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First, calculate the area of the sector : . Next, find the length of in the right-angled triangle : . The area of triangle is . Shaded area = . Correct to 3 significant figures, this is .

M1 for or or 75.4. M1 for or 20.8. M1 for or or 125. A1 for 49.3 (accept 49.3 - 49.4).

Multiply both sides by to clear denominators: . Expand both sides: . Simplify: . Rearrange into standard form: . Apply the quadratic formula: . This gives and . Correct to 3 significant figures, and .

M1 for clearing denominators: . M1 for simplifying to a 3-term quadratic: . M1 for correct substitution into the quadratic formula. A1 for both 2.59 and -3.09.

The multipliers for the three years are , , and respectively. The equation for the final value is . Calculate the combined multiplier: . Solve for : .

M1 for at least one correct multiplier: , , or . M1 for set up of the equation: . M1 for finding combined multiplier 1.08864 (or intermediate step like ). A1 for 12500.

First, use Pythagoras' theorem in right-angled triangle to find : . Next, use the Sine Rule in triangle : . Solve for : . Since is acute: .

M1 for . A1 for (or ). M1 for . A1 for 40.6 (accept 40.5 - 40.7).

The hemisphere has radius . Since the flat face is joined to the cone, the height of the cone is . The volume of the hemisphere is . The volume of the cone is . The total volume is . Correct to 3 significant figures, this is .

M1 for finding the height of the cone: . M1 for . M1 for . A1 for 192 (accept 192 - 193).

Let the total number of people be , so there are men and women. The number of men under 30 is . The number of women under 30 is . The total number of people under 30 is . We are given that . The total number of people is .

M1 for letting men be and women (or using ratio parts). M1 for expressing people under 30 as and (or parts and parts). M1 for (or parts = 210). A1 for 400.

First differences: . Second differences: . The second difference is constant and is , so the coefficient of is . Subtract from the original terms: for , ; for , ; for , ; for , ; for , . The sequence of differences is , which has -th term . Thus, the -th term of the quadratic sequence is .

M1 for finding first differences and second differences . M1 for identifying as part of the formula. M1 for subtracting to get and finding linear rule . A1 for (or equivalent).

The total number of counters is . The probability of choosing two red counters without replacement is . This gives . Factoring this quadratic: . Since must be a positive integer, .

M1 for . M1 for . M1 for (or equivalent quadratic). A1 for 10 (with negative solution rejected).

To find \(n((P \cap Q) \cup (R \cap P'))\), we first list the elements of each set:
\(P = \{ 3, 6, 9, 12, 15, 18, 21, 24 \}\)
\(Q = \{ 4, 8, 12, 16, 20, 24 \}\)
\(R = \{ 1, 2, 3, 4, 6, 8, 12, 24 \}\)

Now, find \(P \cap Q\), which represents elements that are multiples of both 3 and 4:
\(P \cap Q = \{ 12, 24 \}\)

Next, find \(R \cap P'\), which represents factors of 24 that are NOT multiples of 3:
Factors of 24 are \(\{ 1, 2, 3, 4, 6, 8, 12, 24 \}\).
Removing any multiples of 3 (which are 3, 6, 12, 24) leaves:
\(R \cap P' = \{ 1, 2, 4, 8 \}\)

Now, find the union of these two sets:
\((P \cap Q) \cup (R \cap P') = \{ 12, 24 \} \cup \{ 1, 2, 4, 8 \} = \{ 1, 2, 4, 8, 12, 24 \}\)

The number of elements in this union is 6.
Thus, \(n((P \cap Q) \cup (R \cap P')) = 6\).

M1: for listing elements of \(P \cap Q = \{12, 24\}\) or finding the elements of \(R\) and \(P'\).
M1: for finding the elements of \(R \cap P' = \{1, 2, 4, 8\}\).
M1: for listing the complete set of the union: \(\{1, 2, 4, 8, 12, 24\}\).
A1: for the correct final answer of 6.

