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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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

Mathematics
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i did this in a quick check last week to.... x.x
Do you know that answer? Because I got it wrong twice lol that's why there is only two options @Icedragon50
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.

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x=\[\sqrt[3]{10+12}\] or x=\[\sqrt[3]{-10+12}\]
\[y=(x-12)^{3}-10\] is the equation, the other post are the answers @twistnflip
replace y with 0, and solve for x.
(when you are asked for "zeros" it is asking for x-interects....)
intercepts*
so its -10+12? @SolomonZelman
\(\large\color{black}{ \displaystyle y = (x - 12)^3 - 10 }\) 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)
\(\large\color{black}{ \displaystyle \sqrt[3]{10} = x - 12 }\)
\(\large\color{black}{ \displaystyle \sqrt[3]{10}+12 = x }\)
Thank you very much! @SolomonZelman , I gave you a medal
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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