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The sequence 4, 10, 16,... is an arithmetic sequence because there is a common difference between every pair of consecutive terms. To find the common difference, we can subtract any term from the following term. For example: - The common difference between the first and second terms is 10 - 4 = 6. - The common difference between the second and third terms is 16 - 10 = 6. Since the common difference is 6, we can find the nth term of the sequence using the formula: nth term = first term + (n - 1) * common difference The first term is 4 and the common difference is 6, so we get: nth term = 4 + (n - 1) * 6 Simplifying this expression, we get: nth term = 6n - 2 Therefore, the correct answer is (A) 2(3n-1).

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In a direct variation, two variables are related by a constant ratio. This can be expressed mathematically as: x ∝ y which means that x is directly proportional to y, or that x and y vary directly. We can also write this relationship as an equation: x = ky where k is the constant of proportionality. In this problem, we are given that (x + 3) varies directly as y. So we can write: x + 3 = ky where k is some constant of proportionality that we don't know yet. We are also given that x = 3 when y = 12. We can use this information to solve for k: x + 3 = ky 3 + 3 = k(12) 6 = 12k k = 0.5 Now that we know k, we can use the equation x + 3 = ky to find x when y = 8: x + 3 = ky x + 3 = 0.5(8) x + 3 = 4 x = 1 Therefore, when y = 8, x = 1. The answer is (a) 1.

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The expression \(\frac{3x-2}{4x^2+9x-9}\) is undefined when the denominator is equal to zero. Therefore, we need to find the values of x that make the denominator zero. We can factor the denominator as follows: \[4x^2 + 9x - 9 = (4x - 3)(x + 3)\] So, the denominator is equal to zero when: \begin{align*} 4x - 3 &= 0 &\text{or} && x + 3 &= 0 \\ 4x &= 3 &&& x &= -3 \\ x &= \frac{3}{4} \end{align*} Therefore, the expression is undefined when \(x = \frac{3}{4}\) or \(x = -3\). So, the answer is \(\frac{3}{4} \hspace{1mm}or \hspace{1mm}-3\).

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A rational number is any number that can be expressed as a ratio of two integers (where the denominator is not zero). - -5 can be expressed as -5/1, which is a ratio of two integers, so it is rational. - \(\sqrt{4}\) is equal to 2, which can be expressed as the ratio 2/1, so it is rational. - \(3\frac{3}{4}\) is equal to 15/4, which is a ratio of two integers, so it is rational. - \(\sqrt{90}\) cannot be expressed as a ratio of two integers. It is irrational. Therefore, the number that is not rational is \(\sqrt{90}\).

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The formula for the volume of a cone is given by V = (1/3)πr^2h, where r is the radius of the base and h is the height of the cone. Since we are given the base diameter, which is 14cm, we can find the radius by dividing it by 2: radius, r = 14/2 = 7cm Also, we are given the volume of the cone, which is 462 cm^3. Substituting these values into the formula for the volume of a cone, we get: 462 = (1/3)π(7^2)h Simplifying this equation, we get: 22h = 198 h = 198/22 h = 9cm Therefore, the height of the cone is 9cm. Hence, the correct option is (d) 9cm.

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To convert a number from base 2 to base 10, we multiply each digit by its corresponding power of 2 and sum the results. For example, in the binary number 1001, the first digit (from the right) represents 2^0 (1), the second digit represents 2^1 (2), the third digit represents 2^2 (4), and the fourth digit represents 2^3 (8). So, to convert 1001 to base 10, we compute: 1*2^0 + 0*2^1 + 0*2^2 + 1*2^3 = 1 + 0 + 0 + 8 = 9 Using this method, we can convert the binary number 1001011 to base 10: 1*2^0 + 1*2^1 + 0*2^2 + 1*2^3 + 0*2^4 + 0*2^5 + 1*2^6 = 1 + 2 + 0 + 8 + 0 + 0 + 64 = 75 Therefore, M5 in base 10 is 75, and since M5 is in base 10, M is simply 7. So the answer is 7.

