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To solve the simultaneous equations, set the two expressions for \( y \) equal to each other:

\[ x^2 - x - 5 = x + 3 \]

Rearrange the equation to equal zero:

\[ x^2 - 2x - 8 = 0 \]

Factorise the quadratic equation:

\[ (x - 4)(x + 2) = 0 \]

This gives the \( x \)-coordinates of the intersection points:

\[ x = 4 \quad \text{and} \quad x = -2 \]

Now substitute these values back into the linear equation \( y = x + 3 \) to find the corresponding \( y \)-values:

For \( x = 4 \):
\[ y = 4 + 3 = 7 \]

For \( x = -2 \):
\[ y = -2 + 3 = 1 \]

So the solutions are \( (4, 7) \) and \( (-2, 1) \).

M1: Equates the two expressions to form a quadratic equation in \( x \), e.g., \( x^2 - x - 5 = x + 3 \)
M1: Rearranges the quadratic into standard form, \( x^2 - 2x - 8 = 0 \)
M1: Factorises or uses the quadratic formula to find both \( x \)-values: \( x = 4 \) and \( x = -2 \)
A1: Correctly identifies both pairs of solutions: \( (4, 7) \) and \( (-2, 1) \).

The value of the ornament after 3 years is given by the formula for compound depreciation:

\[ V = V_0 \times \left(1 - \frac{r}{100}\right)^t \]

Substitute the given values into the formula:

\[ 5832 = 8000 \times \left(1 - \frac{r}{100}\right)^3 \]

Divide both sides by 8000:

\[ \left(1 - \frac{r}{100}\right)^3 = \frac{5832}{8000} \]
\[ \left(1 - \frac{r}{100}\right)^3 = 0.729 \]

Take the cube root of both sides:

\[ 1 - \frac{r}{100} = \sqrt[3]{0.729} \]
\[ 1 - \frac{r}{100} = 0.9 \]

Solve for \( r \):

\[ \frac{r}{100} = 0.1 \]
\[ r = 10 \]

M1: Sets up a correct algebraic equation for the 3-year depreciation, e.g., \( 8000 \times k^3 = 5832 \)
M1: Solves to find the value of the cube of the multiplier, \( k^3 = 0.729 \) (or equivalent fraction \( \frac{729}{1000} \))
M1: Finds the multiplier \( k = 0.9 \) by taking the cube root
A1: Correctly calculates \( r = 10 \)

First, find the angle of the sector, \( \theta \). The formula for the area of a sector is:

\[ \text{Area} = \frac{\theta}{360} \times \pi r^2 \]

Substitute the given values:

\[ 48\pi = \frac{\theta}{360} \times \pi \times 12^2 \]
\[ 48\pi = \frac{\theta}{360} \times 144\pi \]

Divide both sides by \( \pi \):

\[ 48 = 144 \times \frac{\theta}{360} \]
\[ \frac{\theta}{360} = \frac{48}{144} = \frac{1}{3} \]

Now, calculate the arc length of the sector using the fraction of the circumference:

\[ \text{Arc length} = \frac{1}{3} \times 2\pi r \]
\[ \text{Arc length} = \frac{1}{3} \times 2\pi \times 12 = 8\pi \]

The perimeter of the sector is the sum of the arc length and the two radii:

\[ \text{Perimeter} = \text{Arc length} + 2r \]
\[ \text{Perimeter} = 8\pi + 2(12) = 8\pi + 24 \]

M1: Uses the sector area formula to set up an equation, e.g., \( \frac{\theta}{360} \times 144\pi = 48\pi \)
M1: Determines the fraction of the circle is \( \frac{1}{3} \) (or that the angle \( \theta = 120^\circ \))
M1: Calculates the arc length as \( 8\pi \) using their fraction or angle
A1: Gives the correct final perimeter of \( 8\pi + 24 \)

First, find the critical values by solving the corresponding quadratic equation:

\[ 2x^2 - 5x - 3 = 0 \]

Factorise the quadratic:

\[ (2x + 1)(x - 3) = 0 \]

This gives the critical values:

\[ x = -0.5 \quad \text{and} \quad x = 3 \]

We require the values of \( x \) for which \( 2x^2 - 5x - 3 \) is strictly greater than zero. Since the coefficient of \( x^2 \) is positive, the graph of the quadratic is a U-shaped parabola. It is above the \( x \)-axis (positive) in the regions outside the critical values.

Therefore, the solution is:

\[ x < -0.5 \quad \text{or} \quad x > 3 \]

M1: Factorises the quadratic or uses the quadratic formula to find critical values, e.g., \( (2x + 1)(x - 3) \)
M1: Correctly identifies the critical values as \( x = -0.5 \) (or \( -\frac{1}{2} \)) and \( x = 3 \)
M1: Identifies the correct regions (the outer tails of the quadratic graph)
A1: Writes the final inequalities correctly: \( x < -0.5 \) or \( x > 3 \) (accept equivalent fractions)

The probability of choosing a red bead first is \( \frac{5}{n} \).

Since the selection is without replacement, there are now \( 4 \) red beads and \( n - 1 \) total beads remaining in the bag.

The probability of choosing a second red bead is \( \frac{4}{n - 1} \).

The probability of choosing two red beads is:

\[ P(\text{Red, Red}) = \frac{5}{n} \times \frac{4}{n - 1} = \frac{20}{n(n - 1)} \]

Set this probability equal to \( \frac{5}{33} \):

\[ \frac{20}{n(n - 1)} = \frac{5}{33} \]

Multiply both sides by \( 33n(n - 1) \):

\[ 20 \times 33 = 5n(n - 1) \]
\[ 660 = 5(n^2 - n) \]

Divide by 5:

\[ 132 = n^2 - n \]
\[ n^2 - n - 132 = 0 \]

Now solve this quadratic equation by factorisation:

\[ (n - 12)(n + 11) = 0 \]

This gives \( n = 12 \) or \( n = -11 \). Since the number of beads \( n \) must be positive, we have \( n = 12 \).

M1: Writes down the probability equation for two red beads without replacement: \( \frac{5}{n} \times \frac{4}{n-1} = \frac{5}{33} \)
M1: Eliminates fractions to obtain the given quadratic equation, e.g., showing \( 660 = 5n(n-1) \) leading to \( n^2 - n - 132 = 0 \)
M1: Factorises the quadratic equation to find the solutions: \( (n-12)(n+11) = 0 \)
A1: Selects the positive value, giving final answer of \( n = 12 \)

Since \( y \) is inversely proportional to the square of \( x \), we can write:

\[ y = \frac{k}{x^2} \]

where \( k \) is a constant of proportionality.

Use the given values \( x = 4 \) and \( y = 9 \) to find \( k \):

\[ 9 = \frac{k}{4^2} \]
\[ 9 = \frac{k}{16} \]
\[ k = 9 \times 16 = 144 \]

So the formula relating \( y \) and \( x \) is:

\[ y = \frac{144}{x^2} \]

Now find \( x \) when \( y = 16 \):

\[ 16 = \frac{144}{x^2} \]
\[ 16x^2 = 144 \]
\[ x^2 = \frac{144}{16} = 9 \]

Since \( x > 0 \):

\[ x = \sqrt{9} = 3 \]

M1: Sets up the proportionality equation \( y = \frac{k}{x^2} \)
M1: Substitutes \( x = 4 \) and \( y = 9 \) to calculate \( k = 144 \)
M1: Substitutes \( y = 16 \) into their equation to obtain \( x^2 = 9 \)
A1: Correctly identifies \( x = 3 \) (rejects \( x = -3 \) as \( x > 0 \))

Using the Cosine Rule to find the length of side \( AC \) (which we can denote as side \( b \)):

\[ b^2 = a^2 + c^2 - 2ac\cos(B) \]

Here, \( a = BC = 10\text{ cm} \), \( c = AB = 7\text{ cm} \), and angle \( B = 60^\circ \).

Substitute the values into the formula:

\[ AC^2 = 10^2 + 7^2 - 2 \times 10 \times 7 \times \cos(60^\circ) \]
\[ AC^2 = 100 + 49 - 140 \times 0.5 \]
\[ AC^2 = 149 - 70 \]
\[ AC^2 = 79 \]

Take the square root of both sides:

\[ AC = \sqrt{79} \approx 8.88819... \]

Rounding to 3 significant figures gives \( AC = 8.89\text{ cm} \).

M1: Correctly applies the Cosine Rule formula with values substituted: \( AC^2 = 10^2 + 7^2 - 2(10)(7)\cos(60^\circ) \)
M1: Simplifies the expression using \( \cos(60^\circ) = 0.5 \) to get \( AC^2 = 149 - 70 \)
M1: Finds \( AC^2 = 79 \) and attempts to find the square root
A1: Correctly rounds to 3 significant figures, obtaining \( 8.89 \)

First, find the first and second differences of the sequence:

Sequence: \( 4, \ 11, \ 22, \ 37, \ 56 \)
First differences: \( 7, \ 11, \ 15, \ 19 \)
Second differences: \( 4, \ 4, \ 4 \)

Since the second difference is constant at 4, the coefficient of the \( n^2 \) term is half of this constant:

\[ a = \frac{4}{2} = 2 \]

So the sequence contains a \( 2n^2 \) term. Now subtract \( 2n^2 \) from each term of the original sequence to find the remaining linear sequence:

For \( n=1 \): \( 4 - 2(1^2) = 4 - 2 = 2 \)
For \( n=2 \): \( 11 - 2(2^2) = 11 - 8 = 3 \)
For \( n=3 \): \( 22 - 2(3^2) = 22 - 18 = 4 \)
For \( n=4 \): \( 37 - 2(4^2) = 37 - 32 = 5 \)

The remaining linear sequence is \( 2, \ 3, \ 4, \ 5, \ \dots \).

