If x-3 is directly proportional to the square of y and x=5 when y=2 ,find x when y=6
x - 3 = k y^2
5 - 3 = 4 k ... k = 1/2
x - 3 = 1/2 * 6^2
I don't understand
If x-3 is directly proportional to the square of y and x=5 when y=2, find x when y=6
Well, let's start by understanding what "directly proportional" means. In this case, it means that as y increases or decreases, x will also increase or decrease, but not necessarily in the same ratio.
To find the value of x when y = 6, we can set up a proportion using the information given.
We know that (x - 3) is directly proportional to y^2, which can be written as:
(x - 3) = k * y^2
Where k is the constant of proportionality.
Now, we can substitute the values x = 5 and y = 2 into the equation to find the value of k:
(5 - 3) = k * 2^2
2 = 4k
k = 2/4
k = 1/2
Now, let's use this value of k in the equation to find the value of x when y = 6:
(x - 3) = (1/2) * 6^2
(x - 3) = (1/2)* 36
(x - 3) = 18
x = 18 + 3
x = 21
So, when y = 6, x = 21. Keep in mind that this is just a mathematical calculation and is not proportional to how funny I am.
To solve this problem, we need to use the concept of direct proportionality.
First, let's write down the proportionality equation:
x - 3 = k * y^2
Where k is the constant of proportionality.
We are given that when x = 5, y = 2. We can substitute these values into the equation to find the value of k:
5 - 3 = k * 2^2
2 = 4k
k = 1/2
Now that we have the value of k, we can solve for x when y = 6. Let's substitute these values into the equation:
x - 3 = (1/2) * 6^2
x - 3 = (1/2) * 36
x - 3 = 18
x = 18 + 3
x = 21
Therefore, when y = 6, x = 21.