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To find the value of x in the equation 0.5x + 2.6 = 5x + 0.35, we need to isolate x on one side of the equation. We can start by subtracting 0.5x from both sides to get rid of the x-term on the left side: 0.5x + 2.6 - 0.5x = 5x + 0.35 - 0.5x Simplifying the left side and combining like terms on the right side, we get: 2.6 = 4.5x + 0.35 Subtracting 0.35 from both sides, we get: 2.6 - 0.35 = 4.5x Simplifying and dividing both sides by 4.5, we get: x = (2.6 - 0.35)/4.5 = 0.5 Therefore, the value of x is 0.5.

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The given sketch is a quadratic function graph. From the graph, we can see that the parabola intersects the x-axis at two points which are (-3, 0) and (2, 0). Therefore, the roots or zeros of the quadratic function are -3 and 2. By comparing the options given, we can see that only option B, y = x² - x - 6, has roots of -3 and 2. Thus, option B is the correct answer.

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To find the probability of selecting a prime number from the set Y = {18, 19, 20, ..., 28, 29}, we need to first identify the prime numbers in this set. The prime numbers in this set are 19, 23, and 29. The total number of elements in the set Y is 12. Therefore, the probability of selecting a prime number from the set Y is 3/12 or 1/4. Therefore, the correct option is (a) \(\frac{1}{4}\).

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The correct answer is \(\frac{qp}{t}\). Since q oranges are sold for t Naira, we can say that the cost of one orange is \(\frac{t}{q}\) Naira. To find how many oranges can be bought for p Naira, we need to divide p by the cost of one orange: Number of oranges = \(\frac{p}{\frac{t}{q}}\) Simplifying, we get: Number of oranges = \(\frac{pq}{t}\)

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The equation of a line in slope-intercept form is given by y = mx + b, where m is the slope of the line and b is the y-intercept. In this question, we are given that the point (-3, -4) lies on the line y = mx + 2. This means that when x = -3, y = -4. We can substitute these values into the equation to get: -4 = m(-3) + 2 Simplifying this expression, we get: -4 = -3m + 2 Subtracting 2 from both sides, we get: -6 = -3m Dividing both sides by -3, we get: m = 2 Therefore, the value of m is 2.

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We can use the rule that log a^n = n log a to solve this problem. Since we are given log 2 = 0.3010 and log 2^y = 1.8062, we can use the second equation to get:

log 2^y = 1.8062
y log 2 = 1.8062     (using the above rule)
y = 1.8062 / log 2
y = 5.9889

Rounding to the nearest whole number, we get y = 6. Therefore, the answer is option A.

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To calculate the percentage error, we first need to find the absolute error, which is the difference between the actual value and the approximate value. In this case, the actual value is 0.375, and the approximate value is 0.4. Absolute error = |actual value - approximate value| Absolute error = |0.375 - 0.4| Absolute error = 0.025 To find the percentage error, we divide the absolute error by the actual value and multiply by 100. Percentage error = (absolute error / actual value) x 100 Percentage error = (0.025 / 0.375) x 100 Percentage error = 6.67 Rounding to two significant figures, the percentage error in approximating 0.375 to 0.4 is 6.7%. Therefore, the correct answer is option D.

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A right pyramid is a pyramid in which the apex is directly above the center of the base. In this case, the base is a square and the pyramid is right, so each triangular face of the pyramid is an isosceles right triangle. The slanting side of the pyramid is a hypotenuse of one of the triangular faces. By the Pythagorean theorem, the length of each leg of the right triangle is equal to the length of the base of the square, which is 4cm. Therefore, each leg has length 4cm and the hypotenuse has length \(2\sqrt{3}\) cm. To find the height of the pyramid, we draw a perpendicular line from the apex of the pyramid to the center of the base. This line divides the square base into four congruent right triangles, each with legs of length 2cm and hypotenuse of length \(2\sqrt{2}\) cm. By the Pythagorean theorem, the height of each of these triangles is \(\sqrt{(2\sqrt{2})^{2} - 2^{2}} = \sqrt{8} = 2\sqrt{2}\). Therefore, the height of the pyramid is also 2\(\sqrt{2}\)cm. The volume of the pyramid is given by the formula: \[\frac{1}{3} \times (\text{area of base}) \times (\text{height})\] The area of the square base is \(4^{2}\) cm\(^{2}\) = 16 cm\(^{2}\), and the height is 2\(\sqrt{2}\) cm. Substituting these values into the formula, we get: \[\frac{1}{3} \times (16) \times (2\sqrt{2}) = \frac{32\sqrt{2}}{3} \approx 10.67 \text{ cm}^{3}\] Therefore, the volume of the pyramid is approximately 10.67 cm\(^{3}\). Hence, the correct option is \(\mathbf{(b)}\) \(10\frac{2}{3}\) cm\(^{3}\).

