Answer:
[tex]y = 3\, \sqrt{5}[/tex] when [tex]x = 9[/tex].
Step-by-step explanation:
The question states that [tex]y[/tex] is proportional to [tex]\sqrt{x}[/tex]. In other words, there is a constant [tex]a[/tex] ([tex]a \ne 0[/tex]) such that [tex]y = a\, \sqrt{x}[/tex] for all [tex]x \ge 0[/tex].
The question also states that [tex]y = x[/tex] when [tex]x = 5[/tex]. Make use of this equality to find the value of [tex]a[/tex].
Since [tex]x = 5[/tex] and [tex]y = x[/tex], it must be true that [tex]y = 5[/tex]. Substitute [tex]x = 5\![/tex] and [tex]y = 5\![/tex] into the equation [tex]y = a\, \sqrt{x}[/tex] and solve for [tex]a[/tex]:
[tex]5 = a\, \sqrt{5}[/tex].
[tex]\begin{aligned}a &= \frac{5}{\sqrt{5}} \\ &= \sqrt{5}\end{aligned}[/tex].
Thus, [tex]y = \sqrt{5}\, (\sqrt{x})[/tex] for all [tex]x \ge 0[/tex].
Substitute in the value [tex]x = 9[/tex] to find the corresponding value of [tex]y[/tex]:
[tex]\begin{aligned} y &= \sqrt{5} \, (\sqrt{9}) \\ &= \sqrt{5} \times 3 \\ &= 3\, \sqrt{5}\end{aligned}[/tex].
