By the Mean value theorem; there is a number c such that :
with
Answer:
yards.
Step-by-step explanation:
Please find the attachment.
We have been given that Marisela cut four equivalent lengths of ribbon. Each was 5 eighths of a yard long.
To find yards of ribbon cut by Marisela, we will multiply 4 by 5 eighths of a yard (5/8).
Express answer as the sum of whole number and the remaining fractional units:
Therefore, Marisela cut yards of ribbon.
