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To evaluate \(\frac{0.009}{0.012}\) in standard form, we need to simplify the fraction and express the result in the form of \(a \times 10^b\), where \(1 \leq a < 10\) and \(b\) is an integer. We can simplify the fraction by dividing both the numerator and the denominator by 0.003: \[\frac{0.009}{0.012} = \frac{0.009 \div 0.003}{0.012 \div 0.003} = \frac{3}{4}\] Now we can express 3/4 in standard form: \[\frac{3}{4} = 0.75 = 7.5 \times 10^{-1}\] Therefore, the answer is option (C) 7.5 x 10^-1.

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To find the roots of a quadratic function, we need to set y to zero and solve for x. Therefore, we have: 2x² - x - 6 = 0 We can factorize this quadratic equation by splitting the middle term: 2x² - 4x + 3x - 6 = 0 2x(x - 2) + 3(x - 2) = 0 (2x + 3)(x - 2) = 0 Setting each factor to zero, we get: 2x + 3 = 0 or x - 2 = 0 Solving for x in each equation, we get: x = -3/2 or x = 2 Therefore, the roots of the equation are -3/2 and 2. So the correct answer is (B) -1.5, 2.

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In a circle, the angle subtended by an arc at the center is twice the angle subtended by the same arc at any point on the circumference of the circle. Using this property, we can find the value of ?PRQ: ?POQ = 80^o (given) Therefore, ?PRQ = 1/2 ?POQ = 1/2 x 80^o = 40^o Now, we can use the value of ?PRQ to find the value of x: ?PRQ = 5x (given) Substituting the value of ?PRQ, we get: 40^o = 5x Simplifying this equation, we get: x = 8 Therefore, the value of x is 8. The answer is option B.

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Since 3 raised to what power gives 243, we need to find the exponent that makes this equation true: 3\(^y\) = 243 We can rewrite 243 as 3\(^5\), so the equation becomes: 3\(^y\) = 3\(^5\) Since the bases are equal, the exponents must also be equal. Therefore: y = 5 So the value of y is 5.

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We can use the fact that sin²θ + cos²θ = 1, to find the value of sinθ or cosθ given the other one. Since sinθ = 1/2, we know that sin²θ + cos²θ = 1 becomes: (1/2)² + cos²θ = 1 Simplifying the equation, we get: 1/4 + cos²θ = 1 cos²θ = 3/4 Taking the square root of both sides, we get: cosθ = ±√3/2 Since we know that cosθ = -√3/2, we can conclude that θ = 150^o (or 5π/6 radians) as this is the angle in the fourth quadrant that has a cosine of -√3/2. Therefore, the answer is option E: 150^o.

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To find the percentage of error, we first need to find the absolute error, which is the difference between the measured value and the true value. In this case, the true value is 1.23 cm, and the measured value is 1.32 cm. Absolute error = |measured value - true value| = |1.32 - 1.23| = 0.09 cm Next, we can find the relative error, which is the ratio of the absolute error to the true value. Relative error = (absolute error / true value) x 100% Relative error = (0.09 / 1.23) x 100% = 7.32% Therefore, the percentage of error, correct to one decimal place, is 7.3%. The answer is option B.

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Range of a distribution refers to the difference between the highest and lowest value in a dataset. To find the range of the given Mathematics test scores, we need to determine the highest and lowest values. The lowest score in the test is 2, while the highest score is 9. Therefore, the range of the distribution is 9 - 2 = 7. Hence, the answer is 7.

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We can simplify this expression using the rules of exponents and radicals. First, we can rewrite 9 as \(3^2\) and 27 as \(3^3\): \(\frac{9^{-\frac{1}{2}}}{27^{\frac{2}{3}}} = \frac{(3^2)^{-\frac{1}{2}}}{(3^3)^{\frac{2}{3}}}\) Next, we can simplify the exponents using the product rule of exponents: \(\frac{(3^2)^{-\frac{1}{2}}}{(3^3)^{\frac{2}{3}}} = \frac{3^{-1}}{3^2} = \frac{1}{3^{1+2}} = \frac{1}{3^3} = \frac{1}{27}\) Therefore, the simplified expression is 1/27. The answer is option E.

