WAEC SSCE - General Mathematics - 1992

Bayanin Amsa

In the given diagram, we have a triangle with two sides of equal length, which means that the angles opposite these sides are also equal. Thus, we can conclude that: 4x - 6 = 3x + 12 (using the fact that angles in a triangle add up to 180 degrees) x = 18 Therefore, the value of x is 18 degrees. Answer: (b) 24°.

Bayanin Amsa

The sector of the circle with an angle of 240° forms a cone. To find the radius of the base of the cone, we need to use the formula relating the arc length of the sector and the circumference of the base of the cone. The arc length of the sector is (240/360) * 2π * 8 = (4/3) * π * 8 = (32/3)π cm. The circumference of the base of the cone is equal to the arc length of the sector, so it is (32/3)π cm. We know that the circumference of a circle is equal to 2πr, where r is the radius of the circle. Therefore, we can write: (32/3)π = 2πr Simplifying the equation, we get: r = (16/3) cm So, the radius of the base of the cone is 16/3 cm. Therefore, the correct option is: 16/3cm

Bayanin Amsa

To find the mode of the given set of data, we need to identify the value that appears most frequently. From the given set of measurements, we can see that 72^o appears 4 times, which is more than any other value. Therefore, the mode of the data set is 72^o. So, the correct option is (e) 72^o.

Bayanin Amsa

Bayanin Amsa

To find the median of a set of scores, we need to arrange the scores in order of magnitude from smallest to largest. The scores in this set are: 48, 55, 65, 72, 78 There are five scores in this set, so the median will be the middle score. Since the scores are already in order, we can simply find the middle score, which is 65. Therefore, the median of this set of scores is 65.

Bayanin Amsa

Bayanin Amsa

Bayanin Amsa

The total number of outcomes when rolling a fair die is 6 since there are six sides on the die. The numbers that are greater than 2 are 3, 4, 5, and 6. So, there are four possible outcomes that are greater than 2. The probability of getting a number greater than 2 with a single fair die is the number of favorable outcomes divided by the total number of possible outcomes, which is 4/6 or 2/3. Therefore, the answer is (d) 2/3.

Bayanin Amsa

Since TR = TS and ?TRS = 60o, then ?TSR = ?TRS = 60o. Therefore, the angles of triangle TSP add up to 180o: ?TPS + ?TSP + ?TSR = 180o. Substituting the known values: x + 60 + 60 = 180. Simplifying: x = 60. Hence, the value of x is 60.

Bayanin Amsa

The exterior angles of a polygon add up to 360 degrees. If each exterior angle of a regular polygon is 18 degrees, then the polygon has 360/18 = 20 exterior angles. Since a regular polygon has the same number of sides as exterior angles, the polygon has 20 sides. Therefore, the answer is (d) 20.

Bayanin Amsa

To simplify this expression, we need to follow the order of operations, which is Parentheses, Exponents, Multiplication and Division, and Addition and Subtraction (PEMDAS). First, we can simplify the expression inside the parentheses: (³/₄ + ¹/₃) x 4¹/₃ + 3¹/₄ = (⁹/₁₂ + ⁴/₁₂) x 4¹/₃ + 3¹/₄ = ¹³/₁₂ x 4¹/₃ + 3¹/₄ Next, we can simplify the exponent 4¹/₃. To do this, we can convert the mixed number to an improper fraction: 4¹/₃ = ^{(4 x 3) + 1}/₃ = ¹³/₃ Now we can substitute the value of 4¹/₃ into the expression: ¹³/₁₂ x 4¹/₃ + 3¹/₄ = ¹³/₁₂ x ¹³/₃ + ¹³/₄ To simplify this, we can first simplify the multiplication of fractions: ¹³/₁₂ x ¹³/₃ = ¹⁶⁹/₃₆ Now we can add the two fractions: ¹⁶⁹/₃₆ + ¹³/₄ = ¹⁶⁹/₃₆ + ³⁹/₃₆ = ²⁰⁸/₃₆ = ^13}{9 Therefore, the simplified expression is ^13}{9, which corresponds to option (E).

Bayanin Amsa

Bayanin Amsa

Bayanin Amsa

To find the value of tan 315°, we can use the identity: tan (θ + 180°) = tan θ Since tan (315° + 180°) = tan 495°, and 495° is coterminal with 135°, we have: tan 315° = tan (135° + 180°) = tan 135° Using the fact that tan 45° = 1, and tan (θ + 90°) = -cot θ, we have: tan 135° = tan (45° + 90°) = -cot 45° = -1 Therefore, the value of tan 315° is -1.

