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 \( \$ \square \).

**Step 1: Identify the formula** We are given the function for the amount in the account as \[ A(t) = 450 + 12t, \] where \( t \) is the number of years after 2006. **Step 2: Calculate the amount for 2007** For the year 2007, \[ t = 2007 - 2006 = 1. \] Plug \( t = 1 \) into the function: \[ A(1) = 450 + 12(1) = 450 + 12 = 462. \] Thus, the amount in the account in 2007 is \$462. **Step 3: Calculate the amount for 2018** For the year 2018, \[ t = 2018 - 2006 = 12. \] Plug \( t = 12 \) into the function: \[ A(12) = 450 + 12(12) = 450 + 144 = 594. \] Thus, the amount in the account in 2018 is \$594. a) The amount in the account in 2007 will be \$462. b) The amount in the account in 2018 will be \$594.