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xxTIMURxx [149]
3 years ago
5

Lines CD and DE are tangent to circle A shown below: Lines CD and DE are tangent to circle A and intersect at point D. Arc CE me

asures 112 degrees. Point B lies on circle A. If Arc CE is 112°, what is the measure of ∠CDE? (4 points) Question 5 options: 1) 124° 2) 136° 3) 68° 4) 56°
Mathematics
1 answer:
Verdich [7]3 years ago
7 0

Answer:

Option (3)

Step-by-step explanation:

By the theorem, angle between two tangents intersecting each other outside the circle is half of the difference between intercepted arcs.

From the figure attached,

Two tangents CD and DE are intersecting each other at point D outside the circle A.

Two intercepted arcs are minor arc CE and major arc CBE,

Measure of arc CE = 112°

Therefore, measure of major arc CBE = 360° - 112°

                                                               = 248°

m(∠CDE) = \frac{1}{2}(\text{arcCBE}-\text{arcCE})

                 = \frac{1}{2}(248-112)

                 = \frac{1}{2}\times 136

                 = 68°

Option (3) will be the correct option.

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3 years ago
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Answer:

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8 0
3 years ago
Of great importance to residents of central Florida is the amount of radioactive material present in the soul of reclaimed phosp
alukav5142 [94]

Answer:

See figure attached and explanation below.

Step-by-step explanation:

For this case we have the following dataset given:

0.74, 0.32, 1.66, 3.59, 4.55, 6.47, 9.99, 0.7, 0.37, 0.76, 1.9, 1.77, 2.42, 1.09, 2.03, 2.69,2.41, 0.54, 8.32, 5.70, 0.75, 1.96, 3.36, 4.06, 12.48

And for this case we can use the followinf R code to create the frequency histogram.

> x<-c(0.74, 0.32, 1.66, 3.59, 4.55, 6.47, 9.99,0.7, 0.37, 0.76, 1.9, 1.77, 2.42, 1.09, 2.03, 2.69,2.41, 0.54, 8.32, 5.70, 0.75, 1.96, 3.36, 4.06,12.48)

> length(x)

[1] 25

> hist(x, prob=TRUE)

And for this case we have the histogram on the figure attached. For the number of classes we use the formula of sturges:

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And for this case we have approximately 6 classes. And that's what we can see on the figure attached

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<span>Thank you for posting your question. I hope you found you were after. Please feel free to ask me another.</span>

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