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).