WAEC SSCE - General Mathematics - 1993

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To factorize 2x\(^2\) - 21x + 45, we need to find two numbers that multiply to give 2\(\times\)45 = 90 and add to give -21. Let's factorize 90 to find such numbers: 90 = 1\(\times\)90 = 2\(\times\)45 = 3\(\times\)30 = 5\(\times\)18 = 6\(\times\)15 Out of these, the pair that adds up to -21 is 6 and 15. So, we can rewrite the expression as: 2x\(^2\) - 21x + 45 = 2x\(^2\) - 6x - 15x + 45 Now, we can group the first two terms and the last two terms separately and factorize them using the distributive law. That gives us: 2x\(^2\) - 6x - 15x + 45 = 2x(x - 3) - 15(x - 3) Notice that we have a common factor of (x - 3) in both terms. We can factor it out to get the final answer: 2x\(^2\) - 21x + 45 = (x - 3)(2x - 15) Therefore, the correct answer is (x - 3)(2x - 15).

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To solve the equation x\(^2\) - 2x - 3 = 0, we can use the quadratic formula, which is given by: x = (-b ± sqrt(b\^2 - 4ac)) / 2a where a, b, and c are the coefficients of the quadratic equation ax\^2 + bx + c = 0. Using the coefficients of the given equation, a=1, b=-2, and c=-3, we get: x = (-(-2) ± sqrt((-2)\^2 - 4(1)(-3))) / 2(1) x = (2 ± sqrt(16)) / 2 x = (2 ± 4) / 2 So the solutions of the equation are: x = 3 or x = -1 Therefore, the correct answer is: (-1, 3).

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To find the median, we need to first arrange the set of numbers in order from least to greatest: 12, 12, 12, 13, 14, 15 The median is the middle number in the set. Since there are six numbers in this set, the middle number will be the average of the two middle numbers. In this case, the two middle numbers are 12 and 13, so their average is: (12 + 13) / 2 = 12.5 Therefore, the median of this set of numbers is 12.5. Option (B) is the correct answer.

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To find the total surface area of a cone, we need to add the area of the base and the curved surface area. Given that the base radius is 5cm, we can find the base area by using the formula for the area of a circle: Area of base = πr² = π(5)² = 25π To find the curved surface area, we need to first find the slant height of the cone. We can use the Pythagorean theorem to find this value: l = √(r² + h²) = √(5² + 12²) = √169 = 13cm Now, we can find the curved surface area using the formula: Curved surface area = πrl Curved surface area = π(5)(13) = 65π Therefore, the total surface area of the cone is: Total surface area = Base area + Curved surface area Total surface area = 25π + 65π = 90π Using the approximation π = 22/7, we get: Total surface area = 90π = 90(22/7) = 282.8571... cm² Rounding off to one decimal place, we get: Total surface area ≈ 282.9 cm² Therefore, the correct option is 4) 282 6/7cm².

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We know that sin x = opposite/hypotenuse = 12/13, therefore, the adjacent side of the right-angled triangle is √(13^2 - 12^2) = 5. Now, we need to find the value of 1 - cos^2x. Using the identity sin^2x + cos^2x = 1, we can write: cos^2x = 1 - sin^2x Substituting sin x = 12/13, we get: cos^2x = 1 - (12/13)^2 cos^2x = 1 - 144/169 cos^2x = 25/169 Therefore, 1 - cos^2x = 1 - 25/169 = 144/169. Hence, the answer is (d) 144/169.

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A fair die has six possible outcomes, which are the numbers 1, 2, 3, 4, 5, and 6. Since we want to find the probability of obtaining a number less than 3, we need to count how many of these outcomes satisfy this condition. There are only two possible outcomes that satisfy this condition: 1 and 2. Therefore, the probability of obtaining a number less than 3 is 2 out of 6 possible outcomes or 2/6, which simplifies to 1/3. So, the answer is: - Probability = 1/3

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To find the probability of obtaining a number less than 3, we need to add up the number of times the die rolled a 1 or a 2, since these are the numbers less than 3. From the table, we see that the number of times the die rolled a 1 is 25 and the number of times it rolled a 2 is 30. So the total number of times the die rolled a number less than 3 is: 25 + 30 = 55 Therefore, the probability of obtaining a number less than 3 is: 55/200 = 0.275 So the answer is option C: 0.275.

