Find the constant of variation for the relation and use it to write an equation for the statement. Then solve the equation.
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Answer: b.
Step-by-step explanation:
Based on the information given, you can write the following expression:
[tex]y=\frac{k}{x^2}[/tex]
Where k is the the constant of variation
If y=4/63 when x=3, then you can substitute these values into the expression and solve for k:
[tex]\frac{4}{63}=\frac{k}{3^2}\\k=9*\frac{4}{63}\\k=\frac{4}{7}[/tex]
Substitute k into the expression. Then the equation is:
[tex]y=\frac{4}{7x^{2}}[/tex]
Substitute x=5 into the equation. Then, y is:
[tex]y=\frac{4}{7(5)^{2}}=\frac{4}{175}}[/tex]
Answer:
Final answer is choice B.[tex]y=\frac{4}{7x^2}[/tex], [tex]y(5)=\frac{4}{175}[/tex]
Step-by-step explanation:
Given that if y varies inversely as square of x, and [tex]y=\frac{4}{63}[/tex] when x=3, then we need to find out y-value when x=5.
We also need to find the constant of variation and the equation.
Since y varies inversely as square of x, so we can write equation
[tex]y=\frac{k}{x^2} [/tex]
where k is constant of variation.
Plug given values [tex]y=\frac{4}{63}[/tex] and x=3
[tex]y=\frac{k}{x^2} [/tex]
[tex]\frac{4}{63} =\frac{k}{3^2}[/tex]
[tex]\frac{4}{63} =\frac{k}{9}[/tex]
[tex]\frac{4}{63}*9 =k[/tex]
[tex]\frac{4}{7} =k[/tex]
Hence constant of variation is [tex]k=\frac{4}{7}[/tex]
Now plug the value of k into formula
we get required equation as [tex]y=\frac{4}{7x^2}[/tex]
Now plug the value of x=5 into above formula
[tex]y=\frac{4}{7x^2}[/tex]
[tex]y=\frac{4}{7(5)^2}[/tex]
[tex]y=\frac{4}{7(25)}[/tex]
[tex]y=\frac{4}{175}[/tex]
Hence final answer is choice B.[tex]y=\frac{4}{7x^2}[/tex], [tex]y(5)=\frac{4}{175}[/tex]