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**Question 1**
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What value of p will make (x2 ${}^{2}$ - 4x + p) a perfect square?

**Answer Details**

(x2 ${}^{\mathrm{}}$ - 4x + p)

Use the coefficient of the middle variable(-4x)

= (−42

)2
${}^{\mathrm{}}$

= (-2)2 ${}^{\mathrm{}}$

= 4

**Question 2**
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If √24 + √96 - √600 = y√6, find the value of y

**Answer Details**

To find the value of y in the given expression, let's simplify it step by step: √24 + √96 - √600 = y√6 First, let's simplify the square roots inside the expression: √24 = √(4 * 6) = √4 * √6 = 2√6 √96 = √(16 * 6) = √16 * √6 = 4√6 √600 = √(100 * 6) = √100 * √6 = 10√6 Now, we substitute the simplified values back into the expression: 2√6 + 4√6 - 10√6 = y√6 Combining like terms: (2 + 4 - 10)√6 = y√6 Simplifying further: -4√6 = y√6 Since the coefficient of √6 on both sides of the equation is the same, we can conclude that the value of y must be -4. Therefore, y = -4. So, the correct answer is -4.

**Question 3**
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In the diagram, MNR is a tangent to the circle centre O at N and ∠NOS = 108°. Find ∠OSN

**Answer Details**

The sum of angles in a triangle = 180º

(180 - 108)º = 72º

Isosceles triangle has both two equal sides and two equal angles.

∠OSN = ∠SON = 722 $\frac{\mathrm{}}{}$

= 36º

**Question 4**
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Make t the subject of k = mt−pr−−−√

**Answer Details**

square both sides to remove the square root

k2 ${}^{2}$ = m2 ${}^{2}$ t−pr $\frac{}{}$

k2rm2 $\frac{{}^{}}{}$ = t - p

t = k2rm2 $\frac{{}^{}}{}$ + p

t = k2r+pm2m2 $\frac{{}^{}}{}$

**Question 5**
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A ladder 6m long leans against a vertical wall at an angle 53º to the horizontal. How high up the wall does the ladder reach?

**Answer Details**

To find how high up the wall the ladder reaches, we can use trigonometry, specifically the sine function. Given: Length of the ladder = 6m Angle between the ladder and the horizontal = 53º We want to find the height of the ladder on the wall. Using the trigonometric relationship for right triangles, we can use the sine function to relate the angle and the sides of the triangle. sin(angle) = opposite / hypotenuse In this case, the height of the ladder on the wall is the opposite side, and the length of the ladder is the hypotenuse. sin(53º) = height / 6 To find the height, we rearrange the equation: height = sin(53º) * 6 Using a calculator, we can evaluate sin(53º) ≈ 0.7986. height ≈ 0.7986 * 6 ≈ 4.7916 Therefore, the height up the wall that the ladder reaches is approximately 4.7916m. So, the correct answer is 4.792m.

**Question 6**
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The length of a rectangle is 10 cm. If its perimeter is 28 cm, find the area

**Answer Details**

Let's start by using the formula for the perimeter of a rectangle: P = 2L + 2W, where L is the length and W is the width. We are given that the length L is 10 cm and the perimeter P is 28 cm. So we can substitute these values into the formula and solve for the width W: P = 2L + 2W 28 = 2(10) + 2W 28 = 20 + 2W 2W = 8 W = 4 Now that we know the width of the rectangle is 4 cm, we can use the formula for the area of a rectangle: A = L x W, where L is the length and W is the width. We already know L is 10 cm and we just found that W is 4 cm. So we can substitute these values into the formula and solve for the area A: A = L x W A = 10 x 4 A = 40 Therefore, the area of the rectangle is 40 cm2.

**Question 7**
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If the equations x2 ${}^{2}$ - 5x + 6 = 0 and x + px + 6 = 0 have the same roots, find the value of p.

