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Given [tex][tex]$y \propto \frac{1}{\sqrt{x}}$[/tex][/tex] and [tex][tex]$y=8$[/tex][/tex] when [tex][tex]$x=4$[/tex][/tex], find [tex][tex]$y$[/tex][/tex] when [tex][tex]$x=49$[/tex][/tex].

Sagot :

Sure! Let's solve this step-by-step.

First, we are given the relationship [tex]\( y \propto \frac{1}{\sqrt{x}} \)[/tex]. This means that [tex]\( y \)[/tex] is directly proportional to the inverse of the square root of [tex]\( x \)[/tex], and we can express this as:

[tex]\[ y = \frac{k}{\sqrt{x}} \][/tex]

where [tex]\( k \)[/tex] is the constant of proportionality.

We are given that [tex]\( y = 8 \)[/tex] when [tex]\( x = 4 \)[/tex]. We can use this information to find the constant [tex]\( k \)[/tex].

1. Substitute [tex]\( y = 8 \)[/tex] and [tex]\( x = 4 \)[/tex] into the equation:

[tex]\[ 8 = \frac{k}{\sqrt{4}} \][/tex]

2. Calculate [tex]\(\sqrt{4}\)[/tex] which is 2:

[tex]\[ 8 = \frac{k}{2} \][/tex]

3. Solve for [tex]\( k \)[/tex]:

[tex]\[ k = 8 \times 2 = 16 \][/tex]

Now that we have the constant [tex]\( k = 16 \)[/tex], we can find [tex]\( y \)[/tex] when [tex]\( x = 49 \)[/tex].

4. Substitute [tex]\( k = 16 \)[/tex] and [tex]\( x = 49 \)[/tex] into the equation [tex]\( y = \frac{k}{\sqrt{x}} \)[/tex]:

[tex]\[ y = \frac{16}{\sqrt{49}} \][/tex]

5. Calculate [tex]\(\sqrt{49}\)[/tex] which is 7:

[tex]\[ y = \frac{16}{7} \][/tex]

6. Simplify, if possible:

[tex]\[ y \approx 2.2857142857142856 \][/tex]

So, when [tex]\( x = 49 \)[/tex], [tex]\( y \approx 2.2857142857142856 \)[/tex].