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2 years ago

9

How can you find the y-coordinates of the midpoint of a vertical line segment with endpoints at (0,0) and (0,-12)? Check all tha

t apply.
A. add the endpoints
B. divide 12 by 2
C. divide -12 by 2
D. multiply -12 by 2

1 answer:

Andru [333]2 years ago

6 0

A add the endpoint xA+xB/2 , Ya +Y1/2

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Jana uses 20 cards to play a memory game the cards are either animal cards or fruit cards. The ratio of animal cards to fruit ca

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Given that,

Total no of cards = 20

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Solution,

Let there are 2x animal cards and 3x fruit cards. ATQ,

2x+3x = 20

5x = 20

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It is also mentioned that, of the cards, 1/3 have bananas on them. It means,

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What is 15% of $150?'

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$22.50

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3 years ago

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A bacteria culture starts with 400 bacteria and grows at a rate proportional to its size. After 4 hours, there are 9000 bacteria

Kaylis [27]

Answer:

A) The expression for the number of bacteria is $P(t) = 400e^{0.7783t}$ .

B) After 5 hours there will be 19593 bacteria.

C) After 5.55 hours the population of bacteria will reach 30000.

Step-by-step explanation:

A) Here we have a problem with differential equations. Recall that we can interpret the rate of change of a magnitude as its derivative. So, as the rate change proportionally to the size of the population, we have

$P' = kP$

where $P$ stands for the population of bacteria.

Writing $P'$ as $\frac{dP}{dt}$ , we get

$\frac{dP}{dt} = kP$ .

Notice that this is a separable equation, so

$\frac{dP}{P} = kdt$ .

Then, integrating in both sides of the equality:

$\int\frac{dP}{P} = \int kdt$ .

We have,

$\ln P = kt+C$ .

Now, taking exponential

$P(t) = Ce^{kt}$ .

The next step is to find the value for the constant $C$ . We do this using the initial condition $P(0)=400$ . Recall that this is the initial population of bacteria. So,

$400 = P(0) = Ce^{k0}=C$ .

Hence, the expression becomes

$P(t) = 400e^{kt}$ .

Now, we find the value for $k$ . We are going to use that $P(4)=9000$ . Notice that

$9000 = 400e^{k4}$ .

Then,

$\frac{90}{4} = e^{4k}$ .

Taking logarithm

$\ln\frac{90}{4} = 4k$ , so $\frac{1}{4}\ln\frac{90}{4} = k$ .

So, $k=0.7783788273$ , and approximating to the fourth decimal place we can take $k=0.7783$ . Hence,

$P(t) = 400e^{0.7783t}$ .

B) To find the number of bacteria after 5 hours, we only need to evaluate the expression we have obtained in the previous exercise:

$P(5) =400e^{0.7783*5} = 19593.723 \approx 19593$ .

C) In this case we want to do the reverse operation: we want to find the value of t such that

$30000 = 400e^{0.7783t}$ .

This expression is equivalent to

$75 = e^{0.7783t}$ .

Now, taking logarithm we have

$\ln 75 = 0.7783t$ .

Finally,

$t = \frac{\ln 75}{0.7783} \approx 5.55$ .

So, after 5.55 hours the population of bacteria will reach 30000.

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OLEGan [10]

The answer is -1,0 1,1 3,2

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