Bayanin Amsa

There are 10 numbers in the set (1, 2, 3, ..., 9, 10). Out of these, the numbers that are greater than or equal to 7 are 7, 8, 9, and 10. So, there are 4 numbers that satisfy the given condition. Therefore, the probability of choosing a number that is greater than or equal to 7 is the number of numbers that satisfy the condition divided by the total number of numbers in the set, which is: 4/10 = 2/5 Hence, the probability that the number chosen at random from the set (1, 2, 3, ..., 9, 10) is greater than or equal to 7 is 2/5.

Bayanin Amsa

We can use the following trigonometric identity: \[\tan \theta = \frac{\sin \theta}{\cos \theta}\] We are given that $\cos \theta = 5/13$. To find $\sin \theta$, we can use the Pythagorean identity: \[\sin^2 \theta + \cos^2 \theta = 1\] Substituting $\cos \theta = 5/13$, we get: \[\sin^2 \theta + \left(\frac{5}{13}\right)^2 = 1\] Simplifying, we get: \[\sin^2 \theta = 1 - \frac{25}{169} = \frac{144}{169}\] Taking the square root of both sides, we get: \[\sin \theta = \frac{12}{13}\] Substituting into the formula for $\tan \theta$, we get: \[\tan \theta = \frac{\sin \theta}{\cos \theta} = \frac{\frac{12}{13}}{\frac{5}{13}} = \boxed{\frac{12}{5}}\]

Bayanin Amsa

Bayanin Amsa

We find the modal score by identifying the score that occurs most frequently in the data. Looking at the frequency distribution, we can see that the score 4 appears most frequently, with a frequency of 5. Therefore, the modal score is 4.

Bayanin Amsa

Bayanin Amsa

Bayanin Amsa

The total number of workers in the company is 24+27+12+15+9+3+18 = 108. The probability of an absent worker being a bricklayer is the ratio of the number of bricklayers to the total number of workers, which is 18/108 = 1/6. Therefore, the probability that the absent worker is a bricklayer is 1/6. Answer: 1/6.

Bayanin Amsa

Bayanin Amsa

First, we simplify both sides of the equation by getting rid of the fractions. To do this, we can multiply both sides by the least common multiple (LCM) of the denominators, which is 12. So we have: \begin{align*} \frac{m}{3} + \frac{1}{2} &= \frac{3}{4} + \frac{m}{4}\\ 4 \cdot \left(\frac{m}{3} + \frac{1}{2}\right) &= 4 \cdot \left(\frac{3}{4} + \frac{m}{4}\right)\\ \frac{4m}{3} + 2 &= \frac{9}{4} + \frac{m}{4} \end{align*} Next, we can simplify further by moving all the terms involving m to one side of the equation, and all the constant terms to the other side: \begin{align*} \frac{4m}{3} - \frac{m}{4} &= \frac{9}{4} - 2\\ \frac{16m}{12} - \frac{3m}{12} &= \frac{9}{4} - \frac{8}{4}\\ \frac{13m}{12} &= \frac{1}{4}\\ 13m &= 3\\ m &= \frac{3}{13} \end{align*} Therefore, the solution to the equation is \(m = \frac{3}{13}\). So the correct answer is 3.

Bayanin Amsa

To calculate the probability that a retrenched worker is a plumber or plasterer, we need to first find the total number of workers in the company, and then find the number of workers who are plumbers or plasterers. The total number of workers in the company is: 24 (Carpenters) + 27 (Labourers) + 12 (Plumbers) + 15 (Plasterers) + 9 (Painters) + 3 (Messengers) + 18 (Bricklayers) = 108 The number of workers who are plumbers or plasterers is: 12 (Plumbers) + 15 (Plasterers) = 27 Therefore, the probability that a retrenched worker is a plumber or plasterer is: 27 (number of plumbers and plasterers) / 108 (total number of workers) = 1/4 So the correct answer is option D: 1/4.

Bayanin Amsa

Bayanin Amsa

The probability of drawing one red ball out of 18 balls is 5/18. Similarly, the probability of drawing one white ball is 9/18 (or 1/2) and the probability of drawing one blue ball is 4/18 (or 2/9). Since we are drawing three balls, we can find the probability of getting one red, one white, and one blue by multiplying the probabilities of each event happening together. Since we are drawing with replacement, the probabilities do not change from draw to draw. Thus, the probability of getting one red, one white, and one blue is: (5/18) x (1/2) x (2/9) = 5/162 Therefore, the answer is option C: 5/162.

