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Question 1 Report
If \(\frac{6x + k}{2x^2 + 7x - 15}\) = \(\frac{4}{x + 5} - \frac{2}{2x - 3}\). Find the value of k.
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Question 2 Report
Find the inverse of \(\begin{pmatrix} 3 & 5 \\ 1 & 2 \end{pmatrix}\)
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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) \(\begin{pmatrix} 2 & -5 \\ -1 & 3 \end{pmatrix}\).
Question 3 Report
If the sum of the roots of 2x\(^2\) + 5mx + n = 0 is 5, find the value of m.
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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.
Question 4 Report
A bag contains 5 red and 5 blue identical balls. Three balls are selected at random without replacement. Determine the probability of selecting balls alternating in color.
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Question 5 Report
Calculate, correct to two decimal places, the area enclosed by the line 3x - 5y + 4 = 0 and the axes.
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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.
Question 6 Report
Find the unit vector in the direction opposite to the resultant of forces. F\(_1\) = (-2i - 3j) and F\(_2\) = (5i - j)
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Question 9 Report
Determine the coefficient of x\(^3\) in the binomial expansion of ( 1 + \(\frac{1}{2}\)x)
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Question 10 Report
In what interval is the function f : x -> 2x - x\(^2\) increasing?
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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.
Question 11 Report
Find the nth term of the linear sequence (A.P) (5y + 1), ( 2y + 1), (1- y),...
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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.
Question 12 Report
The distance(s) in metres covered by a particle in motion at any time, t seconds, is given by S =120t - 16t\(^2\). Find in metres, the distance covered by the body before coming to rest.
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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.
Question 13 Report
A function f defined by f : x -> x\(^2\) + px + q is such that f(3) = 6 and f(3) = 0. Find the value of q.
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Question 14 Report
P(3,4) and Q(-3, -4) are two points in a plane. Find the gradient of the line that is normal to the line PQ.
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Question 15 Report
If \(\int^3_0(px^2 + 16)dx\) = 129. Find the value of p.
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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.
Question 16 Report
In which of the following series can be the formula S = \(\frac{a}{1 - r}\) where a is the first term and r is the common ratio, be used to find the sum of all the terms?
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Question 17 Report
If cos x = -0.7133, find the values of x between 0\(^o\) and 360\(^o\)
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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°].
Question 18 Report
Given that P = {x : 1 \(\geq\) x \(\geq\) 6} and Q = {x : 2 < x < 10}. Where x are integers, find n(p \(\cap\) Q)
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Question 20 Report
A binary operation * is defined on the set of real number, R, by \(x*y = x^2 - y^2 + xy\), where \(x, \in R\). Evaluate \((\sqrt{3})*(\sqrt{2})\)
\({\color{red}2x} \times 3\)
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Question 21 Report
If log 5(\(\frac{125x^3}{\sqrt[ 3 ] {y}}\) is expressed in the values of p, q and k respectively.
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Question 23 Report
The variables x and y are such that y =2x\(^3\) - 2x\(^2\) - 5x + 5. Calculate the corresponding change in y and x changes from 2.00 to 2.05.
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Question 24 Report
A force of 230N acts in its direction 065\(^o\). Find its horizontal component.
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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.
Question 25 Report
If V = plog\(_x\), (M + N), express N in terms of X, P, M and V
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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.
Question 26 Report
If \(\begin{pmatrix} p+q & 1\\ 0 & p-q \end {pmatrix}\) = \(\begin{pmatrix} 2 & 1 \\ 0 & 8 \end{pmatrix}\)
Find the values of p and q
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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.
Question 27 Report
Find the median of the numbers 9,7, 5, 2, 12,9,9, 2, 10, 10, and 18.
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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.
Question 28 Report
If the binomial expansion of (1 + 3x)\(^6\) is used to evaluate (0.97)\(^6\), find the value of x.
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Question 29 Report
A circle with centre (5,-4) passes through the point (5, 0). Find its equation.
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Question 30 Report
Calculate the probability that the product of two numbers selected at random with replacement from the set {-5,-2,4, 8} is positive
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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 \(\frac{1}{2}\).
Question 31 Report
Given that F = 3i - 12j, R = 7i + 5j and N = pi + qj are forces acting on a body, if the body is in equilibrium. find the values of p and q.
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Question 32 Report
Given that 2x + 3y - 10 and 3x = 2y - 11, calculate the value of (x - y).
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Question 33 Report
A stone was dropped from the top of a building 40m high. Find, correct to one decimal place, the time it took the stone to reach the ground. [Take g = 9.8ms\(^{-2}\)]
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Question 34 Report
Given that X : R \(\to\) R is defined by x = \(\frac{y + 1}{5 - y}\) , y \(\in\) R, find the domain of x.
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Question 35 Report
Calculate the variance of \(\sqrt{2}\), (1 + \(\sqrt{2}\)) and (2 + \(\sqrt{2}\))
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Question 36 Report
A three-digit odd number less than 500 is to be formed from 1,2,3,4 and 5. If repetition of digits is allowed, in how many ways can this be done?
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Question 37 Report
Consider the statements:
x: Birds fly
y: The sky is blue
Which of the following statements can be represented as x \(\to\) y?
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Question 38 Report
If X = \(\frac{3}{5}\) and cos y = \(\frac{24}{25}\), where X and Y are acute, find the value of cos (X + Y).
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Question 39 Report
Find the angle between i + 5j and 5i - J
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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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