WAEC SSCE - General Mathematics - 2012
Ajụjụ 1 Ripọtì
Given that the mean of the scores 15, 21, 17, 26, 18 and 29 is 21, calculate the standard deviation of the scores
Akọwa Nkọwa
To calculate the standard deviation of the scores, we first need to find the variance, which is the average of the squared deviations from the mean. 1. Find the mean of the scores: Mean = (15 + 21 + 17 + 26 + 18 + 29)/6 = 126/6 = 21 2. Calculate the deviations from the mean for each score: 15 - 21 = -6 21 - 21 = 0 17 - 21 = -4 26 - 21 = 5 18 - 21 = -3 29 - 21 = 8 3. Square each deviation: (-6)^2 = 36 0^2 = 0 (-4)^2 = 16 5^2 = 25 (-3)^2 = 9 8^2 = 64 4. Find the average of the squared deviations: (36 + 0 + 16 + 25 + 9 + 64)/6 = 150/6 = 25 5. Take the square root of the variance to get the standard deviation: Standard deviation = sqrt(25) = 5 Therefore, the standard deviation of the scores is 5. The answer is 5.
Ajụjụ 2 Ripọtì
In the diagram, O is a circle centre of the circle PQRS and < PSR = 86o. If < PQR = xo, find x
Akọwa Nkọwa
Ajụjụ 4 Ripọtì
Factorise completely: 32x2y - 48x3y3
Akọwa Nkọwa
We can begin by factoring out the greatest common factor of the two terms, which is 16x2y. This gives: 32x2y - 48x3y3 = 16x2y(2 - 3xy2) So the correct option is 16x2y(2 - 3xy2).
Ajụjụ 5 Ripọtì
If x and y are variables and k is a constant, which of the following describes an inverse relationship between x and y?
Akọwa Nkọwa
The equation that describes an inverse relationship between x and y is y = k/x, which is the second option. An inverse relationship between two variables means that as one variable increases, the other variable decreases in a proportional manner. In other words, if we double the value of x, the value of y would be halved, and vice versa. In the equation y = k/x, as x increases, y decreases, and as x decreases, y increases. The constant k represents the scale of the relationship between x and y. The other options do not describe an inverse relationship between x and y. is a direct proportion where y increases as x increases. is a power relationship where y increases as the square root of x increases. is a linear relationship where y increases at a constant rate as x increases.
Ajụjụ 6 Ripọtì
The lengths of the minor and major arcs 54cm and 126cm respectively. Calculate the angle of the major sector
Akọwa Nkọwa
Ajụjụ 7 Ripọtì
The volume of a cuboid is 54cm3. If the length, width and height of the cuboid are in the ratio 2:1:1 respectively, find its total surface area
Akọwa Nkọwa
To solve this problem, we first need to find the dimensions of the cuboid. Let the length, width and height be 2x, x and x respectively. Then we have: Volume of cuboid = length x width x height 54 = 2x * x * x 54 = 2x^3 x^3 = 27 x = 3 Therefore, the length of the cuboid is 2x = 6cm, the width is x = 3cm, and the height is x = 3cm. To find the total surface area, we need to find the area of each face and add them up. The total surface area is given by: Total surface area = 2lw + 2lh + 2wh = 2(6*3) + 2(6*3) + 2(3*3) = 36 + 36 + 18 = 90 cm^2 Therefore, the total surface area of the cuboid is 90cm^2. The correct answer is option B.
Ajụjụ 8 Ripọtì
The sum of the interior angles of regular polygon is 1800o. How many sides has the polygon?
Akọwa Nkọwa
The sum of the interior angles of a polygon can be found by the formula: sum of interior angles = (n-2) x 180 degrees where n is the number of sides in the polygon. In this problem, we are given that the sum of the interior angles of the polygon is 1800 degrees. So we can set up an equation: 1800 = (n-2) x 180 Simplifying the equation: 10 = n - 2 n = 12 Therefore, the polygon has 12 sides.
Ajụjụ 10 Ripọtì
If x + 0.4y = 3 and y = \(\frac{1}{2}\)x, find the value of (x + y)
Akọwa Nkọwa
We are given that x + 0.4y = 3 and y = \(\frac{1}{2}\)x. Substitute y in the first equation: x + 0.4 * (\(\frac{1}{2}\)x) = 3 x + 0.2x = 3 1.2x = 3 x = 2.5 Substitute x in the equation for y: y = \(\frac{1}{2}\) * 2.5 = 1.25 Then, x + y = 2.5 + 1.25 = 3.75. Therefore, the value of (x + y) is 3\(\frac{3}{4}\). Hence, the correct option is (c) 3\(\frac{3}{4}\).
