Answer
a. $\sqrt 43$;
b. 6;
c. $\sqrt 19$;
d. $\sqrt 11$
Work Step by Step
Since $f(x,y) = \sqrt x^{2} + 2y^{2}$
a. $h(5,3) = \sqrt 5^{2} + 2(3)^{2}=\sqrt 43$
b. $h(2,4) = \sqrt 2^{2} + 2(4)^{2}=6$
c. $h(-1,-3) = \sqrt (-1)^{2} + 2(-3)^{2}=\sqrt 19$
d. $h(-3,-1) = \sqrt (-3)^{2} + 2(-1)^{2}=\sqrt 11$