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In a triangle, an altitude is a line segment drawn from a vertex perpendicular to the opposite side. In the given diagram, \(\bar{PS}\) and \(\bar{QT}\) are two altitudes of triangle PQR. By definition, the altitude \(\bar{QT}\) is perpendicular to the side \(\bar{PR}\). Therefore, ∠QTR is a right angle. Also, we know that the sum of the angles in a triangle is 180 degrees. So, we can find the measure of ∠RQT as follows: ∠RQT = 180° - ∠QTR - ∠QRT We don't know the value of either ∠QTR or ∠QRT, but we can find one of them using the fact that \(\bar{PS}\) is an altitude. Since \(\bar{PS}\) is perpendicular to \(\bar{QR}\), ∠QPS is a right angle. Therefore, ∠QTR = ∠QTP + ∠RTP Since ∠QTP and ∠RTP are both complementary to ∠QPR, which is a known angle in the diagram, we can find their values: ∠QTP = ∠RTP = 90° - ∠QPR Now we can substitute these values into our equation for ∠RQT: ∠RQT = 180° - (90° - ∠QPR) - (90° - ∠QPR) Simplifying this expression gives: ∠RQT = 2∠QPR - 90° Therefore, the answer is option (B) ?SRP, since ∠QPR and ∠SRP are corresponding angles and thus equal to each other.

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To arrive on schedule, the train needs to cover a distance of 60 km at a speed of 72 km/hr. However, the train starts 10 minutes late, which means it has lost 10 minutes of travel time. We need to find the speed the train needs to travel to make up for the lost time and arrive on schedule. We can use the formula: distance = speed x time Let's first calculate the time taken to cover the distance of 60 km at the speed of 72 km/hr. time = distance / speed time = 60 km / 72 km/hr time = 0.8333 hours or 50 minutes (rounded to the nearest minute) Since the train is starting 10 minutes late, it has only 40 minutes to cover the distance of 60 km to arrive on schedule. We can use the same formula to find the speed the train needs to travel to cover the distance in 40 minutes. speed = distance / time speed = 60 km / (40/60) hours speed = 90 km/hr Therefore, the correct answer is 90 km/hr.

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We can start by simplifying each term separately before adding them up: \(3\sqrt{12} = 3\sqrt{4\cdot3} = 3\cdot2\sqrt{3} = 6\sqrt{3}\) \(10\sqrt{3}\) is already in its simplest form. \(\frac{6}{\sqrt{3}} = \frac{6}{\sqrt{3}} \cdot \frac{\sqrt{3}}{\sqrt{3}} = \frac{6\sqrt{3}}{3} = 2\sqrt{3}\) Now we can substitute these values back into the original expression: \(3\sqrt{12} + 10\sqrt{3} - \frac{6}{\sqrt{3}} = 6\sqrt{3} + 10\sqrt{3} - 2\sqrt{3} = 14\sqrt{3}\) Therefore, the simplified form of the expression is \(14\sqrt{3}\), which is.

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The formula for the area of a trapezium is given as: Area = 1/2 × (sum of the parallel sides) × (perpendicular distance between them) In this case, we have the lengths of the parallel sides as 9 cm and 12 cm, and the area as 105 cm². Substituting these values in the formula, we get: 105 = 1/2 × (9 + 12) × (perpendicular distance) 105 = 1/2 × 21 × (perpendicular distance) 105 = 10.5 × (perpendicular distance) Dividing both sides by 10.5, we get: Perpendicular distance = 105/10.5 Perpendicular distance = 10 Therefore, the perpendicular distance between the parallel sides is 10 cm. Hence, the correct option is (c) 10cm.

