Answer:
Is there supposed to be a photo? Or attachment?
Step-by-step explanation:
For this case what we must do is find a quadratic function that is already factored.
This is because in the factored quadratic equations, it is easier to observe the zeros of the function.
In this case, the zeros of the function represent the time at which the company did not make any profit.
We have the following equation:
p (t) = 40 (t - 3) (t + 2) (t - 5) (t + 3)
We observed that there was no gain in:
t = 3
t = 5
The other roots are discarded because they are negative
Answer:
a.p (t) = 40 (t - 3) (t + 2) (t - 5) (t + 3)
Answer:
Horizontal asymptote of the graph of the function f(x) = (8x^3+2)/(2x^3+x) is at y=4
Step-by-step explanation:
I attached the graph of the function.
Graphically, it can be seen that the horizontal asymptote of the graph of the function is at y=4. There is also a <em>vertical </em>asymptote at x=0
When denominator's degree (3) is the same as the nominator's degree (3) then the horizontal asymptote is at (numerator's leading coefficient (8) divided by denominator's lading coefficient (2)) 
Assuming independence,
prob=P(late,early)+P(early,late)=(1/10)(2/5)+(2/5)(1/10)