m(t)=1200*((100-12)/100)t
Just as an explanation for where the formula comes from above (you don't have to do all this working, it's just more for
why we have the general form for these functions).
Let's let the initial mass be
, then at time
,
=0)
.
Let's let
be the proportion of the mass that is left after time
, that is if the mass is reduced by
, then we will have
of the original mass left (as done in the post above). That is proportion wise we will have
of the mass left.
e.g. After 1 second we have
=m_{0}\times a^{1})
, which is effectively finding the fraction of the original mass that is left.
If we were to do the same thing again, that is find the fraction that the mass decays to after another second, we would again multiply by the proportion
.
=m{0}\times m(1)=m_{0}\times a^{1}\times a=m_{0}\times a^{2})
And again for the third second.
=m_{0}\times m(2)=m_{0}\times a^{2}\timesa=m_{0}\times a^{3})
.
Now we can just double check that our initial condition holds,
,
=m_{0}\times a^{0}=m_{0})
So we can see the pattern that forms the general form of the equation,
=m_{0}\times a^{t})
.
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