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Serjik [45]
3 years ago
10

Can someone help with this question

Mathematics
1 answer:
KatRina [158]3 years ago
4 0
I think it’s 7 Becuase the Ae and DE are congruent
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Step-by-step explanation:

3x^2-2x-1=0\\\\3(x^2-\frac{2}{3}x-\frac{1}{3})=0\\ \\3(x^2-\frac{2}{3}x-\frac{1}{3}+\frac{4}{9})=3(\frac{4}{9})\\ \\ 3(x^2-\frac{2}{3}x+\frac{1}{9})=\frac{4}{3}\\ \\ 3(x-\frac{1}{3})^2=\frac{4}{3}\\ \\(x-\frac{1}{3})^2=\frac{4}{9}\\ \\x-\frac{1}{3}=\pm\frac{2}{3}

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