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.

step-by-step explanation:

in the absence of parentheses, i will have to make some assumptions:   i interpret your (3x+5/y^6)(4y^3/5)​ to mean:

3x + 5     4y^(3/5)       (3x + 5)(4)

* =

    y^6             1               y^(6-3/5)

if this is not what you expected, repost your question using parentheses to indicate whether you meant

4y^3

    or   4y^(3/5)

    5

step-by-step explanation:

this is not possible to solve with the info given.   you need at least an one angle somewhere.