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3 years ago

Find the area of the shape.

Mathematics

2 answers:

yanalaym [24]3 years ago

6 0

Answer:

The area is 50.24

Step-by-step explanation:

Here is the equation to find the area of a circle (π=3.14):

A=πr²

A=3.14·<u>4²</u>

A=<u>3.14·16</u>

A=50.24

nikklg [1K]3 years ago

3 0

Your question has been heard loud and clear.

Alright , so basically , the area of a circle formula = pi * R squared

R = radius = 4 units

Pi = 3.14

So , Area = 3.14 into 4 = 12.56 units

So Area of the circle = 12.56 units

Thank you

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Triangle RST is translated 2 units left and then reflected over the y-axis.

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Answer:

A and B

Step-by-step explanation:

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3 years ago

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In Buffalo, New York the temperature was -14 F in the morning. If the temperature dropped 7 F, what is the temperature now?

Nata [24]

It would be -14-7 = -21 F

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3 years ago

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A sample of 4 different calculators is randomly selected from a group containing 12 that are defective and 33 that have no defec

Aleks04 [339]

The probability that at least one of the calculators is defective is : 0.725

<h3>What is probability?</h3>

Probability is the likelihood of an event happening or not.

if there is certainty that the event would happen probability is 1.

Analysis:

probability = required outcome/possible outcome

for 33 good calculators and 12 defective calculators, total calculators = 45

4 calculators were chosen, possible outcome = 45C4 = 148995

Required outcome = 33C3 . 12C1 + 33C2 . 12C2 + 33C1 . 12C3 + 33C0 . 12C4

= 65472 + 34848 + 7260 + 495 = 108075

Probability = 108075/148995 = 0.725

Learn more about Probability: brainly.com/question/25870256

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7 0

2 years ago

Yuppp , i need help

TEA [102]

Answer:

Correct answer B

Step-by-step explanation:

Choose one point of the original rectangle In this case I chose E. I translated it 6 to the left and three down and I ended up at point M. Rotating the rectangle, the original rectangle will match up with the second rectangle.

7 0

3 years ago

Population Growth A lake is stocked with 500 fish, and their population increases according to the logistic curve where t is mea

Alexus [3.1K]

Answer:

a) Figure attached

b) For this case we just need to see what is the value of the function when x tnd to infinity. As we can see in our original function if x goes to infinity out function tend to 1000 and thats our limiting size.

c) $p'(t) =\frac{19000 e^{-\frac{t}{5}}}{5 (1+19e^{-\frac{t}{5}})^2}$

And if we find the derivate when t=1 we got this:

$p'(t=1) =\frac{38000 e^{-\frac{1}{5}}}{(1+19e^{-\frac{1}{5}})^2}=113.506 \approx 114$

And if we replace t=10 we got:

$p'(t=10) =\frac{38000 e^{-\frac{10}{5}}}{(1+19e^{-\frac{10}{5}})^2}=403.204 \approx 404$

d) $0 = \frac{7600 e^{-\frac{t}{5}} (19e^{-\frac{t}{5}} -1)}{(1+19e^{-\frac{t}{5}})^3}$

And then:

$0 = 7600 e^{-\frac{t}{5}} (19e^{-\frac{t}{5}} -1)$

$0 =19e^{-\frac{t}{5}} -1$

$ln(\frac{1}{19}) = -\frac{t}{5}$

$t = -5 ln (\frac{1}{19}) =14.722$

Step-by-step explanation:

Assuming this complete problem: "A lake is stocked with 500 fish, and the population increases according to the logistic curve p(t) = 10000 / 1 + 19e^-t/5 where t is measured in months. (a) Use a graphing utility to graph the function. (b) What is the limiting size of the fish population? (c) At what rates is the fish population changing at the end of 1 month and at the end of 10 months? (d) After how many months is the population increasing most rapidly?"

Solution to the problem

We have the following function

$P(t)=\frac{10000}{1 +19e^{-\frac{t}{5}}}$

(a) Use a graphing utility to graph the function.

If we use desmos we got the figure attached.

(b) What is the limiting size of the fish population?

For this case we just need to see what is the value of the function when x tnd to infinity. As we can see in our original function if x goes to infinity out function tend to 1000 and thats our limiting size.

(c) At what rates is the fish population changing at the end of 1 month and at the end of 10 months?

For this case we need to calculate the derivate of the function. And we need to use the derivate of a quotient and we got this:

$p'(t) = \frac{0 - 10000 *(-\frac{19}{5}) e^{-\frac{t}{5}}}{(1+e^{-\frac{t}{5}})^2}$

And if we simplify we got this:

$p'(t) =\frac{19000 e^{-\frac{t}{5}}}{5 (1+19e^{-\frac{t}{5}})^2}$

And if we simplify we got:

$p'(t) =\frac{38000 e^{-\frac{t}{5}}}{(1+19e^{-\frac{t}{5}})^2}$

And if we find the derivate when t=1 we got this:

$p'(t=1) =\frac{38000 e^{-\frac{1}{5}}}{(1+19e^{-\frac{1}{5}})^2}=113.506 \approx 114$

And if we replace t=10 we got:

$p'(t=10) =\frac{38000 e^{-\frac{10}{5}}}{(1+19e^{-\frac{10}{5}})^2}=403.204 \approx 404$

(d) After how many months is the population increasing most rapidly?

For this case we need to find the second derivate, set equal to 0 and then solve for t. The second derivate is given by:

$p''(t) = \frac{7600 e^{-\frac{t}{5}} (19e^{-\frac{t}{5}} -1)}{(1+19e^{-\frac{t}{5}})^3}$

And if we set equal to 0 we got:

$0 = \frac{7600 e^{-\frac{t}{5}} (19e^{-\frac{t}{5}} -1)}{(1+19e^{-\frac{t}{5}})^3}$

And then:

$0 = 7600 e^{-\frac{t}{5}} (19e^{-\frac{t}{5}} -1)$

$0 =19e^{-\frac{t}{5}} -1$

$ln(\frac{1}{19}) = -\frac{t}{5}$

$t = -5 ln (\frac{1}{19}) =14.722$

7 0

3 years ago