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To evaluate this expression using logarithm tables, we can use the following steps: 1. Take the logarithm (base 10) of each of the numbers in the expression. 2. Add the logarithms of the two factors in the numerator. 3. Subtract the logarithm of the denominator from the result in step 2. 4. Take the antilogarithm (base 10) of the result in step 3. Using these steps, we get: log(5.34) = 0.727 log(67.4) = 1.829 log(2.7) = 0.431 log(5.34 x 67.4) = log(5.34) + log(67.4) = 0.727 + 1.829 = 2.556 log(5.34 x 67.4) - log(2.7) = 2.556 - 0.431 = 2.125 antilog(2.125) = 133.2 Therefore, the value of the expression is 133.2. The answer is option C.

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We know that log_ax = p This means that a^p = x So, we can express x in terms of a and p as x = a^p. Therefore, the answer is x = a^p.

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We are given the equation: log p = 2 log x + 3 log q. We know that log a + log b = log(ab) and log a - log b = log(a/b). Using these properties, we can rewrite the given equation as follows: log p = log x^2 + log q^3 log p = log (x^2q^3) Now we can write the exponential form of this equation as: p = x^2q^3 Therefore, the expression for p in terms of x and q is p = x^2q^3. Hence, the correct answer is "p = x^2q^3".

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We know that ABDE and FCDE are parallelograms which means their opposite sides are equal and parallel. Therefore, |AB| = |DE| and |AE| = |BD|. We can calculate the area of parallelogram ABDE as follows: Area of ABDE = |AB| x |PC| We need to calculate |AB| and we can do this by finding the length of |DE|. In parallelogram FCDE, the opposite sides |DE| and |FC| are equal. Therefore, |DE| = |FC| = 12.2cm. Also, we know that FCDE is a parallelogram, so the height |PC| is equal to the height of parallelogram ABDE. Now, we can find the length of |AB| using the Pythagorean theorem. |AB|² = |AE|² + |DE|² |AB|² = (4.0cm)² + (12.2cm)² |AB|² = 149.24cm² |AB| = 12.2cm Therefore, the area of ABDE is: Area of ABDE = |AB| x |PC| = 12.2cm x 4.0cm = 48.8cm² Therefore, the answer is 48.8cm².

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The mode of a set of data is the value that appears most frequently. To find the mode of the scores in the Mathematics test, we need to determine which score occurs the most number of times. From the given data, we can see that the score "7" appears 6 times, which is more than any other score. Therefore, the mode of the score is 7. So the answer is (e) 7.

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We are given the equation: x² + 2x + 1 = 25 To solve for x, we first simplify the left-hand side of the equation by subtracting 24 from both sides: x² + 2x - 24 = 0 We can then factor this quadratic equation as follows: (x + 6)(x - 4) = 0 Using the zero product property, we know that this equation is true if either (x + 6) = 0 or (x - 4) = 0. Solving for x in each case, we get: x + 6 = 0 or x - 4 = 0 x = -6 or x = 4 Therefore, the solutions to the original equation are x = -6 and x = 4. So the answer is: -6, 4

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The union of sets A and B (A ∪ B) is the set of elements that are in either set A, set B or both. Similarly, the union of sets A and C (A ∪ C) is the set of elements that are in either set A, set C or both. The intersection of these two sets ((A ∪ B) ∩ (A ∪ C)) is the set of elements that are common to both sets. Therefore, we can first find the union of A and B which gives us {a, b, c, d, e}. Then, we can find the union of A and C which gives us {a, b, c, d, e, f}. The intersection of these two sets is the set of elements that are in both sets, which is {a, b, c, d, e}. Therefore, (A∪B)∩(A∪C) = {a, b, c, d, e}. So the correct answer is: {a,b,c,d,e}.

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Since sin x = cos x, we have: sin x = cos x Dividing both sides by cos x, we get: tan x = 1 Taking the inverse tangent (tan^-1) of both sides, we get: x = 45^o Therefore, the value of x for which sin x = cos x is 45^o. Therefore, the correct option is (d) 45^o.

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The problem is asking us to find the point of intersection of the two graphs y = x2 and y = x + 6. This can be done by solving the equations simultaneously. We need to find the values of x and y that satisfy both equations. We can do this by substituting y = x + 6 for y in the equation y = x2, giving us: x + 6 = x2 Rearranging this equation gives us: x2 - x - 6 = 0 We can factor this quadratic equation to obtain: (x - 3)(x + 2) = 0 Thus, the solutions are x = 3 or x = -2. To find the corresponding values of y, we can substitute these values of x into either of the original equations. For example, if we use y = x + 6, we get: When x = 3, y = 3 + 6 = 9, giving us the point (3, 9). When x = -2, y = -2 + 6 = 4, giving us the point (-2, 4). Therefore, the point of intersection of the two graphs is (3, 9) and (-2, 4).

