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Bingel [31]
3 years ago
9

I need help solving and checking the answer before my test tomorrow. ‍♀️im gonna fail so if you know anything please solve it an

d show how to check the answer ​

Mathematics
2 answers:
Anna11 [10]3 years ago
5 0

Step-by-step explanation:

-3+5=-10

−3x+5−5=−10−5

−3x=−15

−3x÷−3=−15÷-3

x=5

(-2 4/5)(1 4/5)

=-14/5×19/5

=-14×9/5×5

=-126/25

=-5 1/25

Fudgin [204]3 years ago
5 0

Answer:

3. x = 5

4. -5  1/25

Step-by-step explanation:

Question 3 Solving:

your equation is -3x + 5 = -10

subtract 5 from both sides to get the term with x alone: -3x = -15

divide both sides by -3 to isolate x (remember that a minus divided by minus is plus): x = 5

To check:

plug in 5 to x and see if you get -10 on both sides

-3(5) + 5 = -10

first multiply -3 and 5:

-15 + 5 = -10

then add 5 to -15:

-10 = -10

if the number are equal then you solved the equation correctly

Question 4:

multiply the whole number by the denominator, then add the numerator to the product, and that number will be your new numerator ( make sure to leave the negative sign out of this computation)

so multiply 2 and 5 to get 10, then add 4 to get 14

your new fraction is -14/5

practice doing the same thing to the other mixed fraction

you should get a new fraction of 9/5

multiply the numerators and denominators

so multiply -14 and 9, and multiply 5 and 5

you now have -126/25

divide 126 by 25 to get your whole number of the mixed fraction

you should have -5  ?/25

to find the numerator, multiply your whole number by your denominator (25 x 5 = 125), then subtract this from your numerator of the improper fraction

126-125 = 1

now you have an answer of -5  1/25

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Step-by-step explanation:

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So the the probability that first die rolled 2 is 1/6 because there appears only on 2 .

and the probability that the second dice rolled a number greater than 2 is 4/6 or 2/3 because there are 4 numbers greater than 2 out of the 6 total .

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The Hudson Valley Stadium Renegades has a capacity of maximum 4,505 people. During a the peak of its popularity, the tickets sol
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219 x 4505 = 986,595


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Name the property of real numbers illustrated by the following equation.
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Answer:

Distributive property

Step-by-step explanation:

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