Hailey ran a few laps. D(n)D(n)D, (, n, )models the duration (in seconds) of the time it took for Hailey to run her n^{th}n th n, start superscript, t, h, end superscript lap. nnn 333 777 999 D(n)D(n)D, (, n, )858585 999999 110110110 When did the lap duration increase faster

Question:

Hailey ran a few laps. D(n), models the duration (in seconds) of the time it took for Hailey to run her nth lap. When did the lap duration increase faster?

A. Between the 3rd and 7th lap

B. Between the 7th and 9th lap

C. The lap duration increased at the same rate over both intervals

B. Between the 7th and 9th lap

Step-by-step explanation:

Given

See attachment for table

Required

Determine when the lap increases faster

To do this, we simply calculate the rate of change between each lap.

From the attachment, we have:

Rate of change is calculated using:

For D(3) to D(7), the rate is:

For D(7) to D(9), the rate is:

The rate of change between the 7th and 9th lap  is greater than the rate of change between then 3rd and 7th lap .

Hence, (b) is correct

B) Between the 7th lap and the 9th lap

Step-by-step explanation:

the 8 in 8,420 is in the thousands place so it's value is 8 thousand.

the 8 in 6,870 is in the hundreds place so it's value is 8 hundred.

divide 8000 by 800 to find the answer:

8000 / 800 = 10

the 8 in 8,420 is 10 times greater than the 8 in 6,870

38.6°

step-by-step explanation:

resultant is 10 pound

smaller force = 8 pound

angle formed = tan⁻¹(

                        =tan⁻¹(

                          =tan⁻¹(0.8)

                        =38.6°