**Answer Details**

The question presents two quadratic equations: x^2 - 5x + 6 = 0 and x + px + 6 = 0. Both equations are said to have the same roots. To find the value of p, we need to determine when two quadratic equations have the same roots. For two quadratic equations, ax^2 + bx + c = 0 and dx^2 + ex + f = 0, to have the same roots, the following conditions must be met: 1. The discriminant of both equations must be zero. 2. The ratios of the corresponding coefficients must be equal. Let's apply these conditions to the given equations: Equation 1: x^2 - 5x + 6 = 0 Equation 2: x + px + 6 = 0 1. Discriminant condition: The discriminant of Equation 1 is given by Δ1 = b1^2 - 4ac = (-5)^2 - 4(1)(6) = 25 - 24 = 1. The discriminant of Equation 2 is given by Δ2 = e^2 - 4df = p^2 - 4(1)(6) = p^2 - 24. Since both equations have the same roots, their discriminants must be equal: Δ1 = Δ2 1 = p^2 - 24 p^2 = 25 p = ±5 2. Ratio of coefficients condition: Comparing the coefficients of the corresponding terms: For Equation 1: a1 = 1, b1 = -5, c1 = 6 For Equation 2: a2 = 1, b2 = p, c2 = 6 The ratio of the coefficients must be equal: a1/a2 = b1/b2 = c1/c2 From a1/a2 = 1/1 = 1 and c1/c2 = 6/6 = 1, we can conclude that: b1/b2 = -5/p = 1 -5/p = 1 p = -5 Therefore, the value of p that satisfies both conditions and makes the equations have the same roots is p = -5.

**Question 8**
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An exterior angle of a regular polygon is 22.5°. Find the number of sides.

**Answer Details**

For a regular polygon with n sides, each exterior angle is equal to 360°/n. Given that the exterior angle of the regular polygon is 22.5°, we can set up an equation as follows: 360°/n = 22.5° Multiplying both sides by n, we get: 360° = 22.5°n Dividing both sides by 22.5°, we get: n = 360°/22.5° = 16 Therefore, the regular polygon has 16 sides. Hence, the answer is (d) 16.

**Question 9**
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The age (years) of some members in a singing group are: 12, 47, 49, 15, 43, 41, 13, 39, 43, 41 and 36.

Find the lower quartile

**Answer Details**

To find the lower quartile, we need to first arrange the ages in ascending order. Here are the ages provided: 12, 47, 49, 15, 43, 41, 13, 39, 43, 41, and 36. Next, we divide the data set into four equal parts, with each part containing an equal number of values. The lower quartile is the median of the first half of the data set. Let's arrange the ages in ascending order: 12, 13, 15, 36, 39, 41, 41, 43, 43, 47, 49 We can see that there are 11 ages in total, which means the first half contains 11/2 = 5.5 ages. Since we can't have a fraction of an age, we round down to the nearest whole number, which is 5. Now, let's look at the five ages in the first half of the data set: 12, 13, 15, 36, 39. To find the lower quartile, we need to find the median of this subset of ages. Since there are an odd number of ages (5), the median is the middle value. In this case, the middle value is 15. Therefore, the lower quartile is 15. So, the correct answer is 15.

**Question 10**
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A rectangle with width 34

cm and area 3 38 $\frac{\mathrm{}}{}$cm2 ${}^{\mathrm{}}$. Find the length

cm and area 3 38 $\frac{\mathrm{}}{}$cm2 ${}^{\mathrm{}}$. Find the length

**Answer Details**

The area of a rectangle = length ✕ width

: Length = area ÷ width → 3 38 $\frac{\mathrm{}}{}$ ÷ 34 $\frac{\mathrm{}}{}$

= 278 $\frac{\mathrm{}}{}$ * 43 $\frac{\mathrm{}}{}$

= 412 $\frac{\mathrm{}}{}$cm

**Question 11**
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In the diagram, ∠ZWZY and WYX are right angles. Find the perimeter of WXYZ.

**Answer Details**

In ΔWYZ:

hyp2 ${}^{2}$ = adj2 ${}^{\mathrm{}}$ + opp2 ${}^{\mathrm{}}$

hyp2 ${}^{\mathrm{}}$ =32 ${}^{\mathrm{}}$ + 42 ${}^{\mathrm{}}$ → 9 + 16

hyp2 ${}^{\mathrm{}}$ = 25

hyp = 5

In ΔWXY:

hyp2 ${}^{\mathrm{}}$ = adj2 ${}^{\mathrm{}}$ + opp2 ${}^{\mathrm{}}$

hyp2 ${}^{\mathrm{}}$ = 122 ${}^{\mathrm{}}$ + 52 ${}^{\mathrm{}}$ = 144 +25

hyp2 ${}^{\mathrm{}}$ = 169

hyp = 13

the perimeter of WXYZ. = 3+4+12+13 → 32cm

**Question 12**
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Simplify 2−18m21+3m

**Answer Details**

2−18m21+3m $\frac{\mathrm{}}{}$ = 2[1−9m2]1+3m $\frac{\mathrm{}}{}$

2[1−3m][1+3m]1+3m $\frac{\mathrm{}}{}$

2[1-3m]

