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ale4655 [162]
3 years ago
14

BC is tangent to circle A at B and to circle D at C. AB= 10 BC= 21 DC= 8 find AD to the nearest tenth

Mathematics
1 answer:
Andrei [34K]3 years ago
4 0
The attached diagram is an interpretation of the question.
If there are other details, please post.

The distance AD is found by Pythagoras theorem, with 
AD=sqrt(BC^2+(AB-DC)^2)=sqrt(21^2+2^2)=sqrt(441+4)=sqrt(445)
=21.095 (to the third decimal place)

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Answer: First option.


Step-by-step explanation:

 1. To solve this problem you can applly the quadratic formula, which is shown below:

 x=\frac{-b+/-\sqrt{b^{2}-4ac}}{2a}

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x^{2}-4x-9-29=0\\x^{2}-4x-38=0

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4. Therefore, when you substitute these values into the quadratic formula, you obtain the following result:

x=\frac{-(-4)+/-\sqrt{(-4)^{2}-4(1)(-38)}}{2(1)}

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Step-by-step explanation:

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