Bayanin Amsa

Bayanin Amsa

To make the expression x\(^2\) - 18x a perfect square, we need to add a constant term. The constant term should be half of the coefficient of x squared, which is (18/2)\(^2\) = 81. Therefore, we need to add 81 to the expression to make it a perfect square. Adding 81 to the expression gives: x\(^2\) - 18x + 81 = (x - 9)\(^2\) Thus, the answer is 81.

Bayanin Amsa

In set theory, the symbol ∩ denotes the intersection of two sets, meaning the set of elements that are in both sets. Therefore, P∩Q denotes the set of elements that are in both P and Q, which is {2}. The complement of a set A, denoted by A', is the set of elements that are not in A but are in the universal set U. Therefore, (P∩Q)' denotes the set of elements that are in U but are not in P∩Q, which is {1, 3, 4}. Thus, the correct answer is: (1, 3, 4).

Bayanin Amsa

Bayanin Amsa

Bayanin Amsa

Since QOR is a diameter, it divides the circle into two equal halves, which means that ?QOR is a right angle (90°). Thus, we have: ?PQR = 180° - ?QOR (sum of angles in a triangle) ?PQR = 180° - 90° (since ?QOR is a right angle) ?PQR = 90° Now, we have a right triangle PQR with right angle at Q. Since we know that ?PSR is 37°, we can use the fact that the sum of angles in a triangle is 180° to find ?PRQ: ?PQR + ?QPR + ?PRQ = 180° 90° + ?QPR + 37° = 180° ?QPR = 180° - 90° - 37° ?QPR = 53° Therefore, the answer is 53°.

Bayanin Amsa

To simplify \((\frac{16}{81})^{\frac{1}{4}}\), we need to take the fourth root of \(\frac{16}{81}\). The fourth root of a number is the number that, when multiplied by itself four times, gives that number. So, to find the fourth root of \(\frac{16}{81}\), we need to find a number that, when multiplied by itself four times, gives \(\frac{16}{81}\). We can rewrite \(\frac{16}{81}\) as \((\frac{2}{3})^4\), since \(2^4=16\) and \(3^4=81\). Therefore, \((\frac{16}{81})^{\frac{1}{4}} = ((\frac{2}{3})^4)^{\frac{1}{4}} = \frac{2}{3}\), which is.

Bayanin Amsa

Bayanin Amsa

Using the Pythagorean theorem, we can find that the length of the hypotenuse of the right triangle on the left is: $$\sqrt{5^2+12^2} = 13$$ Since the triangle on the right is similar to the triangle on the left, we can set up the following proportion: $$\frac{5}{12} = \frac{t}{13}$$ Cross-multiplying, we get: $$12t = 65$$ Therefore: $$t = \frac{65}{12} \approx 5.4$$ So the answer is closest to 5.6cm. Therefore the correct option is (A).

Bayanin Amsa

When you roll a fair die, there are six possible outcomes, each with an equal chance of occurring. These outcomes are the numbers 1, 2, 3, 4, 5, and 6. To find the probability of obtaining 4 or 6, we need to count how many of the outcomes correspond to these numbers. In this case, there are two outcomes that satisfy the condition, namely 4 and 6. Therefore, the probability of obtaining 4 or 6 is 2 out of 6, or 2/6. We can simplify this fraction by dividing both the numerator and denominator by their greatest common factor, which is 2. This gives us 1/3. Therefore, the probability of obtaining 4 or 6 when rolling a fair die is 1/3.

Bayanin Amsa

We can use the following property of logarithms: - \(\log_{a}b + \log_{a}c = \log_{a}(bc)\) - \(\log_{a}b - \log_{a}c = \log_{a}\left(\dfrac{b}{c}\right)\) Using this property, we can rewrite the expression as: \begin{align*} \log_{10} 25 + \log_{10} 32 - \log_{10} 8 &= \log_{10}(25\cdot 32) - \log_{10} 8 \\ &= \log_{10}(800) - \log_{10} 8 \\ &= \log_{10}\left(\frac{800}{8}\right) \\ &= \log_{10}(100) \\ &= 2 \end{align*} Therefore, the value of the expression is 2. Thus, the answer is (2).

