Question

10 owl bucks if you help me

Answer 1

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.

Part A: Write functions to represent Cory and Roger's collections throughout the years. (4 points)
Part B: How many cars does Cory have after 6 years? How many does Roger have after the same number of years? (2 points)
Part C: After approximately how many years is the number of cars that Cory and Roger have the same? Justify your answer mathematically. (4 points)

Answer 2

Hi again!

Answer 3

So for part A, Roger's function is relatively easy to figure out

Answer 4

Ok ^_^ i closed my question cause i know u'll help me

Answer 5

So since he is constantly adding one more and he already starts with 40, we know it will be y=x+40
Where 40 is the constant (he starts with this many)
x is the rate at which he adds cars (he adds one so its 1x, but thats the same as x)

Answer 6

thats part A?

Answer 7

not all, we need cory's too

Answer 8

hold on, I'm just gonna figure this out, cory's is a bit weird

Answer 9

can u give me direct answers? :/ sorry im going to get screamed at and probably get punished physically if i dont get it done

Answer 10

double check the question, because cory's is a bit weird

Answer 11

sorta hard to explain whats weird

Answer 12

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.

Part A: Write functions to represent Cory and Roger's collections throughout the years. (4 points)
Part B: How many cars does Cory have after 6 years? How many does Roger have after the same number of years? (2 points)
Part C: After approximately how many years is the number of cars that Cory and Roger have the same? Justify your answer mathematically. (4 points)

Answer 13

im copying and pasting it

Answer 14

sorry nix, thats wrong

Answer 15

because think about it, if x is zero, then the whole thing would be zero. But he starts out with 15 cars so it has to be something else

Answer 16

Cory:

\(y = 15(1.2)^x\)

Answer 17

The question is weird because his collection increases by 20% every year. So its increasing by 20% then increase 20% of that 20%.

Essentialy its something like (15x1.2)1.2 etc.

Answer 18

YES thats genius

Answer 19

mathstudent is right

Answer 20

so A is for cory, y=15(1.2)^x
and for roger, y=x+40

Answer 21

For B, just plug in 6 for x

Answer 22

for both equations. So for roger it would be 46, and for cory 44.78976

Answer 23

but then we can't have a bit of a car.... so i guess 44?

Answer 24

wait which is part a?

Answer 25

cory, y=15(1.2)^x
and for roger, y=x+40

Answer 26

its asking for the functions in A

Answer 27

ahh so those are the functions correct?

Answer 28

yea

Answer 29

and for C

Answer 30

part b?

Answer 31

For B just plug in x

Answer 32

i mean 6 for x

Answer 33

so I put "plug in x"

Answer 34

Here is the explanation of Cory's function.

He starts with 15 cars.
His cars go up by 20% at the end of each year.
After 1 year, he has \(1.2 \times 15\)
At the begining of the second year, he starts with \(1.2 \times 15\) cars.
At the end of year 2, he now has \(1.2 \times (1.2 \times 15) = 15 \times (1.2)^2\)
At the end of year 3, he has \(1.2 \times (1.2 \times(1.2 \times 15)) = 15 \times 1.2^3\)
That pattern leads to the equation:

After x years, he has y cars

\(y = 15(1.2)^x\)

Answer 35

NONONONON0

Answer 36

oh so put "6 for x" in part b?

Answer 37

For B you plug in 6 for x yourself, and thats the answers

Answer 38

we definitely know rogers, which is 46
y=6+40

Answer 39

y = 15(1.2)^x

Answer 40

44.78976

Answer 41

But, idk what your instructor wants, because you can't have like 2/3 of a car

Answer 42

idk im just gonna submit only part a

Answer 43

thanks anyways please medal @mathstudent55

Answer 44

No problem, sorry could be that big of a help

Answer 45

it's alright

Answer 46

let's name "x" the number of years.

we know that 15 +15*1.20^{x} =40+x

let's try after 5 years: Roger would have 45 cars and Cory 52 ( i just guessed 5, and then I calculated 15 +15*1.20^{5} ): so, no, it has to be less than 5.

let's try 4:

Roger will have 44, and cory 15+31=45, so that's quite good asnwer!

Let's try with 3:

Roger will have 43, and Cory 15+ 25=40,

so 4 years are better: after approximately 4 years.

(perhaps a better strategy would be to just calculate after each year starting at 1 and going up...)

Answer 47

message me back if i got it right