Ajụjụ 11 Ripọtì
Convert 3510 to number in base 2
Akọwa Nkọwa
To convert a number from base 10 to base 2 (binary), we need to divide the decimal number by 2 repeatedly until the quotient is zero. Then, we write the remainders in reverse order. Let's convert 3510 to binary using this method: 35 / 2 = 17 remainder 1 17 / 2 = 8 remainder 1 8 / 2 = 4 remainder 0 4 / 2 = 2 remainder 0 2 / 2 = 1 remainder 0 1 / 2 = 0 remainder 1 The remainders, in reverse order, are 100011, which is the binary equivalent of 3510. Therefore, the correct answer is option (C) 100011. Each digit in a binary number represents a power of 2, starting from the rightmost digit, which represents 2^0 = 1. The next digit represents 2^1 = 2, the next represents 2^2 = 4, and so on. To convert a binary number to decimal, we multiply each digit by its corresponding power of 2 and add up the results.
Ajụjụ 12 Ripọtì
In 1995, the enrollment of two schools X and Y were 1,050 and 1,190 respectively. Find the ration of the enrollments of X and Y
Akọwa Nkọwa
To find the ratio of enrollments of schools X and Y, we need to divide the enrollment of school X by the enrollment of school Y. Ratio of enrollments of X and Y = Enrollment of X / Enrollment of Y Enrollment of X = 1,050 Enrollment of Y = 1,190 Ratio of enrollments of X and Y = 1,050 / 1,190 Simplifying this fraction by dividing both numerator and denominator by 10 gives: Ratio of enrollments of X and Y = 105 / 119 Further simplifying by dividing both numerator and denominator by 7 gives: Ratio of enrollments of X and Y = 15 / 17 Therefore, the ratio of the enrollments of X and Y is 15:17. Answer option (B) is correct.
Ajụjụ 13 Ripọtì
Express 3 - [\(\frac{x - y}{y}\)] as a single fraction
Akọwa Nkọwa
To express 3 - [\(\frac{x - y}{y}\)] as a single fraction, we first need to simplify the expression inside the square brackets: 3 - [\(\frac{x - y}{y}\)] = 3 - (\(\frac{x}{y}\) - \(\frac{y}{y}\)) = 3 - \(\frac{x}{y}\) + 1 = 4 - \(\frac{x}{y}\) Therefore, 3 - [\(\frac{x - y}{y}\)] can be expressed as a single fraction: = 4 - \(\frac{x}{y}\) = \(\frac{4y - x}{y}\) Hence, the answer is \(\frac{4y - x}{y}\).
Ajụjụ 14 Ripọtì
In the diagram, |QR| = 10m, |SR| = 8m
< QPS = 30o, < QRP = 90o and |PS| = x, Find x
Akọwa Nkọwa
Ajụjụ 15 Ripọtì
In the diagram, MQ//RS, < TUV = 70o and < RLV = 30o. Find the value of x
Ajụjụ 16 Ripọtì
The graph is that of y = 2x2 - 5x - 3. For what value of x will y be negative? What is the gradient of y = 2x2 - 5x - 3 at the point x = 4?
Akọwa Nkọwa
Ajụjụ 17 Ripọtì
Is the diagram, MN, PQ and RS are three intersecting straight lines. Which of the following statements is/are true? i. t = y ii. x + y + z + m = 180o ii. x + m + n = 180o iv. x + n = m + z
Ajụjụ 19 Ripọtì
If N2,500.00 amounted to N3,50.00 in 4 years at simple interest, find the rate at which the interest was charged
Akọwa Nkọwa
Simple interest is calculated by multiplying the principal amount, the interest rate, and the time in years. The formula for simple interest is given as: Simple Interest = (Principal x Rate x Time) / 100 In this problem, we know that the principal amount (P) is N2,500.00, the amount (A) after 4 years is N3,500.00. We need to find the rate (R) at which the interest was charged. Using the formula for simple interest, we can write: A = P + I where I is the interest amount. We can rearrange the formula to solve for I: I = A - P = N3,500.00 - N2,500.00 = N1,000.00 Now, we can substitute the values we know into the formula for simple interest and solve for R: I = (P x R x T) / 100 N1,000.00 = (N2,500.00 x R x 4) / 100 Simplifying, we get: R = (N1,000.00 x 100) / (N2,500.00 x 4) = 10% Therefore, the rate at which the interest was charged is 10%.
Ajụjụ 20 Ripọtì
In the diagram, MN//PO, < PMN = 112o, < PNO = 129oo and < MPN = yo. Find the value of y
Ajụjụ 21 Ripọtì
If x and y are variables and k is a constant, which of the following describes an inverse relationship between x and y?
Akọwa Nkọwa
An inverse relationship between two variables means that as one variable increases, the other variable decreases. Mathematically, an inverse relationship can be represented by an equation where one variable is multiplied or divided by a constant. Out of the given equations, the equation that represents an inverse relationship between x and y is y = \(\frac{k}{x}\). To see why this is the case, let's consider what happens to y as x increases. Suppose x doubles in value. Then, according to the equation y = \(\frac{k}{x}\), y will be halved in value. Similarly, if x triples in value, y will be divided by 3. This means that as x increases, y decreases, which is the characteristic of an inverse relationship. On the other hand, in the equation y = kx, as x increases, y also increases. This is not an inverse relationship but a direct relationship. The same applies to y = k\(\sqrt{x}\) and y = x + k. Therefore, the equation that describes an inverse relationship between x and y is y = \(\frac{k}{x}\).
Ajụjụ 23 Ripọtì
Express 302.10495 correct to five significant figures
Ajụjụ 24 Ripọtì
Mr. Manu travelled from Accra to Pamfokromb a distance of 720km in 8 hours. What will be his speed in m/s?
Akọwa Nkọwa
To find the speed in meters per second (m/s), we need to convert the distance and time to meters and seconds respectively. Distance is given as 720km. 1km is equal to 1000m, so 720km = 720 x 1000m = 720,000m. Time taken is given as 8 hours. 1 hour is equal to 3600 seconds, so 8 hours = 8 x 3600 seconds = 28,800 seconds. Speed = Distance ÷ Time = 720,000m ÷ 28,800s = 25m/s. Therefore, Mr. Manu's speed is 25m/s. (25m/s) is the correct answer.
Ajụjụ 25 Ripọtì
A kite flies on a taut string of length 50m inclined at tan angle 54o to the horizontal ground. The height of the kite above the ground is
Ajụjụ 26 Ripọtì
If p = {prime factors of 210} and Q = {prime less than 10}, find p \(\cap\) Q
Akọwa Nkọwa
The prime factors of 210 are 2, 3, 5, and 7. The prime numbers less than 10 are 2, 3, 5, and 7. The intersection (p ∩ Q) is the set of elements that are in both p and Q. Therefore, p ∩ Q is {2, 3, 5, 7}. So, the correct option is: - {2,3,5,7}
Ajụjụ 27 Ripọtì
Solve for x in the equation; \(\frac{1}{x} + \frac{2}{3x} = \frac{1}{3}\)
Akọwa Nkọwa
To solve the equation \(\frac{1}{x} + \frac{2}{3x} = \frac{1}{3}\), we can use the following steps: Step 1: Find the common denominator. The denominators in this equation are x and 3x. The common denominator is 3x. Step 2: Multiply both sides of the equation by the common denominator. Multiplying both sides by 3x gives: 3 + 2 = x Step 3: Simplify and solve for x. Simplifying the left side gives: 5 = x Therefore, the solution to the equation is x = 5. x = 5, is the correct answer. Options 2, 3, and 4 are incorrect because they do not equal 5.
Ajụjụ 30 Ripọtì
The nth term of a sequence is Tn = 5 + (n - 1)2. Evaluate T4 - T6
Akọwa Nkọwa
To evaluate T4 - T6, we first need to find the values of T4 and T6. T4 = 5 + (4 - 1)2 = 5 + 9 = 14 T6 = 5 + (6 - 1)2 = 5 + 25 = 30 Therefore, T4 - T6 = 14 - 30 = -16 Hence, the answer is -16.
Ajụjụ 34 Ripọtì
Solve the inequality: \(\frac{-m}{2} - \frac{5}{4} \geq \frac{5m}{12} - \frac{7}{6}\)
Akọwa Nkọwa
Ajụjụ 35 Ripọtì
Solve (\(\frac{27}{125}\))-\(\frac{1}{3}\) x (\(\frac{4}{9}\))\(\frac{1}{2}\)
Akọwa Nkọwa
Ajụjụ 36 Ripọtì
The graph is that of y = 2x2 - 5x - 3. For what value of x will y be negative? For what value of x will y be negative?
Akọwa Nkọwa
To find when y is negative, we need to find the values of x for which 2x^2 - 5x - 3 is negative. One way to do this is to use the quadratic formula: x = (-b ± sqrt(b^2 - 4ac)) / 2a where a = 2, b = -5, and c = -3. Plugging these values in, we get: x = (-(-5) ± sqrt((-5)^2 - 4(2)(-3))) / 2(2) x = (5 ± sqrt(49)) / 4 x = (5 ± 7) / 4 So the solutions are x = -3/2 and x = 2. We can check that when x < -3/2 or x > 2, y is positive, and when -3/2 < x < 2, y is negative. Therefore, the answer is option (C) -\(\frac{1}{2} < x < 3\).
Ajụjụ 37 Ripọtì
The curved surface area of a cylindrical tin is 704cm2. If the radius of its base is 8cm, find the height. [Take \(\pi = \frac{22}{7}\)]`
Ajụjụ 38 Ripọtì
The sum of 12 and one third of n is 1 more than twice n. Express the statement in the form of an equation
Akọwa Nkọwa
Ajụjụ 39 Ripọtì
Simplify: \(\frac{54k^2 - 6}{3k + 1}\)
Akọwa Nkọwa
To simplify the expression \(\frac{54k^2 - 6}{3k + 1}\), we can start by factoring out the numerator: \(\frac{54k^2 - 6}{3k + 1} = \frac{6(9k^2 - 1)}{3k + 1}\) We can further simplify the numerator by recognizing that \(9k^2 - 1\) is a difference of squares, and can be factored as \((3k + 1)(3k - 1)\): \(\frac{6(9k^2 - 1)}{3k + 1} = \frac{6(3k + 1)(3k - 1)}{3k + 1}\) We can then cancel out the common factor of \(3k + 1\) in the numerator and denominator, which gives: \(\frac{6(3k - 1)}{1} = 6(3k - 1)\) Therefore, the simplified expression is \(6(3k - 1)\). The correct option is: - 6(3k - 1)
Ajụjụ 41 Ripọtì
Simplify; \(\frac{3\sqrt{5} \times 4\sqrt{6}}{2 \sqrt{2} \times 3\sqrt{2}}\)
Akọwa Nkọwa
Ajụjụ 42 Ripọtì
In \(\bigtriangleup\) XYZ, /XY/ = 8cm, /YZ/ = 10cm and /XZ/ = 6cm. Which of these relation is true?
Ajụjụ 43 Ripọtì
The diagram is a circle with centre P. PRST are points on the circle. Find the value of < PRS
Akọwa Nkọwa
Ajụjụ 44 Ripọtì
The bar chart shows the frequency distribution of marks scored by students in a class test. What is the median of the distribution?
Akọwa Nkọwa
Ajụjụ 45 Ripọtì
If x + y = 2y - x + 1 = 5, find the value of x
Akọwa Nkọwa
We are given two equations, x + y = 5 and 2y - x + 1 = 5, and we need to find the value of x. From the first equation, we can rearrange it to get x = 5 - y. Substituting this value of x into the second equation, we get: 2y - (5 - y) + 1 = 5 Simplifying the equation, we get: 3y - 4 = 5 Adding 4 to both sides, we get: 3y = 9 Dividing both sides by 3, we get: y = 3 Substituting this value of y into x + y = 5, we get: x + 3 = 5 Subtracting 3 from both sides, we get: x = 2 Therefore, the value of x is 2.
Ajụjụ 46 Ripọtì
Find the coefficient of m in the expression of (\(\frac{m}{2} - 1 \frac{1}{2}\)) (m + \(\frac{2}{3}\))
Akọwa Nkọwa
Ajụjụ 47 Ripọtì
The position of three ships P,Q and R at sea are illustrated in the diagram. The arrows indicated the North direction. The bearing of Q from P is 050o and < PQR = 72o. Calculate the bearing of R and Q
Ajụjụ 48 Ripọtì
The bar chart shows the frequency distribution of marks scored by students in a class test. Calculate the mean of the distribution.
Akọwa Nkọwa
To calculate the mean of a frequency distribution, we need to find the sum of all the values multiplied by their respective frequencies and divide by the total number of values. Looking at the bar chart, we can see that there are 5 values: 1, 2, 3, 4, and 5. The corresponding frequencies are 1, 2, 5, 6, and 1, respectively. To calculate the mean, we first need to calculate the sum of all the values multiplied by their respective frequencies: (1 x 1) + (2 x 2) + (3 x 5) + (4 x 6) + (5 x 1) = 1 + 4 + 15 + 24 + 5 = 49 Next, we need to calculate the total number of values, which is the sum of all the frequencies: 1 + 2 + 5 + 6 + 1 = 15 Finally, we divide the sum of the values multiplied by their frequencies by the total number of values: 49 / 15 = 3.27 (rounded to two decimal places) Therefore, the mean of the distribution is approximately 3.27. The correct answer is not provided in the options, but the closest one is option (C) 2.4.
Ajụjụ 49 Ripọtì
Alfred spent \(\frac{1}{4}\) of his money on food, \(\frac{1}{3}\) on clothing and save the rest. If he saved N72,20.00, how much did he spend on food?
Akọwa Nkọwa
Let A be the total amount of money that Alfred has. From the given information, he spent \(\frac{1}{4}\) of A on food and \(\frac{1}{3}\) of A on clothing, and he saved the rest. Therefore, we have: Amount spent on food = \(\frac{1}{4}\)A Amount spent on clothing = \(\frac{1}{3}\)A Amount saved = A - (\(\frac{1}{4}\)A + \(\frac{1}{3}\)A) Amount saved = A - \(\frac{7}{12}\)A Amount saved = \(\frac{5}{12}\)A We know that Alfred saved N72,20.00, so we can write: \(\frac{5}{12}\)A = N72,20.00 Multiplying both sides by \(\frac{12}{5}\), we get: A = N172,80.00 Therefore, the amount spent on food is: \(\frac{1}{4}\)A = \(\frac{1}{4}\) x N172,80.00 = N43,20.00 Hence, Alfred spent N43,20.00 on food. Answer is correct.
Ajụjụ 50 Ripọtì
(a) A box contains 40 identical discs which are either red or white. If the probability of picking a red disc is \(\frac{1}{4}\); Calculate the number of (i) white discs ; (ii) red discs that should be added such that the probability of picking a red disc will be \(\frac{1}{3}\).
(b) A salesman bought some plates at N50.00 each. If he sold all of them for N600.00 and made a profit of 20% on the transaction, how many plates did he buy?
Akọwa Nkọwa
None
Ajụjụ 51 Ripọtì
Three towns X, Y and Z are such that Y is 20 km from X and 22 km from Z. Town X is 18 km from Z. A health centre is to be built by the government to serve the three towns. The centre is to be located such that patients from X and Y travel equal distance to access the health centre while patients from Z will travel exactly 10 km to reach the Health centre.
(a) Using a scale of 1 cm to 2 km, find the construction, using a pair of compasses and ruler only, the possible positions the Health centre can be located.
(b) In how many possible locations can the Health centre be built?
(c) Measure and record the distances of the location from town X.
(d) Which of these locations would be convenient for all three towns?
None
Akọwa Nkọwa
None
Ajụjụ 52 Ripọtì
| Class Interval |
Frequency |
| 60 - 64 | 2 |
| 65 - 69 | 3 |
| 70 - 74 | 6 |
| 75 - 79 | 11 |
| 80 - 84 | 8 |
| 85 - 89 | 7 |
| 90 - 94 | 2 |
| 95 - 99 | 1 |
The table shows the distribution of marks scored by students in an examination. Calculate, correct to 2 decimal places, the
(a) mean ; (b) standard deviation of the distribution.
Akọwa Nkọwa
None
Ajụjụ 53 Ripọtì
A point H is 20 m away from the foot of a tower on the same horizontal ground. From the point H, the angle of elevation of the point P on the tower and the top (T) of the tower are 30° and 50° respectively. Calculate, correct to 3 significant figures :
(a) /PT/; (b) the distance between H and the top of the tower
(c) The position of H if the angle of depression of H from the top of the tower is to be 40°.
Akọwa Nkọwa
None
Ajụjụ 54 Ripọtì
(a) 
In the diagram, TU is tangent to the circle. < RVU = 100° and < URS = 36°. Calculate the value of angle STU.
(b) In triangle XYZ, |XY| = 5 cm, |YZ| = 8 cm and |XZ| = 6 cm. P is a point on the side XY such that |XP| = 2 cm and the line through P, parallel to YZ meets XZ at Q. Calculate |QZ|.
Ajụjụ 55 Ripọtì
Sonny is twice as old as Wale. Four years ago, he was four times as old as Wale. When will the sum of their ages be 66?
Akọwa Nkọwa
None
Ajụjụ 56 Ripọtì
(a) A boy had M Dalasis (D). He spent D15 and shared the remainder equally with his sister. If the sister's share was equal to \(\frac{1}{3}\) of M, find the value of M.
(b) A number of tourists were interviewed on their choice of means of travel. Two- thirds said that they travelled by road, \(\frac{13}{30}\) by air and \(\frac{4}{15}\) by both air and road. If 20 tourists did not travel by either air or road ; (i) represent the information on a Venn diagram ; (ii) how many tourists (1) were interviewed ; (2) travelled by air only?
Akọwa Nkọwa
None
Ajụjụ 57 Ripọtì
(a) Copy and complete the table of values for \(y = 1 - 4\cos x\).
| x | 0° | 30° | 60° | 90° | 120° | 150° | 180° | 210° | 240° | 270° | 300° |
| y | -3.0 | 1.0 | 4.5 | -1.0 |
(b) Using a scale of 2cm to 30° on the x- axis and 2cm to 1 unit on the y- axis, draw the graph of \(y = 1 - 4\cos x\) for \(0° \leq x \leq 360°\).
(c) Use the graph to : (i) solve the equation \(1 - 4\cos x = 0\) ; (ii) find the value of y when x = 105° ; (iii) find x when y = 1.5.
Akọwa Nkọwa
None
Ajụjụ 58 Ripọtì
(a) (i) Using a scale of 2 cm to 1 unit on both axes, on the same graph sheet, draw the graphs of \(y - \frac{3x}{4} = 3\) and \(y + 2x = 6\).
(ii) From your graph, find the coordinates of the point of intersection of the two graphs.
(iii) Show, on the graph sheet, the region satisfied by the inequality \(y - \frac{3}{4}x \geq 3\).
(b) Given that \(x^{2} + bx + 18\) is factorized as \((x + 2)(x + c)\). Find the values of c and b.
Ajụjụ 59 Ripọtì
(a)
In the diagram, ABCD is a rectangular garden (3n - 1)m long and (2n + 1)m wide. A wire mesh 135m long is used to mark its boundary and to divide it into 8 equal plots. Find the value of n.
(b) A cylinder with base radius 14 cm has the same volume as a cube of side 22 cm. Calculate the ratio of the total surface area of the cylinder to that of the cube. [Take \(\pi = \frac{22}{7}\)]
None
Akọwa Nkọwa
None
Ajụjụ 60 Ripọtì
In the diagram, O is the centre of the circleand XY is a chord. If the radius is 5 cm and /XY/ = 6 cm, calculate, correct to 2 decimal places, the :
(a) angle which XY subtends at the centre O ;
(b) area of the shaded portion.
Akọwa Nkọwa
None
Ajụjụ 61 Ripọtì
(a) 
In the diagram, /PQ/ = 6 cm, /QR/ = 13 cm, /RS/ = 5 cm and < RSQ is a right- angled triangle. Calculate, correct to one decimal place, /PS/.
(b) 
The diagram show a wooden structure in the form of a cone mounted on a hemispherical base. The vertical height of the cone is 24 cm and the base radius 7 cm. Calculate, correct to 3 significant figures, the surface area of the structure. [Take \(\pi = \frac{22}{7}\)].
Akọwa Nkọwa
None
Ajụjụ 62 Ripọtì
(a) Simplify : \(\frac{1\frac{1}{4} + \frac{7}{9}}{1\frac{4}{9} - 2\frac{2}{3} \times \frac{9}{64}}\)
(b) Given that \(\sin x = \frac{2}{3}\), evaluate, leaving your answer in surd form and without using tables or calculator, \(\tan x - \cos x\).
Akọwa Nkọwa
None