Bayanin Amsa

The area of a parallelogram is given by the formula: Area = base × height In this problem, the area and height of the parallelogram are given as 513cm\(^2\) and 19cm respectively. We can rearrange the formula to solve for the base: base = Area ÷ height Substituting the given values, we get: base = 513cm\(^2\) ÷ 19cm = 27cm Therefore, the base of the parallelogram is 27cm. The answer is, 27cm.

Bayanin Amsa

To find the distance between points X and Y, we can use the Pythagorean theorem, which states that for a right triangle, the square of the length of the hypotenuse (the side opposite the right angle) is equal to the sum of the squares of the lengths of the other two sides. In this case, we can consider points X, Y, and Z as the vertices of a right triangle, where the right angle is at point Z. Then, XY is the hypotenuse, and we know that XZ = 12m and YZ = 5m. Therefore, we have: XY^2 = XZ^2 + YZ^2 XY^2 = 12^2 + 5^2 XY^2 = 144 + 25 XY^2 = 169 XY = √169 XY = 13m Therefore, the distance between points X and Y is 13m. So the answer is option (C) 13m.

Bayanin Amsa

Bayanin Amsa

Bayanin Amsa

Bayanin Amsa

Bayanin Amsa

To find the percentage error, we need to find the absolute difference between the measured value and actual value, divide it by the actual value, and then multiply by 100 to get a percentage. Absolute difference = |actual value - measured value| = |4.25 - 4.10| = 0.15 Percentage error = (|actual value - measured value| / actual value) x 100% Percentage error = (0.15 / 4.25) x 100% = 3.529% (rounded to 3 decimal places) Therefore, the percentage error is approximately 3.529%, which corresponds to 3 9/17%.

Bayanin Amsa

Bayanin Amsa

Bayanin Amsa

To add the fractions \(\frac{1}{1-x}\) and \(\frac{2}{1+x}\), we need to have a common denominator. We can obtain this by multiplying the first fraction by \((1+x)/(1+x)\) and the second fraction by \((1-x)/(1-x)\). This gives us: $$ \frac{1}{1-x} \cdot \frac{1+x}{1+x} + \frac{2}{1+x} \cdot \frac{1-x}{1-x} = \frac{1+x}{1-x^2} + \frac{2(1-x)}{1-x^2} $$ Combining the two fractions gives us: $$ \frac{1+x+2(1-x)}{1-x^2} = \frac{3-x}{1-x^2} $$ Therefore, the single fraction is \(\boxed{\frac{3-x}{1-x^2}}\).

Bayanin Amsa

Bayanin Amsa

(a) Cumulative frequency table and curve (ogive)

The cumulative frequency is plotted against the upper class boundary of each class. The boundaries are obtained by adding \(0.5\) to each upper class limit.

Starting the curve at the lower boundary \((0.5,\,0)\) and plotting each point \((\text{upper boundary},\ \text{cumulative frequency})\), then joining them with a smooth curve, gives the ogive below. Here \(N=50\).

(b) Estimates from the graph

(i) Median. The median corresponds to a cumulative frequency of \(\dfrac{N}{2}=\dfrac{50}{2}=25\). A horizontal line from cumulative frequency \(25\) meets the curve, and the vertical line down to the profit axis gives the median.

Reading from the graph (confirmed by interpolation between the points \((30.5,\,24)\) and \((40.5,\,35)\)):

(ii) 80th percentile. This corresponds to a cumulative frequency of \(\dfrac{80}{100}\times 50 = 40\). A horizontal line from \(40\) meets the curve, and the vertical drop gives the 80th percentile.

Reading from the graph (confirmed by interpolation between \((40.5,\,35)\) and \((50.5,\,45)\)):

(c) Modal weekly profit

The modal class is the class with the highest frequency, i.e. 21 - 30 (frequency \(12\)). Using

where \(L=20.5\) (lower boundary of the modal class), \(f=12\), \(f_1=6\) (frequency of the class before), \(f_2=11\) (frequency of the class after) and \(c=10\):

Bayanin Amsa

None

Bayanin Amsa

(a) \(2x+5y=\tfrac{13}{2}\) and \(5x-2y=9\). Multiply the first by \(2\) and the second by \(5\): \[4x+10y=13,\qquad25x-10y=45.\] Adding: \(29x=58\Rightarrow x=2\). Then \(2(2)+5y=6\tfrac12\Rightarrow5y=2\tfrac12\Rightarrow y=\tfrac12\). So \(x=2,\;y=\tfrac12\).

