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AfilCa [17]
3 years ago
12

Michael is constructing a circle circumscribed about a triangle. He has partially completed the construction, as shown below. Wh

at should be his next step in the construction?
Triangle DEF is shown with two sets of intersecting arc markings on either side of side EF.

Connect the arc markings to complete the perpendicular bisector
Find the perpendicular bisectors for side DF and DE
Connect three arc markings together to form the triangle
Use the intersection of arcs to determine the center of the circle

Is the answer A or something else??
Mathematics
1 answer:
OlgaM077 [116]3 years ago
8 0

Answer:

Connect the arc markings to complete the perpendicular bisector .

I took the test and got it right.

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The greatest common factor( GCF) of 36 and 84 is 12.
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How can i solve this equation? please help<br> -3&lt;3x+3&lt;12
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Then move all terms not containing x from the center section of the inequality. 
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Janine is packing carrots. Each large box holds 20 2-pound bags of carrots. If janine has 800 bags of carrots, how many boxes ca
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Write an equation in slope-intercept form that has a slope of 3 and a y-intercept of 6
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6 0
3 years ago
A ball of radius 15 has a round hole of radius 5 drilled through its center. Find the volume of the resulting solid. Hint: The u
OverLord2011 [107]

Answer:

The volume of the ball with the drilled hole is:

\displaystyle\frac{8000\pi\sqrt{2}}{3}

Step-by-step explanation:

See attached a sketch of the region that is revolved about the y-axis to produce the upper half of the ball. Notice the function y is the equation of a circle centered at the origin with radius 15:

x^2+y^2=15^2\to y=\sqrt{225-x^2}

Then we set the integral for the volume by using shell method:

\displaystyle\int_5^{15}2\pi x\sqrt{225-x^2}dx

That can be solved by substitution:

u=225-x^2\to du=-2xdx

The limits of integration also change:

For x=5: u=225-5^2=200

For x=15: u=225-15^2=0

So the integral becomes:

\displaystyle -\int_{200}^{0}\pi \sqrt{u}du

If we flip the limits we also get rid of the minus in front, and writing the root as an exponent we get:

\displaystyle \int_{0}^{200}\pi u^{1/2}du

Then applying the basic rule we get:

\displaystyle\frac{2\pi}{3}u^{3/2}\Bigg|_0^{200}=\frac{2\pi(200\sqrt{200})}{3}=\frac{400\pi(10)\sqrt{2}}{3}=\frac{4000\pi\sqrt{2}}{3}

Since that is just half of the solid, we multiply by 2 to get the complete volume:

\displaystyle\frac{2\cdot4000\pi\sqrt{2}}{3}

=\displaystyle\frac{8000\pi\sqrt{2}}{3}

5 0
4 years ago
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