Certainly! Let's solve the problem step by step for [tex]\(x = 4\)[/tex] and [tex]\(y = 6\)[/tex].
We need to find the value of [tex]\(5x + \frac{1}{2} xy^2\)[/tex].
First, we will calculate [tex]\(5x\)[/tex]:
[tex]\[ 5x = 5 \cdot 4 = 20 \][/tex]
Next, we will calculate [tex]\(\frac{1}{2} xy^2\)[/tex]:
[tex]\[ xy^2 = 4 \cdot 6^2 \][/tex]
[tex]\[ 6^2 = 36 \][/tex]
[tex]\[ xy^2 = 4 \cdot 36 = 144 \][/tex]
[tex]\[ \frac{1}{2} xy^2 = \frac{1}{2} \cdot 144 = 72 \][/tex]
Now, we add the two parts together:
[tex]\[ 5x + \frac{1}{2} xy^2 = 20 + 72 \][/tex]
Thus, the final value is:
[tex]\[ 20 + 72 = 92 \][/tex]
So, the value of [tex]\(5x + \frac{1}{2} xy^2\)[/tex] when [tex]\(x = 4\)[/tex] and [tex]\(y = 6\)[/tex] is [tex]\(92\)[/tex].
