Explain how -2 and 3/2 are the roots of the equation 2x2 + x - 6 = 0.

Step 1: Verify the roots by substitution (Question 21). The given quadratic equation is $2x^2 + x - 6 = 0$. We need to explain how $-2$ and $\frac{3}{2}$ are its roots. A number is a root of an equation if substituting it into the equation makes the equation true (i.e., the expression equals zero).

For $x = -2$: Substitute $x = -2$ into the equation: $2(-2)^2 + (-2) - 6$ $2(4) - 2 - 6$ $8 - 2 - 6$ $6 - 6 = 0$ Since the expression equals $0$, $x = -2$ is a root of the equation.

For $x = \frac{3}{2}$: Substitute $x = \frac{3}{2}$ into the equation: $2\left(\frac{3}{2}\right)^2 + \left(\frac{3}{2}\right) - 6$ $2\left(\frac{9}{4}\right) + \frac{3}{2} - 6$ $\frac{18}{4} + \frac{3}{2} - 6$ $\frac{9}{2} + \frac{3}{2} - 6$ $\frac{12}{2} - 6$ $6 - 6 = 0$ Since the expression equals $0$, $x = \frac{3}{2}$ is a root of the equation. Thus, both $-2$ and $\frac{3}{2}$ satisfy the equation, confirming they are its roots.

Step 2: Write the formula for the $n^{th}$ term of an AP and find the $10^{th}$ term (Question 22). The formula to calculate the $n^{th}$ term of an Arithmetic Progression (AP) is: $a_n = a + (n-1)d$ where: • $a_n$ is the $n^{th}$ term. • $a$ is the first term. • $n$ is the term number. • $d$ is the common difference.

Given AP: $3, 7, 11, \dots$ Here, the first term $a = 3$. The common difference $d = 7 - 3 = 4$. We need to find the $10^{th}$ term, so $n = 10$. Substitute these values into the formula: $a_{10} = 3 + (10-1)4$ $a_{10} = 3 + (9)4$ $a_{10} = 3 + 36$ $a_{10} = 39$ The $10^{th}$ term of the AP is: $\boxed{39}$

Step 3: Express trigonometric ratios in terms of $\sin A$ (Question 23). We use the fundamental trigonometric identities. Assume $A$ is an acute angle, so $\cos A > 0$.

a) Express $\cos A$ in terms of $\sin A$: We know the identity $\sin^2 A + \cos^2 A = 1$. Rearrange to solve for $\cos A$: $\cos^2 A = 1 - \sin^2 A$ $\cos A = \sqrt{1 - \sin^2 A}$ $\boxed{\cos A = \sqrt{1 - \sin^2 A}}$

b) Express $\tan A$ in terms of $\sin A$: We know $\tan A = \frac{\sin A}{\cos A}$. Substitute the expression for $\cos A$ from part (a): $\tan A = \frac{\sin A}{\sqrt{1 - \sin^2 A}}$ $\boxed{\tan A = \frac{\sin A}{\sqrt{1 - \sin^2 A}}}$

c) Express $\csc A$ in terms of $\sin A$: We know $\csc A = \frac{1}{\sin A}$. $\boxed{\csc A = \frac{1}{\sin A}}$

d) Express $\sec A$ in terms of $\sin A$: We know $\sec A = \frac{1}{\cos A}$. Substitute the expression for $\cos A$ from part (a): $\sec A = \frac{1}{\sqrt{1 - \sin^2 A}}$ $\boxed{\sec A = \frac{1}{\sqrt{1 - \sin^2 A}}}$

Step 4: Prove that a parallelogram circumscribing a circle is a rhombus (Question 24). Let ABCD be a parallelogram circumscribing a circle with center O. Let the sides AB, BC, CD, and DA touch the circle at points P, Q, R, and S respectively.

According to the property that tangents drawn from an external point to a circle are equal in length: • From vertex A: $AP = AS$ (1) • From vertex B: $BP = BQ$ (2) • From vertex C: $CQ = CR$ (3) • From vertex D: $DR = DS$ (4)

Add equations (1), (2), (3), and (4): $AP + BP + CR + DR = AS + BQ + CQ + DS$ Rearrange the terms: $(AP + BP) + (CR + DR) = (AS + DS) + (BQ + CQ)$ From the figure, we can see that: $AP + BP = AB$ $CR + DR = CD$ $AS + DS = AD$ $BQ + CQ = BC$ So, the equation becomes: $AB + CD = AD + BC$ (5)

Since ABCD is a parallelogram, we know that opposite sides are equal in length: $AB = CD$ (6) $BC = AD$ (7)

Substitute (6) and (7) into (5): $AB + AB = BC + BC$ $2AB = 2BC$ $AB = BC$

Since $AB = BC$, and we know that in a parallelogram, opposite sides are equal ($AB=CD$ and $BC=AD$), this implies that all four sides are equal: $AB = BC = CD = DA$. A parallelogram with all four sides equal is a rhombus. Therefore, a parallelogram circumscribing a circle is a rhombus.

