Which of the following is a singleton set?
2 answers:
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The picture in the attached figure
let
AB=x
we know that
perimeter of the figure=10*x
perimeter=42.5 cm
so
42.5=10*x
x=42.5/10
x=4.25 cm
area of the figure=area of rectangle +area of square
area of rectangle=4.25*(4.25*3)----> 54.1875 cm²
area of square=4.25²----> 18.0625 cm²
area of the figure=54.1875+18.0625-----> 72.25 cm²
the answer is
72.25 cm²
let
AB=x
we know that
perimeter of the figure=10*x
perimeter=42.5 cm
so
42.5=10*x
x=42.5/10
x=4.25 cm
area of the figure=area of rectangle +area of square
area of rectangle=4.25*(4.25*3)----> 54.1875 cm²
area of square=4.25²----> 18.0625 cm²
area of the figure=54.1875+18.0625-----> 72.25 cm²
the answer is
72.25 cm²
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Answer:
- relative minimum -6√3 at x = -√3
- relative maximum 6√3 at x = √3
- decreasing on x < -√3 and x > √3
- increasing on -√3 < x < √3
- see below for a graph
Step-by-step explanation:
I find it convenient to draw the graph first when looking for relative extrema.
The function can be differentiated to get ...
f'(x) = -3x^2 +9
This is zero when ...
-3x^2 +9 = 0
x^2 = 3
x = ±√3 . . . . . x-values of relative extrema
Then the extreme values are ...
f(±√3) = x(9 -x^2) = (±√3)(9 -3) = ±6√3
The lower extreme (minimum) corresponds to the lower value of x (-√3), so the extrema are ...
(x, y) = (-√3, -6√3) and (√3, 6√3)
__
Since the leading coefficient is negative and the degree is odd, the function is decreasing for values of x below the minimum and above the maximum. It is increasing for values of x between the minimum and the maximum.
decreasing: x < -√3, and √3 < x
increasing: -√3 < x < √3