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The probability that 3 customers waiting in a bank will be served in the sequence of their arrival at the bank can be calculated using the permutation formula. Since there are three customers waiting in line, the total number of possible arrangements is 3! = 6. However, only one arrangement results in the customers being served in the sequence of their arrival at the bank. Therefore, the probability of this happening is 1/6. Hence, the correct answer is 1/6.

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In this problem, we can use trigonometry to determine the height of the building. Let h be the height of the building and d be the distance from the point on the ground to the foot of the building. We can see that the angle of depression is the angle between the line of sight from the top of the building to the point on the ground and the horizontal line. Therefore, we can draw a right-angled triangle with the height of the building, the distance from the point on the ground to the foot of the building, and the line of sight from the top of the building to the point on the ground as the sides. From the triangle, we can use the tangent function since we have the opposite and adjacent sides. Therefore: tan(20.3°) = h/40 h = 40 tan(20.3°) ≈ 14.8m Therefore, the height of the building is approximately 14.8m. The correct answer is 14.8m.

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Given: \(d=60t-10t^2\), where \(d\) is in metres and \(t\) is in seconds.

Using the stated scales, plot the following points and join them with a smooth curve.

\(t\) (s)	1	2	3	4	5
\(d=60t-10t^2\) (m)	50	80	90	80	50

Scale: \(2\text{ cm}\) represents \(1\text{ s}\) on the \(t\)-axis and \(2\text{ cm}\) represents \(20\text{ m}\) on the \(d\)-axis.

(a)(i) From the graph, \(d=70\) m at about \(t=1.6\) s on the upward journey. Therefore, it takes 1.6 s to reach a height of 70 m. (It passes through 70 m again at about \(4.4\) s while descending.)

(a)(ii) At \(t=5\), the graph gives \(d=50\). Hence, the height after 5 seconds is 50 m.

(a)(iii) The maximum point of the curve is \((3,90)\). Therefore, the stone reaches its maximum height after 3.0 s.

(b) Draw a tangent to the curve at \(t=4\). Using two points on the tangent, for example \((3.2,96)\) and \((4.8,64)\),

\[\text{slope}=\frac{64-96}{4.8-3.2}=\frac{-32}{1.6}=-20\text{ m s}^{-1}.\]

Therefore, the slope of the graph when \(t=4\) s is \(-20\text{ m s}^{-1}\).

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(a) If the roots are \(\tfrac{1}{2}\) and \(-7\), the equation is

Multiply the first factor by 2 to clear the fraction (using \((2x-1)\)):

This is the smallest equation with integer coefficients.

(b) Let the daughter's present age be \(d\) and the father's present age be \(f\).

"Three years ago the father was four times as old as the daughter is now":

"The product of their present ages is 430":

Using the quadratic formula, \(\;d = \dfrac{-3 \pm \sqrt{3^2 + 4(4)(430)}}{2(4)} = \dfrac{-3 \pm \sqrt{6889}}{8} = \dfrac{-3 \pm 83}{8}.\)

Taking the positive root, \(d = \dfrac{80}{8} = 10\), so \(f = 4(10) + 3 = 43\).

The daughter is 10 years old and the father is 43 years old (product \(= 430\)).

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(a)

Since triangle \(PQR\) is right-angled at \(Q\), by Pythagoras' theorem,

\[PR^2=PQ^2+QR^2=(3a)^2+(4a)^2=9a^2+16a^2=25a^2.\]

\[\therefore\ PR=5a\text{ cm}.\]

(b)(i) Diagram

The diagram below is drawn to scale in the correct relative positions. \(XY=24\text{ km}\) is on a bearing of \(060^\circ\), \(YZ=18\text{ km}\), and \(XZ=30\text{ km}\).

(ii) Bearing of \(Z\) from \(Y\)

\[24^2+18^2=576+324=900=30^2.\]

Therefore, \(\angle XYZ=90^\circ\). The bearing of \(X\) from \(Y\) is \(240^\circ\). Hence,

\[\text{bearing of }Z\text{ from }Y=240^\circ-90^\circ=150^\circ.\]

Bearing of \(Z\) from \(Y\) = \(150^\circ\).

(iii) Bearing of \(X\) from \(Z\)

Using the cosine rule at \(Z\),

\[\cos \angle XZY=\frac{30^2+18^2-24^2}{2(30)(18)}=\frac{648}{1080}=0.6.\]

\[\angle XZY=\cos^{-1}(0.6)=53.13^\circ.\]

Since the bearing of \(Y\) from \(Z\) is \(150^\circ+180^\circ=330^\circ\),

\[\text{bearing of }X\text{ from }Z=330^\circ-53.13^\circ=276.87^\circ\approx277^\circ.\]

Bearing of \(X\) from \(Z\) = \(277^\circ\).

