a) Compute x+y and \(\sqrt{x^2+y^2}\) when x=5 and y=12 b) When is
\(x+y=\sqrt{x^2+y^2}?\) When is
\(x+y\neq\sqrt{x^2+y^2}\)
(a) If you know the Pythagorean triple (5, 12, 13), the answer is 13. Thus, \(\sqrt{25+144}=\sqrt{169}=\boxed{13}.\)
b. the first part is one, the second part is any number aside from one.
a) x + y = 5 + 12 = 17
Tertre's answer for sqrt (x^2 + y^2) is good
b) x + y = sqrt (x^2 + y^2) square both sides
x^2 + 2xy + y^2 = x^2 + y^2
2xy = 0
xy = 0
So....this is true when either x or y (or both) = 0
c) this will not be true when xy ≠ 0
