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i only need help doing #3 @jim_thompson5910
A(t) = 39,145[(1+ 0.03/12)1/4]^(4)(12)(-45.5) what i have so far
the pic doesn't match the problem

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sooo im completely wrong lol just start over then?
did you post the wrong pic?
oops i did....
you meant to say this right? \[\Large A = P\left[ \left(1+\frac{r}{n}\right)^{\frac{1}{c}} \right]^{cnt}\] \[\Large A = 39145\left[ \left(1+\frac{0.03}{12}\right)^{\frac{1}{4}} \right]^{4*12*t}\]
yes and for t i had -45.5
t is time in years how is time negative?
idk but my teacher said that answer was right..
my biggest prob is solving 1.00025^1/4 because i dont have a calculator
he said that a negative time value is possible?
how strange
oh nvm I didn't read #2 fully lol
you can use this free online calculator http://web2.0calc.com/
if you want to evaluate 1.00025^(1/4), make sure you type that in and hit the equal sign don't forget the parenthesis
idk what im not doing right
I'm guessing #3 wants you to solve for r?
solve this for r maybe? \[\Large 1+\frac{r}{n} = P\left[ \left(1+\frac{r}{n}\right)^{\frac{1}{c}} \right]^{cnt}\]
hmm not 100% sure
\[a(t)=39,145[(1+\frac{ .003 }{ 12 })^{(1/4)}]\] ^(4)(12)(-45.5)
39,145[(1.00025)^1/4]^(-2184) -39,145=1^-2184 -39,145=1 39,145
where did i go wrong
which variable are you trying to solve for?
or you're just trying to evaluate 39,145[(1.00025)^1/4]^(-2184) ?
i guess just evaluate that cause the question asks to solve for the adjusted intrest rate
i feel like my equations right and i dont know why im getting it wrong ):
I'm stuck on the fact that they bring in c = 4, but the 'c' terms cancel (when they divide). This problem is just odd. Hmm
its making me mad cause i know im doing it right
are there any logs in here?
this is what I get http://web2.0calc.com/#39145((1.00025)^(1/4))^(-2184)
are there any logs in here?
yeah thats whatt i got to
there's the log button at the bottom
i mean to solve this problem
sorry I don't know how this works.

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