11. Define a linear polynomial. Give an example. A linear polynomial is a polynomial of degree one. This means the highest power of the variable in the polynomial is 1. Example: $P(x) = 3x + 5$ 12. Find the value of $(32)^{\frac{2}{5}}$. Step 1: Express the base 32 as a power of 2. $$32 = 2^5$$ Step 2: Substitute this into the expression. $$(32)^{\frac{2}{5}} = (2^5)^{\frac{2}{5}}$$ Step 3: Apply the exponent rule $(a^m)^n = a^{m \times n}$. $$(2^5)^{\frac{2}{5}} = 2^{5 \times \frac{2}{5}}$$ $$ = 2^2$$ Step 4: Calculate the final value. $$2^2 = 4$$ The value is $\boxed{4}$. 13. Represent the equation $5x = 2y + 2$ in the form of $ax + by + c = 0$. Step 1: Move all terms to one side of the equation to set it equal to zero. $$5x - 2y - 2 = 0$$ Step 2: Compare with the standard form $ax + by + c = 0$. Here, $a=5$, $b=-2$, and $c=-2$. The equation in the required form is $\boxed{5x - 2y - 2 = 0}$. 14. In the figure ABCD is a parallelogram if $\angle DAB = 75^\circ$ and $\angle DBC = 60^\circ$, find $\angle CDB$ and $\angle ADB$. Step 1: Use the property of alternate interior angles for parallel lines. In a parallelogram ABCD, side $AD \parallel BC$. When $BD$ is a transversal, $\angle ADB = \angle DBC$ (alternate interior angles). Given $\angle DBC = 60^\circ$. Therefore, $\angle ADB = 60^\circ$. Step 2: Use the property of adjacent angles in a parallelogram. Adjacent angles in a parallelogram are supplementary. $$\angle DAB + \angle ABC = 180^\circ$$ Given $\angle DAB = 75^\circ$. $$75^\circ + \angle ABC = 180^\circ$$ $$\angle ABC = 180^\circ - 75^\circ$$ $$\angle ABC = 105^\circ$$ Step 3: Find $\angle ABD$. We know that $\angle ABC = \angle ABD + \angle DBC$. $$105^\circ = \angle ABD + 60^\circ$$ $$\angle ABD = 105^\circ - 60^\circ$$ $$\angle ABD = 45^\circ$$ Step 4: Use the property of alternate interior angles again. In a parallelogram ABCD, side $AB \parallel DC$. When $BD$ is a transversal, $\angle CDB = \angle ABD$ (alternate interior angles). Since $\angle ABD = 45^\circ$. Therefore, $\angle CDB = 45^\circ$. The values are $\boxed{\angle CDB = 45^\circ}$ and $\boxed{\angle ADB = 60^\circ}$. 15. The surface area of a sphere is $100\pi \text{ cm}^2$. Find its radius. Step 1: Write the formula for the surface area of a sphere. The surface area $A$ of a sphere with radius $r$ is given by: $$A = 4\pi r^2$$ Step 2: Substitute the given surface area into the formula. $$100\pi \text{ cm}^2 = 4\pi r^2$$ Step 3: Solve for $r^2$. Divide both sides by $4\pi$: $$\frac{100\pi}{4\pi} = r^2$$ $$25 = r^2$$ Step 4: Solve for $r$. $$r = \sqrt{25}$$ $$r = 5 \text{ cm}$$ The radius of the sphere is $\boxed{5 \text{ cm}}$. 16. Verify whether $x = \frac{4}{5}$ is a zero of a polynomial $p(x) = 5x - \pi$. Step 1: Substitute the given value of $x$ into the polynomial. $$p\left(\frac{4}{5}\right) = 5\left(\frac{4}{5}\right) - \pi$$ Step 2: Simplify the expression. $$p\left(\frac{4}{5}\right) = 4 - \pi$$ Step 3: Check if the result is zero. Since $4 - \pi \neq 0$ (as $\pi \approx 3.14159$), $x = \frac{4}{5}$ is not a zero of the polynomial. $\boxed{\text{No, } x = \frac{4}{5} \text{ is not a zero of } p(x) = 5x - \pi}$. 17. Express $1.\overline{27}$ in the form of $\frac{p}{q}$. Step 1: Let $x$ be the given repeating decimal. $$x = 1.272727... \quad (1)$$ Step 2: Since two digits are repeating, multiply equation (1) by $10^2 = 100$. $$100x = 127.272727... \quad (2)$$ Step 3: Subtract equation (1) from equation (2). $$100x - x = 127.272727... - 1.272727...$$ $$99x = 126$$ Step 4: Solve for $x$. $$x = \frac{126}{99}$$ Step 5: Simplify the fraction by dividing the numerator and denominator by their greatest common divisor, which is 9. $$x = \frac{126 \div 9}{99 \div 9}$$ $$x = \frac{14}{11}$$ The value of $1.\overline{27}$ in the form $\frac{p}{q}$ is $\boxed{\frac{14}{11}}$. 18. In the figure, O is the centre of the circle. If $\angle BOC = 100^\circ$, then find the value of $x$. Step 1: Identify the relationship between the angle at the center and the angle at the circumference. The angle subtended by an arc at the center is double the angle subtended by it at any point on the remaining part of the circle. Here, arc BC subtends $\angle BOC$ at the center O and $\angle BDC$ (which is $x$) at point D on the circumference. Step 2: Apply the theorem. $$\angle BDC = \frac{1}{2} \angle BOC$$ Step 3: Substitute the given value of $\angle BOC$. $$x = \frac{1}{2} \times 100^\circ$$ $$x = 50^\circ$$ The value of $x$ is $\boxed{50^\circ}$. 19. Verify which of the following are solutions of the equation $x + 2y = 6$. (a) $(2, 2)$ (b) $(0, 3)$ To verify, substitute the $x$ and $y$ values from each option into the equation $x + 2y = 6$. *a) For the point $(2, 2)$: Substitute $x=2$ and $y=2$ into the equation: $$2 + 2(2) = 6$$ $$2 + 4 = 6$$ $$6 = 6$$ Since the left side equals the right side, $(2, 2)$ is a solution. *b) For the point $(0, 3)$: Substitute $x=0$ and $y=3$ into the equation: $$0 + 2(3) = 6$$ $$0 + 6 = 6$$ $$6 = 6$$ Since the left side equals the right side, $(0, 3)$ is a solution. Both (a) and (b) are solutions to the equation $x + 2y = 6$. $$\boxed{\text{Both (a) } (2, 2) \text{ and (b) } (0, 3) \text{ are solutions.}}$$