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Which organism is used in the production of Humulin (human insulin)?
Recombinant E. coli (bacteria) or yeast are engineered to produce human insulin (Humulin). Bacteria are most common for large-scale production.
Let $P=\begin{pmatrix}1&0&0\\3&1&0\\9&3&1\end{pmatrix}$ and $Q=[q_{ij}]$ be two $3\times3$ matrices satisfying $Q - P^5 = I_3$. Find $\dfrac{q_{21}+q_{31}}{q_{32}}$.
Notice $P = I + N$ where $N=\begin{pmatrix}0&0&0\\3&0&0\\9&3&0\end{pmatrix}$ (strictly lower triangular, so $N^3=0$).
$P^5 = (I+N)^5 = I + 5N + 10N^2$ (since $N^3=0$).
$N^2 = \begin{pmatrix}0&0&0\\0&0&0\\9&0&0\end{pmatrix}$... computing: $(N^2)_{31}=3\cdot3=9$, others 0.
$P^5 = I + 5N + 10N^2$. So $Q = P^5 + I = 2I + 5N + 10N^2$.
$q_{21} = 5\cdot3 = 15$, $q_{31} = 5\cdot9+10\cdot9=45+90=135$... wait: $q_{31}=(5N)_{31}+(10N^2)_{31}=5\cdot9+10\cdot9=45+90=135$. Wait $N_{31}=9, N^2_{31}=9$: $q_{31}=5(9)+10(9)=135$. $q_{32}=(5N)_{32}=5\cdot3=15$.
$\dfrac{q_{21}+q_{31}}{q_{32}} = \dfrac{15+135}{15} = \mathbf{10}$
A sales ledger control account included: (1) Contra with purchases ledger: $500 debit; (2) Discount allowed: $600 debit; (3) Irrecoverable debts written off: $1,200 debit. Which entries must be corrected?
Option A (all three) is correct.
- Contra entry should be a credit (reduces receivables), not debit.
- Discount allowed should be a credit (reduces amount owed by customers), not debit.
- Irrecoverable debts written off should be a credit (reduces receivable), not debit.
Using $l = L + \frac{TL}{AY} = L + kT$:
- $l_1 = L + kT_1$ and $l_2 = L + kT_2$ [cite: 506]
- Solving for $L$ by eliminating $k$: $L = \frac{T_2l_1 - T_1l_2}{T_2 - T_1}$[cite: 515].
The practice of Flextime is best defined as:
Flextime is the practice of permitting employees to choose, with certain limitations, their own working hours. This helps organizations attract qualified individuals by allowing both employment and family needs to be addressed.
Using $\vec{s} = \vec{r}_0 + \vec{u}t + \frac{1}{2}\vec{a}t^2$:
- $x = 2 + 5(2) + \frac{1}{2}(4)(2^2) = 2 + 10 + 8 = 20$
- $y = 4 + 4(2) + \frac{1}{2}(4)(2^2) = 4 + 8 + 8 = 20$
- Distance from origin $D = \sqrt{20^2 + 20^2} = 20\sqrt{2}\text{ m}$.
A ball is dropped from rest at $t = 0$. After 6 s, another ball is thrown downward from the same point with speed $v$. Both balls meet at $t = 18\text{ s}$. Find $v$ ($g = 10\text{ m/s}^2$).
Ball 1 falls for 18 s: $h_1 = \dfrac{1}{2}(10)(18)^2 = 1620\text{ m}$
Ball 2 falls for $18 - 6 = 12\text{ s}$: $h_2 = 12v + \dfrac{1}{2}(10)(12)^2 = 12v + 720$
Setting $h_1 = h_2$: $12v + 720 = 1620 \Rightarrow 12v = 900 \Rightarrow v = \mathbf{75\text{ m/s}}$
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