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In a right-angled triangle, the hypotenuse is the longest side, and it is opposite to the right angle. Using the Pythagorean theorem, we know that the sum of the squares of the two shorter sides is equal to the square of the hypotenuse. So, we have: \begin{align*} x^2 + (x+2)^2 &= (x+1)^2 \\ x^2 + x^2 + 4x + 4 &= x^2 + 2x + 1 \\ x^2 + 2x^2 + 4x + 4 &= x^2 + 2x + 1 \\ 3x^2 + 4x + 4 &= x^2 + 2x + 1 \\ 2x^2 + 2x + 3 &= 0 \end{align*} We can solve this quadratic equation using the quadratic formula: \begin{align*} x &= \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \\ &= \frac{-2 \pm \sqrt{2^2 - 4(2)(3)}}{2(2)} \\ &= \frac{-2 \pm \sqrt{-8}}{4} \\ &= \frac{-1}{2} \pm \frac{\sqrt{2}}{2}i \end{align*} Since x must be greater than 0, the only valid solution is: \begin{align*} x &= \frac{-1}{2} + \frac{\sqrt{2}}{2}i \end{align*} However, we are asked to find the length of the hypotenuse, which is given by: \begin{align*} \sqrt{x^2 + (x+2)^2} &= \sqrt{\left(\frac{-1}{2} + \frac{\sqrt{2}}{2}i\right)^2 + \left(\frac{1}{2} + \frac{\sqrt{2}}{2}i\right)^2} \\ &= \sqrt{\frac{1}{4} - \frac{1}{2}\sqrt{2}i - \frac{1}{4} - \frac{1}{2}\sqrt{2}i + \frac{1}{4} + \frac{\sqrt{2}}{2}i + \frac{1}{4} - \frac{\sqrt{2}}{2}i} \\ &= \sqrt{\frac{1}{2}} \\ &= \frac{\sqrt{2}}{2} \approx 0.707 \end{align*} Therefore, the length of the hypotenuse is approximately 0.707 cm, which is closest to 5 cm.

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The curved surface area of a cone is given by the formula: \[\text{Curved surface area} = \pi rl,\] where $r$ is the radius of the base, $l$ is the slant height, and $\pi$ is the constant pi (approximately 3.14). In this case, the height of the cone is 4 cm and the radius of the base is 3 cm. To find the slant height, we can use the Pythagorean theorem: \[l^2 = r^2 + h^2.\] Plugging in the values we get, \[l^2 = 3^2 + 4^2 = 9 + 16 = 25.\] So, $l = 5$ cm. Now, we can use the formula for curved surface area, substituting $r=3$ cm and $l=5$ cm: \[\text{Curved surface area} = \pi \cdot 3 \cdot 5 = 15\pi \text{ cm}^2.\] Therefore, the answer is (B) $15\pi cm^2$.

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The given function is \(y = 6 + x - x^2\). We are required to find the value of x when y is equal to \(3\frac{1}{3}\), which is also equal to \(\frac{10}{3}\). Substituting \(y = \frac{10}{3}\) in the given equation, we get \[\frac{10}{3} = 6 + x - x^2\] Rearranging the terms, we get \[x^2 - x + \frac{8}{3} = 0\] Solving the quadratic equation, we get two solutions for x, which are \(\frac{4}{3}\) and \(1\frac{1}{2}\). However, we are asked to find the positive value of x, which is \(\boxed{2\frac{1}{5}}\).

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We are given the universal set \(\varepsilon = \{1, 2, 3, \dots, 15\}\) and the three subsets

\[A = \{1, 3, 5, 7, 9, 11\}, \quad B = \{2, 3, 5, 7, 11, 15\}, \quad C = \{3, 6, 9, 12, 15\}.\]

(a) Building the Venn diagram. Work out each region before drawing so that every element sits in exactly one place.

• In all three sets: \(A \cap B \cap C = \{3\}\).
• In \(A\) and \(B\) only: \(\{5, 7, 11\}\).
• In \(A\) and \(C\) only: \(\{9\}\).
• In \(B\) and \(C\) only: \(\{15\}\).
• In \(A\) only: \(\{1\}\).
• In \(B\) only: \(\{2\}\).
• In \(C\) only: \(\{6, 12\}\).
• Outside all three sets: \(\varepsilon \setminus (A \cup B \cup C) = \{4, 8, 10, 13, 14\}\).

