Differentiating an integral removes the integral.
f(x) = integral of dt/sqrt(t^3 + 2)
f'(x) = 1/sqrt(x^3 + 2)
f'(1) = 1/sqrt(1^3 + 2)
f'(1) = 1/sqrt(3) = sqrt(3)/3.
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Sounds to me as tho you are to graph 3x+5y<10, and that after doing so you are to restrict the shaded answer area created by the "constraint" inequality x≤y+1. OR x-1 ≤ y OR y≥x-1. If this is the correct assumption, then please finish the last part of y our problem statement by typing {x-y<=1}.
First graph 3x+5y = 10, using a dashed line instead of a solid line.
x-intercept will be 10/3 and y-intercept will be 2. Now, because of the < symbol, shade the coordinate plane BELOW this dashed line.
Next, graph y=x-1. y-intercept is -1 and x intercept is 1. Shade the graph area ABOVE this solid line.
The 2 lines intersect at (1.875, 0.875). To the LEFT of this point is a wedge-shaped area bounded by the 2 lines mentioned. That wedge-shaped area is the solution set for this problem.
Answer:
-8 I believe
Step-by-step explanation:
Substitute the x values back into the equation
(1, 22)
(2, 18)
(3, 14)
(6, 2)
Answer:
(5x+1)(x - 2)
Step-by-step explanation:
