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mr Goodwill [35]
3 years ago
12

In circle K , diameter CD is drawn and point E is located on the circle such that CE=16 and DE=30 .What is the length of the rad

ius of circle K?
Mathematics
1 answer:
egoroff_w [7]3 years ago
7 0

Answer:

The length of the radius of the circle K is 17\ units

Step-by-step explanation:

we know that

The triangle CDE is a right triangle with the 90 degree angle at point E

so

Applying the Pythagoras Theorem

Find the length of CD (diameter of the circle K)

CD^{2}=CE^{2}+DE^{2}

substitute the values

CD^{2}=16^{2}+30^{2}

CD^{2}=1,156

CD=34\ units

Find the radius

Remember that the radius is half the diameter

so

r=34/2=17\ units

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Simplify seven square root of three end root minus four square root of six end root plus square root of forty eight end root min
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<u>Step-by-step explanation:</u>

We need to Simplify seven square root of three end root minus four square root of six end root plus square root of forty eight end root minus square root of fifty four. Which is equivalent to 7\sqrt{3}- 4\sqrt{6} + \sqrt{48} - \sqrt{54} :

7\sqrt{3}- 4\sqrt{6} + \sqrt{48} - \sqrt{54}

⇒ 7\sqrt{3}- 4\sqrt{6} + \sqrt{48} - \sqrt{44}

⇒ 7\sqrt{3}- 4\sqrt{6} + \sqrt{16(3)} - \sqrt{9(6)}

⇒ 7\sqrt{3}- 4\sqrt{6} + 4\sqrt{(3)} - 3\sqrt{(6)}

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⇒ 11(1.723)- 7(2.449)

⇒ 1.909

Therefore, 7\sqrt{3}- 4\sqrt{6} + \sqrt{48} - \sqrt{54} simplified as 11\sqrt{3}- 7\sqrt{6} or 1.909 .

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Answer:

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Step-by-step explanation:

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