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Bella plans to put $200 into a savings account. She can place her money into an account represented by p(x) = 3x + 200, or into another account represented by n(x) = 200(1.04)x. Which account has the highest value in 4 years? Which account has the highest in 15 years?

Respuesta :

p(x) = 3x + 200
x=4 ⇒ p(4) = 3(4)+200=12+200=212
x=15 ⇒ p(15) = 3(15)+20=45+200=245

n(x) = 200(1.04)^xx
=4 ⇒ n(4) = 200 (1.04)^4 = 233.97
x=15⇒n(15)=200(1.04)^15 = 360.19

Then n(x) has the highest value in 4 years; n(x) has the highest value in 15 years  

Answer:

n(x) has the highest value in 4 years

and n(x) has the highest value in 15 years  

Step-by-step explanation:

p(x) = 3x + 200

Putting x=4

⇒ p(4) = 3×4+200

           =12+200

            =212

Putting x=15

⇒ p(15) = 3×15+200

             =45+200

             =245

n(x) = [tex]200\times (1.04)^x[/tex]

Putting x=4

⇒ n(4) =  [tex]200\times (1.04)^4[/tex]

         = 233.97

Putting x=15

⇒n(15)= [tex]200\times (1.04)^{15}[/tex]

          = 360.19

On comparing the values of n(x) and p(x) at x=4 and x=15,we get

n(x) has the highest value in 4 years

and n(x) has the highest value in 15 years  

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