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Likurg_2 [28]
3 years ago
13

Find the value of x.

Mathematics
2 answers:
Shalnov [3]3 years ago
8 0

Answer:

74 degrees

Step-by-step explanation:

The arc degree measurement of the arc that is colored in orange is given as 254 degrees.

The remaining part is the part in green.

The green part plus the orange part should equal 360 degrees because that would make a full rotation around the circle.

360-254=106.

The green arc has measurement 106 degrees.

We can find x by computing half the difference of the arcs there.

That is,

x=\frac{1}{2}(254-106)

x=\frac{1}{2}(148)

x=74

shutvik [7]3 years ago
6 0

Answer:

  74°

Step-by-step explanation:

There are at least a couple of ways to get there.

1) Find the central angle between the radii to the points of tangency. It will be ...

  360° - 254° = 106°

The measure of x is the supplement of this:

  x = 180° -106° = 74°

__

2) Find half the difference of the subtended arcs. The arc shown is 254°; the one not shown is 360° -254° = 106°. Half the difference of these is ...

  (254°-106°)/2 = 148°/2 = 74°

__

3) Perhaps more simply, we can find this measure symbolically. Let y represent the large arc shown as 254°. Then the smaller one is 360°-y and their difference is ...

   difference = y - (360° -y) = 2y -360°

Half this value, the value of angle x is ...

  (2y -360°)/2 = y -180°

For y=254°, this is ...

  x = 254° -180° = 74°

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3 years ago
A square napkin is folded in half on the diagonal and placed on the diameter of a round plate (see diagram below). If the folded
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Answer:

A=81(\pi-1)\ in^2

Step-by-step explanation:

step 1

Find the area of the plate

The area of a circle is given by the formula

A=\pi r^{2}

we have

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substitute

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step 2

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we know that

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Applying Pythagorean theorem

D^2=2b^2

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Divide by 2

162/2=81\ in^2

step 3

Find the area of the space on the plate that is NOT covered by the napkin

we know that

The  area of the space on the plate that is NOT covered by the napkin, is equal to subtract the area of the square napkin folded (is a half of the area of the square napkin) from the area of the plate

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A=(81\pi-81)\ in^2

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A=81(\pi-1)\ in^2

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