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inn [45]
3 years ago
9

12 is 60% of what number?

Mathematics
1 answer:
goldenfox [79]3 years ago
8 0

Answer: 20

<u>Step-by-step explanation:</u>

12 is 60% of what number

12 = 0.60 * n

20 = n  

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<h3>1. Inscribe a circle ΔJKL. Identify the point of concurrency that is the center of the circle you drew. </h3>

To inscribe a circle in the attached triangle, we must follow the following steps. The resultant figure is the first one below.

1. Draw the angle bisector of angle JKL. The straight line in the green one.

2. Draw the angle bisector of angle KLJ. The straight line in the blue one.

3. Draw point D at these two lines.

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To circumscribe a circle in the attached triangle, we must follow the following steps:

1. Draw the perpendicular bisector of the side FG. The straight line in the red one. The resultant figure is the second one below.

2. Draw the perpendicular bisector of the side GH. The straight line in the blue one.

3. The center of the circle is the intersection of these two lines

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<h3>3. Construct the two lines tangents to circle</h3>

A tangent to a circle is a straight line that touches the circle at an only point. If a point A is outside a circle, we can draw two lines that pass through this point and are tangent to the circle at two different points each. So this is shown in the third figure. Lines red and blue are tangent to the circle at two different points.

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Find the area of the unshaded reason for these two problems​
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9514 1404 393

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Step-by-step explanation:

The area of the unshaded region is the difference between the overall area of the rectangle and the area of the shaded region. It can also be found directly using the dimensions of the unshaded region.

a) We have the height of the right triangle, but need to know its base. The Pythagorean theorem can help with that. Let the base of the triangle be represented by x. Then ...

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The area of the unshaded triangle is given by the formula ...

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__

b) The shaded area at upper left is a square 100 ft on a side. Its area is ...

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7 0
3 years ago
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