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lorasvet [3.4K]
3 years ago
15

Help finding the length of arc RS using 3.14 for pi

Mathematics
2 answers:
jek_recluse [69]3 years ago
8 0

Answer:

the answer is 17

Step-by-step explanation:

katen-ka-za [31]3 years ago
5 0

Answer:

8.48in

Step-by-step explanation:

The length of arc RS is given by:

l=\frac{\theta}{360}\times 2\pi r

From the diagram the radius is r=6.48 in and the central angle of sector RS is \theta=150\degree

We use \pi=3.14 and substitute the radius and central angle to obtain:

l=\frac{150}{360}\times 2\times3.14\times  6.48in

We simplify to get:

l=\frac{5}{12}\times 2\times 6.48in

l=\frac{5}{6}\times 3.14\times  6.48in^2

l=8.48in

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(x/16)-(x+2)/8))=2<br><br><br><br> PLS HELP ME BRAINLIEST TO THE FIRST CORRECT ANSWER
ArbitrLikvidat [17]

Simplify both sides of the equation

x/16−(x+2/8) = 2x/16 + −1/8x + −1/4 = 2

Distribute

1/16x + −1/8x + −1/4 = 2

(1/16x + −1/8x)+(−1/4) = 2

Combine Like Terms

−1/16x + −1/4 = 2

Add 1/4 to both sides.

−1/16x + −1/4 + 1/4 = 2 + 1/4

−1/16x = 9/4

Multiply both sides by 16/(-1).

(16/−1)*(−1/16x)=(16/−1)*(9/4)

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Please help ASAP! Thanks! For the figures below, assume they are made of semicircles, quarter circles and squares. For each shap
bazaltina [42]

Answer:

Part 1) A=36(\pi-2)\ cm^2

Part 2) P=6(\pi+2\sqrt{2})\ cm

Step-by-step explanation:

Part 1) Find the area

we know that

The area of the shape is equal to the area of a quarter of circle minus the area of an isosceles right triangle

so

A=\frac{1}{4}\pi r^{2}-\frac{1}{2}(b)(h)

we have that the base and the height of triangle is equal to the radius of the circle

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substitute

A=\frac{1}{4}\pi (12)^{2}-\frac{1}{2}(12)(12)

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simplify

Factor 36

A=36(\pi-2)\ cm^2

Part 2) Find the perimeter

The perimeter of the figure is equal to the circumference of a quarter of circle plus the hypotenuse of the right triangle

<em>The circumference of a quarter of circle is equal to</em>

C=\frac{1}{4}(2\pi r)=\frac{1}{2}\pi r

substitute the given values

C=\frac{1}{2}\pi (12)

C=6\pi\ cm

The hypotenuse of right triangle is equal to (applying the Pythagorean Theorem)

AC=\sqrt{12^2+12^2}

AC=\sqrt{288}\ cm

simplify

AC=12\sqrt{2}\ cm

Find the perimeter

P=(6\pi+12\sqrt{2})\ cm

simplify

Factor 6

P=6(\pi+2\sqrt{2})\ cm

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