What is the set of values for which 12 - x^2 > 8 and 2x + 3 ≥ 5?
A. 1 ≤ x
B. 1 < x ≤ 2
C. 1 ≤ x < 2
D. 2 < x
E. -1 ≤ x < 2
To solve this question we will have two parts.
-
part:
12-x^2>8 \to 12>8 +x^2 \to 12-8 > x^2 \to 4>x^2
Or better typed: x^2<4. This means that -2<x<2. -
part:
2x +3 \ge 5 \to 2x \ge 5-3 \to 2x \ge 2. Dividing the whole equation with 2 gives us the final form: x\ge 1
Now, when both parts are finished, we need to see which areas collide. The best way to do this is to draw a numeric line:
The area that collides is the solution: (C)1\le x<2
TIP: Put the circles on the numeric line to indicate that the x can take that value, so you don’t exclude it.
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What does 8 and 2x +3 mean???
Lmao now I get it, I’m such a dumbass
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