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In a single toss of a fair die, there are six possible outcomes: 1, 2, 3, 4, 5, and 6. Out of these outcomes, three of them (2, 4, and 6) are even numbers. Therefore, the probability of getting an even number in a single toss of a fair die is 3 out of 6 or 1 out of 2. This can be expressed as a fraction of 1/2 or a decimal of 0.5. So, the correct answer is 1/2.

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The volume of a cone is given by the formula: V = (1/3)πr²h, where r is the radius of the base and h is the height of the cone. The cone in the question has height 14 cm and base radius 4.5 cm. Therefore, its volume is: V = (1/3) × (22/7) × (4.5)² × 14 = 346.5 cm³ Half of this volume is: V/2 = 346.5/2 = 173.25 cm³ Therefore, the volume of water that is exactly half the volume of the cone is 173.25 cm³. The answer closest to this value is 148.5 cm³, so the correct option is: - 148.5cm³

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To find the angle subtended by Ondo in the pie chart, we need to first calculate the total number of students in the Faculty of Science. Total number of students = 45 + 410 + 105 + 126 + 154 = 840 Next, we need to find the percentage of students from Ondo: Percentage of students from Ondo = (Number of students from Ondo / Total number of students) x 100 = (126/840) x 100 = 15 Since there are five states in total, and the pie chart represents 360 degrees, we can find the angle subtended by Ondo as follows: Angle subtended by Ondo = (Percentage of students from Ondo / 100) x 360 = (15/100) x 360 = 54 degrees Therefore, the answer is (e) 54^o.

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To find the percentage error, we need to first calculate the absolute error which is the difference between the measured value and the actual value. Absolute error = |1.80 - 1.75| = 0.05m Then, we can calculate the percentage error using the formula: Percentage error = (Absolute error / Actual value) x 100% Percentage error = (0.05 / 1.75) x 100% = 2.86% Therefore, the percentage error in the boy's measurement is approximately 2.86%. The closest option to this value is 2⁶/₇%.

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The first equation can be rearranged as x = y/3. Then substituting this value of x into the second equation, we get: 4y - 5(y/3) = 14 Multiplying both sides by 3, we get: 12y - 5y = 42 7y = 42 y = 6 Substituting y = 6 into the first equation, we get: x = y/3 = 2 Therefore, the solution to the simultaneous equations is x = 2 and y = 6. So the answer is 2,6.

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The area of a rhombus is half the product of the lengths of its diagonals. Therefore, if we know the area of the rhombus and one of its diagonals, we can use this formula to find the length of the other diagonal. Let's call the length of the first diagonal 12 cm, and let's call the length of the other diagonal d cm. The area of the rhombus is given as 60 cm\(^2\). We can write the formula for the area of a rhombus as: Area = (1/2) x (diagonal 1) x (diagonal 2) Substituting the values we know, we get: 60 = (1/2) x 12 x d Simplifying this equation, we get: 60 = 6d Dividing both sides by 6, we get: d = 10 Therefore, the length of the other diagonal is 10 cm. So, the answer is option C - 10cm.

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Since sin x = cos 90 - x for any angle x, we can write: sin x = cos (90 - x) Given sin x = cos 50°, we can substitute to get: cos (90 - x) = cos 50° Taking the inverse cosine of both sides, we get: 90 - x = 50° Solving for x, we get: x = 40° Therefore, the answer is option A: 40°.

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The sum of the interior angles of a polygon with n sides is given by the formula (n - 2) x 180 degrees. Since a right angle is equal to 90 degrees, 30 right angles is equal to 30 x 90 = 2700 degrees. Thus, we can set up the equation: (n - 2) x 180 = 2700 Solving for n: n - 2 = 15 n = 17 Therefore, the polygon has 17 sides, so the answer is (d) 17 sides.

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To find the probability of obtaining 2, we need to look at the frequency of occurrence of 2 in the outcomes. From the table, we see that 2 appears 30 times out of 200. Therefore, the probability of obtaining 2 on a single roll of the die is: Probability of obtaining 2 = (number of times 2 appears) / (total number of rolls) Probability of obtaining 2 = 30 / 200 Probability of obtaining 2 = 0.15 Therefore, the probability of obtaining 2 is 0.15, which is.

