1. **Problem b:** Determine the equation of the parabola after transformation $(x,y) \to (x+4, \frac{1}{4}y)$ applied to $y=x^2$. 2. The original parabola is $y=x^2$. 3. The transformation changes $x$ to $x+4$ and $y$ to $\frac{1}{4}y$. 4. Substitute $x$ with $x+4$ in the original equation: $y = (x+4)^2$. 5. Since $y$ is scaled by $\frac{1}{4}$, the new $y$ is $\frac{1}{4}y$, so: $$\frac{1}{4}y = (x+4)^2 \implies y = 4(x+4)^2$$ --- 1. **Problem c:** Show that $\sin 60^\circ = \frac{2 \tan 30^\circ}{1 + \tan^2 30^\circ}$. 2. Recall the double angle formula for tangent: $$\tan 2\theta = \frac{2 \tan \theta}{1 - \tan^2 \theta}$$ 3. But here, the expression resembles the sine double angle formula rewritten in terms of tangent. 4. Calculate $\tan 30^\circ = \frac{1}{\sqrt{3}}$. 5. Substitute into the right side: $$\frac{2 \times \frac{1}{\sqrt{3}}}{1 + \left(\frac{1}{\sqrt{3}}\right)^2} = \frac{\frac{2}{\sqrt{3}}}{1 + \frac{1}{3}} = \frac{\frac{2}{\sqrt{3}}}{\frac{4}{3}} = \frac{2}{\sqrt{3}} \times \frac{3}{4} = \frac{3}{2\sqrt{3}} = \frac{\sqrt{3}}{2}$$ 6. Since $\sin 60^\circ = \frac{\sqrt{3}}{2}$, the equality holds. --- 1. **Problem d i:** On a 100 chart, an L shape is formed by shading four numbers. Write a function for the sum of values in the L in terms of the greatest value $g$ in the L. 2. The L shape includes the greatest value $g$, the number above it $g-10$, and two numbers to the left $g-1$ and $g-2$. 3. Sum of values in L: $$S = g + (g-10) + (g-1) + (g-2) = 4g - 13$$ --- 1. **Problem d ii:** Given sum $S=71$, find greatest value $g$. 2. From $S=4g - 13$, solve for $g$: $$71 = 4g - 13 \implies 4g = 84 \implies g = 21$$ --- 1. **Problem 4 a:** Calculate $n$ if $(\sqrt{6} + \sqrt{8})(\sqrt{n} - \sqrt{8}) = 0$. 2. For product to be zero, either factor is zero. 3. $\sqrt{6} + \sqrt{8} \neq 0$, so: $$\sqrt{n} - \sqrt{8} = 0 \implies \sqrt{n} = \sqrt{8} \implies n = 8$$ --- 1. **Problem 4 b i:** Dechen Choden buys shares at 28% discount, face value 100, with 25000 money. 2. Discounted price per share: $$100 - 0.28 \times 100 = 72$$ 3. Number of shares: $$\frac{25000}{72} \approx 347.22 \implies 347 \text{ shares (whole shares)}$$ --- 1. **Problem 4 b ii:** Dividend at 17% on face value 100 per share. 2. Dividend per share: $$0.17 \times 100 = 17$$ 3. Total dividend: $$347 \times 17 = 5899$$ --- 1. **Problem 4 b iii:** Yield percentage = (Dividend / Amount invested) \times 100 2. Amount invested: $$347 \times 72 = 24984$$ 3. Yield: $$\frac{5899}{24984} \times 100 \approx 23.6\%$$ --- 1. **Problem 4 c i:** For $2t + 5m = 4$, write $m$ as a function of $t$. 2. Solve for $m$: $$5m = 4 - 2t \implies m = \frac{4 - 2t}{5}$$ --- 1. **Problem 4 c ii:** Write $t$ as a function of $m$. 2. Solve for $t$: $$2t = 4 - 5m \implies t = \frac{4 - 5m}{2}$$ --- 1. **Problem 4 d:** Find surface area of paper to make cone-shaped cup with diameter 6 cm and depth 4 cm. 2. Radius $r = \frac{6}{2} = 3$ cm. 3. Slant height $l = \sqrt{r^2 + h^2} = \sqrt{3^2 + 4^2} = 5$ cm. 4. Lateral surface area (paper needed): $$\pi r l = \pi \times 3 \times 5 = 15\pi \approx 47.12 \text{ cm}^2$$ --- 1. **Problem 5 a i:** Lobzang works two jobs paying 600 and 500 per hour, wants to earn 4500. 2. Let hours at job 1 be $x$, job 2 be $y$. 3. Income equation: $$600x + 500y = 4500$$ --- 1. **Problem 5 a ii:** Total hours worked is 8. 2. Hours equation: $$x + y = 8$$ --- 1. **Problem 5 a iii:** Solve system: $$600x + 500y = 4500$$ $$x + y = 8$$ 2. From second, $y = 8 - x$. 3. Substitute: $$600x + 500(8 - x) = 4500 \implies 600x + 4000 - 500x = 4500 \implies 100x = 500 \implies x = 5$$ 4. Then $y = 8 - 5 = 3$. --- 1. **Problem 5 b:** Calculate $3A + B$ for matrices $$A = \begin{bmatrix}4 & 1 & 0 \\ 5 & 10 & 2\end{bmatrix}, B = \begin{bmatrix}2 & 4 & 3 \\ 1 & 3 & 4\end{bmatrix}$$ 2. Calculate $3A$: $$3A = \begin{bmatrix}12 & 3 & 0 \\ 15 & 30 & 6\end{bmatrix}$$ 3. Add $3A + B$: $$\begin{bmatrix}12+2 & 3+4 & 0+3 \\ 15+1 & 30+3 & 6+4\end{bmatrix} = \begin{bmatrix}14 & 7 & 3 \\ 16 & 33 & 10\end{bmatrix}$$