The picture is not there.
Are you given the initial population of the city at ?
The population of a city is P(t) at any one time. The rate of decline in population is proportional to the population P(t), that is, dP(t)/dt = -KP(t).
What will the percentage rate of decline in population be after 10 years?
(k = ln(0.9)/-4
Last edited by jadg; 26 May 2018 at 6:49 PM.
The picture is not there.
Are you given the initial population of the city at ?
HSC 2018 - [English Adv.] • [Maths Ext. 1] • [Maths Ext. 2] • [Chemistry] • [Software Design and Development]
_{1}∫^{(3√3)} t^{2} dt cos(3π/9) = log(^{3}√e) | Integral t^{2} dt, From 1 to the cube root of 3. Times the cosine, of three pi over nine, Equals log of the cube root of e.
Sorry it wont upload
that is the whole first bit of the question, it then asks
a) Show that P(t)=P(t(subscript 0))e^-kt is a solution of the differential equation dP(t)/dt = -KP(t)
and
b) What percentage decline will there be after 10 years, given a 10% decline in 4 years? (answer = 23%)
and then c) as above, no initial value
Assuming you have done a) and b), you would have
Then substitute and you will get the population after 10 years as a percentage of the original population .
Last edited by fan96; 27 May 2018 at 2:46 PM.
HSC 2018 - [English Adv.] • [Maths Ext. 1] • [Maths Ext. 2] • [Chemistry] • [Software Design and Development]
_{1}∫^{(3√3)} t^{2} dt cos(3π/9) = log(^{3}√e) | Integral t^{2} dt, From 1 to the cube root of 3. Times the cosine, of three pi over nine, Equals log of the cube root of e.
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