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Mrac [35]
3 years ago
15

|x-2| + |2x+1| ≥ 3 How to solve this inequality? HELP

Mathematics
1 answer:
Phoenix [80]3 years ago
4 0

Answer: \begin{bmatrix}\mathrm{Solution:}\:&\:x\le \:-\frac{2}{3}\quad \mathrm{or}\quad \:x\ge \:0\:\\ \:\mathrm{Decimal:}&\:x\le \:-0.66666\dots \quad \mathrm{or}\quad \:x\ge \:0\\ \:\mathrm{Interval\:Notation:}&\:(-\infty \:,\:-\frac{2}{3}]\cup \:[0,\:\infty \:)\end{bmatrix}

Step-by-step explanation:

\left|x-2\right|+\left|2x+1\right|\ge \:3

x

x\le \:-\frac{2}{3}\quad \mathrm{or}\quad \:0\le \:x

x\le \:-\frac{2}{3}\quad \mathrm{or}\quad \:x\ge \:0

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To find f(3), simplify insert [x = 3] in the given equation:

f(3) = 5 × 2³

f(3) = 5 × 2 × 2 × 2

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4 0
2 years ago
Write an inequality that represents the missing dimension x
forsale [732]

For this case we have that by definition, the area of a rectangle is given by:

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Also we have:

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3 0
3 years ago
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Gwar [14]

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$0.89/apple

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7 0
2 years ago
What is the solution to the system of equations?
Mrac [35]

Answer:

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Substitute that in 2/3x - 1/3y = -2 gives:

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8 0
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