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To convert a decimal number to a binary number, we need to repeatedly divide the decimal number by 2 and take note of the remainder. The binary number is obtained by arranging the remainders in the reverse order. Here's the process for converting 89\(_{10}\) to binary: 1. Divide 89 by 2. The quotient is 44 and the remainder is 1. 2. Divide 44 by 2. The quotient is 22 and the remainder is 0. 3. Divide 22 by 2. The quotient is 11 and the remainder is 0. 4. Divide 11 by 2. The quotient is 5 and the remainder is 1. 5. Divide 5 by 2. The quotient is 2 and the remainder is 1. 6. Divide 2 by 2. The quotient is 1 and the remainder is 0. 7. Divide 1 by 2. The quotient is 0 and the remainder is 1. The remainders, taken in reverse order, are 1011001, which is the binary equivalent of 89\(_{10}\). Therefore, the correct answer is option (B) 1011001.

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We can use the trigonometric ratio tangent to solve this problem. Let h be the height of the tank. Then, we have: tan(45°) = h/500 Since tan(45°) = 1, we can simplify to: h/500 = 1 h = 500 Therefore, the height of the tank is 500 meters. Note that the answer is not in the options provided.

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To add or subtract fractions, we need to have a common denominator. Here, we have denominators of 6r and 4r. The least common multiple (LCM) of 6r and 4r is 12r. So, we need to convert each fraction so that it has a denominator of 12r. \(\frac{5}{6r} - \frac{3}{4r} = \frac{5 \cdot 4}{6r \cdot 4} - \frac{3 \cdot 3}{4r \cdot 3}\) Simplifying the resulting fractions, we get: \(\frac{20}{24r} - \frac{9}{12r} = \frac{20}{24r} - \frac{18}{24r}\) Now, we can combine the fractions by subtracting the numerators and keeping the same denominator: \(\frac{20}{24r} - \frac{18}{24r} = \frac{2}{24r}\) Finally, we can simplify the resulting fraction by dividing the numerator and the denominator by 2: \(\frac{2}{24r} = \frac{1}{12r}\) Therefore, the answer is:

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The given problem involves set theory and requires finding the union and intersection of sets. We are given three sets: S, T, and R. Set S contains six elements, {1, 2, 3, 4, 5, 6}, set T contains four elements, {2, 4, 5, 7}, and set R contains three elements, {1, 4, 5}. First, we need to find the intersection of sets S and T, denoted as S∩T, which represents the elements that are common to both sets. The elements common to sets S and T are 2, 4, and 5. Next, we need to find the union of the intersection of sets S and T with set R, denoted as (S∩T) ∪ R, which represents all the elements that are in either set. The elements in (S∩T) ∪ R are {1, 2, 4, 5}. Therefore, the answer is {1, 2, 4, 5}.

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The set of integers from 20 to 30 is {20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30}. We need to find the probability of selecting a prime number from this set. Prime numbers are the numbers that are only divisible by 1 and themselves. The prime numbers in this set are 23 and 29. So, out of the 11 numbers in the set, only 2 are prime. Therefore, the probability of selecting a prime number from the set is 2/11. Hence, the answer is (a) 2/11.

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The inequality 3m + 3 > 9 can be solved by isolating m on one side of the inequality sign. First, subtract 3 from both sides to get 3m > 6. Then, divide both sides by 3 to get m > 2. Therefore, the solution to the inequality is m > 2, which means any value of m that is greater than 2 would make the inequality true. The correct label for the solution is "m > 2".

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To find the area of a sector, we need to use the formula: Area of sector = (θ/360) x πr² where θ is the angle subtended at the centre of the circle, r is the radius of the circle, and π is a mathematical constant approximately equal to 22/7. In this case, the radius is given as 9cm and the angle is given as 120°. So, substituting these values into the formula, we get: Area of sector = (120/360) x (22/7) x 9² = (1/3) x (22/7) x 81 = 754/7 ≈ 107.71 cm² (rounded to the nearest cm²) Therefore, the correct answer is: 85 cm² (option C).

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When a fair die is rolled, there are 6 possible outcomes, each with an equal probability of 1/6. The only way to obtain a sum of 7 is by rolling a combination of (1,6), (2,5), (3,4), (4,3), (5,2), or (6,1). That means there are 6 possible ways to get a sum of 7 out of 36 possible outcomes. Therefore, the probability of obtaining a sum of 7 is 6/36, which simplifies to 1/6. So the answer is: 1/6.

