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3 years ago

9

When evaluating (6x+9)^2 - 5 for x = 1, the given value of 1 is substituted for what part of the expression?

2 answers:

Veronika [31]3 years ago

7 0

The answer is C, I took the test.

Naya [18.7K]3 years ago

5 0

The correct answer would be C. The variable

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Biologists stocked a lake with 80 fish and estimated the carrying capacity (the maximal population for the fish of that species

kotegsom [21]

Answer:

$P(t) = \frac{160000e^{1.36t}}{2000 + 80(e^{1.36t} - 1)}$

Step-by-step explanation:

The logistic equation is the following one:

$P(t) = \frac{KP(0)e^{rt}}{K + P(0)(e^{rt} - 1)}$

In which P(t) is the size of the population after t years, K is the carrying capacity of the population, r is the decimal growth rate of the population and P(0) is the initial population of the lake.

In this problem, we have that:

Biologists stocked a lake with 80 fish and estimated the carrying capacity (the maximal population for the fish of that species in that lake) to be 2,000. This means that $P(0) = 80, K = 2000$ .

The number of fish tripled in the first year. This means that $P(1) = 3P(0) = 3(80) = 240$ .

Using the equation for P(1), that is, P(t) when $t = 1$ , we find the value of r.

$P(t) = \frac{KP(0)e^{rt}}{K + P(0)(e^{rt} - 1)}$

$240 = \frac{2000*80e^{r}}{2000 + 80(e^{r} - 1)}$

$280*(2000 + 80(e^{r} - 1)) = 160000e^{r}$

$280*(2000 + 80e^{r} - 80) = 160000e^{r}$

$280*(1920 + 80e^{r}) = 160000e^{r}$

$537600 + 22400e^{r} = 160000e^{r}$

$137600e^{r} = 537600$

$e^{r} = \frac{537600}{137600}$

$e^{r} = 3.91$

Applying ln to both sides.

$\ln{e^{r}} = \ln{3.91}$

$r = 1.36$

This means that the expression for the size of the population after t years is:

$P(t) = \frac{160000e^{1.36t}}{2000 + 80(e^{1.36t} - 1)}$

4 0

3 years ago

I need the answer pls reply

just olya [345]

Answer:

2 is the answer I did this once in class

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2 years ago