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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The inequality is

write 7 as 7b/b to have all the expressions in common denominator:






here b=1 is a root and b=0 is not in the domain of the expression, but it still has an effect in the sign of the expression.
the sign table of

is :
+++++++[0] --------[1] +++++
this means that for values of b to the left of 0 and to the right of 1, the expression is positive, and for values of b in (0, 1), the expression is negative.
that is

for b∈(0, 1)
Answer: (0, 1)
Answer:5.088
.5÷786=6.36
6.36×0.8=5.088
I hope this is good enough:
Answer:
The point that the graph does not pass through is (3,1)
Step-by-step explanation:
we have
(0, 5), (2, 2), (3, 1), and (4, -1)
Draw the points, and join them to plot the linear equation and determine which point does not belong to the line.
using a graphing tool
see the attached figure
therefore
The point that the graph does not pass through is (3,1)