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anonymous

  • one year ago

if the point(x, sqrt 3/2) is on the unit circle, what is x

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  1. jim_thompson5910
    • one year ago
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    The point is \[\Large \left(x, \frac{\sqrt{3}}{2}\right)\] right?

  2. anonymous
    • one year ago
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    yes

  3. jim_thompson5910
    • one year ago
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    Notice how that point is the form (x,y) where \[\Large y = \frac{\sqrt{3}}{2}\] Do you agree with this statement?

  4. anonymous
    • one year ago
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    yes

  5. jim_thompson5910
    • one year ago
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    Now plug this y value into the unit circle equation and solve for x \[\Large x^2+y^2 = 1\] \[\Large x^2+\left( \frac{\sqrt{3}}{2}\right)^2 = 1\] \[\Large x^2+ \frac{\left(\sqrt{3}\right)^2}{2^2} = 1\] \[\Large x^2+ \frac{3}{4} = 1\] Do you see how to solve for x from here?

  6. anonymous
    • one year ago
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    no

  7. jim_thompson5910
    • one year ago
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    how would you move the 3/4 to the other side? what must you do to both sides?

  8. jim_thompson5910
    • one year ago
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    hint: you need to undo the addition

  9. anonymous
    • one year ago
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    subtract 3/4 from each side

  10. jim_thompson5910
    • one year ago
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    yes, that gives you \[\Large x^2+ \frac{3}{4} = 1\] \[\Large x^2+ \frac{3}{4}\color{red}{- \frac{3}{4}} = 1\color{red}{- \frac{3}{4}}\] \[\Large x^2 = 1- \frac{3}{4}\] \[\Large x^2 = \frac{4}{4}- \frac{3}{4}\] \[\Large x^2 = \frac{4-3}{4}\] \[\Large x^2 = \frac{1}{4}\]

  11. jim_thompson5910
    • one year ago
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    Now if \[\Large x^2 = \frac{1}{4}\] then what must x be? How would you isolate x?

  12. anonymous
    • one year ago
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    1/2

  13. jim_thompson5910
    • one year ago
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    x = 1/2 is one solution what is the other?

  14. anonymous
    • one year ago
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    i don't know

  15. jim_thompson5910
    • one year ago
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    hint: take a number like 10 and square it to get 100. You can also square -10 to get 100 as well since (-10)^2 = (-10)*(-10) = 100 so that's why the solutions to x^2 = 100 are x = -10 or x = 10

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