Question

Solve :

1) $2000(1+.08/4)^4(18)

Answer 1

2000(1+.08/4)^4(18)

2000(1+0.02)^72

2000(1.02)^72

2000(4.161140375)

8,322.28075

8,322.28

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So

\[\Large 2000\left(1+\frac{0.08}{4}\right)^{4(18)} = 8,322.28\]

Answer 2

Thanks! I went back and did it and that's what I got , miscalculated it . @jim_thompson5910

Answer 3

ok great, glad you figured out where you went wrong

Answer 4

8000=22000(1-.0125/12)12t

A=1500e.075(5)

Answer 5

@jim_thompson5910 ^^

Answer 6

ok so let's solve 8000=22000(1-.0125/12)^(12t) for t first

Answer 7

wait a second, is that really 22,000 or is it supposed to be 2,000?

Answer 8

It's correct.

Answer 9

so it's 22,000 ?

Answer 10

Yes.

Answer 11

ok thx

Answer 12

8000=22000(1-.0125/12)^(12t)

8000/22000 = (1-.0125/12)^(12t)

0.363636 = (1-.0125/12)^(12t)

ln(0.363636) = ln((1-.0125/12)^(12t))

-1.01160191 = 12t*ln(1-.0125/12)

-1.01160191 = 12t*(-0.0010422)

-1.01160191/(-0.0010422) = 12t

970.640865 = 12t

12t = 970.640865

t = 970.640865/12

t = 80.88673875

Answer 13

For the last one

A=1500e^(.075(5))

A=1500e^(0.375)

A=1500(1.454991)

A=2182.4865

A=2182.49

Answer 14

So the question says: You buy a new car for $22,000. The value decreases by 12.5% each year. When will the car have a value of $8000? What's the answer to that? I'm lost.

Answer 15

oh it would just be

8000=22000(1-0.125)^t

Answer 16

solve that for t

Answer 17

Okay good so that was the answer I got for the third one the second time around, thanks!

Answer 18

So how would you solve for t? I was stuck on that.And so then the equation isn't setup like I originally had it?

Answer 19

follow the same idea to solve 8000=22000(1-0.125)^t

Answer 20

so you would divide both sides by 22000

then take the Ln of both sides

that will allow you to pull down the exponent and solve for t

Answer 21

Can you show me in equation format?

Answer 22

one sec

Answer 23

8000 = 22000(1-0.125)^t

8000/22000 = (1-0.125)^t

0.363636 = (1-0.125)^t

ln(0.363636) = ln((1-0.125)^t)

ln(0.363636) = t*ln(0.875)

-1.011599 = t*ln(0.875)

-1.011599 = t*(-0.133531)

-1.011599/(-0.133531) = t

7.575753 = t

t = 7.575753

Answer 24

So it will take roughly 7 and a half years for the car to go from $22,000 in value to $8,000 in value

Answer 25

Thank you so much for your help. :)

Answer 26

np