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Question 1 Report
How many numbers greater than 200 can be formed from the digits 1,2,3,4, 5 if no digit is to be repeated in any particular number?
Answer Details
Question 2 Report
If P = \(\begin {pmatrix} 2 & 3\\ -4 & 1 \end {pmatrix}\), Q = \(\begin{pmatrix} 6 \\ 8 \end {pmatrix}\) and PQ = k \(\begin {pmatrix} 45\\ -20 \end {pmatrix}\). Find the value of k.
Answer Details
Given matrices are: $$P = \begin {pmatrix} 2 & 3\\ -4 & 1 \end {pmatrix}, \quad Q = \begin{pmatrix} 6 \\ 8 \end {pmatrix}, \quad PQ = k \begin {pmatrix} 45\\-20 \end {pmatrix}$$ We know that for two matrices to be multiplied, the number of columns of the first matrix should be equal to the number of rows of the second matrix. In this case, the number of columns of $P$ is 2 and the number of rows of $Q$ is 2, so we can multiply them. $$PQ = \begin {pmatrix} 2 & 3\\ -4 & 1 \end {pmatrix} \begin{pmatrix} 6 \\ 8 \end {pmatrix} = \begin {pmatrix} 2(6)+3(8)\\-4(6)+1(8) \end {pmatrix} = \begin {pmatrix} 45\\-20 \end {pmatrix}$$ Comparing the above equation with the given equation, we get: $$k = \frac{\text{corresponding elements in } PQ}{\text{corresponding elements in } \begin {pmatrix} 45\\-20 \end {pmatrix}} = \frac{-20}{45} = -\frac{4}{9}$$ Therefore, the value of $k$ is -\(\frac{4}{5}\). ()
Question 3 Report
Point X and Y are on the same horizontal base as the foot of a building such that X is 96m due east of the building and Y is due west. If the angle of elevation of the top of that building from X is 30\(^o\) and that of Y is 50\(^o\), calculate the distance of Y from the base of the building.
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Question 4 Report
If the mean of 2, 5, (x + 1), (x + 2), 7 and 9 is 6, find the median.
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Question 5 Report
Which of the following vectors is perpendicular to \(\begin{pmatrix} -1 & 3 \end{pmatrix}\)?
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Question 6 Report
The probabilities that John and Jane will pass an examination are 0.9 and 0.7 respectively. Find the probability that at least one of them will pass the examination.
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Question 8 Report
A 35 N force acts on a body of mass 5 kg for 2 seconds. Calculate the change in momentum of the body.
Answer Details
To calculate the change in momentum of the body, we can use the formula: Change in momentum = Force × time Here, the force acting on the body is 35 N and the time for which it acts is 2 seconds. The mass of the body is given as 5 kg. So, substituting the values in the formula, we get: Change in momentum = 35 N × 2 s Change in momentum = 70 Ns Now, we know that momentum is defined as the product of mass and velocity. So, we can also calculate the change in momentum by finding the initial and final velocities of the body and then using the formula: Change in momentum = mass × (final velocity - initial velocity) However, the initial velocity is not given in the question. So, we cannot use this method to find the change in momentum. Therefore, the answer is 70 kg ms\(^{-1}\).
Question 9 Report
Find the area between line y = x + 1 and the x-axis from x = -2 to x = 0.
Answer Details
To find the area between the line y = x + 1 and the x-axis from x = -2 to x = 0, we need to integrate the equation of the line with respect to x over the interval [-2, 0] and take the absolute value of the result. The equation of the line y = x + 1 can be rewritten as x = y - 1, which gives us a different way to represent the line. Integrating this expression with respect to x over the interval [-2, 0] gives: ∫[-2,0] (y - 1) dx = [xy - x] from -2 to 0 = (0-0) - (-2*(-1)) = 2 Taking the absolute value of this result gives us an area of 2 square units. Therefore, the answer is 2 square units.
Question 10 Report
A uniform beam, PQ. is 100 m long and weighs 35 N. It is placed on a support at a point 40 cm from P. If weights of 54 N and FN are attached at P and Q respectively in order to keep it in a horizontal position, calculate, correct to the nearest whole number, the value of F.
Question 12 Report
The second and fourth terms of an exponential sequence (G.P) are \(\frac{2}{9}\) and \(\frac{8}{81}\) respectively. Find the sixth term of the sequence
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Question 13 Report
Given that X and Y are independent events such that P(X) = 0.5, P(Y) = m and P(X U Y) = 0.75, find the value of m.
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Question 14 Report
If g : r \(\to\) 5 - 2r, r is a real number, find the image of -3
Answer Details
The given function is g : r \(\to\) 5 - 2r, where r is a real number. To find the image of -3, we need to substitute -3 for r in the function and simplify: g(-3) = 5 - 2(-3) g(-3) = 5 + 6 g(-3) = 11 Therefore, the image of -3 under the function g is 11.