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(a) A.P. \(5, 9, 13, \dots\) with \(a = 5,\; d = 4\).

\[T_{25} = a + 24d = 5 + 24(4) = 5 + 96 = 101\]

(b) G.P. \(\dfrac{8}{9},\, -\dfrac{4}{3},\, 2, \dots\) with \(a = \dfrac{8}{9}\) and common ratio

\[r = \frac{-4/3}{8/9} = -\frac{4}{3}\times\frac{9}{8} = -\frac{3}{2}\]

\[T_5 = ar^4 = \frac{8}{9}\left(-\frac{3}{2}\right)^4 = \frac{8}{9}\times\frac{81}{16} = \frac{9}{2} = 4\tfrac12\]

(c) For the G.P., \(T_3 = ar^2 = 48\) and \(T_6 = ar^5 = 14\tfrac29 = \dfrac{128}{9}\). Dividing:

\[r^3 = \frac{ar^5}{ar^2} = \frac{128/9}{48} = \frac{8}{27} \;\Rightarrow\; r = \frac{2}{3}\]

\[ar^2 = a\left(\tfrac49\right) = 48 \;\Rightarrow\; a = 108\]

The first four terms are \(108,\; 72,\; 48,\; 32\).

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Let the foot of the perpendicular from the peak \(P\) to the horizontal level of \(X\) and \(Y\) be \(N\), with \(Y\) nearer the mountain. Let \(|YN| = d\) and let \(h\) be the height of \(P\) above the level \(XY\).

From \(Y\) (angle of elevation \(35^\circ\)):

\[\tan 35^\circ = \frac{h}{d} \;\Rightarrow\; h = 0.7d\]

From \(X\) (angle of elevation \(15^\circ\), and \(|XN| = d + 4000\)):

\[\tan 15^\circ = \frac{h}{d + 4000} \;\Rightarrow\; h = 0.3(d + 4000)\]

Equating:

\[0.7d = 0.3d + 1200 \;\Rightarrow\; 0.4d = 1200 \;\Rightarrow\; d = 3000\text{ m}\]

\[h = 0.7(3000) = 2100\text{ m}\]

This \(h\) is the height of the peak above the level of \(X\) and \(Y\). Since \(X\) (and \(Y\)) are \(5200\) m above sea level, the height of the mountain above sea level is:

\[5200 + 2100 = 7300\text{ m}\]

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(a) Common denominator \((m+2n)(m-3n)\):

\[\frac{3}{m+2n} - \frac{2}{m-3n} = \frac{3(m-3n) - 2(m+2n)}{(m+2n)(m-3n)}\]

\[= \frac{3m - 9n - 2m - 4n}{(m+2n)(m-3n)} = \frac{m - 13n}{(m+2n)(m-3n)}\]

(b) Let the tens digit be \(t\) and the units digit be \(u\). The number is \(10t + u\).

Sum of digits: \(t + u = 11\).

Interchanging the digits gives \(10u + t\), which exceeds the original by 9:

\[10u + t = (10t + u) + 9 \;\Rightarrow\; 9u - 9t = 9 \;\Rightarrow\; u - t = 1\]

Solving \(t + u = 11\) and \(u - t = 1\): \(u = 6,\; t = 5\).

The number is \(\mathbf{56}\).

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(a)(i) Write \(45 = 9 \times 5 = 3^2 \times 5\):

\[\log_{10}45 = 2\log_{10}3 + \log_{10}5 = 2(0.477) + 0.699 = 0.954 + 0.699 = 1.653\]

(ii) Solve \(x^{0.8265} = 45\). Take logarithms of both sides:

\[0.8265\,\log_{10}x = \log_{10}45 = 1.653\]

\[\log_{10}x = \frac{1.653}{0.8265} = 2.000 \;\Rightarrow\; x = 10^{2} = 100\]

(b) Evaluate \(\sqrt{\dfrac{2.067}{0.0348 \times 0.538}}\) using logarithms.

Denominator log \(= \bar{2}.5416 + \bar{1}.7308 = \bar{2}.2724\). Then

\[\log(\text{fraction}) = 0.3155 - \bar{2}.2724 = 2.0431\]

\[\log(\text{answer}) = \tfrac12(2.0431) = 1.0216 \;\Rightarrow\; \text{answer} = 10.51\]

Hence \(\sqrt{\dfrac{2.067}{0.0348 \times 0.538}} \approx 10.51\).

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(a) Completing the table for \(y = 3\sin 2\theta - \cos\theta\). Substitute each value of \(\theta\), remembering that the angle inside the sine is \(2\theta\).

