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- just_one_last_goodbye

10 owl bucks if you help me

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- just_one_last_goodbye

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)

- anonymous

Hi again!

- anonymous

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

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- just_one_last_goodbye

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

- anonymous

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)

- just_one_last_goodbye

thats part A?

- anonymous

not all, we need cory's too

- anonymous

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

- just_one_last_goodbye

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

- anonymous

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

- anonymous

sorta hard to explain whats weird

- just_one_last_goodbye

- just_one_last_goodbye

im copying and pasting it

- anonymous

sorry nix, thats wrong

- anonymous

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

- mathstudent55

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

- anonymous

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.

- anonymous

YES thats genius

- anonymous

mathstudent is right

- anonymous

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

- anonymous

For B, just plug in 6 for x

- anonymous

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

- anonymous

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

- just_one_last_goodbye

wait which is part a?

- anonymous

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

- anonymous

its asking for the functions in A

- just_one_last_goodbye

ahh so those are the functions correct?

- anonymous

yea

- anonymous

and for C

- just_one_last_goodbye

part b?

- anonymous

For B just plug in x

- anonymous

i mean 6 for x

- just_one_last_goodbye

so I put "plug in x"

- mathstudent55

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

- anonymous

NONONONON0

- just_one_last_goodbye

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

- anonymous

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

- anonymous

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

- just_one_last_goodbye

y = 15(1.2)^x

- anonymous

44.78976

- anonymous

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

- just_one_last_goodbye

idk im just gonna submit only part a

- just_one_last_goodbye

thanks anyways please medal @mathstudent55

- anonymous

No problem, sorry could be that big of a help

- just_one_last_goodbye

it's alright

- nonopro

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...)

- nonopro

message me back if i got it right

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