Step 5: Find the total surface area of the toy (Question 25). The toy is a cone mounted on a hemisphere of the same radius. Given: Radius of hemisphere ($r$) = $3.5$ cm Radius of cone ($r$) = $3.5$ cm Total height of the toy ($H$) = $15.5$ cm

The height of the hemispherical part is equal to its radius, $r = 3.5$ cm. The height of the conical part ($h_c$) is the total height minus the height of the hemisphere: $h_c = H - r$ $h_c = 15.5\, cm - 3.5\, \text{cm}$ $h_c = 12\, \text{cm}$

Now, calculate the slant height ($l$) of the cone using the Pythagorean theorem: $l = \sqrt{r^2 + h_c^2}$ $l = \sqrt{(3.5\, cm)^2 + (12\, cm)^2}$ $l = \sqrt{12.25\, cm^2 + 144\, cm^2}$ $l = \sqrt{156.25\, cm^2}$ $l = 12.5\, \text{cm}$

The total surface area of the toy is the sum of the curved surface area of the cone and the curved surface area of the hemisphere. Curved surface area of cone ($CSA_{cone}$) = $\pi r l$ Curved surface area of hemisphere ($CSA_{hemisphere}$) = $2\pi r^2$

Total Surface Area ($TSA$) = $CSA_{cone} + CSA_{hemisphere}$ $TSA = \pi r l + 2\pi r^2$ Factor out $\pi r$: $TSA = \pi r (l + 2r)$ Substitute the values ($r = 3.5$, $l = 12.5$, $\pi \approx \frac{22}{7}$): $TSA = \frac{22}{7} \times 3.5\, cm \times (12.5\, cm + 2 \times 3.5\, cm)$ $TSA = \frac{22}{7} \times 3.5\, cm \times (12.5\, cm + 7\, cm)$ $TSA = \frac{22}{7} \times 3.5\, cm \times 19.5\, \text{cm}$ $TSA = 22 \times 0.5\, cm \times 19.5\, \text{cm}$ $TSA = 11\, cm \times 19.5\, \text{cm}$ $TSA = 214.5\, cm^2$ The total surface area of the toy is: $\boxed{214.5\, cm^2}$

Step 6: Write the formula for mean by step deviation method and explain its terms (Question 26). The formula to find the mean ($\bar{x}$) by the step deviation method for grouped data is: $\bar{x} = A + \left(\frac{\sum f_i u_i}{\sum f_i}\right)h$ Where: • $\bar{x}$ is the mean of the data. • $A$ is the assumed mean, which is a value chosen from the midpoints of the class intervals (usually the midpoint of the middle class interval). • $f_i$ is the frequency of the $i^{th}$ class interval. • $u_i$ is the step deviation for the $i^{th}$ class interval, calculated as $u_i = \frac{x_i - A}{h}$. • $x_i$ is the class mark (midpoint) of the $i^{th}$ class interval. • $h$ is the class size (or class width), which is the difference between the upper and lower limits of a class interval. • $\sum f_i u_i$ is the sum of the products of frequencies and corresponding step deviations. • $\sum f_i$ is the sum of all frequencies (total number of observations).

Step 7: Find the probability that the sum of two numbers appearing on the top of the dice is 8 (Question 27). When two dice are thrown, the total number of possible outcomes is $6 \times 6 = 36$. The possible outcomes are pairs $(d_1, d_2)$ where $d_1$ is the result of the first die and $d_2$ is the result of the second die.

We need to find the outcomes where the sum of the two numbers is 8. These are: • $(2, 6)$ • $(3, 5)$ • $(4, 4)$ • $(5, 3)$ • $(6, 2)$ There are 5 favorable outcomes.

The probability $P(sum is 8)$ is given by: $P(sum is 8) = \frac{Number of favorable outcomes}{Total number of outcomes}$ $P(sum is 8) = \frac{5}{36}$ The probability that the sum of the two numbers appearing on the top of the dice is 8 is: $\boxed{\frac{5}{36}}$