The future value that accrues when $500 is invested at 7%, compounded continuously, is s(t) = 500e0.07t where t is the number of years. (round your answers to the nearest cent.) (a) at what rate is the money in this account growing when t = 6? $ per year (b) at what rate is it growing when t = 12? $ per year

  (a) $53.27 per year

  (b) $81.07 per year

Step-by-step explanation:

The derivative of the function is ...

  S'(t) = 0.07·500e^(0.07t) = 35e^(0.07t)

(a) The derivative evaluated at t=6 is ...

  S'(6) = 35·e^0.42 ≈ 53.27

At t=6, the account is growing at the rate of $53.27 per year.

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(b) The derivative evaluated at t=12 is ...

  S'(12) = 35·e^0.84 ≈ 81.07

At t=12, the account is growing at the rate of $81.07 per year.

income. 10x+1300

expenses 3x+(1800+1500)

expenses 3x+3300

to break even:

income = expense,

10x + 1300 = 3x + 3300,

solve for x.

step-by-step explanation:

mark brainliest