Question 20: In an arithmetic progression, the first term is 2, and the sum of the 1st and 6th terms is 16.5. What is the 4th term? Question 21: The angle of a sector of a circle of diameter 8cm is $135^{\circ}$. Find the area of the sector. (Take $\pi = \frac{22}{7}$). Question 22: Simplify $36^{\frac{1}{2}} \times 64^{-\frac{1}{3}} \times 5^0$. Question 23: The number 186047 was corrected to 186,000. Which of the following can correctly describe the degree of approximation mode? I. To the nearest hundred II. To the nearest thousand III. To 3 significant figures IV. To 4 significant figures Question 24: A student is told to draw the graph of $y = x^2 + 4x - 6$. He is then told to draw a linear graph on the same axis such that the intersection of the two graphs will give the solutions to the equation $x^2 + 4x - 7 = 0$. What is the... (The rest of the question is cut off).

1. Problem Description

Question 20: In an arithmetic progression, the first term is 2, and the sum of the 1st and 6th terms is 16.

5. What is the 4th term?

Question 21: The angle of a sector of a circle of diameter 8cm is $135^{\circ}$. Find the area of the sector. (Take $\pi = \frac{22}{7}$).

Question 22: Simplify $36^{\frac{1}{2}} \times 64^{-\frac{1}{3}} \times 5^0$.

Question 23: The number 186047 was corrected to 186,

0

0

0. Which of the following can correctly describe the degree of approximation mode?

I. To the nearest hundred

II. To the nearest thousand

III. To 3 significant figures

IV. To 4 significant figures

Question 24: A student is told to draw the graph of $y = x^2 + 4x - 6$. He is then told to draw a linear graph on the same axis such that the intersection of the two graphs will give the solutions to the equation $x^2 + 4x - 7 = 0$. What is the... (The rest of the question is cut off).

2. Solution Steps

Question 20:

Let $a_1$ be the first term and $a_n$ be the nth term of an arithmetic progression. Let $d$ be the common difference. We are given that $a_1 = 2$ and $a_1 + a_6 = 16.5$.

The nth term of an arithmetic progression is given by $a_n = a_1 + (n-1)d$. Thus, $a_6 = a_1 + 5d = 2 + 5d$.

We have $a_1 + a_6 = 2 + (2 + 5d) = 4 + 5d = 16.5$.

So, $5d = 16.5 - 4 = 12.5$.

Then, $d = \frac{12.5}{5} = 2.5$.

We want to find the 4th term, $a_4 = a_1 + 3d = 2 + 3(2.5) = 2 + 7.5 = 9.5$.

Question 21:

The radius of the circle is half the diameter, so $r = \frac{8}{2} = 4$ cm.

The area of the sector is given by the formula:

Area $= \frac{\theta}{360^{\circ}} \times \pi r^2$, where $\theta$ is the angle of the sector.

Area $= \frac{135^{\circ}}{360^{\circ}} \times \frac{22}{7} \times 4^2 = \frac{3}{8} \times \frac{22}{7} \times 16 = \frac{3 \times 22 \times 2}{7} = \frac{132}{7} = 18 \frac{6}{7} \text{ cm}^2$.

Question 22:

$36^{\frac{1}{2}} \times 64^{-\frac{1}{3}} \times 5^0 = \sqrt{36} \times \frac{1}{\sqrt[3]{64}} \times 1 = 6 \times \frac{1}{4} \times 1 = \frac{6}{4} = \frac{3}{2} = 1 \frac{1}{2}$.

Question 23:

186047 corrected to

1

8

6

0

0

0. I. Nearest hundred: 186047 rounded to the nearest hundred is

1

8

6

0

0

0. True.

II. Nearest thousand: 186047 rounded to the nearest thousand is

1

8

6

0

0

0. True.

III. 3 significant figures: 186047 rounded to 3 significant figures is

1

8

6

0

0

0. True.

IV. 4 significant figures: 186047 rounded to 4 significant figures is

1

8

6

0

0

0. True.

All statements are correct.

Question 24:

We are given $y = x^2 + 4x - 6$ and $x^2 + 4x - 7 = 0$.

The equation $x^2 + 4x - 7 = 0$ can be obtained by subtracting 1 from the equation $y = x^2 + 4x - 6$, so $y - 1 = x^2 + 4x - 7 = 0$.

Thus, $y = 1$. The linear graph is $y = 1$.

3. Final Answer

Question 20: B. 9.5 (Note: This was not an option, therefore there may have been an error in the question.) This number does not match any given answer, but is closest to B. 9.

2.

Question 21: C. $18 \frac{6}{7} \text{ cm}^2$

Question 22: D. $1 \frac{1}{2}$

Question 23: A. All of them

Question 24: The linear graph is y =

1.