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Anna11 [10]
3 years ago
12

It is predicted that garden gnome value will go up by 2% each year, forming a geometric progression. Jean-Claude has a garden gn

ome currently valued at £80,000. If the rate of inflation is correct:
a) What will Jean-Claude's gnome be worth after 1 year?


b) What is the common ratio of the geometric progression?


c) What will Jean-Claude's gnome be worth after 10 years?


d) It will take k years for the value of Jean-Claude's gnome to exceed £120,000. Find k.
Mathematics
1 answer:
yawa3891 [41]3 years ago
8 0

Answer:

Step-by-step explanation:

a ) The worth of Jean-Claude's gnome after 1 year

= 80000 x 1.02

=  £81600

b ) common ratio of geometric progression

= 81600 / 80000

= 1.02

c )  worth after 10 years

= 80000 x 1.02¹⁰

= 80000 x 1.21899

= £97519.55

d )

120000 = 80000 x 1.02^k

1.5 = 1.02^k

ln1.5 = k ln 1.02

k = ln 1.5 / ln 1.02

= .4054 / .0198

= 20.47 years .

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Step-by-step explanation:

Assuming that the differential equation is

\frac{dP}{dt} = 0.04P\left(1-\frac{P}{500}\right).

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First of all, this is an example of the logistic equation, which has the general form

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In order to make the calculation easier we are going to solve the general equation, and later substitute the values of the constants, notice that k=0.04 and K=500 and the initial condition P(0)=100.

Notice that this equation is separable, then

\frac{dP}{P(1-P/K)} = kdt.

Now, intagrating in both sides of the equation

\int\frac{dP}{P(1-P/K)} = \int kdt = kt +C.

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\frac{1}{P(1-P/K)} = \frac{1}{P} - \frac{1}{K-P}.

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We have obtained that:

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which is equivalent to

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Taking exponentials in both hands:

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Hence,

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The next step is to substitute the given values in the statement of the problem:

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We calculate the value of A using the initial condition P(0)=100, substituting t=0:

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