If x varies inversely as (y2 - 1) and is equal to 24 when y = 10, the value of x when y = 5 will be ?
Question
If x varies inversely as (y<sup>2</sup> - 1) and is equal to 24 when y = 10, the value of x when y = 5 will be ?
Solution
The problem states that x varies inversely as (y^2 - 1). This can be written as:
x = k / (y^2 - 1)
where k is the constant of variation.
We know that x = 24 when y = 10. We can substitute these values into the equation to find k:
24 = k / (10^2 - 1) 24 = k / 99 k = 24 * 99 k = 2376
Now that we know k, we can find x when y = 5:
x = 2376 / (5^2 - 1) x = 2376 / 24 x = 99
So, the value of x when y = 5 is 99.
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