Answer:
(i) The value of m when t = 30 is 13.2
(ii) The value of t when the mass is half of its value at t=0 is 34.7
(iii) The rate of the mass when t=50 is -0.18
Step-by-step explanation:
(i) The m value when t = 30 is:
[tex] m = 24e^{-0.02t} = 24e^{-0.02*30} = 13.2 [/tex]
Then, the value of m when t = 30 is 13.2
(ii) The value of the mass when t=0 is:
[tex] m_{0} = 24e^{-0.02t} = 24e^{-0.02*0} = 24 [/tex]
Now, the value of t is:
[tex] ln(\frac{m_{0}/2}{24}) = -0.02t [/tex]
[tex] t = -\frac{ln(\frac{24}{2*24})}{0.02} = 34.7 [/tex]
Hence, the value of t when the mass is half of its value at t=0 is 34.7
(iii) Finally, the rate at which the mass is decreasing when t=50 is:
[tex] \frac{dm}{dt} = \frac{d}{dt}(24e^{-0.02t}) = 24(e^{-0.02t})*(-0.02) = -0.48* (e^{-0.02*50}) = -0.18 [/tex]
Therefore, the rate of the mass when t=50 is -0.18.
I hope it helps you!