• \(\theta = 30^\circ:\ 3\sin 60^\circ - \cos 30^\circ = 2.598 - 0.866 = 1.732 \approx 1.7\)
• \(\theta = 60^\circ:\ 3\sin 120^\circ - \cos 60^\circ = 2.598 - 0.500 = 2.098 \approx 2.1\)
• \(\theta = 120^\circ:\ 3\sin 240^\circ - \cos 120^\circ = -2.598 + 0.500 = -2.098 \approx -2.1\)
• \(\theta = 150^\circ:\ 3\sin 300^\circ - \cos 150^\circ = -2.598 + 0.866 = -1.732 \approx -1.7\)

(b) Graph of \(y = 3\sin 2\theta - \cos\theta\) for \(0^\circ \le \theta \le 180^\circ\). The seven table points are plotted and joined with a smooth curve; the curve rises above the tabulated points to a peak near \(\theta = 49^\circ\) and dips to a trough near \(\theta = 131^\circ\).

(c)(i) Solution of \(3\sin 2\theta - \cos\theta = 0\). These are the values of \(\theta\) where the curve crosses the \(\theta\)-axis. Reading the three crossings from the graph gives

This is confirmed algebraically, since \(3\sin 2\theta - \cos\theta = \cos\theta(6\sin\theta - 1) = 0\) gives \(\cos\theta = 0\) (so \(\theta = 90^\circ\)) or \(\sin\theta = \tfrac{1}{6}\) (so \(\theta \approx 10^\circ\) and \(\theta \approx 170^\circ\)).

(c)(ii) Maximum value of \(y\). The highest point of the curve lies between the tabulated points, near \(\theta = 49^\circ\), not at \(\theta = 60^\circ\). Reading the peak gives

Examination note: the greatest value of \(y\) is read from the top of the smooth curve, which rises higher than any tabulated point. Taking the largest table value (\(2.1\) at \(\theta = 60^\circ\)) as the maximum is a common error; the true maximum of about \(2.3\) occurs near \(\theta = 49^\circ\).

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The universal set \(\varepsilon\) is the set of all integers. Listing the given sets over the relevant range:

• \(P = \{x : x < 0\} = \{\dots, -3, -2, -1\}\) (negative integers)
• \(Q = \{\dots, -5, -3, -1, 1, 3, 5, \dots\}\) (odd integers)
• \(R = \{x : -2 \le x < 7\} = \{-2, -1, 0, 1, 2, 3, 4, 5, 6\}\)

(a) \(Q \cap R\) = odd integers lying in \(R\):

\[Q \cap R = \{-1,\, 1,\, 3,\, 5\}\]

(b) \(R'\) = all integers not in \(R\):

\[R' = \{\dots, -4, -3\} \cup \{7, 8, 9, \dots\} = \{x : x \le -3 \text{ or } x \ge 7\}\]

(c) \(P' = \{0, 1, 2, 3, \dots\}\) (non-negative integers). Then

\[P' \cup R' = \{0,1,2,\dots\} \cup \{\dots,-4,-3\}\cup\{7,8,\dots\} = \varepsilon \setminus \{-2,\,-1\}\]

i.e. every integer except \(-1\) and \(-2\).

(d) \(P \cap Q\) = negative odd integers \(= \{\dots, -5, -3, -1\}\). Its complement is every integer that is not a negative odd integer:

\[(P \cap Q)' = \{\dots, -6, -4, -2,\, 0, 1, 2, 3, 4, \dots\}\]

that is, all non-negative integers together with all negative even integers.

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(a) Theorem: the angle an arc subtends at the centre is twice the angle it subtends at the circumference.

Let \(O\) be the centre, and let arc \(AB\) subtend \(\angle AOB\) at the centre and \(\angle ACB\) at a point \(C\) on the remaining circumference. Join \(CO\) and produce it to \(X\).

Triangle \(OAC\) is isosceles because \(OA = OC\) (radii), so \(\angle OAC = \angle OCA\). The exterior angle equals the sum of the interior opposite angles:

Likewise, from isosceles triangle \(OBC\), \(\angle BOX = 2\,\angle OCB\). Adding,

Hence the central angle is twice the inscribed angle on the same arc. (Q.E.D.)

(b) \(O\) is the centre of circle \(ACDB\), with \(\angle CAO = 26^\circ\) and \(\angle AOB = 130^\circ\); \(AD\) is the diameter through \(O\) (the dotted line \(A\!-\!O\!-\!D\)).