Let the initial number of red, blue, and yellow counters be \(3x\), \(5x\), and \(8x\) respectively.

We are given that there are 40 more yellow counters than red counters:
\(8x - 3x = 40\)
\(5x = 40\)
\(x = 8\)

Now we can calculate the initial number of each color:
Red = \(3 \times 8 = 24\)
Blue = \(5 \times 8 = 40\)
Yellow = \(8 \times 8 = 64\)

Let \(y\) be the number of red counters added to the bag. The new ratio of Red : Blue : Yellow is \(5 : 5 : 8\).
Since the number of blue and yellow counters remains unchanged, we have:
Blue = 40
Yellow = 64

The ratio of Blue to Yellow is \(40 : 64 = 5 : 8\), which matches the new ratio parts.
The new ratio specifies 5 parts for red counters. Since 5 parts corresponds to 40 counters:
New number of Red counters = 40.

The number of red counters added is:
\(y = 40 - 24 = 16\).

M1: for forming the equation \(8x - 3x = 40\) or finding \(x = 8\).
A1: for finding the initial number of red counters as 24, or blue as 40, or yellow as 64.
M1: for using the new ratio to set up a correct equation, e.g., \(\frac{24+y}{40} = \frac{5}{5}\) or \(\frac{24+y}{64} = \frac{5}{8}\).
A1: for 16.

The total height of the shape is the sum of the height of the rectangle and the radius of the semicircle:
\(h + r = 14 \implies h = 14 - r\)

The width of the rectangle is \(w = 2r\).
The area of the rectangle is:
\(\text{Area} = w \times h = 2r \times h = 80 \text{ cm}^2\)

Substitute \(h = 14 - r\) into the area equation:
\(2r(14 - r) = 80\)
\(28r - 2r^2 = 80\)
\(r^2 - 14r + 40 = 0\)

Solve the quadratic equation by factoring:
\((r - 10)(r - 4) = 0\)
So, \(r = 10\) or \(r = 4\).

If \(r = 10\):
\(w = 2(10) = 20\) cm
\(h = 14 - 10 = 4\) cm
This contradicts \(h > w\), so we reject \(r = 10\).

If \(r = 4\):
\(w = 2(4) = 8\) cm
\(h = 14 - 4 = 10\) cm
This satisfies \(h > w\) (since 10 > 8), so \(r = 4\) and \(h = 10\).

The perimeter of the shape consists of the bottom edge of the rectangle (\(w = 8\) cm), two vertical sides of the rectangle (\(2h = 20\) cm), and the curved boundary of the semicircle (\(\pi r = 4\pi\) cm):
\(\text{Perimeter} = w + 2h + \pi r = 8 + 20 + 4\pi = 28 + 4\pi \approx 40.566 \text{ cm}\)

Correct to 3 significant figures, the perimeter is 40.6 cm.

M1: for setting up simultaneous equations, e.g., \(h + r = 14\) and \(2r \times h = 80\).
M1: for formulating and solving the quadratic equation to get \(r = 4\) and \(h = 10\) (with rejection of \(r = 10\)).
M1: for a correct perimeter expression, e.g., \(w + 2h + \pi r\) or \(8 + 2(10) + 4\pi\).
A1: for 40.6 (accept answers in the range 40.5 to 40.6).

Let \(a\) be the cost of an adult ticket (in £) and \(c\) be the cost of a child ticket (in £).

