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givi [52]
3 years ago
6

Regular pentagon abcde is inscribed in a circle. find arch measure ab. if the radius is 6, find ab.​

Mathematics
2 answers:
SSSSS [86.1K]3 years ago
4 0

a. Your answer and reasoning are correct.

b. If O is the center of the circle, then angle ABO has measure 72º, so that by the law of cosines

AB^2=6^2+6^2-2\cdot6\cdot6\cos72^\circ\implies AB=3\sqrt{10-2\sqrt5}

OleMash [197]3 years ago
4 0

just to add to the superb reply above by @LammettHash

\bf \textit{arc's length}\\\\ s=\cfrac{\pi \theta r}{180}~~ \begin{cases} r=radius\\ \theta =angle~in\\ \qquad degrees\\ \cline{1-1} r=6\\ \theta =72 \end{cases}\implies s=\cfrac{\pi (72)(6)}{180}\implies s=\cfrac{12\pi }{5}\implies s\approx 7.54

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Find the next two terms of the following sequence: 14, 38, 74, 122, 182, 254...
Bezzdna [24]
Let's find the difference between each number.
38-14=24
74-38=36
122-74=48
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As you can tell, you're adding 12 to each difference before. The next difference would be 72+12=84. Let's add.
254+84=338
Now the next difference would be 84+12=96.
338+96=434
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5 0
4 years ago
In a shipment of 71 vials, only 13 do not have hairline cracks. If you randomly select 2 vials from the shipment, what is the pr
yanalaym [24]

Answer:

0.0313

Step-by-step explanation:

Given that in the shipment of 71 vials, only 13 do not have hairline cracks

Probability of not having a hairline cracks = \frac{13}{71}

If you randomly select 2 vials from the shipment

(Probability of not having a hairline cracks)' = \frac{13-1}{71-1} = \frac{12}{70}

what is the probability that none of the 2 vials have hairline cracks.

i.e (\frac{13}{71}) * (\frac{12}{70})

= 0.183 × 0.171

= 0.0313

Hence, the probability that none of the 2 vials have hairline cracks = 0.0313

6 0
3 years ago
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What are the numbers that are supposed to be plugged in?

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