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3 years ago

The area of sector AOB is what fractional part of the area of circle O?

Mathematics

2 answers:

svetoff [14.1K]3 years ago

6 0

I had to look for the figure and here is my answer:

Given the circle attached to this question, the area of the sector AOB is the 60 of the fractional part of the area of the circle O. Since the whole circle is 360, this would be 60/360. To simplify this, we would arrive to the final answer of 1/6. Hope this helps.

qaws [65]3 years ago

4 0

Answer: 1/6 is the answer

Step-by-step explanation:

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RSB [31]

Answer:

49 - 9π

Step-by-step explanation:

Area of white portion = area of triangle - area of circle

To find the area of the triangle:

The area of a triangle can be calculated using this formula

$A=\frac{bh}{2}$ where b = base length and h = height

The height given is 7m and the base length given is 14m

That being said we plug in the given values into the formula

$\frac{14*7}{2} \\14*7=98\\\frac{98}{2} =49$

Hence, the area of the triangle is 49 sq meters.

To find the area of circle

( note that the area is in terms of pi; shown by the answer choices. )

Formula for area of a circle = $A=\pi r^2$

We are given the diameter ( 6m )

However, we need the radius

To convert diameter to radius we divide by 2

6/2 = 3 so the radius = 3m

Now we plug in 3 into the formula

$A=\pi 3^2\\3^2=9\\A=9\pi$

Remember we do not apply pi because the answer should be in terms of pi.

Now we subtract the two

Area of white portion = 49 - 9π

Hence, the answer is A

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3 years ago

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Answer:

0.45 you divide them

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Answer:

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Step-by-step explanation:

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3 years ago

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Gnoma [55]

Answer:

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Step-by-step explanation:

To find when 30m + 80 = 40m, isolate m.

30m + 80 = 40m

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6 0

3 years ago

Find the volume of the solid of revolution formed by rotating about the x--axis the region bounded by the given curves.

amm1812

Answer:

$V = \frac{8\pi}{3}$

Step-by-step explanation:

For this case we are interested on the region shaded on the figure attached.

And we can find the volume with the method of rings.

The area on this case is given by:

$A(x) = \pi [f(x)]^2 = \pi r^2 = \pi [3x]^2 = 9\pi x^2$

And the volume is given by the following formula:

$V= \int_{a}^b A(x) dx$

For our case our limits are x=0 and x=2 so we have this:

$V = \pi \int_{0}^{2} x^2 dx$

And if we solve the integral we got this:

$V= \pi [\frac{x^3}{3}]\Big|_0^{2}$

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$V=\pi [(\frac{8}{3} )-(\frac{0}{3} )]$

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3 years ago