Question

&

Answer 1

i only need help doing #3
@jim_thompson5910

Answer 2

A(t) = 39,145[(1+ 0.03/12)1/4]^(4)(12)(-45.5)
what i have so far

Answer 3

the pic doesn't match the problem

Answer 4

sooo im completely wrong lol just start over then?

Answer 5

did you post the wrong pic?

Answer 6

oops i did....

Answer 7

you meant to say this right?
\[\Large A = P\left[ \left(1+\frac{r}{n}\right)^{\frac{1}{c}} \right]^{cnt}\]
\[\Large A = 39145\left[ \left(1+\frac{0.03}{12}\right)^{\frac{1}{4}} \right]^{4*12*t}\]

Answer 8

yes and for t i had -45.5

Answer 9

t is time in years

how is time negative?

Answer 10

idk but my teacher said that answer was right..

Answer 11

my biggest prob is solving 1.00025^1/4 because i dont have a calculator

Answer 12

he said that a negative time value is possible?

Answer 13

how strange

Answer 14

oh nvm I didn't read #2 fully lol

Answer 15

you can use this free online calculator

http://web2.0calc.com/

Answer 16

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

Answer 17

idk what im not doing right

Answer 18

I'm guessing #3 wants you to solve for r?

Answer 19

solve this for r maybe?

\[\Large 1+\frac{r}{n} = P\left[ \left(1+\frac{r}{n}\right)^{\frac{1}{c}} \right]^{cnt}\]

Answer 20

hmm not 100% sure

Answer 21

\[a(t)=39,145[(1+\frac{ .003 }{ 12 })^{(1/4)}]\] ^(4)(12)(-45.5)

Answer 22

39,145[(1.00025)^1/4]^(-2184)

-39,145=1^-2184

-39,145=1

39,145

Answer 23

where did i go wrong

Answer 24

which variable are you trying to solve for?

Answer 25

or you're just trying to evaluate 39,145[(1.00025)^1/4]^(-2184) ?

Answer 26

i guess just evaluate that cause the question asks to solve for the adjusted intrest rate

Answer 27

i feel like my equations right and i dont know why im getting it wrong ):

Answer 28

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

Answer 29

its making me mad cause i know im doing it right

Answer 30

are there any logs in here?

Answer 31

this is what I get

http://web2.0calc.com/#39145((1.00025)^(1/4))^(-2184)

Answer 32

are there any logs in here?

Answer 33

yeah thats whatt i got to

Answer 34

there's the log button at the bottom

Answer 35

@UsukiDoll

Answer 36

@UsukiDoll

Answer 37

i mean to solve this problem

Answer 38

sorry I don't know how this works.