Answer:
(a) x² - 4x > 3:
First, move everything to one side to get x² - 4x - 3 > 0. This factors to (x - 3)(x + 1) > 0. The solution is x < -1 or x > 3.
(d) 4x - x² < 0:
Factor the left side to get x(4 - x) < 0. This inequality holds when 0 < x < 4.
(e) 2r² + 10 ≤ 5 - 2³:
Simplify the equation to get 2r² + 10 ≤ -3. Subtract 10 from both sides: 2r² ≤ -13. Since the coefficient of r² is positive, there are no real solutions.
(h) x² - 4x + 12 ≤ x + 3:
First, move everything to one side to get x² - 5x + 9 ≤ 0. This quadratic does not have real solutions, so there are no solutions.
(c) 2x² < 9x + 5:
Move everything to one side to get 2x² - 9x - 5 < 0. This factors to (2x + 1)(x - 5) < 0. The solution is -1/2 < x < 5.
(f) x² - x + 1 > 0:
This quadratic is always positive since its discriminant is negative. So, the solution is all real numbers.
(i) r² + 19 ≤ 5(x - 1):
Simplify the equation to get r² + 19 ≤ 5x - 5. Subtract 19 from both sides: r² ≤ 5x - 24. There are no restrictions on r for this inequality.
(c) x + 2 < 2x + 5:
Subtract x from both sides: 2 < x + 5. Subtract 5 from both sides: -3 < x.
(f) |x + 3| > 12:
This inequality holds when x + 3 > 12 or x + 3 < -12. The solution is x > 9 or x < -15.
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