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anonymous
 5 years ago
find the domain
Q(y)=sqrt (y^2+1)  cuberoot (1y)
anonymous
 5 years ago
find the domain Q(y)=sqrt (y^2+1)  cuberoot (1y)

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Domain of y is all real numbers. Explanation: since the quantity y^2 is always positive ((y^2)+1)^(1/2) is always defined and since the a cube root is defined along all reals y can take any value. There does not exist a value of y that will make the function undefined

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0and what about in this case 4 / (x9)  sqrt (x^3 36)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0x definitely can't equal 9 (since you would be dividing by 0) and x^3 must be greater than 36 (since you can't have a negative in a square root so x > 36^(1/3) and x \[\neq\] 9
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