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From the diagram, \(O\) is the centre, radius \(r = 3.2\ \text{cm}\), and \(\angle PRQ = 42^\circ\) is an angle at the circumference standing on chord \(PQ\).

Angle at the centre. The angle subtended at the centre is twice the angle at the circumference on the same arc:

\[ \angle POQ = 2 \times 42^\circ = 84^\circ \]

(a)(i) Area of minor sector POQ.

\[ A_{\text{sector}} = \frac{\theta}{360^\circ}\,\pi r^2 = \frac{84}{360}\times \pi \times (3.2)^2 \]

\[ = \frac{84}{360}\times \pi \times 10.24 = 0.23333 \times 32.1699 = 7.51\ \text{cm}^2 \]

(a)(ii) Area of the shaded part. The shaded region is the minor segment cut off by chord \(PQ\); it equals the sector minus triangle \(POQ\).

\[ A_{\triangle POQ} = \tfrac{1}{2} r^2 \sin\theta = \tfrac{1}{2}\times 10.24 \times \sin 84^\circ = 5.12 \times 0.99452 = 5.0920\ \text{cm}^2 \]

\[ A_{\text{shaded}} = 7.5063 - 5.0920 = 2.41\ \text{cm}^2 \]

(b) Base radius of the cone. When the sector is rolled into a cone, its arc length becomes the circumference of the base, while the sector radius becomes the slant height.

Arc length of sector:

\[ \ell = \frac{\theta}{360^\circ}\times 2\pi r = \frac{84}{360}\times 2\pi \times 3.2 = 4.6915\ \text{cm} \]

Set equal to base circumference \(2\pi R\):

\[ 2\pi R = 4.6915 \;\Rightarrow\; R = \frac{4.6915}{2\pi} = 0.7467 \approx 0.7\ \text{cm} \]

Answers: (i) \(7.51\ \text{cm}^2\); (ii) \(2.41\ \text{cm}^2\); (b) base radius \(\approx 0.7\ \text{cm}\).

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None

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Surface area of the cone-on-hemisphere structure.

From the diagram, the common radius (from centre \(O\) to the base edge) is \(r=7\text{ cm}\), and the vertical height of the cone is \(h=24\text{ cm}\). The exposed surface is the curved surface of the cone plus the curved surface of the hemisphere; the flat circular join is internal and is not counted.

Slant height of the cone:

Curved surface area of the cone:

Curved surface area of the hemisphere:

Total surface area:

Correct to three significant figures, the surface area is \(\mathbf{858\text{ cm}^{2}}\).

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(a) \(6\log(x+4) = \log 64\)

\(\log(x+4)^6 = \log 64\)

\((x+4)^6 = 64 = 2^6\)

Taking the positive sixth root, \(x + 4 = 2\), so \(\mathbf{x = -2}\). (Check: \(x+4 = 2 > 0\), so the logarithm is defined.)

(b) \(U = \{1,2,3,4,5,6,7,8,9,10\}\), \(X = \{1,2,4,6,7,8,9\}\), \(Y = \{1,2,3,4,7,9\}\), \(Z = \{2,3,4,7,9\}\).

First, \(Z' = U \setminus Z = \{1,5,6,8,10\}\).

Next, \(X \cap Y = \{1,2,4,7,9\}\).

Then \((X \cap Y) \cap Z' = \{1,2,4,7,9\} \cap \{1,5,6,8,10\} = \mathbf{\{1\}}\).

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(a)(i) \(\cos\alpha = 0.6717\). From the cosine tables, \(\alpha = \cos^{-1}(0.6717) \approx \mathbf{47.8°}\).

(ii) Using \(\sin^2\alpha + \cos^2\alpha = 1\):

\(\sin\alpha = \sqrt{1 - (0.6717)^2} = \sqrt{1 - 0.4512} = \sqrt{0.5488} = \mathbf{0.7408}\)

(Equivalently, \(\sin 47.8° = 0.7408\).)

(b) The angle of depression of \(P\) from the top \(T\) equals the angle of elevation of \(T\) from \(P\) (alternate angles), \(= 23.6°\). The horizontal distance from the foot of the building to \(P\) is \(50\text{ m}\).