From the given information, we can write the simultaneous equations:
1) \(3a + 5c = 41.50\)
2) \(4a + 2c = 39.00\)

To eliminate \(c\), multiply equation (1) by 2 and equation (2) by 5:
\(6a + 10c = 83.00\)
\(20a + 10c = 195.00\)

Subtract the first of these new equations from the second:
\(14a = 112.00\)
\(a = 8.00\)

Substitute \(a = 8.00\) back into equation (2):
\(4(8.00) + 2c = 39.00\)
\(32.00 + 2c = 39.00\)
\(2c = 7.00\)
\(c = 3.50\)

Now, calculate the total cost of 2 adult tickets and 3 child tickets:
\(\text{Total Cost} = 2a + 3c = 2(8.00) + 3(3.50) = 16.00 + 10.50 = 26.50\)

M1: for a correct algebraic method to eliminate one variable from the two linear equations.
A1: for finding the correct price of one type of ticket: adult ticket = £8.00 or child ticket = £3.50.
M1: for finding the correct price of both types of tickets.
A1: for the correct final answer of 26.50 (or 26.5).

First, express all terms in base 2:
\(8^{2x-3} = (2^3)^{2x-3} = 2^{3(2x-3)} = 2^{6x-9}\)
\(16^{x+1} = (2^4)^{x+1} = 2^{4(x+1)} = 2^{4x+4}\)
\(4^{4x-1} = (2^2)^{4x-1} = 2^{2(4x-1)} = 2^{8x-2}\)
\(64 = 2^6\)

Substitute these expressions back into the equation:
\[\frac{2^{6x-9} \times 2^{4x+4}}{2^{8x-2}} = 2^6\]

Simplify the numerator by adding exponents:
\[2^{(6x-9) + (4x+4)} = 2^{10x-5}\]

Now, divide by subtracting exponents:
\[2^{(10x-5) - (8x-2)} = 2^{2x-3}\]

Set this equal to the right side of the equation:
\[2^{2x-3} = 2^6\]

Since the bases are the same, we equate the indices:
\[2x - 3 = 6\]
\[2x = 9 \implies x = 4.5\]

M1: for expressing at least two of the terms as correct powers of 2 (e.g., \(2^{6x-9}\), \(2^{4x+4}\), or \(2^{8x-2}\)).
M1: for simplifying the numerator to \(2^{10x-5}\) or equivalent.
M1: for obtaining a correct linear equation in \(x\), such as \(2x - 3 = 6\) or equivalent.
A1: for 4.5 (or \(\frac{9}{2}\) or \(4\frac{1}{2}\)).

The total number of pens in the box is \(5 + 3 + 2 = 10\).

For all three pens to be of different colours, we must select exactly one red, one blue, and one green pen in any order.

Let's first find the probability of choosing Red first, Blue second, and Green third:
\(P(\text{Red, Blue, Green}) = \frac{5}{10} \times \frac{3}{9} \times \frac{2}{8} = \frac{30}{720} = \frac{1}{24}\)

There are \(3! = 6\) different orders in which we can draw three distinct colours (RBG, RGB, BRG, BGR, GRB, GBR).
Each of these 6 arrangements has the exact same probability of occurring because the product of the numerators is always \(5 \times 3 \times 2 = 30\) and the product of the denominators is always \(10 \times 9 \times 8 = 720\).

Therefore, the total probability is:
\(\text{Total Probability} = 6 \times \frac{1}{24} = \frac{6}{24} = \frac{1}{4}\).

M1: for calculating the probability of a single valid arrangement of three different colors, e.g. \(\frac{5}{10} \times \frac{3}{9} \times \frac{2}{8}\) or \(\frac{1}{24}\).
M1: for recognizing there are 6 permutations of the three colors (by multiplying by 6 or listing all 6 paths).
M1: for a complete expression, e.g. \(6 \times \frac{30}{720}\).
A1: for the simplified fraction \(\frac{1}{4}\) (accept 0.25).

We can split the shape into two right-angled triangles sharing the side \(AB\): \(ABC\) and \(ABD\).

In right-angled triangle \(ABC\):
\(\tan(40^\circ) = \frac{AB}{BC} = \frac{12}{BC}\)
\(BC = \frac{12}{\tan(40^\circ)} \approx 14.3005\text{ cm}\)

In right-angled triangle \(ABD\):
\(\tan(25^\circ) = \frac{AB}{BD} = \frac{12}{BD}\)
\(BD = \frac{12}{\tan(25^\circ)} \approx 25.7341\text{ cm}\)

Since \(C\) lies on the line \(BD\), the length of \(CD\) is:
\(CD = BD - BC = 25.7341 - 14.3005 = 11.4336\text{ cm}\)

Correct to 3 significant figures, the length of \(CD\) is 11.4 cm.

