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sesenic [268]
3 years ago
13

I need help with this question please!!!

Mathematics
1 answer:
stealth61 [152]3 years ago
5 0

Answer:

Step-by-step explanation:

Each point of function b(x) is shifted by rule

x -----> ( x - 2 )

and

y -----> (y + 3)

( - 1.5, 0) ------> (- 3.5, 3)

(- 1, - 1) -----> ( - 3, 2)

(0, - 2) ------> ( - 2, 1)

(1, - 1) -----> ( - 1, 2)

(2, 3) -----> (0, 6)

(4, 7) -----> (2, 10)

......

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Give me and example

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3 years ago
Help me pleas. And no bots. :)
andreev551 [17]

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otez555 [7]

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98

Step-by-step explanation:

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Tim has L inches of wire. He cuts the wire into pieces. The length of each piece is k inches. How many pieces of wire does Tim h
Yuki888 [10]

Answer:

L/k

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7 0
2 years ago
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
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