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3 years ago

ASAP PLEASE HELP IM FAILING

Mathematics

2 answers:

aniked [119]3 years ago

5 0

The last one is the answer

good luck

Nitella [24]3 years ago

3 0

The answer is choice D

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Can someone please tell me why we write a number as 'difference' in math?

uranmaximum [27]

Answer:

We need to do difference so we can have how many numbers differ between the 2 numbers

Step-by-step explanation:

For example: 5-2=? We want to know what is the difference between the number which is 3 hope I explained it the best I can!!

8 0

3 years ago

4. Determine the following difference. (2x² + 3x - 10) - (x² - x + 111) please explain

SIZIF [17.4K]

Hi there! The answer is A.

(2x² + 3x - 10) - (x² - x + 111)
Let's solve this problem step by step!

First work out the parenthesis. Because we multiply by negative 1, we mustn't forget to change the symbols in front of the factors.
2x² + 3x - 10 - x² + x - 111

Now rearrange our expression, in order to collect the terms later.
2x² - x² + x + 3x - 10 - 111

And finally collect the terms.
x² + 4x -121

The answer is A.
~ Hope this helps you!

8 0

3 years ago

Read 2 more answers

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

Plz helppp I’m dying

Lana71 [14]

Answer:

if you are dying you should call EMS

Step-by-step explanation:

7 0

3 years ago

A 10 ft pole has a support rope that extends from the top of the pole to the ground. The rope and the ground form a 30 degree an

mr Goodwill [35]

Answer:

The length of rope is 20.0 ft . Hence, <u>option (1) </u> is correct.

Step-by-step explanation:

In the figure below AB represents pole having height 10 ft and AC represents the rope that is from the top of pole to the ground. BC represent the ground distance from base of tower to the rope.

The rope and the ground form a 30 degree angle that is the angle between BC and AC is 30°.

In right angled triangle ABC with right angle at B.

Since we have to find the length of rope that is the value of side AC.

Using trigonometric ratios

$\sin C=\frac{\text{perpendicular}}{\text{hypotenuse}}$