What is 6 3/4 times 1 1/2
2 answers:
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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.
Answer:
43.94?
Step-by-step explanation:
I think if you use inverse cos you should get the correct number. So, to find cos, you need to know the adjacent side and the hypotenuse. Here, the reference angle would be the person holding the kite. The distance from the person to the barn is the adjacent side, which is 18. The string of the kite, I assume, would be 25.
cos(∅) =
cos(∅) = 0.72
cos⁻¹(0.72) = 43.94
I don't know if that's what you're looking for, but... If it's not correct, lmk so I can rewrite it. :)