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Kitty [74]
3 years ago
8

James wants to construct a regular hexagon inscribe in a circle with center O and diameter PQ

Mathematics
1 answer:
Brums [2.3K]3 years ago
6 0

Answer:

B. Place the compass at P and draw an arc of length OP across the circle.

Below is the diagram for help.

Step-by-step explanation:

We are given with an already constructed circle of diameter PQ and OP as radius.

To draw a regular hexagon(six-sided), we need to make first cut on the circle of an arc equal to radius of the circle taking any point of the diameter as center

So here, We will consider P as a center and take an arc of length OP and make a cut on the circle.

This would be the first step and James need to repeat the same steps by considering the newly formed cut as center each time.

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At an end of the year sale, Gabriela bought more than 12 bottles of hand soaps and lotions. If x represents the number of hand s
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Step-by-step explanation:

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6 0
3 years ago
Read 2 more answers
Plz someone help me.
satela [25.4K]
Hope this helps you! (and I'm right LOL)


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3 years ago
The Japanese Yen corresponds to the American Dollar. If the exchange rate is 150 Yen for 1 Dollar, then a Japanese car sold in A
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Answer:

125,000

Step-by-step explanation:

Take the Japanese yen and divide by 12

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3 0
3 years ago
Anyone know how to solve this??
Tom [10]

Given : \frac{(x^\frac{2}{5})^9\;.\;(x^\frac{-4}{15})}{x^\frac{1}{3}}

\implies {(x^\frac{18}{5})\;.\; (x^\frac{-4}{15})}\;.\;(x^\frac{-1}{3})

\implies x^(^\frac{18}{5} ^-^\frac{4}{15}^-^\frac{1}{3}^)

\implies x^(^(^\frac{54 - 4}{15}^)^-^\frac{1}{3}^) = x^(^(^\frac{50}{15}^)^-^\frac{1}{3}^) = x^(^(^\frac{10}{3}^)^-^\frac{1}{3}^) = x^(^\frac{9}{3}^) = x^3

\implies x^k = x^3

\implies k = 3

7 0
3 years ago
Greg is in a car at the top of a roller-coaster ride. The distance, d, of the car from the ground as the car descends is determi
ANTONII [103]

Answer:

It takes 3 seconds over the interval [0,3]

Step-by-step explanation:

To find when the roller coaster reaches the ground, find when d=0.

0=144-16t^2

To solve divide each term by 16 and factor:

\frac{0}{16}=\frac{144}{16} -\frac{-16t^2}{16}  \\0= 9 - t^2\\0=(3-t)(3+t)

Solve for t by setting each factor to 0.

t-3=0 so t=3

t+3=0 so t=-3

This means the car is in the air from 0 to 3 second.

8 0
3 years ago
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