Let $a_1$, $a_2$, $\dots$, $a_{12}$ be twelve equally spaced points on a circle with radius 1. find\[(a_1 a_2)^2 + (a_1 a_3)^2 + \dots + (a_{11} a_{12})^2.\](the sum includes the square of the distance between any pair of points, so the sum includes $\binom{12}{2} = 66$ terms.)

The sum evaluates to 144. See the linked question in the comments for details.

so firstly, we need to isolate the y variables to be able to solve these inequalities. firstly, add 0.5x on both sides of the first inequality and subtract 1.5x on both sides of the second inequality:

now since the slope is positive for the first inequality, this means that the line going upwards belongs to the first inequality, and the line going downwards belongs to the second inequality.

next, since y ≥ in the first inequality, this means that the shaded region will be above the first inequality's line, thus shading regions a and b.

next, since y ≤ in the second inequality, this means that the shaded region is going to be below this line, thus shading regions a and c.

since both lines shade region a, this means that region a is the solution.

step-by-step explanation:

mark brainliest!