8 - (-d) = 43
8 + d = 43 <==
<h3>
Answer: 0.6</h3>
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Work Shown:
sin(angle) = opposite/hypotenuse
sin(T) = VU/VT
sin(T) = 3/5
sin(T) = 0.6
P(t)=500(1+4t/(50+t^2 ))
P'(t) = 500 [(50+t^2).4 - 4t.2t]/(50+t^2)^2
by the quotient rule
500 (-4t^2 + 200)/(t^2 + 50)^2
Hence
P'(2) = 500 . (-16 + 200)/54^2 ~= 31.6
Solution:
As region bounded by y-axis, the line y=6, and the line y=1/2 is a line segment of definite length on y-axis.
We consider a line , one dimensional if it's thickness is negligible.
So, Line is two dimensional if it's thickness is not negligible becomes a quadrilateral.
So, Area (region bounded by y-axis, the line y=6, and the line y=1/2 is a line segment of definite length on y-axis)= Area of line segment between [,y=6 and y=1/2.]= 6-1/2=11/2 units if we consider thickness of line as negligible.
What you have written down is correct, x=-3