WAEC - Mathematics (2008 - No. 9)
Given that x = 2 and y = -\(\frac{1}{4}\), evaluate \(\frac{x^2y - 2xy}{5}\)
zero
\(\frac{1}{5}\)
1
2
Giải thích
Given; x = 2; y = \(\frac{-1}{4}\)
= \(\frac{x^2y - 2xy}{5}\)
= \(\frac{2^2(\frac{-1}{4}) - 2(2)(\frac{-1}{4})}{5}\)
= \(\frac{4(\frac{-1}{4}) + 4(\frac{-1}{4})}{5}\)
= \(\frac{1 + 1}{5}\)
= \(\frac{0}{5}\)
= 0
= \(\frac{x^2y - 2xy}{5}\)
= \(\frac{2^2(\frac{-1}{4}) - 2(2)(\frac{-1}{4})}{5}\)
= \(\frac{4(\frac{-1}{4}) + 4(\frac{-1}{4})}{5}\)
= \(\frac{1 + 1}{5}\)
= \(\frac{0}{5}\)
= 0