# The quantity y varies directly with the square of x. If y = 3 when x = 9, find y when x is 11....

## Question:

The quantity {eq}y {/eq} varies directly with the square of {eq}x {/eq}. If {eq}y = 3 {/eq} when {eq}x = 9 {/eq}, find {eq}y {/eq} when {eq}x {/eq} is {eq}11 {/eq}. Round answer to the nearest hundredth.

## Direct Variation:

Direct variation explains a relationship between two variables that change in the same direction. For such variables, their ratio is equal to a constant which is known as the proportionality constant or the constant of variation.

If te variable {eq}y {/eq} varies directly as the square of {eq}x {/eq}, we can write this as:

• {eq}y\propto x^2 {/eq}

Removing the proportionality sign and adding a constant of variation, we have:

• {eq}y = kx^2 {/eq}

Given that {eq}y = 3 {/eq} when {eq}x = 9 {/eq}, we can solve for the constant of variation as:

• {eq}3 = k\times 9^2 {/eq}
• {eq}k = \dfrac{3}{9^2} = \dfrac{3}{81} = \dfrac{1}{27} {/eq}

Thus, an equation that gives the value of y for any given value of x is equal to:

• {eq}y = \dfrac{1}{27}x^2 {/eq}

Using the above equation, the value of y when x = 11 is equal to:

• {eq}y = \dfrac{1}{27}\times 11^2 = \boxed{\dfrac{121}{27}} {/eq} 