M1: for a correct trigonometric ratio to find \(BC\), e.g., \(\tan(40^\circ) = \frac{12}{BC}\) or \(BC = \frac{12}{\tan(40^\circ)}\).
M1: for a correct trigonometric ratio to find \(BD\), e.g., \(\tan(25^\circ) = \frac{12}{BD}\) or \(BD = \frac{12}{\tan(25^\circ)}\).
M1: for subtracting the two calculated lengths: \(BD - BC\).
A1: for 11.4 (accept answers in the range 11.4 to 11.5).

First, list the elements of each set:
\(\mathcal{E} = \{1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12\}\)
\(A = \{2, 3, 5, 7, 11\}\)
\(B = \{3, 6, 9, 12\}\)

Now find the union of \(A\) and \(B\):
\(A \cup B = \{2, 3, 5, 6, 7, 9, 11, 12\}\)

The complement of the union, \((A \cup B)'\), contains all the elements in the universal set \(\mathcal{E}\) that are not in \(A \cup B\):
\((A \cup B)' = \{1, 4, 8, 10\}\).

M1: For listing the elements of at least one of the sets \(A\) or \(B\) correctly, or for a list showing the elements of \(A \cup B\) with no more than two errors.
M1: For a complete method to find the complement, listing at least three correct elements with no incorrect ones, or showing \(A \cup B = \{2, 3, 5, 6, 7, 9, 11, 12\}\).
A1: For the correct set \(\{1, 4, 8, 10\}\) (order of elements does not matter, brackets not strictly required).

Let the normal price be \(x\).
Since the price is reduced by \(15\%\), the sale price is \(100\% - 15\% = 85\%\) of the normal price.

We can write this as:
\(0.85x = 119\)

Solve for \(x\):
\(x = \frac{119}{0.85} = 140\)

Thus, the normal price of the vacuum cleaner is \(\$140\).

M1: For writing a correct expression linking \(119\) with \(85\%\), e.g., \(0.85 \times \text{Normal Price} = 119\) or \(119 \div 85\).
M1: For a complete method to find the normal price, e.g., \(\frac{119}{0.85}\) or \(1.4 \times 100\).
A1: For the correct answer \(140\).

The 10th term of the sequence, \(u_{10}\), can be found using the formula:
\(u_{10} = S_{10} - S_9\)

First, calculate \(S_{10}\):
\(S_{10} = 2(10)^2 + 5(10) = 2(100) + 50 = 250\)

Next, calculate \(S_9\):
\(S_9 = 2(9)^2 + 5(9) = 2(81) + 45 = 162 + 45 = 207\)

Now, subtract \(S_9\) from \(S_{10}\):
\(u_{10} = 250 - 207 = 43\)

M1: For an attempt to substitute \(n=10\) or \(n=9\) into the given formula for \(S_n\), or setting up the expression \(S_{10} - S_9\).
M1: For correctly evaluating both \(S_{10} = 250\) and \(S_9 = 207\), or for correctly determining the general expression for the \(n\)th term, \(u_n = 4n + 3\).
A1: For the correct answer \(43\).

The total probability of choosing any counter from the bag is \(1\).
Therefore, the probability of choosing either a blue counter or a yellow counter is:
\(1 - 0.3 = 0.7\)

We are given the ratio of blue counters to yellow counters as \(2 : 5\).
This means there are \(2 + 5 = 7\) parts in total for the blue and yellow counters.