The \( n \)-th term of this linear sequence has a common difference of 1 and a first term of 2, so its formula is:

\[ n + 1 \]

Combining the quadratic and linear parts, the \( n \)-th term of the original sequence is:

\[ 2n^2 + n + 1 \]

M1: Finds the first and second differences to determine that the second difference is 4
M1: Identifies the coefficient of \( n^2 \) as \( 2 \) (showing the term \( 2n^2 \))
M1: Subtracts \( 2n^2 \) from the terms of the sequence to establish the linear sequence \( 2, 3, 4, 5, \dots \) and finds its general term \( n + 1 \)
A1: Writes the correct final expression: \( 2n^2 + n + 1 \) (or equivalent)

The area of the equilateral triangle is given by \(\frac{1}{2} a b \sin C = \frac{1}{2} \times 8 \times 8 \times \sin(60^\circ) = 16\sqrt{3} \approx 27.7128\text{ cm}^2\). Since the total area of the logo is \(50\text{ cm}^2\), the area of the sector is \(50 - 16\sqrt{3} \approx 22.2872\text{ cm}^2\). The formula for the area of a sector is \(\frac{\theta}{360} \times \pi r^2\). Substituting the radius \(r = 8\) gives \(\frac{\theta}{360} \times \pi \times 8^2 = 22.2872\). Solving for \(\theta\): \(\theta = \frac{22.2872 \times 360}{64\pi} \approx 39.905\). Thus, to 1 decimal place, \(\theta = 39.9\).

M1 for finding the area of the equilateral triangle: \(16\sqrt{3}\) (or \(27.7\)). M1 for finding the area of the sector: \(50 - 16\sqrt{3}\) (or \(22.3\)). M1 for setting up the equation \(\frac{\theta}{360} \times \pi \times 8^2 = \text{sector area}\). A1 for \(\theta = 39.9\) (accept answers in the range 39.8 to 40.0).

For the first class interval \(30 < t \le 40\), the class width is \(10\) minutes, which is represented by \(2\text{ cm}\) on the grid. Therefore, \(1\text{ cm}\) of width represents \(5\) minutes. The frequency density for this interval is \(\text{frequency} \div \text{class width} = 45 \div 10 = 4.5\). This is represented by a height of \(9\text{ cm}\), so \(1\text{ cm}\) of height represents a frequency density of \(4.5 \div 9 = 0.5\). For the second class interval \(40 < t \le 55\), the class width is \(15\) minutes, represented by a width of \(3\text{ cm}\) on the grid. The bar height of \(12\text{ cm}\) represents a frequency density of \(12 \times 0.5 = 6\). Therefore, the frequency for this interval is \(\text{frequency density} \times \text{class width} = 6 \times 15 = 90\).

M1 for finding the frequency density of the first interval (4.5). M1 for establishing the horizontal scale (\(1\text{ cm} = 5\text{ mins}\)) or vertical scale (\(1\text{ cm} = 0.5\text{ frequency density}\)). M1 for calculating the frequency density of the second interval as 6. A1 for 90.

The lawn has dimensions \(x\) and \(2x + 5\). Adding a \(1\text{ m}\) path on all sides increases both dimensions by \(2\text{ m}\). So, the overall width is \(x + 2\) and the overall length is \(2x + 5 + 2 = 2x + 7\). The total area is given by \((x + 2)(2x + 7) = 104\). Expanding the brackets: \(2x^2 + 7x + 4x + 14 = 104 \implies 2x^2 + 11x + 14 = 104\). Rearranging to form a quadratic: \(2x^2 + 11x - 90 = 0\). Factorising the quadratic equation: \((2x - 9)(x + 10) = 0\). This gives the solutions \(x = 4.5\) or \(x = -10\). Since the width must be positive, \(x = 4.5\).

M1 for writing down expressions for the outer dimensions: \(x + 2\) and \(2x + 7\). M1 for setting up the equation \((x + 2)(2x + 7) = 104\). M1 for simplifying to the quadratic equation \(2x^2 + 11x - 90 = 0\). A1 for finding the correct positive solution \(x = 4.5\).

The probability of choosing a green bead first is \(\frac{7}{n}\). Since the first bead is not replaced, the probability of choosing a green bead second is \(\frac{6}{n-1}\). The probability of choosing two green beads is \(\frac{7}{n} \times \frac{6}{n-1} = \frac{42}{n(n-1)}\). We are given that this probability is \(\frac{7}{22}\), so \(\frac{42}{n(n-1)} = \frac{7}{22}\). Dividing both sides by 7: \(\frac{6}{n(n-1)} = \frac{1}{22}\). This gives \(n(n-1) = 132 \implies n^2 - n - 132 = 0\). Factorising the quadratic: \((n - 12)(n + 11) = 0\). Since \(n\) must be positive, \(n = 12\).

M1 for writing the probability of both green beads as \(\frac{7}{n} \times \frac{6}{n-1}\). M1 for forming the equation \(\frac{42}{n(n-1)} = \frac{7}{22}\). M1 for simplifying to \(n^2 - n - 132 = 0\). A1 for solving to get \(n = 12\).

Using the compound interest formula: \(4500 \times \left(1 + \frac{r}{100}\right)^4 = 5200\). Dividing both sides by 4500: \(\left(1 + \frac{r}{100}\right)^4 = \frac{5200}{4500} = \frac{52}{45}\). Taking the fourth root of both sides: \(1 + \frac{r}{100} = \sqrt[4]{\frac{52}{45}} \approx 1.036814\). Subtracting 1: \(\frac{r}{100} \approx 0.036814\). Multiplying by 100: \(r \approx 3.6814\). To 2 decimal places, \(r = 3.68\).

M1 for setting up the equation \(4500 \times \left(1 + \frac{r}{100}\right)^4 = 5200\). M1 for isolating the power: \(\left(1 + \frac{r}{100}\right)^4 = \frac{52}{45}\) (or \(\approx 1.156\)). M1 for taking the fourth root to obtain \(1 + \frac{r}{100} \approx 1.0368\). A1 for \(r = 3.68\) (accept answers in range 3.68 to 3.7).

First, factorise the numerator by finding two numbers that multiply to \(2 \times (-12) = -24\) and add to \(-5\). These numbers are \(-8\) and \(3\). Rewrite and factorise by grouping: \(2x^2 - 8x + 3x - 12 = 2x(x - 4) + 3(x - 4) = (2x + 3)(x - 4)\). Next, factorise the denominator by taking out the highest common factor: \(3x^2 - 12x = 3x(x - 4)\). Write the expression with the factorised forms: \(\frac{(2x + 3)(x - 4)}{3x(x - 4)}\). Cancel the common factor \((x - 4)\) from the numerator and denominator to get \(\frac{2x + 3}{3x}\).

M2 for factorising the numerator to \((2x + 3)(x - 4)\) (M1 for a partial factorisation or a binomial product of the form \((2x+a)(x+b)\) where \(ab=-12\) or \(2b+a=-5\)). M1 for factorising the denominator to \(3x(x - 4)\). A1 for the fully simplified fraction \(\frac{2x + 3}{3x}\) (or equivalent).

Since \(y\) is inversely proportional to the square of \(x\), we can write the formula as \(y = \frac{k}{x^2}\), where \(k\) is a constant. Substitute \(x = 4\) and \(y = 9\) into this equation: \(9 = \frac{k}{4^2} \implies 9 = \frac{k}{16} \implies k = 9 \times 16 = 144\). So, the equation relating \(y\) and \(x\) is \(y = \frac{144}{x^2}\). Now substitute \(y = 16\) to find \(x\): \(16 = \frac{144}{x^2} \implies x^2 = \frac{144}{16} \implies x^2 = 9\). Since \(x > 0\), we take the positive square root: \(x = 3\).

M1 for expressing the relationship as \(y = \frac{k}{x^2}\). M1 for substituting the given values to find \(k = 144\). M1 for setting up \(16 = \frac{144}{x^2}\) and rearranging to \(x^2 = 9\). A1 for \(x = 3\).

First, apply the Cosine Rule to find \(AC\): \(AC^2 = AB^2 + BC^2 - 2 \cdot AB \cdot BC \cdot \cos(ABC)\). Substituting the values: \(AC^2 = 8.4^2 + 11.2^2 - 2 \times 8.4 \times 11.2 \times \cos(64^\circ) = 70.56 + 125.44 - 188.16 \times 0.438371 = 196 - 82.4839 = 113.516\). Therefore, \(AC = \sqrt{113.516} \approx 10.654\text{ cm}\). Next, calculate the area of triangle \(ABC\): \(\text{Area} = \frac{1}{2} \cdot AB \cdot BC \cdot \sin(64^\circ) = \frac{1}{2} \times 8.4 \times 11.2 \times \sin(64^\circ) \approx 42.279\text{ cm}^2\). Since \(BD\) is perpendicular to \(AC\), the area can also be expressed as \(\text{Area} = \frac{1}{2} \cdot AC \cdot BD\). Thus, \(42.279 = \frac{1}{2} \times 10.654 \times BD \implies BD = \frac{2 \times 42.279}{10.654} \approx 7.9365\text{ cm}\). Correct to 3 significant figures, \(BD = 7.94\text{ cm}\).

M1 for using the Cosine Rule to find \(AC^2\) or \(AC\) (e.g. \(AC^2 = 113.5\) or \(AC = 10.7\)). M1 for calculating the area of the triangle: \(42.3\text{ cm}^2\) (or for using the Sine Rule to calculate angle \(A \approx 70.8^\circ\)). M1 for setting up the equation \(\text{Area} = \frac{1}{2} \times AC \times BD\) (or \(\sin A = \frac{BD}{8.4}\)). A1 for \(BD = 7.94\) (accept answers in the range 7.93 to 7.95).

1. State the area of the square:
\(\text{Area of square} = (2x)^2 = 4x^2\).

2. State the area of the four semicircles. Each has a diameter of \(2x\), so the radius \(r = x\):
\(\text{Area of 4 semicircles} = 4 \times \frac{1}{2}\pi x^2 = 2\pi x^2\).

3. Set up the equation for the total area:
\(4x^2 + 2\pi x^2 = 120\).

4. Factorise out \(x^2\):
\(x^2(4 + 2\pi) = 120\).

5. Solve for \(x\):
\(x^2 = \frac{120}{4 + 2\pi} \approx 11.6695\)
\(x = \sqrt{11.6695} \approx 3.416\).

Rounding to 3 significant figures gives \(x = 3.42\).

