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2 years ago

The prime factorization of GCF (18, 30)=

Mathematics

1 answer:

WITCHER [35]2 years ago

8 0

Answer:

We found the factors and prime factorization of 18 and 30. The biggest common factor number is the GCF number. So the greatest common factor 18 and 30 is 6.

Step-by-step explanation:

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35

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Please help me The point D,E,F and G all lie on the same segment, in that order, such that ratio of DE:EF:FG is equal to 4:1:4.

kap26 [50]

Answer:

4+1+4+g=36

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Step-by-step explanation:

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If you sold 14 iphones in 3 hours, how many could you sell in 3 eight hour days? please help!

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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:

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

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2 years ago

Please help me find y and show work/steps. I don't know how to do it

trasher [3.6K]

We will check if y and 60cm are parallel.

If the lengths 28cm, 56cm and 15cm, 30cm are in proportion, then y and the segment 60cm are parallel.

$\dfrac{56}{28}=\dfrac{30}{15}\\\\2=2\qquad CORRECT!$

OK. We know: y and 60cm are parallel. Therefore we have equation:

$\dfrac{y}{28+56}=\dfrac{60}{56}\\\\\dfrac{y}{84}=\dfrac{15}{14}\qquad|\text{cross multiply}\\\\14y=15\cdot84\qquad|\text{divide both sides by 14}\\\\y=90cm$

<h3>Answer: y = 90 cm.</h3>

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2 years ago

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