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What are the limitations of CVP analysis?
- It assumes fixed costs will remain constant.
- It cannot be used if multiple products are produced in a constant mix.
- It ignores the possibility of semi-variable costs.
Option C (1 and 3) is correct.
- Statement 1 (True limitation): Fixed costs can change in steps in reality.
- Statement 2 (False): CVP can be used with multiple products in a constant mix.
- Statement 3 (True limitation): CVP ignores semi-variable costs, treating all costs as purely fixed or variable.
For $x\in\left(0,\dfrac{1}{4}\right)$, the derivative of $\tan^{-1}\!\left(\dfrac{6x\sqrt{x}}{1-9x^3}\right)$ equals $\sqrt{x}\cdot g(x)$. Find $g(x)$.
Write $\frac{6x\sqrt{x}}{1-9x^3} = \frac{2\cdot3x^{3/2}}{1-(3x^{3/2})^2}=\frac{2t}{1-t^2}$ where $t=3x^{3/2}$.
So $\tan^{-1}\!\left(\frac{2t}{1-t^2}\right)=2\tan^{-1}(t)=2\tan^{-1}(3x^{3/2})$
Differentiate: $\frac{d}{dx}[2\tan^{-1}(3x^{3/2})] = \frac{2}{1+9x^3}\cdot 3\cdot\frac{3}{2}x^{1/2} = \frac{9\sqrt{x}}{1+9x^3}$
Since this equals $\sqrt{x}\cdot g(x)$: $g(x)=\dfrac{9}{1+9x^3}$
If $\dfrac{1}{2^{11} \cdot 5^{17}}$ is written as a terminating decimal, how many nonzero digits will it contain?
Rewrite the expression to have a denominator that is a power of 10:
$\dfrac{1}{2^{11} \cdot 5^{17}} = \dfrac{1}{2^{11} \cdot 5^{17}} \times \dfrac{2^6}{2^6} = \dfrac{2^6}{2^{17} \cdot 5^{17}} = \dfrac{64}{10^{17}}$
So the decimal is $\dfrac{64}{10^{17}} = 0.\underbrace{00\ldots0}_{15}64$
The nonzero digits are $6$ and $4$ β that is, two nonzero digits.
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