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If $\alpha$ and $\beta$ are the two roots of $x^2+2x+2=0$, find $\alpha^{15}+\beta^{15}$.
Roots: $x=\frac{-2\pm\sqrt{4-8}}{2}=-1\pm i$. So $\alpha=-1+i, \beta=-1-i$.
In polar form: $|\alpha|=\sqrt{2}$, $\arg(\alpha)=\frac{3\pi}{4}$. So $\alpha=\sqrt{2}\,e^{i3\pi/4}$.
$\alpha^{15}=(\sqrt{2})^{15}e^{i\cdot45\pi/4}=2^{15/2}e^{i\pi/4}$ (since $45\pi/4=11\pi+\pi/4$, so $e^{i45\pi/4}=e^{i\pi/4}\cdot(-1)^{11}=-e^{i\pi/4}$... careful: $45/4=11.25$, $11\pi+\pi/4$, $e^{i(11\pi+\pi/4)}=e^{i\pi}\cdot e^{i\pi/4} \cdot (-1)^{10}... $)
More cleanly: $\alpha^{15}+\beta^{15}=2\,\text{Re}(\alpha^{15})=2(\sqrt{2})^{15}\cos\!\left(\frac{45\pi}{4}\right)$. Since $\cos(45\pi/4)=\cos(\pi/4+11\pi)=-\cos(\pi/4)=-\frac{1}{\sqrt{2}}$: $=2\cdot2^{15/2}\cdot(-\frac{1}{\sqrt{2}})=-2^{15/2+1-1/2}=-2^8=-\mathbf{256}$
Given that $0 < x < y < 1$, compare the two quantities:
Quantity A: $1 - y$
Quantity B: $y - x$
We know $0 < x < y < 1$. Let's test specific values to see if the relationship is fixed.
Example 1: $x = 0.1,\ y = 0.2$.
Quantity A: $1 - 0.2 = 0.8$. Quantity B: $0.2 - 0.1 = 0.1$. A > B.
Example 2: $x = 0.1,\ y = 0.9$.
Quantity A: $1 - 0.9 = 0.1$. Quantity B: $0.9 - 0.1 = 0.8$. B > A.
Since the relationship changes depending on the values of $x$ and $y$, it cannot be determined from the information given.
Bloom's Taxonomy classifies educational objectives into three domains:
- Cognitive domain: Knowledge, comprehension, application, analysis, synthesis, evaluation
- Affective domain: Attitudes, values, emotional responses
- Psychomotor domain: Physical and motor skills β e.g., performing injections, wound dressing, CPR
In nursing education, the psychomotor domain is critical for clinical skill development. A Head Nurse supervising nursing students or staff must assess all three domains for comprehensive competency evaluation.
If $2x = 3y = 4z = 20$, what is the value of $12xyz$?
From $2x = 20$: $x = 10$.
From $3y = 20$: $y = \dfrac{20}{3}$.
From $4z = 20$: $z = 5$.
Now compute $12xyz$:
$12 \times 10 \times \dfrac{20}{3} \times 5 = 12 \times \dfrac{1000}{3} = \dfrac{12000}{3} = 4{,}000$
Motor vehicles costing $530,000 were depreciated at 10% straight-line instead of the correct 25%. After incorrect entries: Motor vehicles at cost $530,000; Provision for depreciation $53,000. Which journal entries will correct the error?
Option C is correct.
Correct depreciation = $530,000 × 25% = $132,500
Already charged = $530,000 × 10% = $53,000
Additional needed = $132,500 − $53,000 = $79,500
Entry: Debit Statement of profit or loss $79,500 | Credit Provision for depreciation $79,500
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