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What are all the real zeroes of y=(x–12)3–7
=
(
–
12
)
3
–
7
?
To find the real zeroes of the equation y = (x-12)^3 - 7, we need to set y equal to zero and solve for x.
0 = (x-12)^3 - 7
Add 7 to both sides:
7 = (x-12)^3
Take the cube root of both sides to eliminate the cube:
∛7 = x-12
Add 12 to both sides:
∛7 + 12 = x
Therefore, the real zero of the equation is x = ∛7 + 12.