Is the relation {(2,-3),(4,6),(3,-1),(6,6), (2,3)} a function? Yes No
1 answer:
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Answer:
n = 59
Step-by-step explanation:
I find it easiest to work problems of this kind using a graphing calculator. That way, extraneous solutions can be avoided. It seems to work well to rewrite the problem, so you're looking for a value of n that makes the result zero. Here, that would mean you want ...
... f(n) = √(n+5) -√(n-10) -1
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The solution by hand involves eliminating the root symbols. You do that by squaring the equation:
... n +5 -2√((n+5)(n-10)) + n -10 = 1
Now, we isolate the remaining root and square again.
... 2n -6 = 2√((n+5)(n-10)) . . . collect terms, add 2√( ) -1
... n -3 = √(n²-5n-50) . . . . . . . divide by 2
... n² -6n +9 = n² -5n -50 . . . . square both sides
... 59 = n . . . . . . . . . . . . . . . . . add 50 +6n -n²