(b) \[\log_{10}(2x+1)-\log_{10}(3x-2)=1\Rightarrow\log_{10}\frac{2x+1}{3x-2}=1\Rightarrow\frac{2x+1}{3x-2}=10.\] \[2x+1=30x-20\Rightarrow28x=21\Rightarrow x=\frac34.\] (Check: \(3x-2=\tfrac14>0\), valid.)

Bayanin Amsa

Setting up. With P as the vertex, the bearing of Q is 075° and the bearing of R is 310°. The angle turned from PQ to PR is \(310^\circ - 75^\circ = 235^\circ\), so the actual angle inside triangle PQR is

\[\angle QPR = 360^\circ - 235^\circ = 125^\circ.\]

(a) Distance QR (cosine rule).

\[QR^2 = PQ^2 + PR^2 - 2(PQ)(PR)\cos\angle QPR\]

\[QR^2 = 50^2 + 90^2 - 2(50)(90)\cos 125^\circ = 2500 + 8100 - 9000(-0.5736)\]

\[QR^2 = 10600 + 5162.2 = 15762.2,\qquad QR = 125.5\text{ km}.\]

(b) Bearing of R from Q. First find \(\angle PQR\) by the sine rule:

\[\frac{\sin\angle PQR}{PR} = \frac{\sin\angle QPR}{QR}\Rightarrow \sin\angle PQR = \frac{90\sin125^\circ}{125.5} = \frac{90(0.8192)}{125.5} = 0.5873.\]

\[\angle PQR = 36.0^\circ.\]

The bearing of P from Q is \(075^\circ + 180^\circ = 255^\circ\). Point R lies to the west of QP, so the bearing of R from Q is

\[255^\circ + 36^\circ = 291^\circ.\]

Answers: (a) \(QR \approx 125.5\text{ km}\); (b) bearing of R from Q \(\approx 291^\circ\).

Bayanin Amsa

(a) Write everything to base \(3\): \(9=3^{2}\), \(81=3^{4}\). \[9^{2x-1}=3^{4x-2},\qquad\frac{81^{x-2}}{3^{x}}=\frac{3^{4x-8}}{3^{x}}=3^{3x-8}.\] So \(4x-2=3x-8\Rightarrow x=-6\).

(b) \[\sqrt{\frac{0.81\times10^{-5}}{2.25\times10^{7}}}=\sqrt{\frac{0.81}{2.25}\times10^{-12}}=\sqrt{0.36\times10^{-12}}=0.6\times10^{-6}=6.0\times10^{-7}.\]

Bayanin Amsa

(a) Pie chart of junior workers

First find the total number of workers:

The angle of each sector is a fraction of the whole circle \(\left(360^\circ\right)\):

The five sector angles add up to \(117^\circ+54^\circ+66^\circ+78^\circ+45^\circ=360^\circ\), confirming the calculation. Drawn to scale with a protractor, the pie chart is:

(b) Mean score of the aptitude test

Multiply each mark \(x\) by its frequency \(f\) to obtain \(fx\):

To the nearest whole number, the mean score is 6 marks.

Bayanin Amsa

(a) \(PQ\) is produced to \(T\), so \(\angle PQR=180^{\circ}-131^{\circ}=49^{\circ}\). \(PR\) is produced to \(S\), so \(\angle PRQ=180^{\circ}-98^{\circ}=82^{\circ}\). The angles of \(\triangle PQR\) sum to \(180^{\circ}\): \[\angle QPR=180^{\circ}-49^{\circ}-82^{\circ}=49^{\circ}.\]

(b) Circumference \(=2\pi r=400\): \[r=\frac{400}{2\times\frac{22}{7}}=\frac{400\times7}{44}=\frac{2800}{44}=63.6\text{ m}\approx64\text{ m}.\]

Bayanin Amsa

None

Bayanin Amsa

(a) Construction (ruler and a pair of compasses only, drawn to scale 1 cm to 1 cm):

Steps.

