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Using $n_1 u_1 = n_2 u_2$:
$128 \times [\text{kg} \cdot \text{m}^{-3}] = n_2 \times [(50 \text{ g}) \cdot (25 \text{ cm})^{-3}]$
$n_2 = 128 \times \left( \frac{1000 \text{ g}}{50 \text{ g}} \right) \times \left( \frac{25 \text{ cm}}{100 \text{ cm}} \right)^3$
$n_2 = 128 \times 20 \times \left( \frac{1}{4} \right)^3 = \frac{128 \times 20}{64} = 2 \times 20 = 40$.
Differentiating high peak vs. high plateau pressure is critical:
| Finding | Cause | Problem |
|---|---|---|
| High Peak + Normal Plateau | Increased airway resistance | ETT kinking/obstruction, secretions, bronchospasm, biting the tube |
| High Peak + High Plateau | Decreased lung compliance | ARDS, pneumothorax, pulmonary edema, auto-PEEP |
Normal compliance \(\approx 50 - 100 \text{ mL/cmH}_2\text{O}\)
Nursing action for high peak, normal plateau:
- Suction the ETT for secretions
- Check for tube kinking or patient biting
- Administer bronchodilator if bronchospasm suspected
- Assess breath sounds bilaterally
Using only principal values of inverse functions, describe the set $A=\left\{x\geq0:\tan^{-1}(2x)+\tan^{-1}(3x)=\dfrac{\pi}{4}\right\}$.
Apply the addition formula: $\tan^{-1}(2x)+\tan^{-1}(3x)=\tan^{-1}\!\left(\dfrac{5x}{1-6x^2}\right)$ when $6x^2<1$.
Setting equal to $\pi/4$: $\dfrac{5x}{1-6x^2}=1 \Rightarrow 5x=1-6x^2 \Rightarrow 6x^2+5x-1=0$
$x=\dfrac{-5\pm\sqrt{25+24}}{12}=\dfrac{-5\pm7}{12}$
$x=\dfrac{2}{12}=\dfrac{1}{6}$ or $x=\dfrac{-12}{12}=-1$.
Since $x\geq0$, only $x=\dfrac{1}{6}$ is valid. Check: $6\cdot(1/6)^2=1/6<1$ ✓. So $A=\{1/6\}$ — a singleton.
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