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anonymous

  • one year ago

I need help with #3. I can't figure out what im suppose to do with it.

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  1. anonymous
    • one year ago
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    just wait im posting it

  2. anonymous
    • one year ago
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  3. anonymous
    • one year ago
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    @Ashleyisakitty

  4. jdoe0001
    • one year ago
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    hmmm looks pretty straightforward what'st he value of "x"? what's the value of "y" in say 3a?

  5. anonymous
    • one year ago
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    x=- Square root of 3 and y= -1

  6. jim_thompson5910
    • one year ago
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    hint: make sure `x^2+y^2 = 1`

  7. anonymous
    • one year ago
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    o.

  8. jdoe0001
    • one year ago
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    ok.. so.. the radius or hypotenuse will be then \(\bf r=\sqrt{x^2+y^2}\implies r=\sqrt{(-\sqrt{3})^2+(-1)^2} \\ \quad \\ r=\sqrt{(-\sqrt{3})(-\sqrt{3})+1} \implies r=\sqrt{(\sqrt{3})^2+1} \\ \quad \\ r=\sqrt{3+1}\implies r=\sqrt{4} \implies r=2\)

  9. jim_thompson5910
    • one year ago
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    jdoe0001 has the most effective method, so go with that

  10. jdoe0001
    • one year ago
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    hmmm shoot

  11. jdoe0001
    • one year ago
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    \(\bf sin(\theta) =\cfrac{y}{r} \qquad % cosine cos(\theta) =\cfrac{x}{r} \qquad % tangent tan(\theta) =\cfrac{y}{x} \\ \quad \\ % cotangent cot(\theta) =\cfrac{x}{y} \qquad % cosecant csc(\theta) =\cfrac{r}{y} \qquad % secant sec(\theta) =\cfrac{r}{x}\)

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spraguer (Moderator)
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is replying to Can someone tell me what button the professor is hitting...

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