M1: For writing a correct expression for the area of the square, e.g., \(4x^2\)
M1: For writing a correct expression for the area of the 4 semicircles, e.g., \(2\pi x^2\)
M1: For setting up the equation \(x^2(4 + 2\pi) = 120\) and attempting to solve for \(x^2\)
A1: For \(x = 3.42\) (or a value in the range \(3.41\) to \(3.42\))

1. Substitute the linear equation into the quadratic equation:
\(x^2 + (2x - 3)^2 = 18\).

2. Expand the bracket:
\(x^2 + 4x^2 - 12x + 9 = 18\).

3. Simplify and set to 0:
\(5x^2 - 12x - 9 = 0\).

4. Factorise the quadratic:
\(5x^2 - 15x + 3x - 9 = 0 \implies 5x(x - 3) + 3(x - 3) = 0 \implies (5x + 3)(x - 3) = 0\).

5. Solve for \(x\):
\(x = 3\) or \(x = -0.6\).

6. Substitute \(x\) back to find \(y\):
If \(x = 3\), \(y = 2(3) - 3 = 3\).
If \(x = -0.6\), \(y = 2(-0.6) - 3 = -4.2\).

Thus, the solutions are \(x = 3, y = 3\) and \(x = -0.6, y = -4.2\).

M1: Substitution of \(2x - 3\) into the quadratic equation to get \(x^2 + (2x - 3)^2 = 18\)
M1: Correct expansion and reduction to a standard quadratic equation, e.g., \(5x^2 - 12x - 9 = 0\)
M1: Solving the quadratic equation to find two values of \(x\) (e.g. \(x = 3\) and \(x = -0.6\))
A1: Both pairs of solutions clearly stated: \(x = 3, y = 3\) and \(x = -0.6, y = -4.2\)

1. Identify the intervals that cover \(t > 25\):
- Part of the interval \(20 < t \le 30\), specifically \(25 < t \le 30\).
- The whole interval \(30 < t \le 50\).

2. Calculate the estimated number of people in \(25 < t \le 30\):
\(\text{Class width} = 30 - 25 = 5\).
\(\text{Frequency} = \text{width} \times \text{frequency density} = 5 \times 2.6 = 13\).

3. Calculate the number of people in \(30 < t \le 50\):
\(\text{Class width} = 50 - 30 = 20\).
\(\text{Frequency} = 20 \times 1.4 = 28\).

4. Total estimated number of people = \(13 + 28 = 41\).

M1: For identifying that the interval \(25 < t \le 30\) has width 5
M1: For calculating the frequency of the interval \(30 < t \le 50\) as \(20 \times 1.4 = 28\)
M1: For finding the partial frequency as \(5 \times 2.6 = 13\)
A1: For the correct total of 41

1. Set up the compound growth equation:
\(1200 \times \left(1 + \frac{x}{100}\right)^6 = 1850\).

2. Divide both sides by 1200:
\(\left(1 + \frac{x}{100}\right)^6 = \frac{1850}{1200} = \frac{37}{24}\).

3. Take the 6th root of both sides:
\(1 + \frac{x}{100} = \left(\frac{37}{24}\right)^{\frac{1}{6}} \approx 1.074811\).

4. Subtract 1:
\(\frac{x}{100} \approx 0.074811\).

5. Multiply by 100:
\(x \approx 7.4811\).

Rounding to 2 decimal places gives \(7.48\).

M1: Set up correct equation, e.g. \(1200 \times (1 + \frac{x}{100})^6 = 1850\)
M1: Attempt to isolate the power term, e.g. \((1 + \frac{x}{100})^6 = 1.5416...\)
M1: Take the 6th root, e.g. \(1 + \frac{x}{100} = 1.0748...\)
A1: For \(7.48\) (or a value in the range \(7.48\) to \(7.50\))

1. Write down the proportionality equation:
\(P = \frac{k q^2}{r}\) where \(k\) is a constant.

2. Substitute the given values to find \(k\):
\(14.4 = \frac{k \times 6^2}{5}\)
\(14.4 = \frac{36 k}{5}\)
\(72 = 36 k \implies k = 2\).

3. Write the formula with the constant:
\(P = \frac{2 q^2}{r}\).

4. Substitute the new values \(q = 10\) and \(r = 8\) to find \(P\):
\(P = \frac{2 \times 10^2}{8} = \frac{200}{8} = 25\).

M1: For establishing the proportional relationship \(P = \frac{k q^2}{r}\)
M1: For substituting \(q = 6\), \(r = 5\), and \(P = 14.4\) to find \(k\)
A1: For finding \(k = 2\)
A1: For \(P = 25\)

1. Use the area of a triangle formula to find angle \(B\):
\(\text{Area} = \frac{1}{2} a c \sin B\)
\(28 = \frac{1}{2} \times 9.5 \times 7.2 \times \sin B\)
\(28 = 34.2 \sin B\)
\(\sin B = \frac{28}{34.2} \approx 0.818713\).

2. Find the obtuse angle \(B\):
\(\sin^{-1}(0.818713) \approx 54.953^\circ\).
Since \(B\) is obtuse, \(B = 180^\circ - 54.953^\circ = 125.047^\circ\).

3. Use the Cosine Rule to find the length of \(AC\):
\(AC^2 = AB^2 + BC^2 - 2 \times AB \times BC \times \cos B\)
\(AC^2 = 7.2^2 + 9.5^2 - 2(7.2)(9.5) \cos(125.047^\circ)\)
\(AC^2 = 51.84 + 90.25 - 136.8 \times (-0.574231)\)
\(AC^2 = 142.09 + 78.555 = 220.645\)
\(AC = \sqrt{220.645} \approx 14.854\).

To 3 significant figures, \(AC = 14.9\text{ cm}\).

M1: For substituting into the area formula: \(28 = \frac{1}{2} \times 7.2 \times 9.5 \times \sin B\)
M1: For finding the obtuse angle \(B \approx 125^\circ\) (condone acute angle \(55^\circ\) if used in the next step)
M1: For substituting their angle and side lengths into the Cosine Rule: \(AC^2 = 7.2^2 + 9.5^2 - 2 \times 7.2 \times 9.5 \times \cos B\)
A1: For \(AC = 14.9\) (accept range \(14.8\) to \(15.0\))

Substitute the expression for \(y\) from the first equation into the second equation:

\(x^2 + (2x - 3)^2 = 18\)

Expand the bracket:

\(x^2 + (4x^2 - 12x + 9) = 18\)

Simplify and rearrange into a standard quadratic form \(ax^2 + bx + c = 0\):

\(5x^2 - 12x - 9 = 0\)

Factorise the quadratic expression:

\((5x + 3)(x - 3) = 0\)

This gives two possible values for \(x\):

\(x = 3\) or \(x = -\frac{3}{5} = -0.6\)

Now find the corresponding \(y\)-values using \(y = 2x - 3\):

- If \(x = 3\), then \(y = 2(3) - 3 = 3\).
- If \(x = -0.6\), then \(y = 2(-0.6) - 3 = -1.2 - 3 = -4.2\).

Therefore, the solutions are \(x = 3, y = 3\) and \(x = -0.6, y = -4.2\).

M1: Substitution of \(y = 2x - 3\) into the quadratic equation to obtain \(x^2 + (2x - 3)^2 = 18\).
M1: Expansion and reduction to the quadratic form \(5x^2 - 12x - 9 = 0\).
M1: Solving the quadratic equation to find both \(x\)-values: \(x = 3\) and \(x = -0.6\) (or equivalent fractions).
A1: For both pairs of fully correct solutions: \(x = 3, y = 3\) and \(x = -0.6, y = -4.2\) (or fractions: \(x = -\frac{3}{5}, y = -\frac{21}{5}\)).

The total number of counters in the bag is \(r + 5\).

The probability of drawing a blue counter first is \(\frac{5}{r + 5}\).

Since the selection is made without replacement, there are now \(r + 4\) counters remaining, of which 4 are blue.

The probability of drawing a second blue counter is \(\frac{4}{r + 4}\).

The probability that both counters are blue is:

\(\frac{5}{r + 5} \times \frac{4}{r + 4} = \frac{20}{(r + 5)(r + 4)}\)

We are given that this probability is \(\frac{2}{9}\):

\(\frac{20}{r^2 + 9r + 20} = \frac{2}{9}\)

Cross-multiplying gives:

\(180 = 2(r^2 + 9r + 20)\)

Divide both sides by 2:

\(90 = r^2 + 9r + 20\)

Rearrange into standard quadratic form:

\(r^2 + 9r - 70 = 0\)

Factorise the quadratic equation:

\((r + 14)(r - 5) = 0\)

This gives \(r = -14\) or \(r = 5\).

Since the number of counters must be positive, \(r = 5\).

M1: Sets up the initial probability product equation: \(\frac{5}{r + 5} \times \frac{4}{r + 4} = \frac{2}{9}\) or equivalent.
M1: Multiplies out the denominators and sets up a cross-multiplied equation, e.g., \(180 = 2(r^2 + 9r + 20)\).
M1: Reduces to the simplified quadratic equation \(r^2 + 9r - 70 = 0\) and attempts to factorise.
A1: Correct value \(r = 5\) (with negative solution discarded).

First, find the area of the entire rectangle:

\(\text{Area}_{\text{rectangle}} = (2x + 3)(x + 4) = 2x^2 + 8x + 3x + 12 = 2x^2 + 11x + 12\)

Next, find the area of the right-angled triangle that was removed:

\(\text{Area}_{\text{triangle}} = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times x \times 2x = x^2\)

The remaining area is the difference between the area of the rectangle and the area of the triangle:

\(\text{Remaining Area} = (2x^2 + 11x + 12) - x^2 = x^2 + 11x + 12\)

We are given that the remaining area is \(38\text{ cm}^2\):

\(x^2 + 11x + 12 = 38\)

Rearrange this into standard quadratic form:

\(x^2 + 11x - 26 = 0\)

Factorise the quadratic:

\((x + 13)(x - 2) = 0\)

This gives \(x = -13\) or \(x = 2\).

Since \(x\) represents physical lengths, it must be positive. Therefore, \(x = 2\).

M1: For expanding the area of the rectangle correctly: \((2x + 3)(x + 4) = 2x^2 + 11x + 12\).
M1: For expressing the area of the triangle as \(x^2\) and subtracting it to get \(x^2 + 11x + 12\).
M1: For setting the expression equal to 38 and rearranging to get a factorisable quadratic: \(x^2 + 11x - 26 = 0\).
A1: For the correct solution \(x = 2\) (negative solution discarded).

