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KatRina [158]
3 years ago
5

A - 2 + 3 = -2 Help me pls

Mathematics
2 answers:
Semmy [17]3 years ago
6 0

Answer:

a = -3

Step-by-step explanation:

a - 2 + 3 = -2 ( cancel equal terms)

a + 3 = 0 (Move the constant to the right)

a = -3

Vika [28.1K]3 years ago
6 0

Answer:

its 3

Step-by-step explanation:

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During optimal conditions, the rate of change of the population of a certain organism is proportional to the population at time
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Answer:

The population is of 500 after 10.22 hours.

Step-by-step explanation:

The rate of change of the population of a certain organism is proportional to the population at time t, in hours.

This means that the population can be modeled by the following differential equation:

\frac{dP}{dt} = Pr

In which r is the growth rate.

Solving by separation of variables, then integrating both sides, we have that:

\frac{dP}{P} = r dt

\int \frac{dP}{P} = \int r dt

\ln{P} = rt + K

Applying the exponential to both sides:

P(t) = Ke^{rt}

In which K is the initial population.

At time t = 0 hours, the population is 300.

This means that K = 300. So

P(t) = 300e^{rt}

At time t = 24 hours, the population is 1000.

This means that P(24) = 1000. We use this to find the growth rate. So

P(t) = 300e^{rt}

1000 = 300e^{24r}

e^{24r} = \frac{1000}{300}

e^{24r} = \frac{10}{3}

\ln{e^{24r}} = \ln{\frac{10}{3}}

24r = \ln{\frac{10}{3}}

r = \frac{\ln{\frac{10}{3}}}{24}

r = 0.05

So

P(t) = 300e^{0.05t}

At what time t is the population 500?

This is t for which P(t) = 500. So

P(t) = 300e^{0.05t}

500 = 300e^{0.05t}

e^{0.05t} = \frac{500}{300}

e^{0.05t} = \frac{5}{3}

\ln{e^{0.05t}} = \ln{\frac{5}{3}}

0.05t = \ln{\frac{5}{3}}

t = \frac{\ln{\frac{5}{3}}}{0.05}

t = 10.22

The population is of 500 after 10.22 hours.

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Answer:

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Step-by-step explanation:

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Multiply and simplify 8 times 3/10
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Answer:

2.4 Hope this helped

Step-by-step explanation:

8/1 x 3/10

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4 years ago
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Meryl spend a total of 68.82 for 2 pairs of sneakers with the same cost . The sale tax is 5.32 . Meryl also use a coupon for 3.0
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$68.82
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A company wanted to estimate the mean lifetime of its new model of lightbulbs. They use a method for testing bulbs that accelera
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