The probability of choosing a yellow counter is \(\frac{5}{7}\) of the combined probability of blue and yellow counters:
\(P(\text{Yellow}) = \frac{5}{7} \times 0.7 = 0.5\)

M1: For calculating the combined probability of blue and yellow counters, e.g., \(1 - 0.3 = 0.7\).
M1: For a complete method to find the probability of a yellow counter, e.g., \(\frac{5}{2+5} \times 0.7\) or establishing the equation \(2x + 5x = 0.7\) to find \(x = 0.1\).
A1: For the correct answer \(0.5\) (or equivalent fraction).

We need to find two numbers that multiply to \(6 \times (-5) = -30\) and add up to \(-7\).
These numbers are \(-10\) and \(3\).

Now, rewrite the middle term of the quadratic expression:
\(6x^2 - 10x + 3x - 5\)

Factorise by grouping terms:
\(2x(3x - 5) + 1(3x - 5)\)

Take out the common binomial factor \((3x - 5)\):
\((2x + 1)(3x - 5)\)

M1: For finding the correct two numbers that multiply to \(-30\) and add to \(-7\) (\(-10\) and \(3\)), or for writing \(6x^2 - 10x + 3x - 5\), or for an initial factorisation attempt resulting in \((ax + b)(cx + d)\) where \(ac = 6\) and \(bd = -5\).
M1: For a partially factorised expression such as \(2x(3x - 5) + 1(3x - 5)\) or a factorised expression with a single sign error, e.g., \((2x - 1)(3x + 5)\).
A1: For the correct fully factorised expression \((2x + 1)(3x - 5)\) (or equivalent order of brackets).

First, address the negative exponent by taking the reciprocal of the fraction:
\(\left(\frac{y^{12}}{64x^6}\right)^{\frac{2}{3}}\)

Next, apply the exponent \(\frac{2}{3}\) to each term in the numerator and the denominator:

For the numerator:
\((y^{12})^{\frac{2}{3}} = y^{12 \times \frac{2}{3}} = y^8\)

For the denominator:
\((64x^6)^{\frac{2}{3}} = 64^{\frac{2}{3}} \times (x^6)^{\frac{2}{3}} = (\sqrt[3]{64})^2 \times x^{6 \times \frac{2}{3}} = 4^2 \times x^4 = 16x^4\)

Combining these results gives:
\(\frac{y^8}{16x^4}\)

M1: For dealing with the negative index by taking the reciprocal of the fraction, or correctly applying the power \(-\frac{2}{3}\) to at least one term inside the brackets (e.g., finding \(64^{-\frac{2}{3}} = \frac{1}{16}\) or \((x^6)^{-\frac{2}{3}} = x^{-4}\)).
M1: For correctly evaluating two of the three components: \(64^{-\frac{2}{3}} = \frac{1}{16}\), \((x^6)^{-\frac{2}{3}} = x^{-4}\), or \((y^{-12})^{-\frac{2}{3}} = y^8\).
A1: For the correct simplified answer \(\frac{y^8}{16x^4}\) (or equivalent, such as \(\frac{1}{16}x^{-4}y^8\)).

The sum of the interior angle and the exterior angle at any vertex of a polygon is always \(180^\circ\).

First, find the size of each exterior angle:
\(\text{Exterior angle} = 180^\circ - 162^\circ = 18^\circ\)

The sum of all the exterior angles of any polygon is \(360^\circ\).
Therefore, the number of sides, \(n\), is:
\(n = \frac{360^\circ}{18^\circ} = 20\)

M1: For a correct method to find the exterior angle, e.g., \(180 - 162\) (implied by an exterior angle of \(18^\circ\)), or for setting up the equation \(\frac{(n - 2) \times 180}{n} = 162\).
M1: For \(\frac{360}{18}\), or for a correct algebraic method to solve the equation, e.g., \(180n - 360 = 162n\) leading to \(18n = 360\).
A1: For the correct answer \(20\).

