Hej!
[tex]y(y + 3) = {y}^{2} + 3y = {( - 4)}^{2} + 3 \times ( - 4) = 16 - 12 = 4 \\ \\ \frac{3}{4} (t - 2)^{2} = \frac{3}{4} ( - 6 - 2)^{2} = \frac{3}{4} \times {( - 8)}^{2} = \frac{3}{4} \times 64 = 3 \times 16 = 48 \\ \\ \frac{4z - 2}{11} = \frac{4 \times ( - 2) - 2}{11} = \frac{ - 8 - 2}{11} = - \frac{10}{11} [/tex]
Pozdrawiam! - Julka
