5:20 and 15:60 equivalent
2 answers:
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so hmmm let's get the area of the whole hexagon, and then get the area of the circle inside it, then <u>subtract the area of the circle from that of the hexagon's</u>, what's leftover is what we didn't subtract, namely the shaded part.
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.