X
3
+
3
x
2
y
+
2
x
y
2
+
y
3
+
x
2
+
3
x
y
+
y
2
A nine can go into a 20 two whole times. After that, there's still
some room left in the 20, but it's only enough room for a 2.
There u go tried my best to work it out!! :))
First you would solve for h(5) by plugging in 5 as your x, then solving it.
h(5) = 5^2 + 1
h(5) = 25 + 1
h(5) = 26
Next you would multiply the 26 by the individual h, which is basically h(1).
h(1) = 1^2 + 1
h(1) = 2
Lastly you multiply your h(1) value by the h(5) value to get your answer.
h(1) • h(5) = 26 • 2
h[h(5)] = 52