**Question 13**
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Find the volume of a cone of radius 3.5cm and vertical height 12cm. [Take π = 22/7]

**Answer Details**

The formula for the volume of a cone is V = 1/3 πr^2h, where r is the radius of the base and h is the height of the cone. Substituting the given values into the formula: V = 1/3 × 22/7 × (3.5cm)^2 × 12cm = 1/3 × 22/7 × 12.25cm^2 × 12cm = 1/3 × 22/7 × 147cm^3 = 22/7 × 49cm^3 = 154cm^3 Therefore, the volume of the cone is 154cm^3. The answer is (d) 154cm3.

**Question 14**
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The mean of two numbers x and y is 4. Find the mean of four numbers x, 2x, y and 2y

**Answer Details**

The mean of two numbers x and y is 4, which means (x + y) / 2 = 4. Solving for x + y, we get x + y = 8. Now we need to find the mean of four numbers x, 2x, y, and 2y. The mean is calculated by adding up all the numbers and dividing by the total number of numbers. So the sum of these four numbers is x + 2x + y + 2y = 3x + 3y, and the total number of numbers is 4. Therefore, the mean is (3x + 3y) / 4. We can substitute 8 for x + y since we know it equals 8, and simplify the expression to get (3x + 3y) / 4 = (3(x+y)) / 4 = (3(8)) / 4 = 6. Therefore, the mean of the four numbers x, 2x, y, and 2y is 6. The answer is option C.

**Question 15**
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Find the correct to two decimal places, the volume of a sphere whose radius is 3cm. [Take π = 227 $\frac{\mathrm{}}{}$]

**Answer Details**

The formula for the volume of a sphere is V = (4/3)πr^3, where r is the radius of the sphere and π is the constant pi (approximately equal to 3.14). Given that the radius of the sphere is 3 cm, we can substitute the value of r into the formula to get: V = (4/3) × 22/7 × 3^3 = (4/3) × 22/7 × 27 ≈ 113.14 cm^3 Therefore, the correct option is 113.14cm3, rounded to two decimal places.

**Question 16**
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Given that log3 ${}_{3}$ 27 = 2x + 1, find the value of x.

**Answer Details**

Recall that: log3 ${}_{3}$ 27 → log3 ${}_{3}$33 ${}^{\mathrm{}}$

3log3 ${}_{3}$3 → 3 * 1

= 3

Then log3 ${}_{3}$ 27 = 2x + 1

→ 3 = 2x + 1

3 - 1 = 2x

2 = 2x

1 = x

**Question 17**
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The straight line y = mx - 4 passes through the point(-4,16). Calculate the gradient of the line

**Answer Details**

To calculate the gradient of the line, we need to find the slope, which is represented by "m" in the equation y = mx - 4. The slope of a line tells us how steep or flat it is. To find the slope, we can use the coordinates of the given point (-4, 16) and the equation of the line. The equation tells us that for any point on the line, the y-coordinate (vertical) is equal to the slope multiplied by the x-coordinate (horizontal) minus 4. Let's substitute the given point's coordinates into the equation: 16 = m(-4) - 4 Now, let's simplify the equation: 16 = -4m - 4 To solve for "m," we need to isolate it on one side of the equation. Let's add 4 to both sides: 16 + 4 = -4m Simplifying further: 20 = -4m To find the value of "m," we divide both sides by -4: 20/-4 = m Simplifying the division: -5 = m Therefore, the gradient or slope of the line is -5.

**Question 18**
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If 5x + 3y=4 and 5x-3y= 2, what is the value of (25x2 ${}^{\mathrm{}}$ -9y2 ${}^{\mathrm{}}$)?

**Answer Details**

5x + 3y=4

5x-3y= 2

Using elimination method

5x + 3y=4 → 5x + 3y=4

-[5x-3y= 2] → -5x +3y= -2

6y = 2

y → 1/3 and x = 3/5

solving (25x2 ${}^{\mathrm{}}$ -9y2 ${}^{\mathrm{}}$)

25 * [3/5]2 ${}^{\mathrm{}}$ -9 * [1/3]2 ${}^{\mathrm{}}$

25 * 925 $\frac{\mathrm{}}{}$ - 9 19

9 - 1 = 8

**Question 19**
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Evaluate, correct to four significant figures, (573.06 x 184.25).

