If y varies inversely with the square of x and y =9 when x=3, find y when x=0.6

Question

answer in integer or decimal

Answer 1

If two quantities vary inversely it means when one goes down the other goes up proportionally so that the product of the two remains constant.

 

We write this as   xy = k where k is the constant and xy is the product (multiplication) of the two quantities.

 

For you question the two quantities are y and x²

 

Therefore our equation is x²y = k

 

When x = 3 and y = 9, x²y = 3²(9) = 81 = k

 

So now our equation is x²y= 81

 

When x = .6, we have .6²y = 81

 

.36y = 81

 

y = 225

Answer 2

Jalesa:

 

You are given that y varies inversely with x².  Therefore, y = k(1/ x²), where k is  a constant, to be determined from other information. We are given that y = 9 when x = 3.  Plugging in, 9 = k (1/ 3²).  That is 9 = k (1/9).  Therefore k = 81..  We have y = 81 (1 / x²) for further use. When  x = 0,6, y = 81 (1/ (0.6)²) = 81 ( 1  /0.36  ) = 81/0.36 = 225.  Final answer.

 

Dr. G.

Answer 3

The basic formula for a "varies inversely" is y = k/x.  (BTW the basic formula for "varies directly" is y = kx).  In this case we have "varies inversely with the square of x", therefore we would write: y = k/(x^2).  This is an equation with 3 variables in it, and in order to do anything, the problem will have to give us more information, which they have.  We know that when y = 9. x = 3.  So we plug in 9 and 3 for y and x and we now  have (9) = k/(3^2).  Simplifying we get 9 = k/9.  Multiplying both sides by 9 we end up with 81 = k, or k = 81.  Now we can rewrite the formula as y = 81/x^2.  Lastly the problem asks us to find y when x = 0.6.  Substituting we get y = 81/(0.6^2) = 81/.36 = 225