First find \(\angle AOC\). In triangle \(OAC\), \(OA = OC\) (radii), so it is isosceles with base angles equal:

Since \(A,O,D\) are collinear (diameter), angles on the straight line at \(O\) give

(ii) \(\angle COB\). With \(C\) above and \(B\) below the diameter, \(\angle COB\) is made up of \(\angle COD\) and \(\angle DOB\):

(i) \(\angle OBC\). Triangle \(OBC\) is isosceles since \(OB = OC\) (radii), so its base angles are equal:

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(a) Cumulative frequency table

Marks (%)	Upper class boundary	Frequency	Cumulative frequency
0–9	9.5	2	2
10–19	19.5	3	5
20–29	29.5	4	9
30–39	39.5	6	15
40–49	49.5	13	28
50–59	59.5	10	38
60–69	69.5	5	43
70–79	79.5	3	46
80–89	89.5	2	48
90–99	99.5	2	50

Plot the cumulative frequencies against the upper class boundaries, beginning with \\((-0.5,0)\\), and join the points with a smooth increasing curve.

(b)(i) Upper quartile

\[\frac{3N}{4}=\frac{3(50)}{4}=37.5.\]

From the curve, the mark corresponding to cumulative frequency \(37.5\) is approximately \(57\).

\[\boxed{Q_3\approx57\text{ marks}}\]

(b)(ii) Pass mark when 60% passed

Number who passed \(=0.60\times50=30\). Hence the number below the pass mark is \(50-30=20\).

From the curve, the mark corresponding to cumulative frequency \(20\) is approximately \(42\).

\[\boxed{\text{Pass mark}\approx42\text{ marks}}\]

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(a) Use \(a\log b = \log b^a\) and combine:

\[\log_{10}\frac{30}{16} - 2\log_{10}\frac{5}{9} + \log_{10}\frac{400}{243} = \log_{10}\frac{30}{16} - \log_{10}\frac{25}{81} + \log_{10}\frac{400}{243}\]

\[= \log_{10}\!\left(\frac{30}{16} \times \frac{81}{25} \times \frac{400}{243}\right)\]

Simplify the product step by step: \(\dfrac{30}{16} = \dfrac{15}{8}\), and \(\dfrac{15}{8}\times\dfrac{81}{25} = \dfrac{243}{40}\), then \(\dfrac{243}{40}\times\dfrac{400}{243} = \dfrac{400}{40} = 10\).

\[= \log_{10} 10 = 1\]

(b) With \(P = 3.6\times10^{-3}\) and \(Q = 2.25\times10^{6}\):

\[\frac{P}{Q} = \frac{3.6\times10^{-3}}{2.25\times10^{6}} = 1.6\times10^{-9} = 16\times10^{-10}\]

\[\sqrt{\frac{P}{Q}} = \sqrt{16\times10^{-10}} = 4\times10^{-5}\]

In standard form, \(\sqrt{\dfrac{P}{Q}} = 4.0\times10^{-5}\).

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(a)

Through C, draw the line RCE parallel to BA and DE.

Since BA is parallel to RC,

\[\angle BCR=52^\circ\]

Also, since RC is parallel to DE,

\[\angle RCD=48^\circ.\]

Hence, using angles about a point,

\[x=360^\circ-(52^\circ+48^\circ)=260^\circ.\]

Therefore, \(x=260^\circ\).

(b)

First draw the boundary lines:

\[y-2x=5\quad\Rightarrow\quad y=2x+5\]

\[2y+x=4\quad\Rightarrow\quad y=2-\frac{x}{2}\]

\[y+2x=10\quad\Rightarrow\quad y=10-2x.\]

Boundary	Points used for plotting	Required side
\(y=2x+5\)	\((0,5),\ (2,9)\)	Below the line; boundary excluded
\(y=2-\frac{x}{2}\)	\((0,2),\ (4,0)\)	On or above the line
\(y=10-2x\)	\((0,10),\ (2,6)\)	On or below the line

The shaded feasible region is shown below. The broken boundary \(y=2x+5\) is not included because the inequality is strict.

The corner points of the region are obtained from intersections of the boundary lines:

\[\left(-\frac65,\frac{13}{5}\right),\qquad \left(\frac54,\frac{15}{2}\right),\qquad \left(\frac{16}{3},-\frac23\right).\]

The points \(\left(-\frac65,\frac{13}{5}\right)\) and \(\left(\frac54,\frac{15}{2}\right)\) are excluded because they lie on \(y-2x=5\).