Let the general formula for the \(n\)-th term of a quadratic sequence be \(an^2 + bn + c\).

Find the first differences between consecutive terms:
- \(10 - 3 = 7\)
- \(21 - 10 = 11\)
- \(36 - 21 = 15\)

Find the second differences:
- \(11 - 7 = 4\)
- \(15 - 11 = 4\)

Since the second difference is constant and equal to 4, we have:

\(2a = 4 \implies a = 2\)

So the sequence contains the term \(2n^2\). Now, subtract \(2n^2\) from each term of the original sequence to find the linear part \(bn + c\):
- For \(n = 1\): \(3 - 2(1)^2 = 3 - 2 = 1\)
- For \(n = 2\): \(10 - 2(2)^2 = 10 - 8 = 2\)
- For \(n = 3\): \(21 - 2(3)^2 = 21 - 18 = 3\)
- For \(n = 4\): \(36 - 2(4)^2 = 36 - 32 = 4\)

The remaining sequence is \(1, 2, 3, 4, \dots\), which has the general formula \(n\) (so \(b = 1\) and \(c = 0\)).

Thus, the formula for the \(n\)-th term of the quadratic sequence is:

\(U_n = 2n^2 + n\)

To find the 20th term, substitute \(n = 20\):

\(U_{20} = 2(20)^2 + 20 = 2(400) + 20 = 800 + 20 = 820\)

M1: For finding the first differences (7, 11, 15) and the second differences (4, 4).
M1: For deducing that \(a = 2\) so the formula begins with \(2n^2\).
M1: For subtracting \(2n^2\) and finding the correct general formula for the \(n\)-th term: \(2n^2 + n\).
A1: For substituting \(n = 20\) to obtain the correct answer of \(820\).

Since \(y\) is inversely proportional to the square of \((x - 1)\), we can write the relationship as:

\(y = \frac{k}{(x - 1)^2}\)

where \(k\) is a constant of proportionality.

Substitute the given values \(x = 3\) and \(y = 9\) to find \(k\):

\(9 = \frac{k}{(3 - 1)^2}\)

\(9 = \frac{k}{2^2}\)

\(9 = \frac{k}{4} \implies k = 36\)

So the equation connecting \(x\) and \(y\) is:

\(y = \frac{36}{(x - 1)^2}\)

Now, substitute \(y = 4\) into this equation to solve for \(x\):

\(4 = \frac{36}{(x - 1)^2}\)

Multiply both sides by \((x - 1)^2\):

\(4(x - 1)^2 = 36\)

Divide by 4:

\((x - 1)^2 = 9\)

Take the square root of both sides:

\(x - 1 = \pm 3\)

This gives two possible equations:
- \(x - 1 = 3 \implies x = 4\)
- \(x - 1 = -3 \implies x = -2\)

Since the question specifies \(x > 1\), we must have \(x = 4\).

M1: Sets up the correct proportional equation: \(y = \frac{k}{(x - 1)^2}\).
M1: Substitutes \(x = 3, y = 9\) correctly to find the value of \(k = 36\).
M1: Substitutes \(y = 4\) to form the equation \(4 = \frac{36}{(x - 1)^2}\) and solves to get \((x - 1)^2 = 9\).
A1: For the final answer \(x = 4\) (rejecting \(x = -2\) as \(x > 1\)).

The height of a bar in a histogram is proportional to the frequency density of the corresponding class interval.

First, calculate the frequency density (FD) of the class interval \(10 < t \le 15\):
- Class Width = \(15 - 10 = 5\)
- Frequency = \(20\)
- \(\text{FD} = \frac{\text{Frequency}}{\text{Class Width}} = \frac{20}{5} = 4\)

We are given that the bar for this interval has a height of 8 cm. Thus, the relationship between the height of the bar and its frequency density is:

\(\text{Height} = \text{FD} \times \text{Scale Factor}\)

\(8 = 4 \times \text{Scale Factor} \implies \text{Scale Factor} = 2\)

Now, find the frequency density of the class interval \(20 < t \le 40\):
- Class Width = \(40 - 20 = 20\)
- Frequency = \(20\)
- \(\text{FD} = \frac{\text{Frequency}}{\text{Class Width}} = \frac{20}{20} = 1\)

Finally, calculate the height of this bar using the scale factor of 2:

\(\text{Height} = 1 \times 2 = 2\text{ cm}\)

M1: For calculating the frequency density of the \(10 < t \le 15\) class interval: \(20 \div 5 = 4\).
M1: For calculating the frequency density of the \(20 < t \le 40\) class interval: \(20 \div 20 = 1\).
M1: For setting up the scaling relationship correctly: \(\text{Height} = \text{FD} \times 2\).
A1: For the correct height of \(2\) cm.

To simplify the algebraic fraction, factorise both the numerator and the denominator.

First, factorise the numerator, \(2x^2 + 5x - 3\):
- We look for two numbers that multiply to \(2 \times (-3) = -6\) and add to \(5\). These numbers are \(6\) and \(-1\).
- Rewrite and factorise by grouping:
\(2x^2 + 6x - x - 3 = 2x(x + 3) - 1(x + 3) = (2x - 1)(x + 3)\)

Next, factorise the denominator, \(4x^2 - 1\):
- This is a difference of two squares:
\(4x^2 - 1 = (2x)^2 - (1)^2 = (2x - 1)(2x + 1)\)

Now, substitute the factorised expressions back into the fraction:

\(\frac{(2x - 1)(x + 3)}{(2x - 1)(2x + 1)}\)

Cancel the common factor \((2x - 1)\) from both the numerator and the denominator:

\(\frac{x + 3}{2x + 1}\)

M1: Factorises numerator to \((2x - 1)(x + 3)\) (or equivalent method).
M1: Factorises denominator to \((2x - 1)(2x + 1)\).
M1: For showing a fraction with both numerator and denominator factorised and a clear attempt to cancel the common factor.
A1: Correct final simplified expression: \(\frac{x + 3}{2x + 1}\).

First, find the turning point of the original curve \(C\) by completing the square for \(y = x^2 - 6x + 5\):

\(y = (x - 3)^2 - 3^2 + 5\)

\(y = (x - 3)^2 - 9 + 5\)

\(y = (x - 3)^2 - 4\)

From this form, we can see that the turning point (minimum) of curve \(C\) is at \((3, -4)\).

The curve \(C\) is translated by the vector \(\begin{pmatrix} -3 \\ 2 \end{pmatrix}\).
This means:
- Every \(x\)-coordinate is shifted by \(-3\) units (left by 3).
- Every \(y\)-coordinate is shifted by \(+2\) units (up by 2).

Apply this transformation to the turning point \((3, -4)\):
- New \(x\)-coordinate: \(3 - 3 = 0\)
- New \(y\)-coordinate: \(-4 + 2 = -2\)

Thus, the turning point of curve \(D\) is \((0, -2)\).

M1: Attempt to complete the square on \(y = x^2 - 6x + 5\), leading to \((x - 3)^2 \pm k\).
M1: Correct completed square form: \(y = (x - 3)^2 - 4\).
M1: Identification of the original turning point as \((3, -4)\) and applying the translation values \(3 + (-3)\) and \(-4 + 2\).
A1: Correct final coordinates: \((0, -2)\).

First, find the frequency density of the interval \(10 < t \le 15\). The class width is \(15 - 10 = 5\), and the frequency is 15. Frequency density = \(15 \div 5 = 3\). The frequency density of the interval \(15 < t \le 25\) is \(3 - 1.2 = 1.8\). The class width of the interval \(15 < t \le 25\) is \(25 - 15 = 10\). Frequency \(x\) = \(\text{class width} \times \text{frequency density} = 10 \times 1.8 = 18\).

M1: For calculating the class width of the interval \(10 < t \le 15\) as 5, or the interval \(15 < t \le 25\) as 10. M1: For calculating the frequency density of the first interval as \(15 \div 5 = 3\). M1: For setting up the calculation for the frequency density of the second interval as \(3 - 1.2 = 1.8\). A1: For 18.

Substitute \(y = 2x - 3\) into the quadratic equation: \(x^2 + (2x - 3)^2 = 18\). Expand the bracket: \(x^2 + 4x^2 - 12x + 9 = 18\). Simplify to form a quadratic equation: \(5x^2 - 12x - 9 = 0\). Factorise the quadratic equation: \((5x + 3)(x - 3) = 0\). This gives \(x = 3\) or \(x = -0.6\). Substitute these back to find \(y\): If \(x = 3\), \(y = 2(3) - 3 = 3\). If \(x = -0.6\), \(y = 2(-0.6) - 3 = -4.2\).

M1: Substitute \(y = 2x - 3\) into the quadratic equation. M1: Rearrange to form a 3-term quadratic equation, e.g., \(5x^2 - 12x - 9 = 0\). M1: Factorise or solve their quadratic equation to find two values of \(x\). A1: Both pairs of correct solutions: \(x = 3, y = 3\) and \(x = -0.6, y = -4.2\) (or equivalents).

The area of a sector is given by \(\frac{\theta}{360} \pi r^2\). Substituting the given values: \(\frac{120}{360} \pi r^2 = 27\pi\), which simplifies to \(\frac{1}{3} r^2 = 27\). Multiplying by 3 gives \(r^2 = 81\), so \(r = 9\text{ cm}\). The arc length of the sector is given by \(\frac{\theta}{360} \times 2\pi r = \frac{120}{360} \times 2\pi \times 9 = 6\pi\text{ cm}\). The perimeter is the sum of the arc length and two radii: \(\text{Perimeter} = 2r + \text{arc length} = 2(9) + 6\pi = 18 + 6\pi\text{ cm}\).

M1: Set up the area equation: \(\frac{120}{360} \pi r^2 = 27\pi\) or equivalent. M1: Solve for the radius to find \(r = 9\). M1: Calculate the arc length: \(\frac{120}{360} \times 2\pi \times 9\) (or \(6\pi\)). A1: Correct perimeter: \(18 + 6\pi\).

