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Sagot :
Let's solve the problem step-by-step.
Given:
- The value of [tex]\( y \)[/tex] varies inversely with [tex]\( x \)[/tex].
- When [tex]\( x = 9 \)[/tex], [tex]\( y = 7 \)[/tex].
We know that for inverse variation, the product [tex]\( k \)[/tex] of [tex]\( x \)[/tex] and [tex]\( y \)[/tex] remains constant:
[tex]\[ k = x \cdot y \][/tex]
First, we will calculate the constant [tex]\( k \)[/tex] using the given values of [tex]\( x \)[/tex] and [tex]\( y \)[/tex]:
[tex]\[ k = 9 \cdot 7 = 63 \][/tex]
Now that we have the constant [tex]\( k = 63 \)[/tex], we use it to find the new value of [tex]\( y \)[/tex] when [tex]\( x = 3 \)[/tex]:
[tex]\[ y = \frac{k}{x} \][/tex]
Substitute [tex]\( k = 63 \)[/tex] and [tex]\( x = 3 \)[/tex] into the equation:
[tex]\[ y = \frac{63}{3} = 21 \][/tex]
So, the value of [tex]\( y \)[/tex] when [tex]\( x = 3 \)[/tex] is:
[tex]\[ y = 21 \][/tex]
Given:
- The value of [tex]\( y \)[/tex] varies inversely with [tex]\( x \)[/tex].
- When [tex]\( x = 9 \)[/tex], [tex]\( y = 7 \)[/tex].
We know that for inverse variation, the product [tex]\( k \)[/tex] of [tex]\( x \)[/tex] and [tex]\( y \)[/tex] remains constant:
[tex]\[ k = x \cdot y \][/tex]
First, we will calculate the constant [tex]\( k \)[/tex] using the given values of [tex]\( x \)[/tex] and [tex]\( y \)[/tex]:
[tex]\[ k = 9 \cdot 7 = 63 \][/tex]
Now that we have the constant [tex]\( k = 63 \)[/tex], we use it to find the new value of [tex]\( y \)[/tex] when [tex]\( x = 3 \)[/tex]:
[tex]\[ y = \frac{k}{x} \][/tex]
Substitute [tex]\( k = 63 \)[/tex] and [tex]\( x = 3 \)[/tex] into the equation:
[tex]\[ y = \frac{63}{3} = 21 \][/tex]
So, the value of [tex]\( y \)[/tex] when [tex]\( x = 3 \)[/tex] is:
[tex]\[ y = 21 \][/tex]
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