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wolverine [178]
3 years ago
7

A circle has 9 times the area of another circle. If the radius of the larger circle is 27 meters, find the radius of the smaller

circle.

Mathematics
1 answer:
hoa [83]3 years ago
5 0

Answer:

Step-by-step explanation:

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All vertical and horizontal cross sections of any one cube are all the same size and shape. What about vertical and horizontal c
svetlana [45]
Horizontal cross sections are the same shape: they are all squares if the pyramid is square or rectangles if the pyramid is a rectangle pyramid.
Vertical cross sections are different shape: At the pyramid top they are triangles, at other places they have four sides.

Horizontal cross sections are different size: the more we go to the top of the pyramind, the smaller they get.
Vertical cross sections are different size: the closer we get to the edges, the smaller they become.
8 0
3 years ago
The question says 5k+6k+4k=90. What does k equal?
alina1380 [7]
5k+6k+4k=90 \\ \\ 15k = 90 \ / \ simplify \\ \\ k =  \frac{90}{15} \ / \ divide \ each \ side \ by \ 15 \\ \\ k = 6 \ / \ simplify \\ \\

The final result is, k = 6.
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Answer:2/16

Step-by-step explanation:

8 0
3 years ago
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Answer:

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7 0
3 years ago
Find the critical numbers of the function f(x) = x6(x − 1)5 what does the first derivative test tell you that the second derivat
patriot [66]
The given function is f(x) = x⁶(x-1)⁵

The first derivative is
f'(x) = 6x⁵(x-1)⁵ + 5x⁶(x-1)⁴
       = x⁵(x-1)⁴(6x - 6 + 5x)
       = x⁵(x-1)⁴(11x - 6)
The critical values are the zeros of f'(x). They are
x = 0, 6/11, and 1.
The critical values indicate that turning points exist for f(x) at the critical points. However, we do not know the nature of the turning points.

Write the first derivative in the form f'(x) = (x-1)⁴(11x⁶ - 6x⁵).
The second derivative is
f''(x) = 4(x-1)³(11x⁶ - 6x⁵) + (x-1)⁴(66x⁵ - 30x⁴)
        = (x-1)³(44x⁶ - 24x⁵ + 66x⁶ - 30x⁵ - 66x⁵ + 30x⁴)
        = (x-1)³(110x⁶ - 120x⁵ + 30x⁴)
        = 10x⁴(x-1)³(11x² - 12x + 3)
The sign of f''(x) at the critical values tell us the nature of the turning point.

f''(0) = 0, therefore a point of inflection exists at x = 0.
f''(6/11) > 0, therefore a local minumum exists at x = 6/11.
f''(1) = 0, therefore a point of inflection exists at x=1.

The graphs shown below confirm these results.

8 0
3 years ago
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