On January 2,2006 , Jorge Sanchez deposited \( \$ 450 \) into a savings account that pays simple interest at a rate of \( \$ 12 \) each year. The amount in the account is a function of time given by A(t) \( =450+12 t \), where t is the number of years after 2006 . Find the amount in the account in a) 2007, b) 2018 . a) The amount in the account in 2007 will be \( \$ \square \). b) The amount in the account in 2018 will be \( \$ \square \).

To find the amounts in the account for the specified years, we simply plug in the values for \( t \) into the function \( A(t) = 450 + 12t \). a) For 2007, \( t = 1 \) (since it's one year after 2006). \[ A(1) = 450 + 12(1) = 450 + 12 = 462 \] Thus, the amount in the account in 2007 will be \( \$ 462 \). b) For 2018, \( t = 12 \) (since it's twelve years after 2006). \[ A(12) = 450 + 12(12) = 450 + 144 = 594 \] Therefore, the amount in the account in 2018 will be \( \$ 594 \).