3 years ago

In which quadrant angle k lies, if tan k<0, an csc k >

Mathematics

1 answer:

Zanzabum3 years ago

7 0

Answer:

A. QII.

Step-by-step explanation:

The tan < 0 in Quadrants 2 and 3.

The csc > 0 in Quadrants 1 and 2.

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Answer: $-4,-15;\ \left(-\dfrac{19}{2},-\dfrac{121}{4}\right)$

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Given

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$t^2+19t+60=0\\\\\Rightarrow t=\dfrac{-19\pm\sqrt{19^2-4\times 1\times 60}}{2\times 1}\\\\\Rightarrow t=\dfrac{-19\pm \sqrt{121}}{2}\\\\\Rightarrow t=\dfrac{-19\pm 11}{2}\\\\\Rightarrow t=-4,-15$

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$y=t^2+2\times \dfrac{19}{2}t+\dfrac{19^2}{2^2}-\dfrac{19^2}{2^2}+60\\\\y=\left(t+\dfrac{19}{2}\right)^2+60-\dfrac{361}{4}\\\\y=\left(t+\dfrac{19}{2}\right)^2-\dfrac{121}{4}\\\\y+\dfrac{121}{4}=\left(t+\dfrac{19}{2}\right)^2\\\\\left(y-(-\dfrac{121}{4}\right)=\left(t-(-\dfrac{19}{2})\right)^2\\\\\text{The vertex is }\left(-\dfrac{19}{2},-\dfrac{121}{4}\right)$

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