The quantity y varies directly with the square of x. If y = 3 when x = 9, find y when x is 11.
Round answer to the nearest hundredth.
y=kx^2
y/x^2 is constant
You want y such that
y/11^2 = 3/9^2
To find the value of y when x is 11, we can use the direct variation equation that states that y varies directly with the square of x. The equation can be written as:
y = kx^2
where k is the constant of variation.
To find the value of k, we can use the given information that when x is 9, y is 3. Plugging these values into the equation, we get:
3 = k * 9^2
Simplifying this equation, we have:
3 = k * 81
To solve for k, divide both sides of the equation by 81:
k = 3/81
Now that we have the value of k, we can use it to find y when x is 11:
y = k * 11^2
Substituting k = 3/81 and x = 11 into the equation, we have:
y = (3/81) * 11^2
Simplifying this equation:
y = (3/81) * 121
y = 3.6173
Therefore, when x is 11, rounded to the nearest hundredth, y is approximately 3.62.