Answer:
h = - r
Step-by-step explanation:
The question requires you to make h the subject of the formula.
S = 2πrh + 2πr²
subtract 2πr² on both sides.
S - 2πr² = 2πrh - 2πr² - 2πr²
S - 2πr² = 2πrh
Dividing both sides by 2πr
(S - 2πr²)/2πr = 2πrh/ 2πr
h = - r
Answer:
The maximum velocity of the object in the time interval [0, 4] is 10 ft/s.
Step-by-step explanation:
Given : The position of an object along a vertical line is given by , where s is measured in feet and t is measured in seconds.
To find : What is the maximum velocity of the object in the time interval [0, 4]?
Solution :
The velocity is rate of change of distance w.r.t time.
Distance in terms of t is given by,
Derivate w.r.t. time,
It is a quadratic function so its maximum is at vertex of the function.
The x point of the function is given by,
Where, a=-3, b=6 and c=7
As 1 lie between interval [0,4]
Substitute t=1 in the function,
Th maximum velocity is 10 ft/s.
Therefore, the maximum velocity of the object in the time interval [0, 4] is 10 ft/s.