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Find the domain of f(x) = (x + 2)/[sqrt{x - 10}].
I see a radical in the denominator of the algebraic fraction. Disregarding the numerator, must I set the radicand (x - 10) to be greater than or equal to 0 and solve for x?
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That sounds like a good way to proceed.
Bob
Children are not defined by school ...........The Fonz
You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you! …………….Bob
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Find the domain of f(x) = (x + 2)/[sqrt{x - 10}].
I see a radical in the denominator of the algebraic fraction. Disregarding the numerator, must I set the radicand (x - 10) to be greater than or equal to 0 and solve for x?
Let me see.
x - 10 >= 0
x >= 10...Domain?
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Yes. Well done!
B
Children are not defined by school ...........The Fonz
You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you! …………….Bob
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Yes. Well done!
B
This is pretty easy stuff. Once I solve the problem, it all comes back to me. This is review. I took precalculus in the Spring 1993 semester at Lehman College in the Bronx, NY.
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