We are given \(\vec{OA} = \mathbf{a}\) and \(\vec{OB} = \mathbf{b}\). Since \(OP : PA = 1 : 2\), we have \(\vec{OP} = \frac{1}{3}\mathbf{a}\). Since \(Q\) is the midpoint of \(OB\), we have \(\vec{OQ} = \frac{1}{2}\mathbf{b}\). The vector \(\vec{AQ}\) can be expressed as \(\vec{AQ} = \vec{AO} + \vec{OQ} = -\mathbf{a} + \frac{1}{2}\mathbf{b}\). Since \(X\) lies on the line \(AQ\), there exists a scalar \(\lambda\) such that \(\vec{OX} = \vec{OA} + \lambda\vec{AQ} = \mathbf{a} + \lambda\left(-\mathbf{a} + \frac{1}{2}\mathbf{b}\right) = (1-\lambda)\mathbf{a} + \frac{1}{2}\lambda\mathbf{b}\). The vector \(\vec{BP}\) can be expressed as \(\vec{BP} = \vec{BO} + \vec{OP} = -\mathbf{b} + \frac{1}{3}\mathbf{a}\). Since \(X\) lies on the line \(BP\), there exists a scalar \(\mu\) such that \(\vec{OX} = \vec{OB} + \mu\vec{BP} = \mathbf{b} + \mu\left(-\mathbf{b} + \frac{1}{3}\mathbf{a}\right) = \frac{1}{3}\mu\mathbf{a} + (1-\mu)\mathbf{b}\). Equating the coefficients of \(\mathbf{a}\) and \(\mathbf{b}\) for the two expressions of \(\vec{OX}\) gives: For \(\mathbf{a}\): \(1-\lambda = \frac{1}{3}\mu\), and for \(\mathbf{b}\): \(\frac{1}{2}\lambda = 1-\mu\). From the second equation, we have \(\mu = 1 - \frac{1}{2}\lambda\). Substituting this into the first equation: \(1-\lambda = \frac{1}{3}\left(1 - \frac{1}{2}\lambda\right)\), which simplifies to \(1-\lambda = \frac{1}{3} - \frac{1}{6}\lambda\). Solving for \(\lambda\) gives \(\frac{2}{3} = \frac{5}{6}\lambda\), hence \(\lambda = \frac{4}{5}\). Substituting \(\lambda = \frac{4}{5}\) back into the expression for \(\vec{OX}\) yields \(\vec{OX} = \left(1-\frac{4}{5}\right)\mathbf{a} + \frac{1}{2}\left(\frac{4}{5}\right)\mathbf{b} = \frac{1}{5}\mathbf{a} + \frac{2}{5}\mathbf{b}\). This completes the proof.

M1: For finding either \(\vec{AQ} = -\mathbf{a} + \frac{1}{2}\mathbf{b}\) or \(\vec{BP} = \frac{1}{3}\mathbf{a} - \mathbf{b}\). M1: For writing a vector expression for \(\vec{OX}\) along \(AQ\), e.g., \(\vec{OX} = (1-\lambda)\mathbf{a} + \frac{1}{2}\lambda\mathbf{b}\). M1: For writing a vector expression for \(\vec{OX}\) along \(BP\), e.g., \(\vec{OX} = \frac{1}{3}\mu\mathbf{a} + (1-\mu)\mathbf{b}\). M1: For equating the coefficients of \(\mathbf{a}\) and \(\mathbf{b}\) to set up two simultaneous equations. M1: For solving the simultaneous equations to find \(\lambda = \frac{4}{5}\) or \(\mu = \frac{3}{5}\). A1: For correctly completing the proof to show \(\vec{OX} = \frac{1}{5}\mathbf{a} + \frac{2}{5}\mathbf{b}\).

