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AlladinOne [14]
3 years ago
10

The profits of mr cash’s company is represented by the equation p(t)=-3t^2+18t-4, where p(t) is the amount of profit in hundreds

of thousands of dollars and t is the number of years of operation. he realizes his company is on the down turn and wishes to sell before he ends up in debt. In what year of operation does Mr. Cash’s business show maximum profit? What is the maximum profit? What time will it be too late to sell business?

Mathematics
1 answer:
eduard3 years ago
4 0
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.

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Answer:

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Step-by-step explanation:

Please see the attachment

From the figure, it is evident that

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Also, there are can be other values depending on the quadrants

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Answer:

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Then

((x+12)-4)+(x+10)=36\\ \\(x+12-4)+(x+10)=36\\ \\(x+8)+(x+10)=36\\ \\x+8+x+10=36\\ \\2x+18=36\\ \\2x+18-18=36-18\\ \\2x=18\\ \\x=9

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MArishka [77]
3b=150 should be the answer.
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ValentinkaMS [17]

Answer: D

Explanation:

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Let's use the y intercept since that would be the easiest. Since x=0, the terms with x cancel, and you will get a leading result of -5

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2 years ago
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I hope this helps you



Area=length.width



343=l.w


7.7.7=l.w


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