These give the Venn diagram below.

(b) Reading answers from the diagram.

(i) \(A'\) is everything outside circle \(A\), so \(C \cap A'\) is the part of circle \(C\) that lies outside \(A\). Reading those regions of \(C\) gives the C-only part \(\{6, 12\}\) together with the \(B\cap C\)-only part \(\{15\}\):

\[C \cap A' = \{6, 12, 15\}.\]

(ii) First form \(B \cup C\) (all of circles \(B\) and \(C\)):

\[B \cup C = \{2, 3, 5, 6, 7, 9, 11, 12, 15\}.\]

Now intersect with \(A'\), keeping only the members of \(B \cup C\) that lie outside circle \(A\). Discarding \(3, 5, 7, 9, 11\) (which are inside \(A\)) leaves

\[A' \cap (B \cup C) = \{2, 6, 12, 15\}.\]

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(a) Evaluate \(\dfrac{3}{\sqrt3}\left(\dfrac{2}{\sqrt3}-\dfrac{\sqrt{12}}{6}\right)\) without tables.

First simplify each surd. \(\dfrac{3}{\sqrt3}=\dfrac{3}{\sqrt3}\times\dfrac{\sqrt3}{\sqrt3}=\dfrac{3\sqrt3}{3}=\sqrt3.\)

Inside the bracket: \(\dfrac{2}{\sqrt3}=\dfrac{2\sqrt3}{3}\) and \(\dfrac{\sqrt{12}}{6}=\dfrac{2\sqrt3}{6}=\dfrac{\sqrt3}{3}.\)

\[\frac{2\sqrt3}{3}-\frac{\sqrt3}{3}=\frac{\sqrt3}{3}.\]

Therefore

\[\sqrt3\times\frac{\sqrt3}{3}=\frac{3}{3}=\boxed{1.}\]

(b) Circle, centre \(O\); \(AB\) produced to \(E\), \(\widehat{ACB}=49^\circ\), \(\widehat{CBE}=68^\circ\).

Since \(A,B,E\) are collinear, \(\widehat{ABC}\) and \(\widehat{CBE}\) are angles on a straight line:

\[\widehat{ABC}=180^\circ-68^\circ=112^\circ.\]

In \(\triangle ABC\):

\[\widehat{BAC}=180^\circ-\widehat{ABC}-\widehat{ACB}=180^\circ-112^\circ-49^\circ=19^\circ.\]

(i) Interior angle \(AOC\). \(\widehat{ABC}=112^\circ\) is the angle at the circumference standing on chord \(AC\); it subtends the major arc \(AC\), whose central angle (reflex \(AOC\)) is \(2\times112^\circ=224^\circ.\) Hence the interior (non-reflex) angle is

\[\widehat{AOC}=360^\circ-224^\circ=\boxed{136^\circ.}\]

(ii) Angle \(BOC\). \(\widehat{BAC}=19^\circ\) at the circumference subtends arc \(BC\); the angle at the centre on the same arc is twice as large:

\[\widehat{BOC}=2\times19^\circ=\boxed{38^\circ.}\]

(Check: \(\widehat{ACB}=49^\circ\Rightarrow\widehat{AOB}=98^\circ,\) and \(98^\circ+38^\circ+224^\circ=360^\circ.\))

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(a) AP with \(a=3,\;d=4,\;n=28\). \[S_{28}=\frac{n}{2}\bigl(2a+(n-1)d\bigr)=14\bigl(6+27\times4\bigr)=14\times114=1596.\]

(b) \(x=\dfrac{2m}{1-m^{2}}=\dfrac{2m}{(1-m)(1+m)}\) and \(y=\dfrac{2m}{1+m}\). \[2x-y=\frac{4m}{(1-m)(1+m)}-\frac{2m}{1+m}=\frac{4m-2m(1-m)}{(1-m)(1+m)}=\frac{2m+2m^{2}}{(1-m)(1+m)}=\frac{2m(1+m)}{(1-m)(1+m)}=\frac{2m}{1-m}.\]

(c) Interior angles of a pentagon sum to \(540^{\circ}\): \[x+2x+3x+2x+(3x-10)=540\Rightarrow 11x-10=540\Rightarrow 11x=550\Rightarrow x=50.\]

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```html

Geometrical Construction

Given: Construct triangle XYZ such that

• |XY| = 8 cm,
• ∠YXZ = 60°,
• ∠XYZ = 30°.

