Question

will medal real quick for cory could you do those functions until 6 years since you had 2 years

Answer 1

C stands for the the cars Cory has. n stands for the year pass Like if n =1, that is year 1. The number of Cars he has C1=1.2C1−1=1.2C0=1.2∗15=18 if n=2, that is year 2, the number of Cars he has is C2=1.2C2−1=1.2C1=1.2∗18=21.6 As we do above 20% of 18 = 3.6 Hence after year 2, he has 18+3.6 =21.6

Answer 2

@Loser66

Answer 3

what is the problem,

Answer 4

Ok, Please help him @DanJS

Answer 5

well what loser66 did was made the function go up 2 to 2 years and i need the function to go up to 6 years

Answer 6

C1=1.2C1−1

are those C(n) cars at n year
and what is the prob you were asked

Answer 7

Write functions to represent Cory and Roger's collections throughout the years. (Cory has 15 die-cast cars in his collection. Each year his collection increases by 20%. Roger has 40 cars in his collection. Each year he collects 1 additional car.

Answer 8

ok, use n=0 as the initial time, the initial amount of cars each have at time 0

Answer 9

huh

Answer 10

nevermind

Answer 11

C(n) - cars at year n

C(0) = 15 starts at 15
C(1) = 1.2*C(0)
C(2) = 1.2*c1 = 1.2^2 * c(0)
C(3) = 1.2*C2 = 1.2^3 * c(0)
..
.
.same form as a general exponential function (y=a*b^x)

Answer 12

C(n) = C(0)*(1.2)^n = 15*(1.2)^n

Answer 13

Rodgers is a linear rate,

R(n) = initial + one added each year n past

R(n) = R(0) + 1*n

Answer 14

Cory collection has another factor of 1.2 multiplied at the end of each year, and starts at n=0 with 15

C(n) = 15*(1.2)^n


Rodger collection just adds 1 after each year, he starts with 40.

R(n) = 40 + n


for n= time passed in years