First, factorise the denominator of the second fraction: \(x^2 - 3x = x(x-3)\). The common denominator is \(x(x-3)\). Multiply the numerator and denominator of the first fraction by \(x\): \(\frac{4x}{x(x-3)} - \frac{2x-1}{x(x-3)}\). Subtract the numerators: \(\frac{4x - (2x - 1)}{x(x-3)} = \frac{4x - 2x + 1}{x(x-3)} = \frac{2x+1}{x(x-3)}\).

M1: Factorise the denominator of the second fraction to \(x(x-3)\). M1: Write both fractions over a common denominator of \(x(x-3)\). M1: Subtract the numerators, correctly expanding and handling the negative sign: \(4x - 2x + 1\). A1: Correct simplified fraction: \(\frac{2x+1}{x(x-3)}\) or \(\frac{2x+1}{x^2-3x}\).

The gradient of the radius from the origin \((0,0)\) to the point \(P(3,6)\) is \(m_{\text{radius}} = \frac{6-0}{3-0} = 2\). Since the tangent is perpendicular to the radius, the gradient of the tangent is \(m_{\text{tangent}} = -\frac{1}{2} = -0.5\). Using the equation of a line with gradient \(-0.5\) passing through \((3,6)\): \(y - 6 = -0.5(x - 3)\). Simplifying this gives \(y - 6 = -0.5x + 1.5\), which rearranges to \(y = -0.5x + 7.5\) (or \(y = -\frac{1}{2}x + \frac{15}{2}\)).

M1: Find the gradient of the radius: \(6/3 = 2\). M1: Find the gradient of the tangent as the negative reciprocal: \(-0.5\) or \(-1/2\). M1: Substitute their perpendicular gradient and point \((3, 6)\) into a line equation. A1: Correct equation in the form \(y = mx + c\), e.g., \(y = -0.5x + 7.5\) or \(y = -\frac{1}{2}x + \frac{15}{2}\).

The total number of counters in the bag is \(x + 3 + 2 = x + 5\). The probability of choosing a blue counter first is \(\frac{x}{x+5}\). Since the counter is not replaced, the probability of choosing a blue counter second is \(\frac{x-1}{x+4}\). The probability of choosing two blue counters is \(\frac{x}{x+5} \times \frac{x-1}{x+4} = \frac{2}{9}\). Expanding this gives \(\frac{x(x-1)}{(x+5)(x+4)} = \frac{2}{9\)}. Cross-multiplying: \(9x(x-1) = 2(x+5)(x+4)\). Expand both sides: \(9x^2 - 9x = 2(x^2 + 9x + 20) = 2x^2 + 18x + 40\). Subtracting terms from both sides gives the required quadratic: \(7x^2 - 27x - 40 = 0\). Factorising this quadratic equation: \((7x+8)(x-5) = 0\). Since the number of counters must be positive, \(x = 5\).

M1: Write the probability equation: \(\frac{x}{x+5} \times \frac{x-1}{x+4} = \frac{2}{9}\). M1: Correct algebraic expansion to get \(9x^2 - 9x = 2x^2 + 18x + 40\). M1: Factorise the quadratic equation: \((7x+8)(x-5) = 0\). A1: Correct value of \(x = 5\) (with negative solution rejected).

A depreciation of \(20\%\) corresponds to a multiplier of \(1 - 0.20 = 0.8\). Let \(V\) be the original value of the car. After 3 years, the value is \(V \times 0.8^3 = 10240\). Since \(0.8^3 = 0.512\), we have \(0.512 V = 10240\). Solving for \(V\): \(V = \frac{10240}{0.512} = \frac{10240000}{512}\). Since \(1024 \div 512 = 2\), the original value of the car is \(\pounds 20,000\).

M1: Identify the multiplier for a \(20\%\) decrease as \(0.8\) or \(\frac{4}{5}\). M1: Formulate the equation \(V \times 0.8^3 = 10240\). M1: Calculate \(0.8^3 = 0.512\) or \(\left(\frac{4}{5}\right)^3 = \frac{64}{125}\). A1: Correct original value of \(20000\) (or \(\pounds 20,000\)).

We are given \(y = k_1 \sqrt{x}\) and \(z = \frac{k_2}{y^3}\). Substituting \(y\) into \(z\) gives \(z = \frac{k_2}{(k_1 \sqrt{x})^3} = \frac{C}{x^{3/2}}\) for some constant \(C\). When \(x = 9\), \(z = 4\): \(4 = \frac{C}{9^{3/2}} = \frac{C}{27}\), which gives \(C = 108\). Thus, \(z = \frac{108}{x^{3/2}}\). When \(x = 36\), \(z = \frac{108}{36^{3/2}} = \frac{108}{216} = 0.5\). Alternatively, when \(x\) increases from 9 to 36 (a factor of 4), \(y\) increases by a factor of \(\sqrt{4} = 2\). Since \(z\) is inversely proportional to \(y^3\), \(z\) is divided by \(2^3 = 8\). Thus, the new value of \(z\) is \(4 \div 8 = 0.5\).

M1: Establish proportional relations, e.g., \(y = k\sqrt{x}\) and \(z = m/y^3\). M1: Combine relations to show \(z\) is inversely proportional to \(x^{3/2}\), or deduce the ratio scale factor of \(y\) when \(x\) changes. M1: Calculate the constant \(C = 108\), or apply the scale factor of \(\frac{1}{8}\) to \(z\). A1: Correct final value of \(0.5\) (or \(\frac{1}{2}\)).

To find the points of intersection, set the two equations equal to each other:

\(2x^2 - 3x - 4 = x + 2\)

Rearrange the equation to equal zero:

\(2x^2 - 4x - 6 = 0\)

Divide the entire equation by 2 to simplify:

\(x^2 - 2x - 3 = 0\)

Factorise the quadratic equation:

\((x - 3)(x + 1) = 0\)

This gives the \(x\)-coordinates of the intersection points:

\(x = 3\) or \(x = -1\)

Substitute these \(x\)-values back into the linear equation \(y = x + 2\) to find the corresponding \(y\)-coordinates:

When \(x = 3\):
\(y = 3 + 2 = 5\) \(\implies (3, 5)\)

When \(x = -1\):
\(y = -1 + 2 = 1\) \(\implies (-1, 1)\)

Thus, the points of intersection are \((3, 5)\) and \((-1, 1)\).

M1: Equating expressions to form an equation in one variable, e.g., \(2x^2 - 3x - 4 = x + 2\)
M1: Simplifying to form a 3-term quadratic equal to zero, e.g., \(2x^2 - 4x - 6 = 0\) or \(x^2 - 2x - 3 = 0\)
M1: Factorising or using formula to find both \(x = 3\) and \(x = -1\)
A1: Both correct coordinate pairs: \((3, 5)\) and \((-1, 1)\)

First, analyze the interval \(10 < t \le 25\):
- Class width = \(25 - 10 = 15\)
- Real width on histogram = 3 cm, so the scale for the horizontal axis is \(15 \div 3 = 5\) minutes per cm.
- Frequency = 15
- Frequency density (FD) = \(\frac{\text{Frequency}}{\text{Class Width}} = \frac{15}{15} = 1\)
- Real height on histogram = 2 cm, so the scale for the vertical axis is 2 cm per unit of frequency density.

Now, analyze the interval \(25 < t \le 30\):
- Class width = \(30 - 25 = 5\)
- Width of the bar on the histogram = \(\frac{5\text{ minutes}}{5\text{ minutes/cm}} = 1\) cm.
- Frequency = 18
- Frequency density (FD) = \(\frac{\text{Frequency}}{\text{Class Width}} = \frac{18}{5} = 3.6\)
- Height of the bar on the histogram = \(3.6 \times 2\text{ cm} = 7.2\) cm.

M1: For finding the horizontal scale, e.g. 1 cm represents 5 minutes, or stating that the width of the second bar is 1 cm.
M1: For calculating the frequency density of both intervals: \(\text{FD} = 1\) for the first interval and \(\text{FD} = 3.6\) for the second interval.
M1: For finding the vertical scale, e.g. FD of 1 represents a height of 2 cm, or setting up a correct ratio for the height: \(\frac{h}{3.6} = \frac{2}{1}\).
A1: Correctly stating both Width = 1 cm and Height = 7.2 cm.

The area of the shaded segment is found by subtracting the area of triangle \(OAB\) from the area of the sector \(OAB\).

1) Area of the sector:
\(\text{Area}_{\text{sector}} = \frac{\theta}{360} \times \pi r^2 = \frac{120}{360} \times \pi \times 6^2\)
\(\text{Area}_{\text{sector}} = \frac{1}{3} \times 36\pi = 12\pi\) \(\text{cm}^2\).

2) Area of triangle \(OAB\):
\(\text{Area}_{\text{triangle}} = \frac{1}{2} a b \sin C = \frac{1}{2} \times 6 \times 6 \times \sin(120^\circ)\)
Since \(\sin(120^\circ) = \sin(60^\circ) = \frac{\sqrt{3}}{2}\):
\(\text{Area}_{\text{triangle}} = 18 \times \frac{\sqrt{3}}{2} = 9\sqrt{3}\) \(\text{cm}^2\).

3) Area of the shaded segment:
\(\text{Area}_{\text{segment}} = 12\pi - 9\sqrt{3}\) \(\text{cm}^2\).

M1: For finding the area of the sector: \(\frac{120}{360} \times \pi \times 6^2 = 12\pi\)
M1: For using the formula for the area of the triangle: \(\frac{1}{2} \times 6 \times 6 \times \sin(120^\circ)\)
M1: For substituting the exact value of \(\sin(120^\circ) = \frac{\sqrt{3}}{2}\) to get \(9\sqrt{3}\)
A1: Correct exact area: \(12\pi - 9\sqrt{3}\) (or any equivalent factorised form like \(3(4\pi - 3\sqrt{3})\))

First, factorise the quadratic expression in the numerator:
\(2x^2 - 5x - 3\)
We seek two numbers that multiply to \(2 \times (-3) = -6\) and add to \(-5\). These are \(-6\) and \(1\).
\(2x^2 - 6x + x - 3 = 2x(x - 3) + 1(x - 3) = (2x + 1)(x - 3)\).

Next, factorise the denominator as a difference of two squares:
\(4x^2 - 1 = (2x - 1)(2x + 1)\).

