A is set of positive multiples of 5 less than 23
1 answer:
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We have that
<span>p(t)=-3t^2+18t-4
using a graphing tool, we can see the maximum of the graph
(see the attached figure)
A) </span><span>In what year of operation does Mr. Cash’s business show maximum profit?
</span>Mr. Cash’s business show maximum profit at year 3 (maximum in the parabole)
<span>B) What is the maximum profit?
23 (hundred of thousand of dollars) = 2.300.000 dollars
</span>c) What time will it be two late?
(This is the time when the graph crosses zero and the profits turn into losses )
5.77 years, or an estimate of about 69 months.
<span>p(t)=-3t^2+18t-4
using a graphing tool, we can see the maximum of the graph
(see the attached figure)
A) </span><span>In what year of operation does Mr. Cash’s business show maximum profit?
</span>Mr. Cash’s business show maximum profit at year 3 (maximum in the parabole)
<span>B) What is the maximum profit?
23 (hundred of thousand of dollars) = 2.300.000 dollars
</span>c) What time will it be two late?
(This is the time when the graph crosses zero and the profits turn into losses )
5.77 years, or an estimate of about 69 months.
Step-by-step explanation:
Take the first derivative
Set the derivative equal to 0.
or
For any number less than -1, the derivative function will have a Positve number thus a Positve slope for f(x).
For any number, between -1 and 1, the derivative slope will have a negative , thus a negative slope.
Since we are going to Positve to negative slope, we have a local max at x=-1
Plug in -1 for x into the original function
So the local max is 2 and occurs at x=-1,
For any number greater than 1, we have a Positve number for the derivative function we have a Positve slope.
Since we are going to decreasing to increasing, we have minimum at x=1,
Plug in 1 for x into original function
So the local min occurs at -2, at x=1