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binarymimicBest ResponseYou've already chosen the best response.0
What equations have you set up to model this problem?
 one year ago

Outkast3r09Best ResponseYou've already chosen the best response.0
I'd first start by deciphering everything you have
 one year ago

binarymimicBest ResponseYou've already chosen the best response.0
If you are unsure where to begin, review exponential functions (http://www.regentsprep.org/Regents/math/ALGEBRA/AE7/ExpDecayL.htm) You should obtain \[50000(1 + 0.07)^{x} = 70000(1  0.04)^{x}\]
 one year ago

Outkast3r09Best ResponseYou've already chosen the best response.0
first lets make 2010 your Pnot or \[P_0=P(0)\]
 one year ago

Outkast3r09Best ResponseYou've already chosen the best response.0
so at \[P(0)=50,000\]
 one year ago

Outkast3r09Best ResponseYou've already chosen the best response.0
for checkerville yes
 one year ago

Outkast3r09Best ResponseYou've already chosen the best response.0
lets rephrase what your equation is \[P=ce^{kt}\] where p is your population , c is a constant unknown atm and k is your rate. t is time. if we put in P(0) you get \[50,000=ce^{0}=c\] \[P=50,000e^{kt}\] since we know the rate at which it increases 7% \[P=50,000e^{.07t}\] next doing for checker ville you get \[P=70,000e^{.04t}\]
 one year ago

Outkast3r09Best ResponseYou've already chosen the best response.0
At P1=P2 is when one will exceed the other and the other will be below soooo \[70,000e^{.04t}=50,000e^{.07t}\] solve for t
 one year ago

satellite73Best ResponseYou've already chosen the best response.1
i am going to make a guess that you are supposed to use \[50000(1 + 0.07)^{x} = 70000(1  0.04)^{x}\]
 one year ago

satellite73Best ResponseYou've already chosen the best response.1
that is \[50000(1.07)^x=70000(.96)^x\]
 one year ago

Outkast3r09Best ResponseYou've already chosen the best response.0
essentially the same thing lol
 one year ago

satellite73Best ResponseYou've already chosen the best response.1
divide by 50000 get \[ 91.07)^x=1.4(.96)^x\]divide by\[ (.96)^x\] get \[\left(\frac{1.07}{.96}\right)^x=1.4\]
 one year ago

satellite73Best ResponseYou've already chosen the best response.1
@Outkast3r09 yeah it is the same except that i am thinking that because it say 7% per year it is using \((1.07)^x\) rather than \(e^{.07x}\) but i could be wrong
 one year ago

satellite73Best ResponseYou've already chosen the best response.1
at any rate, solve via \[x=\frac{\ln(1.4)}{\ln(\frac{1.07}{.96})}\]
 one year ago

satellite73Best ResponseYou've already chosen the best response.1
say this exact same problem yesterday. they really reach for these word problems, don't they?
 one year ago

binarymimicBest ResponseYou've already chosen the best response.0
satellite, your equation is correct and your final solution is also correct
 one year ago

Outkast3r09Best ResponseYou've already chosen the best response.0
nah it's because i meant to put the rate as 1.07 and .96 so you'd get the same thing
 one year ago
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