a circular pool measures 10 ft across. 1 cubic yard of concrete is to be used to create a circular border of uniform width around the pool. if the border's depth is 2 inches, how wide would will it be?

volume of stuff is 2 inches times (area of outer circle - area of inner circle)

22 inches = 1/6 foot
1 yard^3 = 27 ft^3

so
27 = (1/6)(pi)(R^2 - 5^2)

solve that for R
then width of path = R-5 in feet

2 inches = 1/6 foot

To find the width of the border, we need to know its volume first. The volume of the concrete used for the border will be the difference between the volume of the larger circle (pool + border) and the volume of the smaller circle (pool only).

First, let's find the radius of the pool:
The diameter of the pool is given as 10 ft, so the radius is half of that, which is 10 ft / 2 = 5 ft.

Now, let's find the area of the pool:
The area of a circle is calculated using the formula: A = πr^2, where A is the area and r is the radius.
So, the area of the pool is A = π * (5 ft)^2 = 25π square feet.

Next, let's find the radius of the larger circle (pool + border):
Since we want to find the width of the border, let's assume that the radius of the larger circle is (5 ft + x), where x is the width of the border.

Now, let's calculate the area of the larger circle (pool + border):
A = π * (5 ft + x)^2 = 25π + 10πx + πx^2 square feet.

We know that the volume of 1 cubic yard is equal to 27 cubic feet.
Now, let's find the volume of the larger circle:
The volume of a cylinder is calculated using the formula: V = Ah, where V is the volume, A is the base area, and h is the height.
So, the volume of the larger circle is V = (25π + 10πx + πx^2) * (2 inches) = (25π + 10πx + πx^2) * (2/12) feet = (25π + 10πx + πx^2) * (1/6) ft^3.

The volume of the concrete used for the border is the difference between the volume of the larger circle and the volume of the smaller circle:
(25π + 10πx + πx^2) * (1/6) - 25π = 1/3 ft^3.

Now, let's solve the equation to find the value of x, which is the width of the border:
(25π + 10πx + πx^2) * (1/6) - 25π = 1/3
(25π + 10πx + πx^2)/6 - 25π = 1/3
(25π + 10πx + πx^2)/6 = 1/3 + 25π
(25π + 10πx + πx^2)/6 = (1 + 75π)/3
25π + 10πx + πx^2 = 2(1 + 75π)
25π + 10πx + πx^2 = 2 + 150π
πx^2 + 10πx + 25π - 150π - 2 = 0
πx^2 + 10πx - 125π - 2 = 0

Now, we can solve this quadratic equation using the quadratic formula or any other appropriate method to find the value of x, which gives us the width of the border.