# y varies inversely as x square.The difference in the values of y when x=1 and x=2is.find the relation between x and y.(plz help me,am not sure abt the answer,i got y=20/x square..plz..help me)

What is the difference in the values of y when x=1 and x=2?

## To find the difference in the values of y when x=1 and x=2, we need to first determine the relationship between x and y.

The given information states that "y varies inversely as x square." This means that as x increases, y decreases, and as x decreases, y increases, but at an inverse rate proportional to the square of x.

Let's denote the constant of variation as k.

The inverse variation equation can be written as:

y = k / x^2

To solve for k, we can use the given information that when x=1, y is some value, let's denote it as y1.

Plugging in x=1 and y=y1 into the equation, we have:

y1 = k / 1^2

y1 = k

Now, let's use the information that when x=2, y is some value, denoted as y2.

Plugging in x=2 and y=y2 into the equation, we have:

y2 = k / 2^2

y2 = k / 4

Now, to find the difference in the values of y when x=1 and x=2, we simply subtract y1 from y2:

Difference in y = y2 - y1

Difference in y = k/4 - k

Difference in y = -3k/4

So, the relation between x and y is: y = -3k/(4x^2)

Therefore, the answer you got, y = 20/x^2, is incorrect. The correct relation is y = -3k/(4x^2).