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NISA [10]
2 years ago
12

Lines CD and DE are tangent to circle A, as shown below: Lines CD and DE are tangent to circle A and intersect at point D. Arc C

E measures 125 degrees. Point B lies on circle A. If arc CE is 125°, what is the measure of ∠CDE? 55° 62.5° 117.5° 125°
Mathematics
1 answer:
melamori03 [73]2 years ago
5 0

The measure of angle CDE ( ∠CDE) is 55°. The correct option is the first option  55°

<h3>Calculating the measure of the angle formed by two tangents</h3>

From the question, we are to determine the measure of angle CDE.

From one of the circle theorems, we have that

"<em>The measure of the angle formed by two tangents that intersect at a point outside a circle is equal to one-half the positive difference of the measures of the intercepted arcs</em>"

Thus,

m ∠CDE = 1/2(measure of arc CBE - measure of arc CE)

From the given information,

Measure of arc CE = 125°

Then,

Measure of arc CBE = 360° - 125°

Measure of arc CBE = 235°

Therefore,

m ∠CDE = 1/2(235° - 125°)

m ∠CDE = 1/2(110°)

m ∠CDE = 55°

Hence, the measure of ∠CDE is 55°

Learn more on Calculating the measure of the angle formed by two tangents here: brainly.com/question/9143293

#SPJ1

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Answer:

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Step-by-step explanation:

The order in which the extras are ordered is not important. So we use the combinations formula to solve this question.

Combinations formula:

C_{n,x} is the number of different combinations of x objects from a set of n elements, given by the following formula.

C_{n,x} = \frac{n!}{x!(n-x)!}

In this problem:

2 options of hamburger(regular or larger)

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1 hamburger type, from a set of 2.

3 extras, from a set of 6. So

C_{2,1}*C_{6,3} = \frac{2!}{1!(2-1)!}*\frac{6!}{3!(6-3)!} = 2*20 = 40

40 different hamburgers can be ordered with exactly three extras

(b) How many different regular hamburgers can be ordered with exactly three extras?

3 extras, from a set of 6. So

C_{6,3} = \frac{6!}{3!(6-3)!} = 20

20 different regular hamburgers can be ordered with exactly three extras

(c) How many different regular hamburgers can be ordered with at least five extras?

Five extras:

5 extras, from a set of 6. So

C_{6,5} = \frac{6!}{5!(6-5)!} = 6

Six extras:

6 extras, from a set of 6. So

C_{6,6} = \frac{6!}{6!(6-6)!} = 1

6 + 1 = 7

7 different regular hamburgers can be ordered with at least five extras

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Step-by-step explanation:

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Step-by-step explanation:

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iren [92.7K]

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