Найдите значение выражения \( \sqrt{x^2 + y^2} \) при данных \( x \) и \( y \).

Для каждой пары значений \( x \) и \( y \) вычислим \( \sqrt{x^2 + y^2} \): 1. \( x = 4, y = 0 \): \( \sqrt{4^2 + 0^2} = \sqrt{16} = 4 \) 2. \( x = 0, y = -6 \): \( \sqrt{0^2 + (-6)^2} = \sqrt{36} = 6 \) 3. \( x = 5, y = -12 \): \( \sqrt{5^2 + (-12)^2} = \sqrt{25 + 144} = \sqrt{169} = 13 \) 4. \( x = -7, y = 24 \): \( \sqrt{(-7)^2 + 24^2} = \sqrt{49 + 576} = \sqrt{625} = 25 \) 5. \( x = 10, y = -24 \): \( \sqrt{10^2 + (-24)^2} = \sqrt{100 + 576} = \sqrt{676} = 26 \) 6. \( x = \sqrt{21}, y = 2 \): \( \sqrt{(\sqrt{21})^2 + 2^2} = \sqrt{21 + 4} = \sqrt{25} = 5 \) 7. \( x = \sqrt{13}, y = 6 \): \( \sqrt{(\sqrt{13})^2 + 6^2} = \sqrt{13 + 36} = \sqrt{49} = 7 \) 8. \( x = 6\sqrt{2}, y = -6\sqrt{2} \): \( \sqrt{(6\sqrt{2})^2 + (-6\sqrt{2})^2} = \sqrt{72 + 72} = \sqrt{144} = 12 \)