If y varies directly as the square root of w and inversely as the cube of x, by what factor will y change if w is tripled and x is halved?
I hate math
alebra is dum
y = k * √w/x^3
The new y, y' will be
y' = k * √3w/(x/2)^3
= k * √3 * √w / (x^3/8)
= k * √w/x^3 * [√3 * 8]
= y * 8√3
To find the factor by which y will change, we need to understand the relationship between y, w, and x based on the given information.
Let's start by writing the direct variation statement:
y varies directly as the square root of w
We can represent this as an equation:
y = k√w
where k is the constant of variation.
Next, let's write the inverse variation statement:
y varies inversely as the cube of x
We can represent this as an equation too:
y = k/x^3
Now that we have the equations, we can find the unknown constant of variation, k.
To do this, we will consider the initial conditions. Let's assume we have some values for y, w, and x.
Let y₁, w₁, and x₁ represent the initial values, and y₂, w₂, and x₂ represent the new values after the changes are made.
Given that y varies directly as the square root of w, we can write:
y₁ = k√w₁
Given that y varies inversely as the cube of x, we can write:
y₁ = k/x₁^3
Now let's consider the values after the changes are made (y₂, w₂, and x₂):
y₂ = k√w₂
y₂ = k/x₂^3
Given that w is tripled (w₂ = 3w₁) and x is halved (x₂ = 0.5x₁), we can substitute these values into the equations:
y₁ = k√w₁
y₂ = k√(3w₁)
y₁ = k/x₁^3
y₂ = k/(0.5x₁)^3
To find the factor by which y will change, we can divide y₂ by y₁:
(y₂/y₁) = (k√(3w₁))/(k√w₁)
(y₂/y₁) = (k/(0.5x₁)^3)/(k/x₁^3)
We notice that the constant of variation, k, cancels out in both equations:
(y₂/y₁) = (√(3w₁))/(√w₁)
(y₂/y₁) = ((0.5x₁)^3)/(x₁^3)
Simplifying further:
(y₂/y₁) = (√3)
(y₂/y₁) = 0.5^3
Therefore, the factor by which y will change if w is tripled and x is halved is (√3) or approximately 1.732.