1. **Find the total surface area of a cone and explain the terms.** The total surface area $A$ of a cone is given by the formula: $$A=\pi r^2 + \pi r l$$ where: - $r$ is the radius of the base of the cone. - $l$ is the slant height of the cone. - $\pi$ (pi) is approximately 3.14159. The first term $\pi r^2$ is the area of the circular base. The second term $\pi r l$ accounts for the lateral surface area of the cone. 2. **Check whether the polynomial $x^3 + x^2 - (2+\sqrt{2})x - \sqrt{2}$ is divisible by $x + x - 2$ or not.** First, simplify the divisor: $$x + x - 2 = 2x - 2$$ Set $2x - 2 = 0$ to find roots: $$2x = 2 \Rightarrow x = 1$$ Check polynomial at $x=1$: $$p(1) = 1^3 + 1^2 - (2+\sqrt{2})(1) - \sqrt{2} = 1 + 1 - 2 - \sqrt{2} - \sqrt{2} = 2 - 2 - 2\sqrt{2} = -2\sqrt{2} \neq 0$$ Since $p(1) \neq 0$, polynomial is **not divisible** by $2x - 2$. 3. **Find the values of $p(-1), p(0), p(2)$ for $p(x) = x^3 - 2x^2 - 3$ and their mean.** Calculate: $$p(-1) = (-1)^3 - 2(-1)^2 - 3 = -1 - 2 - 3 = -6$$ $$p(0) = 0^3 - 2 \cdot 0^2 - 3 = -3$$ $$p(2) = 2^3 - 2(2)^2 - 3 = 8 - 8 - 3 = -3$$ Mean value: $$\frac{p(-1)+p(0)+p(2)}{3} = \frac{-6 - 3 -3}{3} = \frac{-12}{3} = -4$$ 4. **Find the three angles of a triangle if they are $(2x)^{\circ}$, $(3x + 10)^{\circ}$ and $(5x - 20)^{\circ}$.** Sum of angles in a triangle is $180^{\circ}$: $$2x + (3x+10) + (5x - 20) = 180$$ Simplify: $$2x + 3x + 10 + 5x - 20 = 180$$ $$10x - 10 = 180$$ $$10x = 190$$ $$x = 19$$ Find angles: $$2x = 2 \times 19 = 38^{\circ}$$ $$3x + 10 = 3 \times 19 + 10 = 57 + 10 = 67^{\circ}$$ $$5x - 20 = 5 \times 19 - 20 = 95 - 20 = 75^{\circ}$$ Final answers: 1. Total surface area formula explained. 2. Polynomial not divisible by $2x - 2$. 3. $p(-1) = -6, p(0) = -3, p(2) = -3$, mean = $-4$. 4. Triangle angles: $38^{\circ}, 67^{\circ}, 75^{\circ}$.