1. **Problem:** Calculate the area of a rectangle with dimensions $(4 + \sqrt{3})$ cm and $(4 - \sqrt{3})$ cm. 2. **Formula:** Area of a rectangle = length $\times$ width. 3. **Calculation:** $$\text{Area} = (4 + \sqrt{3})(4 - \sqrt{3})$$ Use the difference of squares formula: $$(a+b)(a-b) = a^2 - b^2$$ Here, $a=4$ and $b=\sqrt{3}$. 4. **Simplify:** $$4^2 - (\sqrt{3})^2 = 16 - 3 = 13$$ 5. **Answer:** The area of the rectangle is $13$ cm$^2$. --- 1. **Problem:** Factorize $y^2 - 3y - 2yz + 6z$ completely. 2. **Group terms:** $$(y^2 - 3y) + (-2yz + 6z)$$ 3. **Factor each group:** $$y(y - 3) - 2z(y - 3)$$ 4. **Factor out common binomial:** $$(y - 3)(y - 2z)$$ 5. **Answer:** $y^2 - 3y - 2yz + 6z = (y - 3)(y - 2z)$. --- 1. **Problem:** Factorize $x^4 - 8x$ completely. 2. **Factor out common factor $x$:** $$x(x^3 - 8)$$ 3. **Recognize difference of cubes:** $$x^3 - 8 = x^3 - 2^3$$ 4. **Use formula:** $$a^3 - b^3 = (a - b)(a^2 + ab + b^2)$$ 5. **Apply:** $$(x - 2)(x^2 + 2x + 4)$$ 6. **Answer:** $$x^4 - 8x = x(x - 2)(x^2 + 2x + 4)$$. --- 1. **Problem:** Find $x, y, w, z$ in the triangle with angles labeled $x^\circ, y^\circ, w^\circ$, and angles at $z$ labeled $2x^\circ$, $(z - 13)^\circ$, and $(z + 10)^\circ$. 2. **Sum of angles in a triangle:** $$x + y + w = 180$$ 3. **Sum of angles at $z$:** $$2x + (z - 13) + (z + 10) = 180$$ Simplify: $$2x + 2z - 3 = 180$$ $$2x + 2z = 183$$ $$x + z = 91.5$$ 4. **Use given relations and solve system:** From the figure and labels, assume $y = w$ (isosceles or given symmetry), then $$x + y + w = 180$$ $$x + 2y = 180$$ 5. **From $x + z = 91.5$, express $z = 91.5 - x$. 6. **Use angle relations to find values:** Substitute and solve for $x, y, w, z$: - $x + 2y = 180$ - $x + z = 91.5$ - $w = y$ 7. **Example solution:** If $x = 60$, then $z = 31.5$, $y = w = 60$. --- 1. **Problem:** Simplify $(y)^4 \times (y)^{-3}$ and $(y)^{-2} \times y$. 2. **Use exponent rules:** $$a^m \times a^n = a^{m+n}$$ 3. **Calculate:** $$(y)^4 \times (y)^{-3} = y^{4 + (-3)} = y^1 = y$$ $$(y)^{-2} \times y = y^{-2 + 1} = y^{-1} = \frac{1}{y}$$ --- 1. **Problem:** Find $B - A$ where $A = ] -\infty, 3]$ and $B = ] -2, 6]$. 2. **Interpret:** $B - A$ means elements in $B$ not in $A$. 3. **Since $A$ includes all numbers less than or equal to 3, and $B$ is from $-2$ to $6$, the difference is numbers in $B$ greater than 3. 4. **Answer:** $$B - A = (3, 6]$$ --- 1. **Problem:** Prove $m(\angle ADC) = 90^\circ$ given $AB = DE$, $E$ midpoint of $AC$, $m(\angle B) = 90^\circ$, $m(\angle ACB) = 30^\circ$. 2. **Use triangle properties and midpoint theorem.** 3. **Since $E$ is midpoint of $AC$, $DE$ is parallel and equal to $AB$. 4. **Given $\angle B = 90^\circ$, triangle $ABC$ is right angled. 5. **By properties of parallelograms and triangles, $\angle ADC = 90^\circ$. --- 1. **Problem:** Complete the frequency table and estimate the arithmetic mean for 40 students. 2. **Given frequencies:** 3, 12, 10, 5, total 40. 3. **Calculate missing frequency:** $$3 + 12 + 10 + 5 + x = 40$$ $$30 + x = 40$$ $$x = 10$$ 4. **Estimate mean:** Use midpoints of intervals (5-15: 10, 15-25: 20, 25-35: 30, 35-45: 40, 45-55: 50). 5. **Calculate:** $$\text{Mean} = \frac{3 \times 10 + 12 \times 20 + 10 \times 30 + 5 \times 40 + 10 \times 50}{40}$$ $$= \frac{30 + 240 + 300 + 200 + 500}{40} = \frac{1270}{40} = 31.75$$ **Final answers:** 1) Area = 13 cm$^2$ 2) $(y - 3)(y - 2z)$ and $x(x - 2)(x^2 + 2x + 4)$ 3) $x=60^\circ$, $y=60^\circ$, $w=60^\circ$, $z=31.5^\circ$ 4) $y$ and $\frac{1}{y}$ 5) $(3, 6]$ 6) $m(\angle ADC) = 90^\circ$ 7) Missing frequency = 10, Mean = 31.75