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Not sure how difficult this would be to solve
asked
Nov 10, 2022
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Not sure how difficult this would be to solve
Mathematics
high-school
Stephen Connolly
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Stephen Connolly
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f (x) = x^2 - 2x + 1
f (x) = (-2)^2 - 2(-2) + 1
= 4 + 4 + 1
= 9
f (x) = ( 0 )^2 - 2 (0) + 1
= 0 - 0 + 1
= 1
0 = x^2 - 2x + 1
x^2 = 2x - 1 = 1
f (2) = 2^2 - 2(2) + 1
= 4 - 4 + 1
= 1
f (3) = 3^2 - 2(3) + 1
= 9 - 6 + 1
= 3 + 1
= 4
Rijk
answered
Nov 16, 2022
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Rijk
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