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3 years ago

The value of x is 40 but how do I get that I need to show work ?

Mathematics

1 answer:

Butoxors [25]3 years ago

6 0

Answer:

120 ÷ 3 = 40

Step-by-step explanation:

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Plz give answer and explanation <br> i will give 24 points

Paladinen [302]

Answer:

please refer to the above attachment.

8 0

3 years ago

Read 2 more answers

2. David determines that 3 sqrt 2 cis(7pi/4) = -3+3i. When he considers the graphical representation of these numbers, he knows

fiasKO [112]

The correct rectangular equivalence of 3sqrt(2)·cis(7pi/4 ) is:

3sqrt(2)·cos( 7pi/4 ) + i·sqrt(2)·sin( 7pi/4 ) = 3 - 3i.

<h3>Where did David go wrong?</h3>

David mistakenly interchanged the Sin function and the Cos function when he was calculating the problem.

Hence the correct rectangular equivalence is:

3sqrt(2)·cos( 7pi/4 ) + i·sqrt(2)·sin( 7pi/4 ) = 3 - 3i.

<h3>What is rectangular equivalence?</h3>

An equation is rectangular in form when it is comprised of Variables like X and Y and can be represented on a Cartesian Plane.

Learn more about rectangular equivalence at:
brainly.com/question/27813225
#SPJ1

5 0

2 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

2 years ago

I NEED HELP PLEASE!!!!!!!!!!!!!!!!!!!!!!!!!

diamong [38]

Geez that is real complicated sorry I’m not sure just go with your gut

6 0

3 years ago

A circle is centered at the point (-3,2) and passes through the point (1,5). What is the radius of the circle?

Akimi4 [234]

Answer:

Step-by-step explanation:

The radius is a distance on a circle from the center to a point on the circle.

We have both of these points describe here in this definition. Using the distance formula will give us the radius.

$\sqrt{(x \text{ difference })^2+(y \text{ difference})^2$

The difference between 1 and -3 is 1-(-3)=4.

The difference between 5 and 2 is 5-2=3.

$\sqrt{4^2+3^2}$

$\sqrt{16+9}$

$\sqrt{25}$

$5$

8 0

3 years ago