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/h^2. 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 Since v'(t) is the acceleration at time t. the acceleration c hours after 2:00 PM is exactly 80 mi/h^2.

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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.

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

true

step-by-step explanation:

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hello from mrbilldoesmath!

answer:

(7 x + 6 y) (49 x^2 - 42 x y + 36 y^2)

discussion:

the trick is to recognize that 343 = 7^3 and 216 = 6^3 and, in general,

a^3 + b^3 =   (a + b) (a^2 - a b + b^2).

in our case

343x^3+216y^3   =   (7x)^3 + (6y)^3

so take a = 7x and b = 6y in the above equation to get the answer.

you,

mrb