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kakasveta [241]
4 years ago
13

Express 356 miles in 8 hours as a unit rate

Mathematics
2 answers:
natulia [17]4 years ago
3 0
356/8 = 44.5

44.5 miles/hour
44.5 miles per hour
Lelechka [254]4 years ago
3 0
The unit is miles per hour
so you would divide 356 by 8 
answer is 44.5 mph which can also be written as 44.5 mi/hr
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Mae-Ying wanted to buy packages of crackers and cheese. Each pack is 35 cents. She has 5 quarters, 3 dimes, and 2 nickels. How m
spin [16.1K]

Answer:

  4

Step-by-step explanation:

She can use a dime and a quarter to buy one package. With 3 dimes and more than 3 quarters, she can buy 3 packages this way.

She can use 2 nickels and a quarter to buy one package. With 2 nickels and the 2 remaining quarters, she can buy 1 package this way.

Mae-Ying can buy 4 packages, and will have 1 quarter left over.

_____

You can add up her money and divide by 35¢ to find the number of packages. Doing that, you get ...

  5·0.25 +3·0.10 +2·0.05 = 1.25 +0.30 +0.10 = 1.65

Then 1.65/0.35 ≈ 4.71, so Mae-Ying can buy 4 packages with the money she has.

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3 years ago
Describe the transformation in this image
bearhunter [10]

Answer:

Rotation

Step-by-step explanation:

4 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
(5x+37)"<br> S<br> W<br> (3x-9)<br> Р<br> T<br> What is the value of x?
svp [43]

Step-by-step explanation:

well, the arcs WS and ST together are the full half-circle and therefore 180°.

that means

5x + 37 + 3x - 9 = 180

8x + 28 = 180

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x = 19

5 0
2 years ago
Dominic owns a small business selling used books. He knows that in the last week
Sav [38]

Answer:

About 9% or 0.09

Step-by-step explanation:

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