## College Algebra 7th Edition

(a) $m(0)=13$ (b) $m(45)=6.62$ kg
We are given: $m(t)=13e^{-0.015t}$ (a) We plug in $t=0$: $m(0)=13e^{-0.015*0}=13*1=13$ (b) We plug in $t=45$: $m(45)=13e^{-0.015*45}=13e^{-0.675}=6.62$ kg