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IRISSAK [1]
3 years ago
12

Bill went to the museum at 11:30 A.M . He stayed for 3 1/2 hours.When did he leave ?

Mathematics
1 answer:
tatuchka [14]3 years ago
5 0
To start, note that an hour is 60 minutes long. A 1/2 hour, or half hour, is then 60/2=30 minutes. Therefore, when we have 11 hours and 30 minutes, we have 11 and a half hours. Adding 3 and a half to that, we get 11.5+3.5=15 (a half can also be expressed as .5, although it's not typically done that way when expressing time - it just might be easier to visualize it this way). Therefore, we are 15 hours into the day. However, we can't just stop there - we have to account for AM and PM. Therefore, we subtract 12 hours from 15. If the number is positive, we are in PM - otherwise, we're in AM. Therefore, as 15-12=3, the time is in PM. The remaining number is the time, so Bill leaves at 3 PM. If we are left with a decimal (e.g. 3.25), we would keep the 3 and multiply the 0.25 (the decimal) by 60 to figure out how many minutes we have, so 3.25 would turn into 3+0.25*60=3:15.

Feel free to ask further questions!
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victus00 [196]

Answer:

a. The critcal points are at

x=0,-5,3

b. Then, x = -5   is a maximum and x=3 is a minimum

c. The absolute minimum is at   x = 3  and the absolute maximum is at  x = -5.

Step-by-step explanation:

(a)

Remember that you need to find the points where

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Therefore you have to solve this equation.

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From that equation you can factor out    20x^2  and you would get

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And from that you would have   20x^2 = 0  , so x = 0.

And you would also have  x^2 +2x-15 = 0.

You can factor that equation as    x^2 +2x -15 = (x+5)(x-3) = 0

Therefore   x=-5 ,   x=3.

So the critcal points are at

x=0,-5,3

b.  

Remember that a function has a maximum at a critical point if the second derivative at that point is negative. Since

f''(x) = 80x^3 + 120x^2 -600x\\f''(-5) = 80(-5)^3 + 120(-5)^2 -600(-5) = -4000 < 0\\\\f''(3) = 80(3)^3 + 120(3)^2 -600(3) = 1440 > 0 \\

Then, x = -5   is a maximum and x=3 is a minimum

c.

The absolute minimum is at   x = 3  and the absolute maximum is at  x = -5.

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