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Y_Kistochka [10]

3 years ago

7

20 POINTS AND BRAINLIEST

1 answer:

emmainna [20.7K]3 years ago

8 0

To solve this equation we must first plug in 5 for x.

$\frac{1}{3}$ <span> (5 - 4x)
</span>↓
$\frac{1}{3}$ (5 - 4 · 5)

The next step would be to multiply 4 by 5 to get rid of the parenthesis within our main parenthesis.

$\frac{1}{3}$ (5 - 4 · 5)

$\frac{1}{3}$ (5 - 20)

The next step would be to subtract both of the numbers within the parenthesis.

$\frac{1}{3}$ (-15)

Now we must multiply.

- $\frac{1}{3}$ · 15 = - $\frac{1×15}{3}$

- $\frac{15}{3}$

Now we simplify our answer to get our final answer.

-5

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In 1859, 24 rabbits were released into the wild in Australia, where they had no natural predators. Their population grew exponen

Mashcka [7]

Answer:

a) P' = P

$P(t) = 24e^{0.693t}$ where t is step of 6 months

b) 7.7 years

c)1064.67 rabbits/year

Step-by-step explanation:

The differential equation describing the population growth is

$\frac{dP}{dt} = P$

Where t is the range of 6 months, or half of a year.

P(t) would have the form of

$P(t) = P_0e^{kt}$

where $P_0 = 24$ is the initial population

After 6 month (t = 1), the population is doubled to 48

$P(1) = 24e^k = 48$

$e^k = 2$

$k = ln(2) = 0.693$

Therefore $P(t) = 24e^{0.693t}$

where t is step of 6 months

b. We can solve for t to get how long it takes to get to a population of 1,000,000:

$24e^{0.693t} = 1000000$

$e^{0.693t} = 1000000 / 24 = 41667$

$0.693t = ln(41667) = 10.64$

$t = 10.64 / 0.693 = 15.35$

So it would take 15.35 * 0.5 = 7.7 years to reach 1000000

c. $P' = P_0ke^{kt}$

We need to resolve for k if t is in the range of 1 year. In half of a year (t = 0.5), the population is 48

$24e^{0.5k) = 48$

$0.5k = ln2 = 0.693$

$k = 1.386$

Therefore, $P' = 1.386*24e^{1.386t}$

At the mid of the 3rd year, where t = 2.5, we can calculate P'

$P' = 1.386*24e^{1.386*2.5} = 1064.67$ rabbits/year

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2 years ago

80% of what how many games is 32 games?

Talja [164]

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Aleksandr-060686 [28]

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Use long division to find each quotient. Show your computation and write your answer as a decimal.

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