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Lesechka [4]
4 years ago
8

A quantity with an initial value of 830 grows exponentially at a rate such that the quantity doubles every 2 weeks. What is the

value of the quantity after 21 day, to the nearest hundredth?
Mathematics
1 answer:
labwork [276]4 years ago
7 0

Answer:

The value of the quantity after 21 days is 2,347.59.

Step-by-step explanation:

The exponential growth function is

A=A_0(1+r)^t

A= The number of quantity after t days

A_0= initial number of quantity

r= rate of growth

t= time in days.

A quantity with an initial value of 830 grows at a rate such that the quantity doubles in 2 weeks = 14 days.

Now A= (2×830)= 1660

A_0 = 830

t = 14 days

r=?

Now plug all value in exponential growth function

1660=830(1+r)^{14}

\Rightarrow \frac{1660}{830}= (1+r)^{14}

\Rightarrow 2= (1+r)^{14}

\Rightarrow (1+r) ^{14}=2

\Rightarrow  (1+r)=\sqrt[14]{2}

\Rightarrow  r=\sqrt[14]{2}-1

Now, to find the quantity after 21 days, we plug A_0 = 830, t= 21 days in exponential function

A=830( 1+\sqrt[14]{2}-1)^{21}

\Rightarrow A=830(\sqrt[14]2)^{21}

\Rightarrow A=830(2)^\frac{21}{14}

\Rightarrow A=830(2)^\frac{3}{2}

\Rightarrow A=2,347.59

The value of the quantity after 21 days is 2,347.59.

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