(a) Construction steps (ruler and compasses only).

1. Draw a line segment XY = 8 cm.
2. At X, construct an angle of 60° with XY. This may be done by constructing an equilateral triangle on XY, since every angle in an equilateral triangle is 60°.
3. At Y, construct an angle of 30° with YX. First construct a 60° angle at Y, then bisect it using the standard compass construction for an angle bisector.
4. Let the 60° ray from X and the 30° ray from Y meet at Z. This gives triangle XYZ.
5. Construct the perpendicular from Z to XY. Let it meet XY at T. Hence, ZT ⟂ XY.
6. The locus of points equidistant from the lines ZY and XY is the angle bisector of ∠ZYX. Construct the internal angle bisector of the 30° angle at Y and call it l₁.
7. Let l₁ meet ZT at S.

(b) Finding |ST|.

Since ∠XYZ = 30° and l<sub>1</sub> bisects this angle,

∠SYT = 15°.

Also, triangle ZXY is right-angled at Z, because

∠XZY = 180° − 60° − 30° = 90°.

In the 30°–60°–90° triangle XYZ, with hypotenuse XY = 8 cm:

XZ = 4 cm,    YZ = 4√3 cm.

The foot of the perpendicular is T, and

YT = 6 cm.

In right triangle YST:

tan 15° = ST / YT.

ST = 6 tan 15° = 6(2 − √3).

