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Denise

  • 5 years ago

How do you solve (x-7)^2+(y-(-11))^2=6^2

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  1. anonymous
    • 5 years ago
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    \[{({x-7})^2}+{({y-(-11)}^2} = {6^2}\] Remember your order of operations: PEMDAS. 1) Simplify the {y-(-11)} first - remember two consecutive negatives is the same as a plus sign. 2) You can multiply out 6^2 on the right at the same time. 3) Lastly Use your rules to multiply a binomial by a binomial (itself). You'll have two of these to multiply out. 4) Finally, combine like terms. Post your progress.

  2. anonymous
    • 5 years ago
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    Oops: \[...(-11))^2\]

  3. anonymous
    • 5 years ago
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    Hi Denise, solve for what? you have 2 variables (x and y) in the equation Ivan

  4. Denise
    • 5 years ago
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    I am honestly lost!

  5. Denise
    • 5 years ago
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    Ivan, I am suppose to write the center-radius form equation of the circle.

  6. anonymous
    • 5 years ago
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    Oops, I was telling you how to expand this equation - not very useful. Remember the basic unit circle equation: \[x^2 + y^2 = (radius)^2 \] -- radius equals 1 for the unit circle. I imagine that your assignment is to find the new center and radius of the circle in the given equation. Correct?

  7. anonymous
    • 5 years ago
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    I just responded, but my answer didnt stick for some reason. I agree with Moss' interpretation. http://www.regentsprep.org/Regents/math/algtrig/ATC1/circlelesson.htm the equation you have, to me, is in the centre- radius form, whereas the multiplied out form is the "general form"

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