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3 years ago

5

Triangle JKL is equilateral. One side of the triangle, JL, is a diameter of circle M. Which is true about line segments JK and K

L?
Both segments are tangent to circle M but are not chords.
One segment is tangent to circle M and one segment is a chord in circle M.
Both segments are chords in circle M but are not tangents.
Neither segment is a chord nor tangent to circle M.

2 answers:

antoniya [11.8K]3 years ago

7 0

It could be concluded that since line JL is a diameter of a circle, therefore, JK and KL are the tangents of the specific circle that JL is part of. By definition, a tangent is "a line that touches a curve at one point without touching it." Both these lines touches the edges of the circle M therefore they are tangents.

Agata [3.3K]3 years ago

4 0

Answer:

Just took the test, it's D!

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2 years ago

Write out the form of the partial fraction decomposition of the function. Do not determine the numerical values of the coefficie

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For part (a), you have

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$\dfrac{x^2}{x^2+x+2}=\dfrac{x^2+x+2-x-2}{x^2+x+2}=1-\dfrac{x+2}{x^2+x+2}$

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$\implies x^2+x+2=\left(x+\dfrac12-\dfrac{\sqrt7}2i\right)\left(x+\dfrac12+\dfrac{\sqrt7}2i\right)$

Then you have

$\dfrac{x+2}{x^2+x+2}=\dfrac a{x+\dfrac12-\dfrac{\sqrt7}2i}+\dfrac b{x+\dfrac12+\dfrac{\sqrt7}2i}$
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When $x=-\dfrac12-\dfrac{\sqrt7}2i$ , you have

$-\dfrac12-\dfrac{\sqrt7}2i+2=b\left(-\dfrac12-\dfrac{\sqrt7}2i+\dfrac12-\dfrac{\sqrt7}2i\right)$
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When $x=-\dfrac12+\dfrac{\sqrt7}2i$ , you have

$-\dfrac12+\dfrac{\sqrt7}2i+2=a\left(-\dfrac12+\dfrac{\sqrt7}2i+\dfrac12+\dfrac{\sqrt7}2i\right)$
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So, you could write

$\dfrac{x^2}{x^2+x+2}=1-\dfrac{x+2}{x^2+x+2}=1-\dfrac {7-3\sqrt7i}{14\left(x+\dfrac12-\dfrac{\sqrt7}2i\right)}-\dfrac {7+3\sqrt7i}{14\left(x+\dfrac12+\dfrac{\sqrt7}2i\right)}$

but that may or may not be considered acceptable by that webpage.

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3 years ago

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