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anonymous

  • one year ago

What are all the real zeros of y = (x - 12)^3 - 10? A. x=3 square-root 10+12 B. x=3 square-root -10+12

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  1. anonymous
    • one year ago
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    i did this in a quick check last week to.... x.x

  2. anonymous
    • one year ago
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    Do you know that answer? Because I got it wrong twice lol that's why there is only two options @Icedragon50

  3. anonymous
    • one year ago
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    Can you use the equation editor to write the answers, I need to seem them clearer to give you the answer choice that is correct.

  4. anonymous
    • one year ago
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    x=\[\sqrt[3]{10+12}\] or x=\[\sqrt[3]{-10+12}\]

  5. anonymous
    • one year ago
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    @twistnflip

  6. anonymous
    • one year ago
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    \[y=(x-12)^{3}-10\] is the equation, the other post are the answers @twistnflip

  7. SolomonZelman
    • one year ago
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    replace y with 0, and solve for x.

  8. SolomonZelman
    • one year ago
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    (when you are asked for "zeros" it is asking for x-interects....)

  9. SolomonZelman
    • one year ago
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    intercepts*

  10. anonymous
    • one year ago
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    so its -10+12? @SolomonZelman

  11. SolomonZelman
    • one year ago
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    \(\large\color{black}{ \displaystyle y = (x - 12)^3 - 10 }\) the very first step (set y=0) \(\large\color{black}{ \displaystyle 0 = (x - 12)^3 - 10 }\) Now, solving for x. \(\large\color{black}{ \displaystyle 10 = (x - 12)^3 }\) (I added 10 to both sides)

  12. SolomonZelman
    • one year ago
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    \(\large\color{black}{ \displaystyle \sqrt[3]{10} = x - 12 }\)

  13. SolomonZelman
    • one year ago
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    \(\large\color{black}{ \displaystyle \sqrt[3]{10}+12 = x }\)

  14. anonymous
    • one year ago
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    Thank you very much! @SolomonZelman , I gave you a medal

  15. SolomonZelman
    • one year ago
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    I don't need medals, my profile looks suspicious because of their multiplicity\(\bf ....\) The cube root, like any odd root is not going to have a ± sign, and this why we ended up with only 1 zero.

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