Question 15 Report
A particle starts from rest and moves in a straight line such that its velocity, V ms\(^{-1}\), at time t second is given by V = 3t\(^2\) - 6t. Calculate the acceleration in the 3rd second.
Answer Details
Acceleration is the rate of change of velocity, so to find the acceleration at any point in time, we need to find the derivative of the velocity function with respect to time. In this case, the velocity function is given by V = 3t\(^2\) - 6t. So, taking the derivative of V with respect to time t, we get: dV/dt = 6t - 6 Now that we have the derivative, we can evaluate it at t = 3 to find the acceleration at the 3rd second. Plugging in t = 3, we get: dV/dt = 6 * 3 - 6 = 12 - 6 = 6 So, the acceleration at the 3rd second is 6 m/s\(^2\).
Question 16 Report
An operation (*) is defined on the set T = {-1, 0, ...., 5} by x * y = x + y - xy. Which of the following operation(s) will give an image which is an element of T?
I. 2(*)5 II. 3(*)5 III. 3(*)4
Answer Details
Question 17 Report
Find correct to the nearest degree,5 the angle between p = 12i - 5j and q = 4i +3j
Answer Details
To find the angle between two vectors, we can use the dot product formula: p \(\cdot\) q = \|p\| \|q\| cos \(\theta\) where p \(\cdot\) q is the dot product of vectors p and q, \|p\| and \|q\| are the magnitudes of vectors p and q respectively, and \(\theta\) is the angle between the two vectors. First, let's calculate the magnitudes of vectors p and q: \|p\| = \(\sqrt{(12)^2 + (-5)^2}\) = \(\sqrt{169}\) = 13 \|q\| = \(\sqrt{(4)^2 + (3)^2}\) = \(\sqrt{25}\) = 5 Next, let's calculate the dot product of vectors p and q: p \(\cdot\) q = (12)(4) + (-5)(3) = 48 - 15 = 33 Substituting the values we obtained into the formula for the dot product, we get: 33 = (13)(5) cos \(\theta\) Solving for cos \(\theta\), we get: cos \(\theta\) = \(\frac{33}{65}\) Using a calculator, we can find that the inverse cosine of \(\frac{33}{65}\) is approximately 59.08\(^o\). Therefore, the angle between vectors p and q is approximately 59 degrees when rounded to the nearest degree. Answer: 59\(^o\).
Question 18 Report
Calculate the mean deviation of 5, 8, 2, 9 and 6
Answer Details
To calculate the mean deviation, we need to first find the mean or average of the given values. Mean or average is calculated by adding up all the values and then dividing the sum by the total number of values. In this case, the sum of the given values is 5 + 8 + 2 + 9 + 6 = 30. Dividing this sum by the total number of values, which is 5 in this case, we get the mean or average as 30/5 = 6. Now, to find the mean deviation, we need to find the deviation of each value from the mean, which is the absolute difference between the value and the mean. For example, the deviation of 5 from the mean is |5 - 6| = 1. Similarly, the deviation of 8 from the mean is |8 - 6| = 2, and so on for all the values. Once we have found the deviation of each value from the mean, we add up all the deviations and divide the sum by the total number of values to get the mean deviation. In this case, the sum of all the deviations is 1 + 2 + 4 + 3 + 0 = 10. Dividing this sum by the total number of values, which is 5, we get the mean deviation as 10/5 = 2. Therefore, the answer is: 2.
Question 19 Report
Solve, correct to three significant figures, (0.3)\(^x\) = (0,5)\(^8\)
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Question 20 Report
Given that \(\frac{1}{x^2 - 4} = \frac{p}{(x + 2)} + \frac{Q}{(x - 2})\)
x \(\neq \pm 2\)
Find the value of (P + Q)
Answer Details
We can start by simplifying the right-hand side of the equation using partial fraction decomposition. To do this, we need to find the values of P and Q. We can use a common denominator on the right-hand side of the equation to get: \[\frac{1}{x^2 - 4} = \frac{p(x - 2) + Q(x + 2)}{(x + 2)(x - 2)}\] Next, we can multiply both sides of the equation by the denominator of the left-hand side to get: \[1 = \frac{p(x - 2) + Q(x + 2)}{(x + 2)(x - 2)}(x^2 - 4)\] Simplifying the right-hand side by multiplying out the terms, we get: \[1 = \frac{(p + Q)x^3 - 4p + 4Q}{(x + 2)(x - 2)}\] Since the left-hand side is just the number 1, the numerator on the right-hand side must also be equal to 1. Therefore, we can set up a system of equations to solve for P and Q: \[p + Q = 0\] \[-4p + 4Q = 1\] Solving for P and Q using the system of equations, we get: \[P = -\frac{1}{4}\] \[Q = \frac{1}{4}\] Therefore, P + Q = 0, which is option (D).
