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Kruka
17 days ago
10

The growth of a species of fish in a lake can be modeled by the function ƒ(x) = 500(1.10)x, where x is time in months since Janu

ary 2019. What does the 500 represent?
Mathematics
1 answer:
Leona [9.2K]17 days ago
6 0

Answer:

f(x) = 500(1.1)^x

We need to determine what the number 500 signifies in this equation. Looking at the general form of an exponential function, we have:

y= ab^x

Where 'a' denotes the constant or initial amount, 'b' is the base, and 'x' is the independent variable (time)

In this specific situation, we have that:

a = 500, b =1.1

Here, 500 indicates the constant or the initial value for the function

Step-by-step explanation:

We have the following function presented:

f(x) = 500(1.1)^x

We need to ascertain what the value 500 represents in the equation. The standard arrangement for an exponential function is:

y= ab^x

Where 'a' stands for the constant or initial value, 'b' as the base, and 'x' signifies the independent variable (time)

In this particular example, we know that:

a = 500, b =1.1

So, 500 represents the constant or starting value for the function

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Solution :

Let t represent the quantity of trucks sold and c represent the cars sold.

The statement indicates that the quantity of cars sold exceeded three times the trucks sold by 39,000.

Thus,

c=3t+39000\\\\t=\dfrac{c-39000}{3}

Plugging the value of c = 216000 into the equation above yields :

t=\dfrac{c-39000}{3}\\\\t=\dfrac{216000-39000}{3}\\\\t=59000

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Answer:

1.21 g/day

Step-by-step explanation:

We start with the fact that

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Where t=Time(in days)

Next, we differentiate with respect to t

P'(t)=5(\frac{1}{1+\frac{t^2}{4}}\times \frac{1}{2})

Using the formula \frac{d(tan^{-1}(x)}{dx}=\frac{1}{1+x^2}

P'(t)=\frac{5}{2}(\frac{4}{4+t^2})

P'(t)=\frac{10}{4+t^2}

We know that P(t)=6

Now, substitute this value

6=2+5tan^{-1}(\frac{t}{2})

5tan^{-1}(\frac{t}{2})=6-2=4

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\frac{t}{2}=tan(\frac{4}{5})

t=2tan(\frac{4}{5})

Insert the given value of t

P'(2tan\frac{4}{5})=\frac{10}{4+4tan^2(\frac{4}{5})}

P'(2tan\frac{4}{5})=\frac{10}{4}\times \frac{1}{1+tan^2(\frac{4}{5})}

We understand that 1+tan^2\theta=sec^2\theta

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P'(2tan(\frac{4}{5})=\frac{5}{2}\times \frac{1}{sec^2(\frac{4}{5})}

P'(2tan\frac{4}{5})=\frac{5}{2}\times cos^2(\frac{4}{5})

By employing cos^2x=\frac{1}{sec^2x}

P'(2tan\frac{4}{5})=\frac{5}{2}\times (0.696)^2=1.21g/day

Consequently, the instantaneous rate at which the mass of the colony changes is=1.21g/day

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