Nina made two investments: investment \text{a}a has a value of \$50$50 at the end of the first year and increases by 8\%8% per year. investment \text{b}b has a value of \$60$60 at the end of the first year and increases by \$3$3 per year. nina checks the value of her investments once a year, at the end of the year. what is the first year in which nina sees that investment \text{a}a's value exceeded investment text{b}b's value?

  year 7

Step-by-step explanation:

If we assume that investment A earns interest compounded annually, its value can be modeled by the equation ...

  A = 50·(1+0.08)^(t-1) . . . . . where t is the year number

The second investment earns $3 per year, so its value can be modeled by the equation ...

  B = 60 + 3(t -1) . . . . . . . . . where t is the year number

We are interested in finding the minimum value of t such that ...

  A > B

  50·1.08^(t-1) > 60 +3(t-1)

This is a mix of exponential and polynomial terms for which no solution method is available using the tools of Algebra. A graphing calculator shows the solution to be ...

  t > 6.552

The value at the end of year 1 is found for t=1, so the values of interest are seen after 6.55 years, in year 7.

the answer is 288..

hope this

use an equation of ratios to determine the amount of gas needed to fill the tank.

    0.2             0.8

  =    

288.75             x

                                                              231 gal

so, 0.2x = (0.8)(288.75 gal), and x = = 1155 gal.

                                                                0.2

it would take 1155 gal. to fill the tank.

the cost would be

1155 gal         $3

* = $3465   (answer)

      1           1 gal

hope this you