\(\tan 23.6° = \dfrac{\text{height}}{50}\)

\(\text{height} = 50 \times \tan 23.6° = 50 \times 0.4369 = 21.85\text{ m}\)

Height of the building \(\approx \mathbf{22\text{ m}}\) (to the nearest metre).

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(a) Two dice give \(6 \times 6 = 36\) equally likely outcomes. A sum of "at least 9" means a sum of 9, 10, 11 or 12.

• Sum 9: (3,6),(4,5),(5,4),(6,3) = 4 ways
• Sum 10: (4,6),(5,5),(6,4) = 3 ways
• Sum 11: (5,6),(6,5) = 2 ways
• Sum 12: (6,6) = 1 way

Favourable outcomes \(= 4+3+2+1 = 10\).

(b) Let \(P(\text{owns a car}) = \frac{1}{6}\), so \(P(\text{does not}) = \frac{5}{6}\). Selections are independent.

(i) Both A and B own a car:

(ii) Of C and D, exactly one owns a car (owns-then-not, or not-then-owns):

(iii) Of X, Y and Z, exactly one owns a car. Choose which one (\(\binom{3}{1}=3\) ways):

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(a) Let the first term be \(a\) and the common difference be \(d\). The \(n\)th term is \(T_n = a + (n-1)d\).

The 8th and 4th terms are \(T_8 = a + 7d\) and \(T_4 = a + 3d\).

(i) Their difference is 20:

The common difference is 5.

(ii) The 8th term is \(1\tfrac{1}{2}\) times the 4th term:

Substitute \(d = 5\):

The first term is 25.

(b) A 5% depreciation each year multiplies the value by \(0.95\) annually. After 9 years:

Now \((0.95)^{9} = 0.6302\) (to 4 d.p.), so

Correct to three significant figures, the value in January 1989 is N6 460.00.

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Construction of Triangle ABC and the Required Loci

Construct triangle ABC using a ruler and pair of compasses only, given that:

• |AB| = 8.4 cm
• |BC| = 6.5 cm
• ∠ABC = 30°

1. Construct triangle ABC

1. Draw a line segment BC = 6.5 cm.
2. At point B, construct a 60° angle using the equilateral-triangle construction: with centre B and any suitable radius, draw an arc; using the same radius from a point on the arc, mark the intersection to form a 60° ray.
3. Bisect the 60° angle to obtain a ray making 30° with BC.
4. With centre B and radius 8.4 cm, draw an arc cutting the 30° ray at A.
5. Join A to C. This completes triangle ABC.

2. Construct locus l<sub>1</sub>

The locus l<sub>1</sub> of points equidistant from the lines AB and BC, and lying within ∠ABC, is the internal angle bisector of ∠ABC.

1. With centre B, draw an arc cutting BA and BC at points D and E.
2. With the same compass radius, draw arcs from D and E to meet at F inside ∠ABC.
3. Draw the line BF. This line is locus l<sub>1</sub>.

3. Construct locus l<sub>2</sub>

The locus l<sub>2</sub> of points equidistant from points B and C is the perpendicular bisector of BC.

1. Set the compass radius to more than half of BC.
2. With centres B and C, draw arcs above and below BC so that they intersect.
3. Join the two arc-intersection points. The resulting line is locus l<sub>2</sub>.

Therefore:

• l<sub>1</sub> is the internal bisector of ∠ABC.
• l<sub>2</sub> is the perpendicular bisector of BC.
• I = l<sub>1</sub> ∩ l<sub>2</sub>.

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(a) Bar chart

(b)(i) Mode

The highest frequency is 4, corresponding to a weight of 61 kg.

\[\boxed{\text{Mode}=61\text{ kg}}\]

(b)(ii) Median

There are \(12\) students. Therefore, the median is the mean of the 6th and 7th observations.

The ordered weights are:

\[55,\ 55,\ 57,\ 59,\ 59,\ \underbrace{61,\ 61}_{\text{6th and 7th}},\ 61,\ 61,\ 63,\ 63,\ 63\]

\[\text{Median}=\frac{61+61}{2}=\boxed{61\text{ kg}}\]

(c) Mean weight

\[\begin{aligned}\sum f&=2+1+2+4+3=12,\\\sum fx&=55(2)+57(1)+59(2)+61(4)+63(3)\\&=110+57+118+244+189\\&=718.\end{aligned}\]

\[\text{Mean}=\frac{\sum fx}{\sum f}=\frac{718}{12}=59.833\ldots\text{ kg}\]

\[\boxed{\text{Mean weight}=60\text{ kg, correct to the nearest kilogram}}\]