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The point (-4, 6) is on the terminal side of theta in standard position. Which is the cosine of theta?

1. - 2sqrt13/13
2. -2/13
3. 2/13
4. 2sqrt13/13

I think it is one of the negative ones, but I'm not sure of which.

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  1. you are correct, since (-4,6) is in QII.

    Recall that cosine is x/r.
    r^2 = 4^2+6^2 = 42
    r = √42 = 2√13

    so, your cos is -4/2√13 = -2/√13

    Or, as they clumsily put it, -2√13/13

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