Answer:
Step-by-step explanation:
If a function has a positive slope, that means the y value is increasing with respect to x. As you go down the x-axis, the y value will continuously increase.
First you want to plot the two sets of coords that they gave you, or else you wont know what the line looks like. Or, you could visually do it in your head. We're going down -4 on the x-axis and down 1 on the y-axis. Then for our second coords, we are going to 0 on the x-axis, then up 3 on the y-axis.
You could plot this for yourself, but im going to do it in my head for simplicity. We also need to find the slope of this function in order for us to find the y-intercept. The slope is change in y divided by the change in x. Subtract the initial position from the final position.
3 - 1 = 2
0 - (-4) = 4
2/4 = 1/2
The new equation is:
Everything is shaded below and to the right of the line. They said the line was solid, so that means less than or equal to,
Let the bushels of wheat is b and weight of the wheat is w.
We can say that more the bushels of wheat more will be the weight of the wheat.
Hence, the quantities vary directly.
Therefore, we have
, where k is the constant of variation.
Now, we have been given that 5 bushels of wheat weigh 136 kg. Thus, we have
Thus, the constant of variation is
Now, we have been given 3.5 bushels of wheat. Hence, we have
Therefore, 3.5 bushels of wheat weigh 95.2 kg
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)
And if we find the derivate when t=1 we got this:
And if we replace t=10 we got:
d)
And then:
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
(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:
And if we simplify we got this:
And if we simplify we got:
And if we find the derivate when t=1 we got this:
And if we replace t=10 we got:
(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:
And if we set equal to 0 we got:
And then:
