WAEC SSCE - Further Mathematics - 2020 (Objective)

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Let's first recall the formula for the sum of the roots of a quadratic equation in the form ax\(^2\) + bx + c = 0: Sum of roots = -b/a In the given equation, 2x\(^2\) + 5mx + n = 0, the coefficient of x\(^2\) is 2, which means a = 2. Therefore, the sum of the roots can be expressed as: Sum of roots = - (5m) / 2 We are given that the sum of the roots is 5, so we can set up an equation: 5 = - (5m) / 2 Multiplying both sides by -2, we get: -10 = 5m Dividing both sides by 5, we obtain: m = -2 Therefore, the value of m is -2. So the answer is -2.0.

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When a force is acting at an angle to the horizontal, we can break it down into two components: the horizontal component and the vertical component. The horizontal component of the force is the part of the force that is acting in the horizontal direction, while the vertical component is the part of the force that is acting in the vertical direction. To find the horizontal component of the force of 230N acting at an angle of 65\(^o\), we use the formula: Horizontal component = Force × cos(angle) where angle is the angle between the force and the horizontal direction. Substituting the values given in the problem, we have: Horizontal component = 230N × cos(65\(^o\)) Using a calculator, we can find that cos(65\(^o\)) is approximately 0.4226. Therefore, the horizontal component of the force is: Horizontal component = 230N × 0.4226 ≈ 97.2N Therefore, the answer is approximately 97.2N.

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To find the area enclosed by the line 3x - 5y + 4 = 0 and the axes, we need to find the points where the line intersects with the x-axis and y-axis. First, let's find the x-intercept. To do this, we set y = 0 and solve for x: 3x - 5(0) + 4 = 0 3x + 4 = 0 3x = -4 x = -4/3 So the x-intercept is (-4/3, 0). Next, let's find the y-intercept. To do this, we set x = 0 and solve for y: 3(0) - 5y + 4 = 0 -5y + 4 = 0 -5y = -4 y = 4/5 So the y-intercept is (0, 4/5). Now we can draw a triangle with vertices at the origin (0,0), the x-intercept (-4/3, 0), and the y-intercept (0, 4/5). The base of the triangle is the x-intercept, which has a length of 4/3. The height of the triangle is the y-intercept, which has a length of 4/5. Therefore, the area of the triangle is: (1/2) * base * height = (1/2) * (4/3) * (4/5) = 8/15 = 0.53 (rounded to two decimal places) Therefore, the answer is option C: 0.53 square units.

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The value of p can be found by solving the definite integral for a given value of p and equating it to the given value of 129. The definite integral of the function px^2 + 16 between the limits 0 and 3 can be calculated using antiderivatives. The antiderivative of px^2 is (1/3)px^3 and the antiderivative of 16 is 16x. So, ∫^3_0(px^2 + 16)dx = (1/3)p * 3^3 + 16 * 3 - (1/3)p * 0^3 - 16 * 0 = (1/3)p * 27 + 48 Now, equating this expression to 129 and solving for p, we get: (1/3)p * 27 + 48 = 129 (1/3)p * 27 = 81 p * 27 = 243 p = 243/27 p = 9 So, the value of p is 9.

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To find the inverse of a matrix, we need to use the formula: \[A^{-1} = \frac{1}{\det(A)}\text{adj}(A)\] where \(\det(A)\) is the determinant of matrix \(A\) and \(\text{adj}(A)\) is the adjugate of matrix \(A\). In this case, we have: \[A = \begin{pmatrix} 3 & 5 \\ 1 & 2 \end{pmatrix}\] The determinant of \(A\) is: \[\det(A) = \begin{vmatrix} 3 & 5 \\ 1 & 2 \end{vmatrix} = (3\times2) - (5\times1) = 1\] The adjugate of \(A\) is: \[\text{adj}(A) = \begin{pmatrix} 2 & -5 \\ -1 & 3 \end{pmatrix}\] Therefore, the inverse of \(A\) is: \[A^{-1} = \frac{1}{\det(A)}\text{adj}(A) = \frac{1}{1}\begin{pmatrix} 2 & -5 \\ -1 & 3 \end{pmatrix} = \begin{pmatrix} 2 & -5 \\ -1 & 3 \end{pmatrix}\] So the answer is (B) \(\begin{pmatrix} 2 & -5 \\ -1 & 3 \end{pmatrix}\).

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A linear sequence is a sequence in which each term differs from the previous term by a constant amount. In other words, if we subtract any term from the term that comes after it, we get the same value. To find the nth term of the linear sequence (5y + 1), (2y + 1), (1 - y), ..., we need to first find the common difference between consecutive terms. We can do this by subtracting the second term from the first term, and then subtracting the third term from the second term: (2y + 1) - (5y + 1) = -3y (1 - y) - (2y + 1) = -3y - 1 Since both subtractions give us the same value of -3y, we know that the common difference between consecutive terms is -3y. Now, we need to find the formula for the nth term of the sequence. We can use the general formula for an arithmetic progression: nth term = a\(_1\) + (n - 1)d where a\(_1\) is the first term, d is the common difference, and n is the term we want to find. In our case, the first term is (5y + 1), the common difference is -3y, and we want to find the nth term. Therefore, we have: nth term = (5y + 1) + (n - 1)(-3y) Simplifying this expression, we get: nth term = 5y + 1 - 3ny + 3y nth term = (8 - 3n)y + 1 Therefore, the nth term of the linear sequence (5y + 1), (2y + 1), (1 - y), ... is (8 - 3n)y + 1.

