just_one_last_goodbye
  • just_one_last_goodbye
10 owl bucks if you help me
Mathematics
jamiebookeater
  • jamiebookeater
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just_one_last_goodbye
  • 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
  • anonymous
Hi again!
anonymous
  • anonymous
So for part A, Roger's function is relatively easy to figure out

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just_one_last_goodbye
  • just_one_last_goodbye
Ok ^_^ i closed my question cause i know u'll help me
anonymous
  • 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
  • just_one_last_goodbye
thats part A?
anonymous
  • anonymous
not all, we need cory's too
anonymous
  • anonymous
hold on, I'm just gonna figure this out, cory's is a bit weird
just_one_last_goodbye
  • 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
  • anonymous
double check the question, because cory's is a bit weird
anonymous
  • anonymous
sorta hard to explain whats weird
just_one_last_goodbye
  • 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)
just_one_last_goodbye
  • just_one_last_goodbye
im copying and pasting it
anonymous
  • anonymous
sorry nix, thats wrong
anonymous
  • 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
  • mathstudent55
Cory: \(y = 15(1.2)^x\)
anonymous
  • 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
  • anonymous
YES thats genius
anonymous
  • anonymous
mathstudent is right
anonymous
  • anonymous
so A is for cory, y=15(1.2)^x and for roger, y=x+40
anonymous
  • anonymous
For B, just plug in 6 for x
anonymous
  • anonymous
for both equations. So for roger it would be 46, and for cory 44.78976
anonymous
  • anonymous
but then we can't have a bit of a car.... so i guess 44?
just_one_last_goodbye
  • just_one_last_goodbye
wait which is part a?
anonymous
  • anonymous
cory, y=15(1.2)^x and for roger, y=x+40
anonymous
  • anonymous
its asking for the functions in A
just_one_last_goodbye
  • just_one_last_goodbye
ahh so those are the functions correct?
anonymous
  • anonymous
yea
anonymous
  • anonymous
and for C
just_one_last_goodbye
  • just_one_last_goodbye
part b?
anonymous
  • anonymous
For B just plug in x
anonymous
  • anonymous
i mean 6 for x
just_one_last_goodbye
  • just_one_last_goodbye
so I put "plug in x"
mathstudent55
  • 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
  • anonymous
NONONONON0
just_one_last_goodbye
  • just_one_last_goodbye
oh so put "6 for x" in part b?
anonymous
  • anonymous
For B you plug in 6 for x yourself, and thats the answers
anonymous
  • anonymous
we definitely know rogers, which is 46 y=6+40
just_one_last_goodbye
  • just_one_last_goodbye
y = 15(1.2)^x
anonymous
  • anonymous
44.78976
anonymous
  • anonymous
But, idk what your instructor wants, because you can't have like 2/3 of a car
just_one_last_goodbye
  • just_one_last_goodbye
idk im just gonna submit only part a
just_one_last_goodbye
  • just_one_last_goodbye
thanks anyways please medal @mathstudent55
anonymous
  • anonymous
No problem, sorry could be that big of a help
just_one_last_goodbye
  • just_one_last_goodbye
it's alright
nonopro
  • 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
  • nonopro
message me back if i got it right

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