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Step-by-step ABG interpretation:
- pH: \(7.28 < 7.35\) → Acidosis
- PaCO\(_2\): \(55 > 45 \text{ mmHg}\) → Respiratory cause (CO\(_2\) is an acid; elevated = acidosis)
- HCO\(_3^-\): \(24 \text{ mEq/L}\) → Normal (22–26 mEq/L) → No metabolic compensation yet
Normal ABG Reference Values:
| Parameter | Normal Range |
|---|---|
| pH | \(7.35 - 7.45\) |
| PaCO\(_2\) | \(35 - 45 \text{ mmHg}\) |
| HCO\(_3^-\) | \(22 - 26 \text{ mEq/L}\) |
| PaO\(_2\) | \(80 - 100 \text{ mmHg}\) |
| SaO\(_2\) | \(95 - 100\%\) |
In the absence of a formal partnership agreement, which of the following rules apply under standard partnership law?
- Partners receive 10% interest per annum on loans to the business.
- Partners receive 5% interest per annum on their capital.
- No interest is charged on partners' drawings.
- Partners are not entitled to a salary.
Under the UK Partnership Act 1890 (often used as the basis for these exams), if no agreement exists: no interest on capital, no salaries, no interest on drawings, and interest on partner loans is usually 5% (not 10%). Thus, only 3 and 4 are correct.
Balkrishna Doshi (B.V. Doshi) was awarded the Pritzker Architecture Prize in 2018 — the first Indian architect to receive this honour.
- Known for: Aranya Community Housing (Indore), CEPT University (Ahmedabad), Sangath (his own studio).
- He was a student/collaborator of both Le Corbusier and Louis Kahn.
- He passed away in January 2023.
Charles Correa received the Pritzker in 2022 (posthumous? — No, the 2022 prize went to Diébédo Francis Kéré). Correa never received the Pritzker.
Let $A=\begin{pmatrix}i&-i\\-i&i\end{pmatrix}$ where $i=\sqrt{-1}$. The system $A^8\begin{pmatrix}x\\y\end{pmatrix}=\begin{pmatrix}8\\64\end{pmatrix}$ has:
$A = \begin{pmatrix}i&-i\\-i&i\end{pmatrix}$. Note $\det(A) = i^2-i^2=0$, so $A$ is singular.
$A^2$: compute $(A)^2$. Row1·Col1: $i(i)+(-i)(-i)=-1+(-1)=-2i$... Let's compute: $A^2_{11}=i\cdot i+(-i)(-i)=i^2+i^2=-1-1=-2$. So $A^2 = -2A$... check: $A^2 = -2\begin{pmatrix}i&-i\\-i&i\end{pmatrix}$.
Therefore $A^8 = (-2)^4 A^4 = 16(A^2)^2=16\cdot4A^2=64\cdot(-2A)=-128A$.
Wait: $A^2=-2A \Rightarrow A^4=4A^2=-8A \Rightarrow A^8=-8A^4... $ Correctly: $A^8=(-2)^7A = -128A$.
$-128A\begin{pmatrix}x\\y\end{pmatrix}=\begin{pmatrix}8\\64\end{pmatrix}$. Since $\det(A)=0$, check consistency. The equations are linearly dependent but RHS $(8,64)$ is not in column space of $A$. Hence no solution.
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