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# Algebra

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can someone help me on these problems? Thank you

Guest Mar 16, 2017
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#1
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I think I can do the first part:

The problem says the max height is 443 feet.

The second height is 443 feet - the radius.

443 feet - 394/2 feet

443 feet - 197 feet = 246 feet

The third height is 443 feet - the diameter.

443 feet - 394 feet = 49 feet

The very bottom height is 0 feet.

hectictar  Mar 16, 2017
#2
+77209
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Note.....the max height - the diameter = starting height [ low point]....so

443 - 394   =   49 ft

1. When

t = 0, h  = 49  [starting point ]

t = 7.5, h =  246  [first quarter point]

t = 15, h = 443    [second quarter point ]

t = 22.5, h = 246  [ third quarter point ]

t = 30, h = 0  [  back to starting point ]

There are probably several trig functons that would work, but here's one possibility:

2. h(t)  = 246 - 197cos[ (pi/15)*t ]    where t is in minutes

3. The vertical shift is 246

4. The amplitude is  l -197 l = 197

5. There is no horizontal shift in this function

6. The period is 2pi / [pi/15]  = 2 * 15  = 30 minutes

The frequency is the reciprocal of the period = 1/30  [it completes one revolution every 30 minutes]

7. See  (2)

CPhill  Mar 16, 2017

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