On a ruler-and-compass construction, the measured value should be approximately 1.6 cm.

```

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(a) Let width \(=w\). Then length \(=2w\) and depth \(=w-4\). An open tank has a base plus four sides: \[\text{Area}=\underbrace{(2w)(w)}_{\text{base}}+\underbrace{2(2w)(w-4)+2(w)(w-4)}_{\text{four sides}}=2w^{2}+6w(w-4).\] Set equal to \(1440\): \[2w^{2}+6w^{2}-24w=1440\Rightarrow 8w^{2}-24w-1440=0\Rightarrow w^{2}-3w-180=0.\] \[(w-15)(w+12)=0\Rightarrow w=15.\] Length \(=2w=\) 30 m.

(b) First arrangement: \(\dfrac{3000x}{100}+\dfrac{2000y}{100}=640\Rightarrow 3x+2y=64.\) Swapped: \(\dfrac{2000x}{100}+\dfrac{3000y}{100}=660\Rightarrow 2x+3y=66.\) Solving: from \(9x+6y=192\) and \(4x+6y=132\), subtract \(\Rightarrow 5x=60\Rightarrow x=12\); then \(3(12)+2y=64\Rightarrow y=14\). So x = 12, y = 14.

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(a) Frequency table. Sorting the 30 lengths into the given class intervals and tallying gives:

Length (cm)	Tally	Frequency \(f\)	Midpoint \(x\)	\(fx\)
45 - 49	|	1	47	47
50 - 54	|||	3	52	156
55 - 59	|||| |	6	57	342
60 - 64	|||| ||	7	62	434
65 - 69	|||| |	6	67	402
70 - 74	||||	4	72	288
75 - 79	||	2	77	154
80 - 84	|	1	82	82
Total		30		1905

(b) Histogram of the distribution. The bars are drawn against the class boundaries \(44.5, 49.5, 54.5, 59.5, 64.5, 69.5, 74.5, 79.5, 84.5\) on the horizontal axis, with heights equal to the class frequencies. Since all class widths are equal, the bar heights are simply the frequencies.

(c) Mean of the distribution.

\[\bar{x}=\frac{\sum fx}{\sum f}=\frac{1905}{30}=63.5\text{ cm}.\]

(d) Probability of selecting a rod in the modal class. The modal class is the one with the highest frequency, \(60 - 64\) (frequency \(7\)). The number of rods in this class is \(7\) out of \(30\), so

\[P(\text{modal class})=\frac{7}{30}.\]

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(a) Simplify \((2a + b)^2 - (b - 2a)^2\). Note that \((b - 2a)^2 = (2a - b)^2\). This is a difference of two squares, \(P^2 - R^2 = (P + R)(P - R)\), with \(P = 2a + b\) and \(R = 2a - b\):

\[(2a + b)^2 - (2a - b)^2 = \big[(2a + b) + (2a - b)\big]\big[(2a + b) - (2a - b)\big].\]

\[= (4a)(2b) = 8ab.\]

(b) Given \(S = K\sqrt{m^2 + n^2}\).

(i) Make \(m\) the subject.

\[\frac{S}{K} = \sqrt{m^2 + n^2} \Rightarrow \left(\frac{S}{K}\right)^2 = m^2 + n^2.\]

\[m^2 = \left(\frac{S}{K}\right)^2 - n^2 \Rightarrow m = \sqrt{\left(\frac{S}{K}\right)^2 - n^2}.\]

(ii) With \(S = 12.2, K = 0.02, n = 1.1\):

\[\frac{S}{K} = \frac{12.2}{0.02} = 610, \qquad \left(\frac{S}{K}\right)^2 = 372100, \qquad n^2 = 1.21.\]

\[m = \sqrt{372100 - 1.21} = \sqrt{372098.79} = 609.999 \approx 610.\]

To the nearest whole number, the positive value of \(m = \mathbf{610}\).

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From the diagram: \(PQ=8\text{ m}\), \(QR=13\text{ m}\), the bearing of Q from P is \(050^\circ\), and the bearing of R from Q is \(130^\circ\).

(a)(i) Length PR. Find \(\angle PQR\). The bearing of P from Q is the back-bearing of \(050^\circ\):

\[050^\circ+180^\circ=230^\circ\]

The bearing of R from Q is \(130^\circ\), so

\[\angle PQR=230^\circ-130^\circ=100^\circ\]

Apply the cosine rule to triangle PQR:

\[PR^2=PQ^2+QR^2-2\,(PQ)(QR)\cos(\angle PQR)\]\[PR^2=8^2+13^2-2(8)(13)\cos100^\circ\]\[PR^2=64+169-208(-0.17365)=269.12\]\[PR=16.4\text{ m (3 s.f.)}\]

(a)(ii) Bearing of R from P. Use the sine rule to find \(\angle QPR\):

\[\frac{\sin\angle QPR}{QR}=\frac{\sin\angle PQR}{PR}\]\[\sin\angle QPR=\frac{13\sin100^\circ}{16.405}=\frac{12.803}{16.405}=0.78040\]\[\angle QPR=51.3^\circ\]

R lies clockwise of Q as seen from P, so the bearing of R from P is

