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KonstantinChe [14]
3 years ago
14

What is the solution set for 4x + 12 ≤ 8?

Mathematics
2 answers:
timurjin [86]3 years ago
7 0

Answer:x

≥

−

7

Step-by-step explanation:

aivan3 [116]3 years ago
5 0

Answer:

Inequality Form: x ≤ -1

Interval Notation: (−∞,−1]

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If g(x) = \dfrac{1}{x}, then g(x+h) = \dfrac{1}{x+h}. It follows that

  \begin{aligned} \\\frac{g(x+h)-g(x)}{h} &= \frac{1}{h} \cdot [g(x+h) - g(x)] \\&= \frac{1}{h} \left( \frac{1}{x+h} - \frac{1}{x} \right)\end{aligned}

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  \begin{aligned} \\\frac{g(x+h)-g(x)}{h} &= \frac{1}{h} \left( \frac{1}{x+h} - \frac{1}{x} \right)\\&=\frac{1}{h} \left(\frac{x}{x(x+h)} - \frac{x+h}{x(x+h)} \right) \\ &=\frac{1}{h} \left(\frac{x-(x+h)}{x(x+h)}\right) \\ &=\frac{1}{h} \left(\frac{x-x-h}{x(x+h)}\right) \\ &=\frac{1}{h} \left(\frac{-h}{x(x+h)}\right) \end{aligned}

Now we can cancel the h in the numerator and denominator under the assumption that h is not 0.

  = \dfrac{-1}{x(x+h)}, h\ne 0

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