The total surface area \(A\) of a closed cylinder is given by \(A = 2\pi r^2 + 2\pi r h\). Given \(A = 150\pi\), we have \(2\pi r^2 + 2\pi r h = 150\pi\). Dividing by \(2\pi\) gives \(r^2 + r h = 75\), so \(h = \frac{75 - r^2}{r}\). The volume \(V\) of the cylinder is given by \(V = \pi r^2 h\). Substituting the expression for \(h\) into the volume formula yields \(V = \pi r^2 \left(\frac{75 - r^2}{r}\right) = 75\pi r - \pi r^3\). To find the stationary value, we differentiate \(V\) with respect to \(r\): \(\frac{\mathrm{d}V}{\mathrm{d}r} = 75\pi - 3\pi r^2\). Setting \(\frac{\mathrm{d}V}{\mathrm{d}r} = 0\) gives \(75\pi - 3\pi r^2 = 0\), which leads to \(r^2 = 25\), so \(r = 5\) (since \(r > 0\)). To prove this stationary value is a maximum, we find the second derivative: \(\frac{\mathrm{d}^2 V}{\mathrm{d}r^2} = -6\pi r\). At \(r = 5\), \(\frac{\mathrm{d}^2 V}{\mathrm{d}r^2} = -30\pi\), which is less than \(0\), confirming a maximum. Substituting \(r = 5\) into the volume formula gives the maximum volume: \(V = 75\pi(5) - \pi(5)^3 = 375\pi - 125\pi = 250\pi\text{ cm}^3\). This completes the proof.

M1: For setting up the surface area equation \(2\pi r^2 + 2\pi r h = 150\pi\). M1: For expressing \(h\) in terms of \(r\), e.g., \(h = \frac{75 - r^2}{r}\). A1: For substituting \(h\) into the volume formula to obtain \(V = 75\pi r - \pi r^3\). M1: For differentiating \(V\) to find \(\frac{\mathrm{d}V}{\mathrm{d}r} = 75\pi - 3\pi r^2\), setting to \(0\), and solving to find \(r = 5\). M1: For showing that the second derivative is negative at \(r = 5\). A1: For substituting \(r = 5\) to correctly obtain the maximum volume of \(250\pi\text{ cm}^3\).

The total volume \(V\) of the solid is the sum of the volumes of the hemisphere and the cylinder: \(V = \frac{2}{3}\pi r^3 + \pi r^2 h\). Given that \(V = 18\pi\), we have \(\frac{2}{3}\pi r^3 + \pi r^2 h = 18\pi\). Dividing both sides by \(\pi\) gives \(\frac{2}{3}r^3 + r^2 h = 18\), which we can rearrange to find \(h\) in terms of \(r\): \(r^2 h = 18 - \frac{2}{3}r^3\), so \(h = \frac{18}{r^2} - \frac{2}{3}r\). The total surface area \(S\) is the sum of the area of the circular base, the curved surface of the cylinder, and the curved surface of the hemisphere: \(S = \pi r^2 + 2\pi r h + 2\pi r^2 = 3\pi r^2 + 2\pi r h\). Substituting our expression for \(h\) into this formula gives: \(S = 3\pi r^2 + 2\pi r \left(\frac{18}{r^2} - \frac{2}{3}r\right)\). Expanding the brackets: \(S = 3\pi r^2 + \frac{36\pi}{r} - \frac{4}{3}\pi r^2\). Combining the \(\pi r^2\) terms: \(S = \left(3 - \frac{4}{3}\right)\pi r^2 + \frac{36\pi}{r} = \frac{5}{3}\pi r^2 + \frac{36\pi}{r}\). This completes the proof.

M1: For the correct volume formula \(\frac{2}{3}\pi r^3 + \pi r^2 h = 18\pi\). M1: For expressing \(h\) in terms of \(r\) to get \(h = \frac{18}{r^2} - \frac{2}{3}r\). M1: For the correct total surface area expression \(S = 3\pi r^2 + 2\pi r h\). A1: For substituting the expression for \(h\) into the surface area formula. M1: For expanding the brackets to get \(\frac{36\pi}{r} - \frac{4}{3}\pi r^2\). A1: For correctly simplifying the expression to show \(S = \frac{5}{3}\pi r^2 + \frac{36\pi}{r}\).