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To solve for p and q, we can equate the corresponding elements of both matrices: p + q = 2 1 = 1 p - q = 8 From the first and third equations, we can obtain: p = 5 q = -3 Therefore, the correct answer is p = 5, q = -3.

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The given equation is V = plog\(_x\), (M + N). We need to express N in terms of X, P, M, and V. We can start by isolating the term containing N: V = plog\(_x\), (M + N) V = p(log\(_x\) M + log\(_x\) N) V - plog\(_x\) M = plog\(_x\) N log\(_x\) N = (V - plog\(_x\) M) / p Now we can solve for N by exponentiating both sides with base x: N = x\(^{\frac{V - plog_x M}{p}}\) Therefore, the expression for N in terms of X, P, M, and V is N = x\(^{\frac{V - plog_x M}{p}}\). Option (A) is the correct answer.

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The given equation represents the distance covered by a particle in motion at any time t seconds. We are asked to find the distance covered by the particle before it comes to rest, which means the particle stops moving. To find the distance covered by the particle before coming to rest, we need to find the time when the particle stops moving. The particle stops moving when its velocity becomes zero. We can find the velocity of the particle at any time t seconds by differentiating the given equation with respect to time: v = dS/dt = 120 - 32t The particle stops moving when v = 0, so we can set the velocity equation to zero and solve for t: 120 - 32t = 0 t = 3.75 seconds Now we know that the particle stops moving after 3.75 seconds. To find the distance covered by the particle before it stops moving, we can substitute this value of t in the original equation: S = 120t - 16t^2 S = 120(3.75) - 16(3.75)^2 S = 225 meters Therefore, the distance covered by the particle before coming to rest is 225 meters. Option (D) is the correct answer.

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To find the median of a set of numbers, we need to arrange the numbers in order from least to greatest. Arranging the given set of numbers in order, we get: 2, 2, 5, 7, 9, 9, 9, 10, 10, 12, 18 There are 11 numbers in the set. Since 11 is an odd number, the median is the middle number when the set is arranged in order. The middle number is the sixth number, which is 9. Therefore, the median of the set of numbers 9, 7, 5, 2, 12, 9, 9, 2, 10, 10, and 18 is 9.

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To calculate the probability that the product of two numbers selected at random with replacement from the set {-5,-2,4,8} is positive, we need to find the total number of ways to select two numbers from the set and the number of ways to select two numbers whose product is positive. There are four numbers in the set, so there are 4 x 4 = 16 possible ways to select two numbers with replacement. For example, we could select -5 and -2, or we could select 4 and 8. To find the number of ways to select two numbers whose product is positive, we need to consider the following cases: 1. Both numbers are positive: There are two positive numbers in the set, so there are 2 x 2 = 4 ways to select two positive numbers. 2. Both numbers are negative: There are two negative numbers in the set, so there are 2 x 2 = 4 ways to select two negative numbers. Therefore, there are a total of 4 + 4 = 8 ways to select two numbers whose product is positive. The probability of selecting two numbers whose product is positive is the number of ways to select two numbers whose product is positive divided by the total number of ways to select two numbers, which is 8/16 or 1/2. Therefore, the answer is, which is \(\frac{1}{2}\).

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To find the angle between the vectors i + 5j and 5i - j, we can use the dot product formula: a · b = |a||b| cos θ where a and b are the two vectors, |a| and |b| are their magnitudes, and θ is the angle between them. First, we need to find the dot product of the two vectors: (i + 5j) · (5i - j) = (1)(5) + (5)(-1) = 0 Next, we need to find the magnitudes of the two vectors: | i + 5j | = √(1² + 5²) = √26 | 5i - j | = √(5² + (-1)²) = √26 Now we can substitute the dot product and magnitudes into the dot product formula to solve for θ: 0 = (√26)(√26) cos θ cos θ = 0 θ = 90° Therefore, the angle between i + 5j and 5i - j is 90 degrees. Option (D) is the correct answer.

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The values of x between 0° and 360° that satisfy cos x = -0.7133 are 135.5° and 224.5°. The cosine function has a range of values between -1 and 1, and for a given value of cosine, there are two angles in the interval [0°, 360°] that satisfy the equation. These angles are found by using the inverse cosine function, also known as arccos. The arccos of -0.7133 is equal to 135.5° and 224.5°, which are the two values of x that satisfy the equation cos x = -0.7133 in the interval [0°, 360°].

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The function f(x) = 2x - x^2 is increasing on an interval if, as x increases, f(x) also increases. To determine this interval, we can first find the critical points of the function, which are the values of x where the derivative of the function is equal to 0 or undefined. The derivative of f(x) = 2x - x^2 is f'(x) = 2 - 2x. Setting f'(x) = 0, we get 2 - 2x = 0, so x = 1. This is the critical point of the function. To determine whether the function is increasing or decreasing at the critical point, we can use the second derivative test. The second derivative of the function f(x) = 2x - x^2 is f''(x) = -2, which is always negative. So, the first derivative (f'(x)) is decreasing at the critical point. Since the first derivative is decreasing at x = 1, the function f(x) = 2x - x^2 is decreasing to the left of x = 1 and increasing to the right of x = 1. So, the function is increasing on the interval x > 1.