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Volgvan
2 years ago
5

Can anyone help me and explain as well

Mathematics
2 answers:
Citrus2011 [14]2 years ago
6 0

Answer:

3x2-3x+9

Step-by-step explanation:

(h-k)(3)

(X2+1-(x-2))(3)

(x2+1-x+2)(3)

3x2+3-3x+6

3x2-3x+9

Alchen [17]2 years ago
6 0

Hello,

h(x)=x²+1

k(x)=x-2

First of all, you must understand that:

(h+k)(x)=h(x)+k(x)

So, (h+k)(3)=h(3)+k(3)=(3²+1)+(3-2)=(9+1)+1=11

3h(2)+2k(3)=3*(2²+1)+2*(3-2)

=3*(4+1)+2*1

=15+2

=17

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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)}

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