1. Draw a base line and mark \(|AB| = 5\text{ cm}\).
2. Construct \(\angle CAB = 120^{\circ}\) at \(A\). With centre \(A\) draw an arc cutting \(AB\); step the radius round the arc twice to obtain \(60^{\circ}\) and then \(120^{\circ}\), and rule the arm \(AX\) through the \(120^{\circ}\) mark.
3. With centre \(A\) and radius \(7.5\text{ cm}\) cut arm \(AX\) at \(C\), so \(|AC| = 7.5\text{ cm}\). Join \(BC\); triangle \(ABC\) is complete.
4. Locus \(l_{1}\): the perpendicular bisector of \(AB\). With centre \(A\), then centre \(B\), and the same radius (a little more than half of \(|AB|\)), draw arcs that meet above and below \(AB\); the line through the two meeting points is \(l_{1}\), the locus of points equidistant from \(A\) and \(B\).
5. Locus \(l_{2}\): the bisector of \(\angle CAB\). With centre \(A\) draw an arc cutting both \(AB\) and \(AC\); from those two cuts draw equal arcs meeting inside the angle, and rule \(l_{2}\) from \(A\) through that meeting point. \(l_{2}\) is the locus of points equidistant from lines \(AB\) and \(AC\).

(b) \(l_{1}\) and \(l_{2}\) intersect at the point marked \(P\) inside the triangle.

(c) Measurement. Measuring directly from the construction, \(|CP| = 6.7\text{ cm}\) (accept \(6.5\text{ cm}\) to \(6.7\text{ cm}\)).

This agrees with a coordinate check. Taking \(A(0,0)\) and \(B(5,0)\), then \(C = 7.5(\cos120^{\circ},\ \sin120^{\circ}) = (-3.75,\ 6.50)\). The perpendicular bisector of \(AB\) is \(x = 2.5\), and the bisector of the \(120^{\circ}\) angle rises at \(60^{\circ}\), so \(P = (2.5,\ 2.5\tan60^{\circ}) = (2.5,\ 4.33)\). Hence

Bayanin Amsa

(a) Completing the table for \(y=\dfrac{5}{2}+x-4x^{2}\).

Each missing value is found by substitution, e.g.

(b) Graph of \(y=\dfrac{5}{2}+x-4x^{2}\) for \(-2.0\le x\le 2.0\). Using a scale of 2 cm to 1 unit on the x-axis and 2 cm to 5 units on the y-axis, the nine points from the table are plotted and joined with a smooth curve (an inverted parabola).

(c) Maximum value of y. The curve turns at its highest point where \(x=-\dfrac{b}{2a}=-\dfrac{1}{2(-4)}=0.125\). Reading the top of the curve from the graph:

So the maximum value of \(y\) is approximately 2.6.

(d) Roots of \(8x^{2}-2x-5=0\). Divide the equation through by \(-2\):

This is exactly \(y=0\), so the required roots are the values of \(x\) where the graph cuts the x-axis. Reading these two intercepts from the curve:

(Check by formula: \(x=\dfrac{1\pm\sqrt{1+40}}{8}=\dfrac{1\pm 6.40}{8}\), giving \(x=-0.68\) or \(x=0.93\), which agrees with the graph.)

Bayanin Amsa

(a) Radius \(r=7\text{ cm}\), angle \(=108^{\circ}\). Arc length \[=\frac{108}{360}\times2\pi r=\frac{108}{360}\times2\times\frac{22}{7}\times7=\frac{3}{10}\times44=13.2\text{ cm}.\] Perimeter of sector \(=\) arc \(+2r=13.2+14=\) 27.2 cm.

(b) Angle of depression \(=30^{\circ}\) equals the angle of elevation of the cliff top from the boat. With the boat \(120\text{ m}\) from the foot: \[\tan30^{\circ}=\frac{h}{120}\Rightarrow h=120\tan30^{\circ}=120\times0.5774=69.28\text{ m}\approx69.3\text{ m}.\]

Bayanin Amsa

(a) Using difference of two squares: \[375^{2}-356^{2}=(375-356)(375+356)=19\times731=13889.\] So \(17x=13889\Rightarrow x=\dfrac{13889}{17}=817\).

(b) \(4^{x}=2^{2x}\) and \(2^{\frac12}\times8=2^{\frac12}\times2^{3}=2^{\frac72}\). Thus \(2x=\tfrac72\Rightarrow x=\tfrac74\).

(c) \(S_{9}=72\Rightarrow\tfrac92(2a+8d)=72\Rightarrow a+4d=8\). Sum of next \(4\) terms \(=S_{13}-S_{9}=71\Rightarrow S_{13}=143\Rightarrow\tfrac{13}{2}(2a+12d)=143\Rightarrow a+6d=11\). Subtracting: \(2d=3\Rightarrow d=1.5\), then \(a=2\). The AP is \[2,\;3\tfrac12,\;5,\;6\tfrac12,\;8,\;\dots\]