Answer
The solutions are
a) $h(2,3,9) = \frac{2}{3}.$
b) $h(1,0,1) = 0.$
c) $h(-2,3,4) = -\frac{3}{2}.$
d) $h(5,4,-6) = -\frac{10}{3}.$
Work Step by Step
We will substitute $x$, $y$ and $z$ from the first, the second and the third coordinate, respectively into the function $h(x,y,z) = \frac{xy}{z}$
a) $h(2,3,9)=\frac{2\times3}{9} = \frac{2}{3}.$
b) $h(1,0,1) = \frac{1\times 0}{1} = 0.$
c) $h(-2,3,4) = \frac{-2\times 3}{4} = -\frac{3}{2}.$
d) $h(5,4,-6) = \frac{5\times4}{-6} = -\frac{10}{3}.$
