Respuesta :

[tex]\bf \qquad \qquad \textit{inverse proportional variation} \\\\ \textit{\underline{y} varies inversely with \underline{x}}\qquad \qquad y=\cfrac{k}{x}\impliedby \begin{array}{llll} k=constant\ of\\ \qquad variation \end{array}\\\\ -------------------------------\\\\ \textit{\underline{y} is inversely proportional to square root of \underline{x}}\qquad y=\cfrac{k}{\sqrt{x}}[/tex]

[tex]\bf \textit{we also know that } \begin{cases} y=24\\ x=100 \end{cases}\implies 24=\cfrac{k}{\sqrt{100}}\implies 24=\cfrac{k}{10} \\\\\\ 240=k\qquad \qquad \boxed{y=\cfrac{240}{\sqrt{x}}}[/tex]

[tex]\bf \textit{when x = 36, what is \underline{y}?}\qquad y=\cfrac{240}{\sqrt{36}}\implies y=\cfrac{240}{6}\implies y=40 \\\\\\ \textit{when y = 600, what is \underline{x}?}\qquad 600=\cfrac{240}{\sqrt{x}}\implies \sqrt{x}=\cfrac{240}{600} \\\\\\ \sqrt{x}=\cfrac{2}{5}\implies x=\left( \cfrac{2}{5} \right)^2\implies x=\left( \cfrac{2^2}{5^2} \right)\implies x=\cfrac{4}{25}[/tex]
y=k/√x,


First way,

y=k/√x
You can find  k.
24=k/√100
24 = k/10
k=24*10, k=240

y=240/√x

When x=36 
y=240/√x
y=240/(√36)=240/(6)= 40,

so 
when x=36, y= 40

When y=600,
y=240/√x
y=600
600=240/√x
√x =240/600 = 24/60=0.4
x=(0.4)²=0.16
When y=600, x=0.16



Second way
y1=k/√x1     (1)
 and y2=k/√x2   (2),
When you divide equation 2 by equation 1, you get

y2/y1 = √x1/√x2.


So,
24/y2= √36/√100
24/y2=6/10
y2=24*10/6=40
For x=36, y=40.

24/600 = √x1/√100
√x1 = 24*√100/600=240/600=0.4
√x1=0.4
x1=0.4²=0.16
x1=0.16

For x=0.16, y =600.


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