Now rewrite the fraction with the factorised terms:
\(\frac{(2x + 1)(x - 3)}{(2x - 1)(2x + 1)}\)

Cancel the common factor of \((2x + 1)\) from the numerator and denominator:
\(\frac{x - 3}{2x - 1}\).

M1: Attempt to factorise the numerator, obtaining \((2x + 1)(x - 3)\) (allow one sign error)
M1: Correctly factorising the denominator as a difference of two squares: \((2x - 1)(2x + 1)\)
M1: Showing a step where common brackets of \((2x + 1)\) are crossed out or cancelled
A1: Final simplified expression of \(\frac{x - 3}{2x - 1}\)

1) Find the gradient of the radius from the origin \((0,0)\) to the point \(P(3, -4)\):
\(m_{\text{radius}} = \frac{-4 - 0}{3 - 0} = -\frac{4}{3}\)

2) The tangent is perpendicular to the radius. Therefore, the gradient of the tangent is the negative reciprocal of \(m_{\text{radius}}\):
\(m_{\text{tangent}} = -\frac{1}{-\frac{4}{3}} = \frac{3}{4}\)

3) Find the equation of the tangent line through \(P(3, -4)\) with gradient \(m = \frac{3}{4}\):
\(y - y_1 = m(x - x_1)\)
\(y - (-4) = \frac{3}{4}(x - 3)\)
\(y + 4 = \frac{3}{4}x - \frac{9}{4}\)

Multiply the entire equation by 4 to remove the fraction:
\(4y + 16 = 3x - 9\)

Rearrange into the form \(ax + by = c\):
\(3x - 4y = 25\) (or \(-3x + 4y = -25\)).

M1: Finding the gradient of the radius: \(-\frac{4}{3}\)
M1: Finding the gradient of the tangent as the perpendicular gradient: \(\frac{3}{4}\) (or negative reciprocal of their radius gradient)
M1: Substituting their tangent gradient and point \((3, -4)\) into a linear equation, e.g., \(y - (-4) = m(x - 3)\)
A1: Correct final equation in the required integer form, e.g., \(3x - 4y = 25\) (or any non-zero integer multiple)

The total number of counters in the bag is \(5 + n\).

The probability of selecting a red counter first is:
\(P(\text{Red}_1) = \frac{5}{5 + n}\)

Since the selection is without replacement, the probability of selecting a red counter second is:
\(P(\text{Red}_2) = \frac{4}{4 + n}\)

The probability that both are red is:
\(P(\text{Red, Red}) = \frac{5}{5 + n} \times \frac{4}{4 + n} = \frac{20}{(5+n)(4+n)}\)

We are given that this probability is \(\frac{5}{14}\):
\(\frac{20}{(5+n)(4+n)} = \frac{5}{14}\)

Divide both sides of the equation by 5:
\(\frac{4}{(5+n)(4+n)} = \frac{1}{14}\)

Cross-multiply to remove the fractions:
\(4 \times 14 = (5+n)(4+n)\)
\(56 = 20 + 9n + n^2\)

Rearrange to form the quadratic equation:
\(n^2 + 9n + 20 - 56 = 0\)
\(n^2 + 9n - 36 = 0\) (as required).

To find \(n\), factorise this quadratic:
\((n + 12)(n - 3) = 0\)

So \(n = -12\) or \(n = 3\).

Since \(n\) represents a number of counters, it must be a positive integer, so we reject \(n = -12\).
Thus, \(n = 3\).

M1: Writing a correct algebraic expression for the probability of getting two red counters, e.g., \(\frac{5}{5+n} \times \frac{4}{4+n}\)
M1: Setting their probability expression equal to \(\frac{5}{14}\) and initiating cross-multiplication
A1: Showing clear algebraic steps to simplify and arrive at the given equation \(n^2 + 9n - 36 = 0\)
A1: Factorising the quadratic to find \(n = 3\) (rejecting the negative solution)

1) Calculate the value of the coin at the end of the first year (\(20\%\) increase on £400):
\(10\%\text{ of } 400 = 40\), so \(20\%\text{ of } 400 = 80\).
Value = \(400 + 80 = 480\) pounds.

2) Calculate the value at the end of the second year (\(10\%\) increase on £480):
\(10\%\text{ of } 480 = 48\).
Value = \(480 + 48 = 528\) pounds.

3) Calculate the value at the end of the third year (\(5\%\) decrease on £528):
\(10\%\text{ of } 528 = 52.80\).
\(5\%\text{ of } 528 = 52.80 \div 2 = 26.40\).
Value = \(528 - 26.40 = 501.60\) pounds.

M1: For calculating the value after Year 1: \(400 \times 1.2 = 480\)
M1: For calculating the value after Year 2: \(480 \times 1.1 = 528\)
M1: For setting up or executing a subtraction of \(5\%\) from 528, e.g. finding \(5\%\text{ of } 528 = 26.40\) or calculating \(528 \times 0.95\)
A1: Correct final value of £501.60 (accept 501.6)

Since \(y\) is inversely proportional to the square of \(x\), we can write the formula:
\(y = \frac{k}{x^2}\)

Substitute the known values \(x = 3\) and \(y = 8\) to solve for the constant of proportionality, \(k\):
\(8 = \frac{k}{3^2}\)
\(8 = \frac{k}{9}\)
\(k = 8 \times 9 = 72\)

So the formula is:
\(y = \frac{72}{x^2}\)

Now substitute \(y = 18\) into the formula to find \(x\):
\(18 = \frac{72}{x^2}\)

Multiply by \(x^2\):
\(18x^2 = 72\)

Divide by 18:
\(x^2 = \frac{72}{18} = 4\)

Since we are looking for the positive value of \(x\):
\(x = \sqrt{4} = 2\).

M1: For writing a correct proportional relationship, e.g., \(y = \frac{k}{x^2}\) or \(y \propto \frac{1}{x^2}\)
M1: Substituting \(x = 3\) and \(y = 8\) to find \(k = 72\)
M1: Setting up the equation \(18 = \frac{72}{x^2}\) and solving to get \(x^2 = 4\)
A1: Correct positive value \(x = 2\)

Since \(y\) is inversely proportional to the square of \(x\), we can write the relationship as:
\[y = \frac{k}{x^2}\]
where \(k\) is a constant. This can also be expressed as \(y x^2 = k\).

Using the first set of values where \(x = d\) and \(y = 18\):
\[18d^2 = k\]

Using the second set of values where \(x = d + 3\) and \(y = 8\):
\[8(d + 3)^2 = k\]

Equating the two expressions for \(k\):
\[18d^2 = 8(d+3)^2\]

Divide both sides by 2:
\[9d^2 = 4(d+3)^2\]

Since \(d > 0\), both sides of the equation are positive. We can take the positive square root of both sides:
\[3d = 2(d+3)\]
\[3d = 2d + 6\]
\[d = 6\]

(Alternatively, expanding the equation gives:
\[9d^2 = 4(d^2 + 6d + 9)\]
\[9d^2 = 4d^2 + 24d + 36\]
\[5d^2 - 24d - 36 = 0\]
Factorising this quadratic equation:
\[(5d + 6)(d - 6) = 0\]
This gives \(d = -1.2\) or \(d = 6\). Since \(d > 0\), the only valid solution is \(d = 6\).)

M1: For establishing a correct algebraic relationship, e.g., \(18d^2 = 8(d+3)^2\) or \(18d^2 = k\) and \(8(d+3)^2 = k\)
M1: For simplifying to a quadratic equation, e.g., \(5d^2 - 24d - 36 = 0\), or for taking the square root of both sides to obtain \(3d = 2(d+3)\)
M1: For a correct method to solve their quadratic equation (e.g., factorising to \((5d+6)(d-6) = 0\)) or solving their linear equation
A1: For \(d = 6\) (with negative solution rejected if quadratic method is used)

Substitute \(y = 2x + 3\) into the second equation: \(x^2 + (2x + 3)^2 = 26\). Expand and simplify: \(x^2 + 4x^2 + 12x + 9 = 26\) which simplifies to \(5x^2 + 12x - 17 = 0\). Solve this quadratic equation: \((5x + 17)(x - 1) = 0\). This gives \(x = 1\) or \(x = -3.4\). Substitute \(x = 1\) into \(y = 2x + 3\) to get \(y = 5\). Substitute \(x = -3.4\) into \(y = 2x + 3\) to get \(y = -3.8\).

M1: For substituting the linear expression into the quadratic. M1: For expanding and simplifying to a correct quadratic in the form \(ax^2 + bx + c = 0\). M1: For a correct method to solve their quadratic equation. A1: For both pairs of correct answers: \(x = 1, y = 5\) and \(x = -3.4, y = -3.8\).

The total frequency is 80, so the median is the 40th value. The cumulative frequency up to \(w = 25\) is \(12 + 24 = 36\). This means the 40th value lies in the class interval \(25 < w \le 40\). We need \(40 - 36 = 4\) more values from this class of frequency 32. The width of this class is \(40 - 25 = 15\). Estimate for median = \(25 + \frac{4}{32} \times 15 = 25 + 1.875 = 26.875\) kg.

M1: Finding that the median is the 40th (or 40.5th) value. M1: Identifying the class interval \(25 < w \le 40\) with cumulative frequency 36. M1: Setting up the interpolation calculation: \(25 + \frac{4}{32} \times 15\). A1: For \(26.875\) kg or \(26.9\) kg (accept \(27.1\) kg if using the 40.5th value).

Let \(O\) be the center of the circle. Using the cosine rule in triangle \(OAB\): \(\cos(\angle AOB) = \frac{8^2 + 8^2 - 12^2}{2 \times 8 \times 8} = \frac{64 + 64 - 144}{128} = -0.125\). Thus, \(\angle AOB = \arccos(-0.125) \approx 97.18^\circ\). Area of sector \(OAB = \frac{97.18}{360} \times \pi \times 8^2 \approx 54.27\text{ cm}^2\). Area of triangle \(OAB = \frac{1}{2} \times 8 \times 8 \times \sin(97.18^\circ) \approx 31.75\text{ cm}^2\). Area of minor segment = Sector area - Triangle area = \(54.27 - 31.75 = 22.52\text{ cm}^2\). Correct to 3 significant figures, the area is \(22.5\text{ cm}^2\).

