Solve for ALL values of x, where 0<x<2Pi. (should be greater than or equal to)

use

>= for greater than or equal
<= for less than or equal

Sierra spent 2/3 of her savings on clothes and 1/5 0n school supplies. She saved the rest. What fraction of her savings did she save?

3/15+10/15=13/15. She saved 2/15.

To solve for all values of x where 0 < x ≤ 2π, we need to consider all numbers between 0 and 2π, including 0 but excluding 2π.

For this problem, we are looking for all values of x in the interval (0, 2π]. It means that x can take any value between 0 and 2π, excluding 0 but including 2π.

To solve this, we can start by expressing the solution in terms of the variable θ instead of x. Usually, θ represents an angle in trigonometric functions. In this case, we want to find values of θ that satisfy the given conditions.

0 < θ ≤ 2π

Next, let's consider the unit circle, which is a circle with a radius of 1 centered at the origin (0, 0) on a coordinate plane. We will use the unit circle to find the solutions for θ.

In the unit circle, the angle formed by the x-axis and the line connecting the origin to a point on the circle represents the angle θ.

Starting from 0 radians (or 0 degrees) on the positive x-axis, we can go clockwise around the unit circle until we reach 2π radians (or 360 degrees), which brings us back to the starting point.

All the angles (values of θ) within the interval (0, 2π] on the unit circle satisfy the given conditions.

So, all the values of x that satisfy 0 < x ≤ 2π are the corresponding angles in radians (or degrees) on the unit circle.

To find these values, we can use the following steps:
1. Starting from 0, go around the unit circle in a clockwise direction until you reach 2π.
2. At each point you pass, write down the corresponding angle in radians or degrees.

The angles (values of x) that satisfy the given conditions will be all the values you recorded.

Therefore, in this case, for 0 < x ≤ 2π, the solution is all the angles on the unit circle between 0 and 2π (excluding 0 but including 2π).