Sheila is looking at some information for the obstacle course she is interested in completing. the x-coordinate is the number of the obstacle, while the y-coordinate is the average time to complete the obstacle, measured in minutes. (1, 8.25), (2, 9.075), (3, 9.9825), (4, 10.98075) sheila use an explicit formula to find the average time she will need for the 8th obstacle.
a. f(8) = 8.25(1.1)^8; f(8) = 17.685
b. f(8) = 8.25(1.1)^7; f(8) = 16.077
c. f(8) = 1.1(8.25)^7; f(8) = 2861345
d. f(8) = 1.1(8.25)^8; f(8) = 23606102

f(8) = 8.25(1.1)^7 ; f(8) = 16.077 ⇒ answer B

Step-by-step explanation:

* Lets explain how to solve the problem

∵ The x-coordinate is the number of the obstacle

∵ The y-coordinate is the average time to complete the obstacle

∵ The order pairs of function are (1 , 8.25) , (2 , 9.075) , (3 , 9.9825) ,

  (4 , 10.98075)

- From these order pairs

# The time to finish the 1st obstacle is 8.25 minutes

# The time to finish the 2nd obstacle is 9.075 minutes

# The time to finish the 3rd obstacle is 9.9825 minutes

# The time to finish the 4th obstacle is 10.98075 minutes

∵ 2nd ÷ 1st = 9.075/8.25 = 1.1

∵ 3rd ÷ 2nd = 9.9825/9.075 = 1.1

∵ 4th ÷ 3rd = 10.98075/9.9825 = 1.1

∴ There is a constant ratio 1.1 between each 2 consecutive terms

∴ The order pairs formed a geometric series

- Any term in the geometric series Un = a r^(n - 1) , where a is the 1st

 term in the series , r is the constant ratio and n is the position of the

 term in the series

∵ a = 8.25 ⇒ the time of the first obstacle

∵ r = 1.1

- Sheila wants to find the average time she will need for the 8th

 obstacle

∴ n = 8

∵ The explicit formula is f(x) = a r^(n - 1)

∴ f(8) = 8.25 (1.1)^(8 - 1)

∴ f(8) = 8.25(1.1)^7

∴ f(8) = 16.076916 ≅ 16.077

* f(8) = 8.25(1.1)^7 ; f(8) = 16.077

hello from mrbilldoesmath!

answer:

25x^2 + 90x + 84

discussion:

mn(x) = m(n(x)) = m( 5x + 9) =

(5x + 9)^2 + 3 =>

25x^2 + 90x + 9^2 + 3 =

25x^2 + 90x + 84                 (as 9^2 = 9*9 = 81)

which is the second choice

regards,  

mrb

p.s.   i'll be on vacation from friday, dec 22 to jan 2, 2019. have a great new year!