This is regarding direct variation.
y varies directly as the square of R. If y = 128 when R = 8, find y when R = 5.
Another one I cannot figure out:
If x varies directly as y, and x = 32 when y = 8, find x when y = 10
Help!
y varies directly as the square of R
becomes ...
y = k √R , where k is a constant
let's use the given information to find k
when y = 128, R = 8, then
128 = k√8
k = 128/√8 or 16√8 or 32√2
so y = 32√(2R)
when R = 5
y = 32√10 = appr. 101.19
for the second
x = ky, try doing it the same way as the first
For the first problem, we are given that y varies directly with the square of R. This can be represented by the equation y = kR^2, where k is the constant of variation.
To find the value of k, we can substitute the given values of y and R into the equation. We have y = 128 and R = 8. Substituting these values, we get:
128 = k * 8^2
128 = k * 64
Dividing both sides of the equation by 64, we find that k = 2.
Now that we have the value of k, we can use it to find y when R = 5. Substituting R = 5 and k = 2 into the equation, we get:
y = 2 * 5^2
y = 2 * 25
y = 50
So, when R = 5, y = 50.
For the second problem, we are given that x varies directly with y. This can be represented by the equation x = ky, where k is the constant of variation.
To find the value of k, we can substitute the given values of x and y into the equation. We have x = 32 and y = 8. Substituting these values, we get:
32 = k * 8
Dividing both sides of the equation by 8, we find that k = 4.
Now that we have the value of k, we can use it to find x when y = 10. Substituting y = 10 and k = 4 into the equation, we get:
x = 4 * 10
x = 40
So, when y = 10, x = 40.
To solve the problems involving direct variation, we can use the formula: y = k*R^2 (for the first question) and x = k*y (for the second question), where k is the constant of variation.
1. First, we need to find the value of k in the first problem. We can do this by substituting the given values into the equation and solving for k:
y = k*R^2
128 = k*8^2
128 = 64k
k = 128/64
k = 2
Now that we have the value of k, we can use it to find y when R = 5:
y = 2*5^2
y = 2*25
y = 50
Therefore, when R = 5, y = 50.
2. Similarly, we can find the value of k in the second problem by substituting the given values into the equation:
x = k*y
32 = k*8
k = 32/8
k = 4
Now that we have the value of k, we can use it to find x when y = 10:
x = 4*10
x = 40
Therefore, when y = 10, x = 40.
I hope this helps!