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Klio2033 [76]
3 years ago
8

Radius OA of circle O has a slope of 2.5. What is the slope of the line tangent to circle O to point A?

Mathematics
2 answers:
mixer [17]3 years ago
5 0
The tangent at A is perpendicular to OA, so has a slope that is the negative reciprocal of that of the radius: -1/2.5 = -2/5.
Mekhanik [1.2K]3 years ago
4 0

Answer:  The required slope of the line tangent to the circle at point A is -0.4.

Step-by-step explanation:  Given that the radius OA of a circle with center O has a slope of 2.5.

We are to find the slope of the line tangent to the circle at the point A.

We know that

The radius of a circle at a point on the circumference of a circle is perpendicular to the tangent line at the same point on the circumference of the circle.

Also, the product of the slopes of two perpendicular lines is -1.

So, if m is the slope of the tangent line at the point A, then we must have

m\times 2.5=-1\\\\\Rightarrow m=-\dfrac{1}{2.5}\\\\\\\Rightarrow m=-\dfrac{10}{25}\\\\\Rightarrow m=-\dfrac{2}{5}\\\\\Rightarrow m=-0.4.

Thus, the required slope of the line tangent to the circle at point A is -0.4.

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Answer:

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Step-by-step explanation:

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a) Exponential growth model is given by N(t) = N_{0} e^{kt} where t is the time in years after 2016.

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therefore  I \times M \times S = 10^{magnitude\hspace{0.1cm}of\hspace{0.1cm}earthquake}

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