M1: For using trigonometry or the cosine rule to find the angle at the center of the circle. A1: For finding the angle as \(97.18^\circ\) or equivalent. M1: For calculating the sector area and the triangle area. A1: For the final answer of \(22.5\) (accept range \(22.4\) to \(22.6\)).

The probability of picking the first lemon sweet is \(\frac{6}{n}\) and the second is \(\frac{5}{n-1}\). Since the probability of both being lemon is \(1/3\): \(\frac{6}{n} \times \frac{5}{n-1} = \frac{1}{3} \Rightarrow \frac{30}{n(n-1)} = \frac{1}{3} \Rightarrow n(n-1) = 90 \Rightarrow n^2 - n - 90 = 0\). Solving the quadratic equation by factoring: \((n - 10)(n + 9) = 0\). Since the number of sweets must be positive, \(n = 10\).

M1: Setting up the probability equation: \(\frac{6}{n} \times \frac{5}{n-1} = \frac{1}{3}\). M1: Rearranging to form the quadratic equation \(n^2 - n - 90 = 0\). M1: Factoring the quadratic to find the solutions \(n = 10\) and \(n = -9\). A1: Rejecting the negative value to give the final answer \(n = 10\).

Write the relationship as \(y = \frac{k}{(x-1)^2}\). Substitute \(y = 5\) and \(x = 4\) to find \(k\): \(5 = \frac{k}{(4-1)^2} \Rightarrow 5 = \frac{k}{9} \Rightarrow k = 45\). Thus, \(y = \frac{45}{(x-1)^2}\). Now, substitute \(y = 20\): \(20 = \frac{45}{(x-1)^2} \Rightarrow (x-1)^2 = \frac{45}{20} = 2.25\). Taking the square root gives \(x - 1 = \pm 1.5\). For a positive value of \(x\), we use \(x - 1 = 1.5 \Rightarrow x = 2.5\).

M1: Writing \(y = \frac{k}{(x-1)^2}\) and substituting \(x = 4, y = 5\). A1: Finding the constant of proportionality \(k = 45\). M1: Substituting \(y = 20\) and solving for \((x-1)^2\). A1: Correctly obtaining the positive solution \(x = 2.5\).

The center of the circle is the origin \(O(0, 0)\). The gradient of the radius \(OP\) is \(m_r = \frac{5 - 0}{3 - 0} = \frac{5}{3}\). Since the tangent is perpendicular to the radius, its gradient is \(m_t = -\frac{3}{5}\). Using the equation of a line with gradient \(-\frac{3}{5}\) passing through \(P(3, 5)\): \(y - 5 = -\frac{3}{5}(x - 3)\). Multiply by 5: \(5(y - 5) = -3(x - 3) \Rightarrow 5y - 25 = -3x + 9\). Rearranging gives \(3x + 5y = 34\).

M1: Correct method to find the gradient of the radius \(OP\). M1: Taking the negative reciprocal of the radius gradient to find the gradient of the tangent. M1: Setting up the straight-line equation using their perpendicular gradient and point \(P\). A1: Correct final equation in the required form (e.g., \(3x + 5y = 34\) or any non-zero integer multiple).

Solve the first inequality: \(3x + 1 > 7 \Rightarrow 3x > 6 \Rightarrow x > 2\). Solve the second inequality: \(x^2 - 2x - 8 \le 0\). Factorizing gives \((x - 4)(x + 2) \le 0\). The critical values are \(x = 4\) and \(x = -2\), so the solution is \(-2 \le x \le 4\). Combining both inequalities gives \(2 < x \le 4\). The integers that satisfy this combined range are \(3\) and \(4\).

M1: Correctly solving the linear inequality to get \(x > 2\). M1: Correctly factorizing the quadratic expression or finding critical values \(-2\) and \(4\). M1: Expressing the quadratic solution as \(-2 \le x \le 4\). A1: Identifying exactly the integers \(3\) and \(4\).

(a) Let \(r\) be the multiplier. After 2 years, \(800 \times r^2 = 648 \Rightarrow r^2 = 0.81 \Rightarrow r = 0.9\). Since the multiplier is 0.9, the depreciation rate is \(10\%\), so \(x = 10\). (b) At the start of 2025, 5 years have passed from the start of 2020. The value is \(800 \times 0.9^5 = 472.392\). To the nearest penny, this is £472.39.

M1: Writing down a correct compound decay equation, e.g., \(800(1 - \frac{x}{100})^2 = 648\). A1: Showing clearly that \(r = 0.9\) or \(x = 10\). M1: Calculating the value for the start of 2025 using \(800 \times 0.9^5\) or \(648 \times 0.9^3\). A1: Correctly rounding the final answer to £472.39.

The area of a sector of a circle is given by \(\frac{\theta}{360} \times \pi r^2\).

For this sector, \(\theta = 120^\circ\), so:
\(\text{Area of sector} = \frac{120}{360} \times \pi r^2 = \frac{1}{3}\pi r^2\).

The area of the circular hole is:
\(\text{Area of hole} = \pi \times 3^2 = 9\pi\).

The remaining area of the metal plate is:
\(\text{Remaining Area} = \text{Area of sector} - \text{Area of hole} = \frac{1}{3}\pi r^2 - 9\pi\).

We are given that this remaining area is \(18\pi\text{ cm}^2\):
\(\frac{1}{3}\pi r^2 - 9\pi = 18\pi\).

Divide both sides of the equation by \(\pi\):
\(\frac{1}{3}r^2 - 9 = 18\).

Add 9 to both sides:
\(\frac{1}{3}r^2 = 27\).

Multiply by 3:
\(r^2 = 81\).

Since the radius must be positive:
\(r = 9\).

M1 for expressing the sector area as \(\frac{1}{3}\pi r^2\)
M1 for subtracting the circular hole area of \(9\pi\)
M1 for equating \(\frac{1}{3}\pi r^2 - 9\pi = 18\pi\) and rearranging to find \(r^2 = 81\)
A1 for \(r = 9\)

The car's value depreciates by \(10\%\) each year.
After 2 years (from the start of 2022 to the start of 2024), the value of the car is:
\(16000 \times (0.90)^2 = 16000 \times 0.81 = £12,960\).

Let the multiplier for the motorbike's depreciation be \(m = 1 - \frac{y}{100}\).
After 2 years, the value of the motorbike is:
\(4000 \times m^2\).

We are given that the car is worth \(5.0625\) times the motorbike:
\(12960 = 5.0625 \times 4000 \times m^2\)
\(12960 = 20250 \times m^2\)
\(m^2 = \frac{12960}{20250} = 0.64\).

Taking the square root (since \(m > 0\)):
\(m = \sqrt{0.64} = 0.8\).

Therefore:
\(1 - \frac{y}{100} = 0.8\)
\(\frac{y}{100} = 0.2 \implies y = 20\).

M1 for finding the value of the car after 2 years: \(16000 \times 0.9^2 = 12960\)
M1 for setting up the equation: \(12960 = 5.0625 \times 4000 \times (1 - y/100)^2\)
M1 for solving to get \((1 - y/100)^2 = 0.64\) and taking the square root to find \(1 - y/100 = 0.8\)
A1 for \(y = 20\)

Find a common denominator for the terms on the left-hand side:
\(\frac{5(x+1) + 2(x-2)}{(x-2)(x+1)} = 2\).

Expand and simplify the numerator:
\(\frac{5x + 5 + 2x - 4}{x^2 - x - 2} = 2\)
\(\frac{7x + 1}{x^2 - x - 2} = 2\).

Multiply both sides by the denominator \(x^2 - x - 2\):
\(7x + 1 = 2(x^2 - x - 2)\)
\(7x + 1 = 2x^2 - 2x - 4\).

Rearrange into a standard quadratic form equal to 0:
\(2x^2 - 9x - 5 = 0\).

Factorise the quadratic expression:
\((2x + 1)(x - 5) = 0\).

This yields two possible solutions:
\(2x + 1 = 0 \implies x = -0.5\)
\(x - 5 = 0 \implies x = 5\).

So, the solutions are \(x = 5\) or \(x = -0.5\).

M1 for putting the fractions over a common denominator: \(\frac{5(x+1)+2(x-2)}{(x-2)(x+1)} = 2\)
M1 for removing the fraction to obtain a quadratic: \(7x + 1 = 2(x^2 - x - 2)\)
M1 for expanding and rearranging to \(2x^2 - 9x - 5 = 0\) and attempting to factorise
A1 for both correct solutions: \(x = 5\) and \(x = -0.5\) (or \(-\frac{1}{2}\))

First, find the number of students who study History but do not study Geography.
Let \(H\) be the set of students studying History and \(G\) be the set of students studying Geography.

The number of students studying at least one of these subjects is:
\(80 - 12 = 68\).

Using the principle of inclusion-exclusion:
\(n(H \cup G) = n(H) + n(G) - n(H \cap G)\)
\(68 = 45 + 40 - n(H \cap G)\)
\(68 = 85 - n(H \cap G) \implies n(H \cap G) = 17\).

The number of students who study History but not Geography (History only) is:
\(n(H \text{ only}) = n(H) - n(H \cap G) = 45 - 17 = 28\).

We select two students at random without replacement:
- The probability that the first student studies History only is \(\frac{28}{80}\).
- The probability that the second student studies History only is \(\frac{27}{79}\).

The probability that both students study History only is:
\(P = \frac{28}{80} \times \frac{27}{79} = \frac{7}{20} \times \frac{27}{79} = \frac{189}{1580}\).

M1 for calculating the number of students studying both subjects: \(45 + 40 - (80 - 12) = 17\)
M1 for calculating the number of students studying History only: \(45 - 17 = 28\)
M1 for a correct product of two probabilities without replacement: \(\frac{28}{80} \times \frac{27}{79}\)
A1 for the simplified fraction \(\frac{189}{1580}\)

We can calculate the frequency of each of the first three intervals using the formula:
\(\text{Frequency} = \text{Frequency density} \times \text{Class width}\).

1) For \(0 < t \le 10\):
\(\text{Frequency} = 15\).

2) For \(10 < t \le 25\):
Class width = \(25 - 10 = 15\).
\(\text{Frequency} = 15 \times 2.4 = 36\).