**Answer Details**

573.06 x 184.25 = 105,586.305

1,05600.00 to four significant figure

What are the Rules for significant figures?

Significant Figures

- All non-zero numbers ARE significant.
- Zeros between two non-zero digits ARE significant.
- Leading zeros are NOT significant.
- Trailing zeros to the right of the decimal ARE significant.
- Trailing zeros in a whole number with the decimal shown ARE significant.

**Question 20**
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If 43x
${}^{\mathrm{}}$ = 16x+1
${}^{}$, find the value of x

**Answer Details**

43x ${}^{\mathrm{}}$ = 42(x+1) ${}^{\mathrm{}}$

3x = 2x + 2

3x - 2x = 2

x = 2

**Question 21**
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Given that sin (5x-28)° = cos (3x-50)", 0°≤ x ≤ 90°, find the value of x.

**Answer Details**

using the trial method by inserting each option in the equation.

Inserting 21º: sin([5 x 21] - 28) = cos([3 x 21] - 50)

sin(105 - 28) = cos (63 - 50)

sin 77º = cos 13º

where:

sin 77º = 0.9744

cos 13º = 0.9744

**Question 22**
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A boy 14 m tall, stood 10m away from a tree of height 12 m. Calculate, correct to the nearest degree, the angle of elevation of the top of the tree from the boy's eyes.

**Answer Details**

The angle of elevation

= Tan θ = oppadj $\frac{}{}$

Tan θ = 12+1410 $\frac{\mathrm{}}{}$

Tan θ = 2610 $\frac{\mathrm{}}{}$

θ = Tan−1 ${}^{}$(2.6)

θ ≈ 70º

**Question 23**
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Consider the statements:

p: Stephen is intelligent

q: Stephen is good at Mathematics

If p⇒q, which of the following is a valid conclusion?

**Answer Details**

The given statements are: - p: Stephen is intelligent - q: Stephen is good at Mathematics The symbol "⇒" means "implies". So, p⇒q means "if Stephen is intelligent, then he is good at Mathematics". , "If Stephen is good at Mathematics, then he is intelligent", is a valid conclusion because it is the contrapositive of p⇒q. In other words, if the original statement is true, then its contrapositive must also be true. The contrapositive of p⇒q is "if Stephen is not good at Mathematics, then he is not intelligent" However, is also valid because the contrapositive of a true statement is always true. , "If Stephen is not intelligent, then he is not good at Mathematics", is the inverse of the original statement and is not necessarily true. In other words, just because Stephen is not intelligent, it does not mean that he is not good at Mathematics. For example, he may have a natural talent for Mathematics even if he is not generally intelligent. , "If Stephen is not good at Mathematics, then he is intelligent", is the converse of the original statement and is also not necessarily true. In other words, just because Stephen is good at Mathematics, it does not mean that he is intelligent. For example, he may be good at Mathematics but struggle with other subjects. So, the valid conclusion is: "If Stephen is good at Mathematics, then he is intelligent".

**Question 24**
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In the diagram, triangle MNR is inscribed in circle MNR and line PQ is a straight line. ∠MRN = 41 and = 141, find ∠QNR

**Answer Details**

We can use the property that an angle inscribed in a circle is half the measure of the arc it intercepts. Since triangle MNR is inscribed in circle MNR, we know that angle MNR is half the measure of arc MR. Similarly, angle MRN is half the measure of arc MN. We also know that arc PQM and arc QNR add up to a full circle, which has a measure of 360 degrees. Using these facts, we can write two equations: - angle MNR = 1/2(arc MR) = 1/2(360 - arc PQM - arc QNR) - angle MRN = 1/2(arc MN) = 1/2(360 - arc PQM) Substituting the given angle measures, we have: - 41 = 180 - arc PQM/2 - arc QNR/2 - 141 = 180 - arc PQM/2 Solving for arc PQM in the second equation gives us arc PQM = 198. Substituting this into the first equation, we have: 41 = 180 - 99 - arc QNR/2 arc QNR/2 = 40 arc QNR = 80 Therefore, angle QNR is half the measure of arc QNR, which is 80 degrees. So, the answer is 80º.

**Question 25**
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