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stepan [7]
3 years ago
10

Circle P has a radius of 5 centimeters. Point Q lies on circle P, and the equation of PQ←→ is y=25x+3. Which equation could be t

he equation of a line tangent to circle P at point Q?
Mathematics
1 answer:
lesya [120]3 years ago
5 0
5x+2y=7 that would be the correct equation
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​

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​

​​

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2 In this case, a=1a=1, b=-2b=−2 and c=-2c=−2.

(x-\frac{2+\sqrt{{(-2)}^{2}-4\times -2}}{2})(x-\frac{2-\sqrt{{(-2)}^{2}-4\times -2}}{2})(x−

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​​

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3 Simplify.

(x-\frac{2+2\sqrt{3}}{2})(x-\frac{2-2\sqrt{3}}{2})(x−

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4 Factor out the common term 22.

(x-\frac{2(1+\sqrt{3})}{2})(x-\frac{2-2\sqrt{3}}{2})(x−

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5 Cancel 22.

(x-(1+\sqrt{3}))(x-\frac{2-2\sqrt{3}}{2})(x−(1+√

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​​ ))(x−

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​

​2−2√

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​

​​

​​ )

6 Simplify brackets.

(x-1-\sqrt{3})(x-\frac{2-2\sqrt{3}}{2})(x−1−√

​3

​

​​ )(x−

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​​

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7 Factor out the common term 22.

(x-1-\sqrt{3})(x-\frac{2(1-\sqrt{3})}{2})(x−1−√

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​2(1−√

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8 Cancel 22.

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7 0
4 years ago
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