3) For \(25 < t \le 40\):
Class width = \(40 - 25 = 15\).
\(\text{Frequency} = 15 \times 1.8 = 27\).

Now find the total frequency of these three intervals:
\(15 + 36 + 27 = 78\).

Since there are 120 people in total, the frequency of the final interval \(40 < t \le 60\) is:
\(120 - 78 = 42\).

The class width of the final interval is:
\(60 - 40 = 20\).

The frequency density for the final interval is:
\(\text{Frequency density} = \frac{42}{20} = 2.1\).

M1 for calculating the frequency of the second interval: \(15 \times 2.4 = 36\)
M1 for calculating the frequency of the third interval: \(15 \times 1.8 = 27\)
M1 for finding the frequency of the final interval: \(120 - (15 + 36 + 27) = 42\)
A1 for the final frequency density \(2.1\)

Apply the Cosine Rule:
\(AC^2 = AB^2 + BC^2 - 2(AB)(BC)\cos(ABC)\).

Substitute the given values into the equation:
\((x+2)^2 = 8^2 + x^2 - 2(8)(x)\cos(60^\circ)\).

Since \(\cos(60^\circ) = 0.5\), we expand and simplify:
\(x^2 + 4x + 4 = 64 + x^2 - 16x(0.5)\)
\(x^2 + 4x + 4 = 64 + x^2 - 8x\).

Subtract \(x^2\) from both sides:
\(4x + 4 = 64 - 8x\).

Add \(8x\) to both sides:
\(12x + 4 = 64\).

Subtract 4 from both sides:
\(12x = 60\).

Divide by 12:
\(x = 5\).

M1 for substituting correctly into the Cosine Rule: \((x+2)^2 = 8^2 + x^2 - 2(8)(x)\cos(60^\circ)\)
M1 for expanding \((x+2)^2\) to \(x^2 + 4x + 4\)
M1 for simplifying the quadratic terms and obtaining the linear equation \(12x = 60\)
A1 for \(x = 5\)

Since \(P\) is inversely proportional to the square root of \(Q\), we can write:
\(P = \frac{k}{\sqrt{Q}}\) where \(k\) is a constant.

Substitute the given values \(Q = 16\) and \(P = 15\) into the formula:
\(15 = \frac{k}{\sqrt{16}}\)
\(15 = \frac{k}{4}\)
\(k = 60\).

So the equation connecting \(P\) and \(Q\) is:
\(P = \frac{60}{\sqrt{Q}}\).

Now, substitute \(P = 12\) to find \(Q\):
\(12 = \frac{60}{\sqrt{Q}}\)
\(\sqrt{Q} = \frac{60}{12}\)
\(\sqrt{Q} = 5\).

Square both sides:
\(Q = 25\).

M1 for setting up the proportion equation: \(P = \frac{k}{\sqrt{Q}}\)
M1 for finding the constant of proportionality \(k = 60\)
M1 for substituting \(P = 12\) into their formula to get \(\sqrt{Q} = 5\)
A1 for \(Q = 25\)

To simplify the algebraic fraction, factorise both the numerator and the denominator.

First, factorise the numerator \(2x^2 - 5x - 3\):
We need two numbers that multiply to \(2 \times (-3) = -6\) and add to \(-5\). These are \(-6\) and \(1\).
\(2x^2 - 6x + x - 3 = 2x(x - 3) + 1(x - 3) = (2x + 1)(x - 3)\).

Next, factorise the denominator \(6x^2 - 18x\):
Taking out the common factor \(6x\):
\(6x(x - 3)\).

Write the fraction with the factored forms:
\(\frac{(2x + 1)(x - 3)}{6x(x - 3)}\).

Cancel the common factor of \((x - 3)\) from both numerator and denominator:
\(\frac{2x + 1}{6x}\).

M1 for factorising the numerator to \((2x + 1)(x - 3)\)
M1 for factorising the denominator to \(6x(x - 3)\)
M1 for showing the cancellation of the common factor \((x - 3)\)
A1 for the fully simplified fraction \(\frac{2x + 1}{6x}\)

Let the multiplier for depreciation be \(y\), where \(y = 1 - \frac{x}{100}\).

After 3 years:
\[40000 \times y^3 = 24565\]

Divide both sides by 40,000:
\[y^3 = \frac{24565}{40000}\]
\[y^3 = 0.614125\]

Take the cube root of both sides:
\[y = \sqrt[3]{0.614125}\]
\[y = 0.85\]

Since \(1 - \frac{x}{100} = 0.85\):
\[\frac{x}{100} = 0.15\]
\[x = 15\]

M1: Sets up the correct equation, e.g. \(40000 \times (1 - \frac{x}{100})^3 = 24565\) or \(40000 \times y^3 = 24565\)
M1: Rearranges to find the value of \(y^3 = 0.614125\) or equivalent fraction
M1: Takes the cube root to find \(y = 0.85\) (or finds \(1 - \frac{x}{100} = 0.85\))
A1: \(15\)

The center of the circle is at the origin \(O(0, 0)\).

First, find the gradient of the radius \(OP\):
\[\text{Gradient of } OP = \frac{6 - 0}{3 - 0} = 2\]

The tangent is perpendicular to the radius. Therefore, the gradient of the tangent, \(m\), is:
\[m = -\frac{1}{2} = -0.5\]

The equation of the tangent passes through \(P(3, 6)\):
\[y - 6 = -0.5(x - 3)\]
\[y - 6 = -0.5x + 1.5\]
\[y = -0.5x + 7.5\]

M1: For finding the gradient of the radius \(OP\) as \(2\)
M1: For finding the perpendicular gradient of the tangent as \(-\frac{1}{2}\) or \(-0.5\)
M1: For substituting the point \((3, 6)\) and their perpendicular gradient into a linear equation form, e.g. \(6 = (-0.5)(3) + c\)
A1: Correct equation, e.g. \(y = -0.5x + 7.5\) or \(y = -\frac{1}{2}x + \frac{15}{2}\)

Substitute \(y = 2x - 3\) into the quadratic equation:
\[x^2 + (2x - 3)^2 = 18\]

Expand the brackets:
\[x^2 + (4x^2 - 12x + 9) = 18\]
\[5x^2 - 12x + 9 = 18\]

Rearrange to equal zero:
\[5x^2 - 12x - 9 = 0\]

Factorise the quadratic equation:
\[(5x + 3)(x - 3) = 0\]

This gives two solutions for \(x\):
\[x = 3 \quad \text{or} \quad x = -0.6\]

Substitute the \(x\) values back into \(y = 2x - 3\) to find the corresponding \(y\) values:
- If \(x = 3\):
\[y = 2(3) - 3 = 3\]
- If \(x = -0.6\):
\[y = 2(-0.6) - 3 = -1.2 - 3 = -4.2\]

So the solutions are \(x = 3, y = 3\) and \(x = -0.6, y = -4.2\).

M1: Correct substitution of linear equation into quadratic, e.g. \(x^2 + (2x - 3)^2 = 18\)
M1: Expansion and simplification to a standard 3-term quadratic, e.g. \(5x^2 - 12x - 9 = 0\)
M1: Correct method to solve their 3-term quadratic equation to find two values of \(x\) (e.g. \(x = 3\) and \(x = -0.6\))
A1: Both pairs of solutions clearly stated: \(x = 3, y = 3\) and \(x = -0.6, y = -4.2\) (or equivalent fractional values)

The area of the segment is found by subtracting the area of the triangle from the area of the sector.

1. **Area of the sector**:
\[\text{Area of sector} = \frac{\theta}{360} \times \pi r^2\]
\[\text{Area of sector} = \frac{135}{360} \times \pi \times 10^2 = 0.375 \times 100\pi = 37.5\pi \approx 117.810\text{ cm}^2\]

2. **Area of the triangle**:
Using the formula \(\text{Area} = \frac{1}{2}ab\sin C\):
\[\text{Area of triangle} = \frac{1}{2} \times 10 \times 10 \times \sin(135^\circ) = 50 \times \sin(135^\circ) \approx 35.355\text{ cm}^2\]

3. **Area of the segment**:
\[\text{Area of segment} = 117.810 - 35.355 = 82.455\text{ cm}^2\]

Rounding to 3 significant figures gives \(82.5\text{ cm}^2\).

M1: For a correct method to find the area of the sector, e.g. \(\frac{135}{360} \times \pi \times 10^2\) (or value \(117.8...\))
M1: For a correct method to find the area of the triangle, e.g. \(\frac{1}{2} \times 10 \times 10 \times \sin(135)\) (or value \(35.3...\) or \(25\sqrt{2}\))
M1: Subtracts their area of the triangle from their area of the sector
A1: Correct final answer of \(82.5\) (allow answers in the range \(82.4\) to \(82.6\))

The total number of counters in the bag is \(x + 4\).

The probability of choosing a green counter first is:
\[P(G_1) = \frac{x}{x+4}\]

Since the selection is without replacement, the probability of choosing a green counter second is:
\[P(G_2) = \frac{x-1}{x+3}\]

The probability that both are green is:
\[\frac{x}{x+4} \times \frac{x-1}{x+3} = \frac{14}{33}\]

Multiply both sides to eliminate the denominators:
\[33x(x - 1) = 14(x + 4)(x + 3)\]

Expand both sides:
\[33x^2 - 33x = 14(x^2 + 7x + 12)\]
\[33x^2 - 33x = 14x^2 + 98x + 168\]

Rearrange into a quadratic equation equal to zero:
\[19x^2 - 131x - 168 = 0\]

Factorise the quadratic:
\[(19x + 21)(x - 8) = 0\]

This gives two solutions for \(x\):
\[x = -\frac{21}{19} \quad \text{or} \quad x = 8\]

Since the number of counters must be a positive integer, \(x = 8\).

M1: Writes a correct probability expression, e.g. \(\frac{x}{x+4} \times \frac{x-1}{x+3} = \frac{14}{33}\)
M1: Eliminates fractions to obtain a simplified expansion, e.g. \(33x(x-1) = 14(x^2 + 7x + 12)\)
M1: Rearranges into a standard quadratic equation of the form \(ax^2 + bx + c = 0\), e.g. \(19x^2 - 131x - 168 = 0\)
A1: Correctly solves the quadratic and selects the positive integer solution \(x = 8\) with working shown