The price of a certain computer stock t days after it is issued for sale is p(t)=100+20t−6t^2 dollars. the price of the stock initially rises, but eventually begins to fall. during what period of time does the stock price rise?
0 < t <
if you owned the stock, after how many days would you sell it? days

0 < t <

After 1.67 days the stocks would be sold out.

Step-by-step explanation:

The price of a certain computer stock after t days is modeled by

p(t) = 100 + 20t - 6t²

Now we will take the derivative of the given function and equate it to zero to find the critical points,

p'(t) = 20 - 12t = 0

t =

t = days

Therefore, there are two intervals in which the given function is defined

(0, ) and (, ∞)

For the interval (0, ),

p'(1) = 20 - 12(1) = 20

For the interval (, ∞),

p'(2) = 20 - 12(2) = -4

Positive value of p'(t) in the interval (0, ) indicates that the function is increasing.

0 < t <

Since at the point t = 1.67 days curve is showing the maximum, so the stocks should be sold after 1.67 days.

687.09 dollars       happy now

step-by-step explanation: 6870

1857 times .63 equals 1169.91 subtract that from 1857 you get 687.09

step-by-step explanation:

hans did not square the 40-unit side; he merely multiplied it by 2.   incorrect.