Question

1. Dylan invested $47,000 in an account paying an interest rate of 4% compounded annually. Assuming no deposits or withdrawals are made, how long would it take, to the nearest tenth of a year, for the value of the account to reach $79,200?
2. x^2 + y^2 = 153 and y = -4x. If (x, y) is a solution to the system of equations above, what is the value of y^3?

Answer 1

Answer:

(1) 13.0 years

(2) $y^{3} =$ {-1728, 1728}

Step-by-step explanation:

(1)

Compound annually:

$Pe^{rt}$ = A

(47000)$e^{(0.04)(t)}$ = 79200

$e^{(0.04)(t)}$ = $\frac{79200}{47000}$

ln($e^{(0.04)(t)}$) = ln($\frac{79200}{47000}$)

ln and e cancel out.

(0.04)(t) = ln($\frac{79200}{47000}$)

t = $\frac{ln(\frac{79200}{47000})}{0.04}$

t = 13.0 years

(2)

$x^{2} +y^{2} = 153\\y = -4x$

Substitute y with -4x.

$x^{2} + (-4x)^{2} = 153$

Solve for x.

x = {-3, 3}

Plug in x values into any equation to find y.

y = -4(-3) and y = -4(3)

y = {-12, 12}

$y^{3}$ = $-12^{3}$ = -1728

$y^{3}$ = $12^{3}$ = 1728

Answer 2

Answer:

13.3    

Step-by-step explanation:

I got it right