For domain 2x sqrt(2+x)>0
x>0,2+x>0,x>-2 combining
we get x>2
f'(x)=[1/{2x sqrt(2+x)}][{2x/(2 sqrt(2+x))}+2 sqrt(2+x)]
See the attached picture.
Answer:
37.04 m/s
Step-by-step explanation:
we can use the equation x = x0 + v0t + 1/2at^2
a = -9.8 m/s^2, the acceleration due to gravity
x0 = 70 m, the starting point of x
v0 = 0 m/s
x = 0
this will allow us to solve for time
0 = 70 + 0 +-4.9t^2
t = 3.779
now we can use the equation v = v0 + at
v = 0 + 9.8(3.77)
v = 37.04 m/s
Answer:
<em>b</em><em> </em><em>=</em><em> </em><em>44</em><em> </em><em>feet</em><em> </em>
Step-by-step explanation:
BY Pythagorus theorem :-
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Answer:
A) -30
Step-by-step explanation:
