Question

*Advance Algebra With Financial Application * [WORK IS SHOWN]
***WHAT GET'S ME IS THAT IM STUCK . I SUBTRACTED THE PURCHASE AND THE DOWN PAYMENT WHICH GOT ME :16740
Fitzgerald purchased a car for $18,980. He made a down payment of $2,240. He applied for a five-year installment loan with an interest rate of 7.4%. What is the total cost of the car after five years?

Answer 1

18,980-2240=16740
M=16740(1+(0.074)/(12))^(12)/(1+(0.074)/(12))^(12)[-1] ?

Answer 2

installment loan? what are the directives for that?

Answer 3

is it monthly payments? or just a wait 5 years payment?

Answer 4

if payments can be calculated, id suggest
\[0=B_o(k)^5-P\frac{1-k^5}{1-k}\]
\[B_o(k)^5=P\frac{1-k^5}{1-k}\]
\[B_o(k)^5\frac{1-k}{1-k^5}=P\]

Answer 5

where k = (1+r/c)

Answer 6

and i spose that 5 needs to be adjusted by c as well

Answer 7

if its monthly payments, that gives me c=12 and, payments of about 4015.68 a month

Answer 8

\[B_o(1+\frac{.074}{12})^{12*5}\frac{-\frac{.075}{12}}{1-(1+\frac{.074}{12})^{12*5}}=P\]

Answer 9

O.o where do i start to solve the formula

Answer 10

lol, im assuming this is just another name for a monthly payment loan that is paid off in 5 years

Answer 11

Bo = 16740, the rest of it is just how i shuffle the weight around

Answer 12

lets say k = 1.00617

Answer 13

\[B_o(1.00617)^{60}~\frac{1-1.00617}{1-1.00617^{60}}=P\]

Answer 14

http://www.wolframalpha.com/input/?i=16740%281.00617%29%5E60+*+-.00617%2F%281-1.00617%5E60%29

so payments each month are about 334.68

Answer 15

if we want to be more exact, 334.65 so not much difference for an approximation

Answer 16

so about 20079 plus the down payment

Answer 17

Is that answer ?

Answer 18

$20,078.40

$22,318.40

$17,978.76

$20,384.52

Answer 19

it looks like 20078.40 is the payment option if we add back in the
2240 down payment
----------
22318.40

Answer 20

@amistre64 Thank youu !!

Answer 21

youre welcome