The spread of a virus is modeled by V (t) = −t 3 + t 2 + 12t, where V (t) is the number of people (in hundreds) with the virus and t is the number of weeks since the first case was observed.
(a) Sketch V (t).
(b) What is a reasonable domain of t for this problem?
(c) Find the average rate of infection from t = 0 to t = 2.
(d) Find the instantaneous rate of infection as a function of t using the limit definition of the derivative.
(e) Find V (2) and V ‘ (2). Write a sentence interpreting V (2) and V ‘ (2) in terms of the number of infected people. Make sure to include units.
(f) Sketch the tangent line to the graph you drew in a. at the point (2, V (2)). State the slope of the tangent line.
(g) Use V (2) and V ‘ (2) to estimate the value of V (2.1).
(h) Find the maximum number of people infected at the same time and when the maximum occurs. Determine the rate of infection at this time.

where is the rest of the

step-by-step explanation:

we need the options

step-by-step explanation:

answer is d-29

step-by-step explanation:

since you didn't include the picture, i can give you the description so you know how to pick the correct answer. the third quartile of a data is always shown in a box plot. in the middle of your graph should be a box. the third quartile is the value that goes with the far right edge of the box.

answer: the price of one pound cabbage = $ 0.8

the point (10,8) represents that the cost of 10 pounds cabbage is $ 8

step-by-step explanation:

since, here is a proportional relation between the number of pounds of cabbage (x) and the price in dollars (y).

therefore, we can write,

⇒ y = k x

where k is the constant of proportionality.

now, point (5,4) is in the line.

thus, it will satisfies the above equation of line.

⇒ 4 = 5 k

⇒ k = 0.8

thus, the equation of the line is,

y = 0.8 x

a) for x = 1,

y = 0.8

therefore, the cost of one pound of cabbage is 0.8 dollars.

b) in the below graph (10,8) represents that the cost of 10 pounds of cabbage is 8 dollars.