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muminat
3 years ago
8

An object is launched upward at 64 ft/sec from a platform that is 80 feet high. What is the objects maximum height if the equati

on of height (h) in terms of time (t) of the object is given by h(t) = -16t2 + 64t + 80?
Mathematics
2 answers:
miss Akunina [59]3 years ago
8 0
Ht + 16t2<span> - 64t - 80 = 0 
</span><span>ht - (((0 - 24t2) + 64t) + 80) = 0
</span><span>3.1 </span>    Solve  <span> <span>ht+16t2-64t-80</span> </span><span> = 0 
</span>theres no solution found honey 
Alenkinab [10]3 years ago
3 0
1) we calculate the first derivative:
h´(x)=-32t+64

2) we equalized to "0" the first derivative, and find out the value of "t".
-32t+64=0
-32t=-64
t=-64/-32
t=2

3) we calculate the second derivative:

h´´=-32<0  ⇒ then , we have a maximum at t=(2)

4) we calculate the height at t=2

h(2)=-16(2)²+64(2)+80=-32+128+80=176.

Answer: the maximum height is 176 m. 
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3 years ago
At the Olympic Games, a runner won the 26.2 mile marathon race in 2 hr 4 min and 1 second. What was his average speed in mph and
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The average speed of the runner is 12.7 mph and 20.4 km/h

Given that the runner ran 26.2 mile in 2hr and 4 minutes, we start of by converting the time from  hours and minutes into minutes and finally hours, since hours is what we need. So, we have

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This means that the runner finished the race in 2.06 hours.

If we are to find the average speed in mile per hour, we have

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for more, check: brainly.com/question/1989219

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3 years ago
SV is a midsegment of △RTU.<br> If TU=y+38 and SV=y–9, what is the value of y?
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3 years ago
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