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just_one_last_goodbye
 one year ago
10 owl bucks if you help me
just_one_last_goodbye
 one year ago
10 owl bucks if you help me

This Question is Closed

just_one_last_goodbye
 one year ago
Best ResponseYou've already chosen the best response.2Cory has 15 diecast 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
 one year ago
Best ResponseYou've already chosen the best response.0So for part A, Roger's function is relatively easy to figure out

just_one_last_goodbye
 one year ago
Best ResponseYou've already chosen the best response.2Ok ^_^ i closed my question cause i know u'll help me

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0So 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
 one year ago
Best ResponseYou've already chosen the best response.2thats part A?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0not all, we need cory's too

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0hold on, I'm just gonna figure this out, cory's is a bit weird

just_one_last_goodbye
 one year ago
Best ResponseYou've already chosen the best response.2can u give me direct answers? :/ sorry im going to get screamed at and probably get punished physically if i dont get it done

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0double check the question, because cory's is a bit weird

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0sorta hard to explain whats weird

just_one_last_goodbye
 one year ago
Best ResponseYou've already chosen the best response.2Cory has 15 diecast 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
 one year ago
Best ResponseYou've already chosen the best response.2im copying and pasting it

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0sorry nix, thats wrong

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0because 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
 one year ago
Best ResponseYou've already chosen the best response.3Cory: \(y = 15(1.2)^x\)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0The 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
 one year ago
Best ResponseYou've already chosen the best response.0mathstudent is right

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so A is for cory, y=15(1.2)^x and for roger, y=x+40

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0For B, just plug in 6 for x

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0for both equations. So for roger it would be 46, and for cory 44.78976

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0but then we can't have a bit of a car.... so i guess 44?

just_one_last_goodbye
 one year ago
Best ResponseYou've already chosen the best response.2wait which is part a?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0cory, y=15(1.2)^x and for roger, y=x+40

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0its asking for the functions in A

just_one_last_goodbye
 one year ago
Best ResponseYou've already chosen the best response.2ahh so those are the functions correct?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0For B just plug in x

just_one_last_goodbye
 one year ago
Best ResponseYou've already chosen the best response.2so I put "plug in x"

mathstudent55
 one year ago
Best ResponseYou've already chosen the best response.3Here 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 \(y = 15(1.2)^x\)

just_one_last_goodbye
 one year ago
Best ResponseYou've already chosen the best response.2oh so put "6 for x" in part b?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0For B you plug in 6 for x yourself, and thats the answers

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0we definitely know rogers, which is 46 y=6+40

just_one_last_goodbye
 one year ago
Best ResponseYou've already chosen the best response.2y = 15(1.2)^x

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0But, idk what your instructor wants, because you can't have like 2/3 of a car

just_one_last_goodbye
 one year ago
Best ResponseYou've already chosen the best response.2idk im just gonna submit only part a

just_one_last_goodbye
 one year ago
Best ResponseYou've already chosen the best response.2thanks anyways please medal @mathstudent55

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0No problem, sorry could be that big of a help

just_one_last_goodbye
 one year ago
Best ResponseYou've already chosen the best response.2it's alright

nonopro
 11 months ago
Best ResponseYou've already chosen the best response.0let'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
 11 months ago
Best ResponseYou've already chosen the best response.0message me back if i got it right
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