we have the equation
[tex]H(t)=180-a(108)^{-t}[/tex]
A) since H is an exponential function 180 represents the maximum posible value of H or the maximum temperature that the center of the cake can reach
b)
we have
H(t)= 22° C
when placed in the oven or t=0
[tex]22=180-a(108)^0[/tex][tex]22=180-a(1)[/tex][tex]a=180-22=158[/tex]
the value of a is 158
C)
H(t)=150°C
we already know a=158
let's solve for t
[tex]150=180-(158)(1.08)^{-t}[/tex][tex]-30=-(158)(1.08)^{-t}[/tex][tex]\frac{30}{158}=1.08^{-t}[/tex][tex]ln(\frac{30}{158})=ln(1.08^{-t)}[/tex][tex]ln(\frac{30}{158})=-t*ln(1.08)[/tex][tex]-t=\frac{ln(\frac{30}{158})}{ln(1.08)}[/tex][tex]t=-\frac{ln(\frac{30}{158})}{ln(1.08)}=21.587[/tex]
now this is the time 29 minutes before taking the cake out of the oven
so the total time is 29+21.587
then the total time the baking thin was in the oven
is 50.587 minutes
