Answer:
(1) 13.0 years
(2) [tex]y^{3} =[/tex] {-1728, 1728}
Step-by-step explanation:
(1)
Compound annually:
[tex]Pe^{rt}[/tex] = A
(47000)[tex]e^{(0.04)(t)}[/tex] = 79200
[tex]e^{(0.04)(t)}[/tex] = [tex]\frac{79200}{47000}[/tex]
ln([tex]e^{(0.04)(t)}[/tex]) = ln([tex]\frac{79200}{47000}[/tex])
ln and e cancel out.
(0.04)(t) = ln([tex]\frac{79200}{47000}[/tex])
t = [tex]\frac{ln(\frac{79200}{47000})}{0.04}[/tex]
t = 13.0 years
(2)
[tex]x^{2} +y^{2} = 153\\y = -4x[/tex]
Substitute y with -4x.
[tex]x^{2} + (-4x)^{2} = 153[/tex]
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}
[tex]y^{3}[/tex] = [tex]-12^{3}[/tex] = -1728
[tex]y^{3}[/tex] = [tex]12^{3}[/tex] = 1728