\[050^\circ+51.3^\circ=101^\circ\ (\text{3 s.f.})\]

(b) Shortest distance from Q to PR, and the area. The area of triangle PQR is

\[\text{Area}=\tfrac12(PQ)(QR)\sin(\angle PQR)=\tfrac12(8)(13)\sin100^\circ\]\[\text{Area}=52(0.98481)=51.2\text{ m}^2\ (\text{3 s.f.})\]

The shortest distance from Q to PR is the perpendicular height h onto base PR. Using \(\text{Area}=\tfrac12(PR)(h)\):

\[h=\frac{2\times\text{Area}}{PR}=\frac{2(51.210)}{16.405}=6.24\text{ m (3 s.f.)}\]

Awọn alaye Idahun

(a) \[\tfrac12\log_{10}25-2\log_{10}3+\log_{10}18=\log_{10}5-\log_{10}9+\log_{10}18=\log_{10}\!\left(\frac{5\times18}{9}\right)=\log_{10}10=1.\]

(b) \(123_{y}=1\cdot y^{2}+2\cdot y+3\). So \[y^{2}+2y+3=83\Rightarrow y^{2}+2y-80=0\Rightarrow(y+10)(y-8)=0.\] Since a base must be positive, \(y=8\).

(c) \[\frac{9^{2x-3}}{3^{x+3}}=1.\] Write \(9=3^{2}\): numerator \(=3^{2(2x-3)}=3^{4x-6}\). Then \[3^{(4x-6)-(x+3)}=3^{0}\Rightarrow 3x-9=0\Rightarrow x=3.\]

Awọn alaye Idahun

Let \(P(A)=\tfrac23,\;P(K)=\tfrac58,\;P(F)=\tfrac34\), so the failure probabilities are \(\tfrac13,\tfrac38,\tfrac14\). The three events are independent, so multiply.

(a) All three pass: \[\frac23\times\frac58\times\frac34=\frac{30}{96}=\frac{5}{16}.\]

(b) None passes: \[\frac13\times\frac38\times\frac14=\frac{3}{96}=\frac{1}{32}.\]

(c) Ade and Kujo only (Fati fails): \[\frac23\times\frac58\times\frac14=\frac{10}{96}=\frac{5}{48}.\]

Awọn alaye Idahun

The relation illustrated is a downward parabola. Reading from the graph, the curve cuts the y-axis at \(y = 6\), cuts the x-axis at \(x = -1.5\) and \(x = 2\), and turns over near \(x = 0.25\). The annotated graph below shows these key features, which we read off to answer each part.

(a) Roots of \(ax^{2}+bx+c=0\)

The roots are where the curve crosses the x-axis (\(y=0\)). From the graph these are:

\[ x = -1.5 \quad \text{or} \quad x = 2. \]

(b) Values of a, b and c, and the relation

The curve cuts the positive y-axis at \(6\), so the intercept gives

\[ c = 6. \]

Substitute the coordinates of the two x-axis crossings, \(Q(2,\,0)\) and \(P(-1.5,\,0)\), into \(y = ax^{2}+bx+c\) with \(y = 0\) and \(c = 6\).

At \(x = 2\):

\[ a(2)^{2} + b(2) + 6 = 0 \implies 4a + 2b = -6 \quad \text{...(1)} \]

At \(x = -1.5\):

\[ a(-1.5)^{2} + b(-1.5) + 6 = 0 \implies \tfrac{9a}{4} - \tfrac{3b}{2} = -6 \implies 9a - 6b = -24 \quad \text{...(2)} \]

Multiply (1) by 3:

\[ 12a + 6b = -18 \quad \text{...(1a)} \]

Add (1a) and (2):

\[ 21a = -42 \implies a = -2. \]

Substitute \(a = -2\) into (1):

\[ 4(-2) + 2b = -6 \implies 2b = 2 \implies b = 1. \]

Therefore \((a,\,b,\,c) = (-2,\,1,\,6)\), and the relation illustrated by the graph is

\[ \boxed{\,y = -2x^{2} + x + 6\,}. \]

(c) Maximum value of y and the corresponding x

The maximum occurs at the turning point. Completing the derivative (or reading the vertex from the graph):

\[ \frac{dy}{dx} = -4x + 1 = 0 \implies x = \tfrac{1}{4} = 0.25. \]

Substituting \(x = 0.25\):

\[ y_{\max} = -2(0.25)^{2} + 0.25 + 6 = -0.125 + 0.25 + 6 = 6.125. \]

So the maximum value is \(y_{\max} \approx 6.13\) (about \(6.1\)), occurring at \(x = 0.25\), which agrees with the peak read from the graph.

(d) Values of x when \(y = 2\)

Draw the horizontal line \(y = 2\) and read where it meets the curve, or solve:

\[ -2x^{2} + x + 6 = 2 \implies -2x^{2} + x + 4 = 0 \implies 2x^{2} - x - 4 = 0. \]

Using the quadratic formula:

\[ x = \frac{-(-1) \pm \sqrt{(-1)^{2} - 4(2)(-4)}}{2(2)} = \frac{1 \pm \sqrt{33}}{4} = \frac{1 \pm 5.745}{4}. \] \[ x = \frac{1 + 5.745}{4} = 1.69 \quad \text{or} \quad x = \frac{1 - 5.745}{4} = -1.19. \]

So when \(y = 2\), \(x \approx -1.2\) or \(x \approx 1.7\), matching the two points where the line \(y = 2\) cuts the curve on the graph.