If y varies inversely as the square of x, and y=0.5 when x=1, find y when x=9
"If y varies inversely as the square of x" ---- >
y = k(1/x^2) , where k is a constant
when x=1, y = .5
sub that in to find k,
write the equation again now that you know k
sub in x=9 and you are done
To find the value of y when x=9 in this inverse variation problem, we can start by setting up the inverse variation equation.
When two variables are inversely proportional, their product remains constant. In this case, y varies inversely as the square of x, so we can write the equation as:
y = k/x^2
where k is the constant of variation.
To find the constant of variation, we can use the given information that when y=0.5, x=1:
0.5 = k/1^2
0.5 = k
So, the equation becomes:
y = 0.5/x^2
Now, we can substitute x=9 into the equation to find the value of y:
y = 0.5/9^2
y = 0.5/81
y = 0.0062 (rounded to four decimal places)
Therefore, when x=9, y is approximately equal to 0.0062.