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Alona [7]
3 years ago
10

Given that the diameter of Circle A is 6 cm , and the radius of Circle B is 18 cm , what can be concluded about the two circles?

Mathematics
1 answer:
ELEN [110]3 years ago
8 0

Answer:

The radius of circle B is 6 times greater than the radius of circle A

The area of circle B is 36 times greater than the area of circle A

Step-by-step explanation:

we have

<em>Circle A</em>

D=6\ cm

The radius of circle A is

r=6/2=3\ cm -----> the radius is half the diameter

<em>Circle B</em>

r=18\ cm

Compare the radius of both circles

3\ cm< 18\ cm

18=6(3)

The radius of circle B is six times greater than the radius of circle A

Remember that , if two figures are similar, then the ratio of its areas is equal to the scale factor squared

All circles are similar

In this problem the scale factor is 6

so

6^{2}=36

therefore

The area of circle B is 36 times greater than the area of circle A

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Answer:

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Step-by-step explanation:

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First, you need to find the average of the bases:

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Then, you multiply that by the height.

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          55 square feet is the answer

8 0
4 years ago
Find the sum of the first 5 terms of the infinite series: 8 + 18 + 28 +
Anna007 [38]

Answer:

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Step-by-step explanation:

3 0
3 years ago
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Hatshy [7]

Answer:

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Step-by-step explanation:

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6 0
3 years ago
How do you find the x intercepts of 3x^(5/3) - 4x^(7/3)
kaheart [24]
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Then either x^(5/3) = 0, or 3 - 4x^(2/3) = 0.

In the latter case, 4x^(2/3) = 3. 

To solve this:  mult. both sides by x^(-2/3).  Then we have

4x^(2/3)x^(-2/3) = 3x^(-2/3),            or 4 = 3x^(-2/3).  It'd be easier to work with this if we rewrote it as

4           3
---  = --------------------
1            x^(+2/3)

Then 

4
---  = x^(-2/3).  Then, x^(2/3) = (3/4), and x = (3/4)^(3/2).  According to my     3                      calculator, that comes out to x = 0.65 (approx.)

Check this result!  subst. 0.65 for x in the given equation.  Is the equation then true?

My method here was a bit roundabout, and longer than it should have been.  Can you think of a more elegant (and shorter) solution?
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Semenov [28]
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