1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
Lera25 [3.4K]
3 years ago
12

Help will give brainliest

Mathematics
1 answer:
mr_godi [17]3 years ago
7 0
The average rate of change is -50x
You might be interested in
Answer please and explain
Andrej [43]
36 pairs
because 2 spots = 1 pair
so 2x=72
and your answer will be 36 pairs
4 0
3 years ago
Juantia will make cookies and the recipce calls for 2/3 cup of flour. She wants to make this recipe four times a month How mcuh
vesna_86 [32]

Answer:

8/3 or 6 2/3 cup

Step-by-step explanation:

2/3 times 4/1 = 8/3 or 6 2/3

3 0
4 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
Fund the separate equation and angles between them
irga5000 [103]
Um, is there angles to look at, or an actual problem?
7 0
3 years ago
Does the equation 3(2x−1)+5=6(x+1) have one, none, or an infinite amount of solutions?
EleoNora [17]

Answer: No solutions

Step-by-step explanation: If you solve the problem all the way, you get 0 = 4 which is not valid so there is simply no solution

8 0
3 years ago
Read 2 more answers
Other questions:
  • Samuel order four did from an online music stores. Each dvd cost 9.99$. He has a 20% discount code and sales tax is 6.75%. What
    6·1 answer
  • PLEASE HELP ME FILL THIS CHART OUT<br> I WILL MARK BRAINIEST
    10·1 answer
  • The price of gold has a mean of $1243 per ounce. The price per ounce has a normal distribution, with a standard deviation of $21
    9·1 answer
  • Please answer correctly
    12·1 answer
  • 98464 rounded to the nearest ten thousand?
    8·1 answer
  • The weight of an object on the moon is
    12·1 answer
  • A dice is a perfect cube. The length of the side of the dice is 1 cm. The density of a dice is 2 g/cm.
    5·1 answer
  • Fine floors can install 15 square yards of carpeting in 4 hours and 30 minutes. At this rate, how long would it take to install
    10·2 answers
  • A baby girl weighed 6 pounds at birth and 8
    5·2 answers
  • SOLVE THIS PROBLEM ASAP PLS
    12·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!