A person invested $440 in an account growing at a rate allowing the money to double every 8 years. How long, to the nearest tenth of a year would it take for the value of the account to reach $2,860?​

9514 1404 393

  21.6 years

Step-by-step explanation:

The first statement tells you the account balance can be modeled by ...

  A = 440(2)^(t/8) . . . . . where t is in years

We want to find t when A = 2860

  2860 = 440(2^(t/8))

  6.5 = 2^(t/8) . . . . . . . . . . divide by 440

  log(6.5) = (t/8)log(2) . . . . take logarithms

  t = 8·log(6.5)/log(2) ≈ 21.604 . . . . . . divide by the coefficient of t; evaluate

It would take about 21.6 years for the value to reach $2,860.

a. -0.3 > -0.33

a reflection across the x-axis , followed by the by a reflection across the y-axis , and then a scale dilation of 0.5 .