Question 21 Report
The function f : x \(\to\) x\(^2\) + px + q has turning point when x = -3 and remainder of -6 when divided by (x + 2). Find the value of q.
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Question 23 Report
Consider the following statements:
p: Birds fly
q: The sky is blue
r: The grass is green
What is the symbolic representation of "If the grass is green and the sky is not blue, then the birds do not fly"?
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Question 24 Report
A particle starts from rest and moves in a straight line such that its velocity, V ms\(^{-1}\), at time t second is given by V = 3t\(^2\) - 6t. Calculate the acceleration in the 3rd second.
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Question 25 Report
If y = (5 - x)\(^{-3}\), and \(\frac{dy}{dx}\)
Answer Details
To find \(\frac{dy}{dx}\), we need to differentiate y with respect to x using the chain rule and the power rule of differentiation. Using the chain rule, we get: \(\frac{dy}{dx}\) = \(\frac{d}{dx}\)(5 - x)\(^{-3}\) = -3(5 - x)\(^{-4}\)\(\frac{d}{dx}\)(5 - x) = -3(5 - x)\(^{-4}\)(-1) = \(\frac{3}{(5 - x)^4}\) Therefore, the correct answer is (c) \(\frac{3}{(5 - x)^4}\).
Question 27 Report
Calculate the distance between points (-2, -5) and (-1, 3)
Answer Details
To calculate the distance between two points, we can use the distance formula: d = \(\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}\) Where (x1, y1) and (x2, y2) are the coordinates of the two points. In this case, the two points are (-2, -5) and (-1, 3). So, we can substitute these values into the distance formula: d = \(\sqrt{(-1 - (-2))^2 + (3 - (-5))^2}\) Simplifying this expression, we get: d = \(\sqrt{(1)^2 + (8)^2}\) d = \(\sqrt{1 + 64}\) d = \(\sqrt{65}\) Therefore, the distance between the points (-2, -5) and (-1, 3) is \(\sqrt{65}\) units. So, option (C) is the correct answer. Note that the distance formula can be used to find the distance between any two points in a two-dimensional plane.
Question 28 Report
Find the coordinates of the centre of the circle 3x\(^2\) + 3y\(^2\) - 6x + 9y - 5 = 0
Answer Details
To find the center of the circle, we need to rewrite the equation in the standard form: (x - h)\(^2\) + (y - k)\(^2\) = r\(^2\) where (h,k) is the center of the circle and r is the radius. Starting with the given equation: 3x\(^2\) + 3y\(^2\) - 6x + 9y - 5 = 0 We can group the x and y terms together: (3x\(^2\) - 6x) + (3y\(^2\) + 9y) = 5 Next, we need to complete the square for both x and y. For the x terms, we can factor out a 3 from the first two terms: 3(x\(^2\) - 2x) To complete the square, we need to add and subtract (\(\frac{2}{2}\))\(^2\) = 1 inside the parenthesis: 3(x\(^2\) - 2x + 1 - 1) Then, we can simplify this expression: 3((x - 1)\(^2\) - 1) = 3(x - 1)\(^2\) - 3 For the y terms, we can follow the same process: 3(y\(^2\) + 3y) = 3(y\(^2\) + 3y + (\(\frac{3}{2}\))\(^2\) - (\(\frac{3}{2}\))\(^2\)) = 3((y + \(\frac{3}{2}\))\(^2\) - \(\frac{9}{4}\)) = 3(y + \(\frac{3}{2}\))\(^2\) - \(\frac{27}{4}\) Now we can substitute these expressions back into the original equation and simplify: 3(x - 1)\(^2\) - 3 + 3(y + \(\frac{3}{2}\))\(^2\) - \(\frac{27}{4}\) = 5 3(x - 1)\(^2\) + 3(y + \(\frac{3}{2}\))\(^2\) = \(\frac{49}{4}\) Dividing both sides by 3, we get: (x - 1)\(^2\) + (y + \(\frac{3}{2}\))\(^2\) = (\(\frac{7}{2}\))\(^2\) Comparing this equation to the standard form: (x - h)\(^2\) + (y - k)\(^2\) = r\(^2\) We can see that the center of the circle is at (1, -\(\frac{3}{2}\)) and the radius is (\(\frac{7}{2}\)). Therefore, the answer is (3). (1, -\(\frac{3}{2}\))
Question 29 Report
Find the value of x for which 6\(\sqrt{4x^2 + 1}\) = 13x, where x > 0
Answer Details
To solve for x in the equation 6\(\sqrt{4x^2 + 1}\) = 13x, we need to isolate x on one side of the equation. First, we can simplify the left-hand side by squaring both sides of the equation: (6\(\sqrt{4x^2 + 1}\))^2 = (13x)^2 Simplifying the left-hand side, we get: 6^2 * (4x^2 + 1) = 13^2 * x^2 Simplifying further: 144x^2 + 36 = 169x^2 Subtracting 144x^2 from both sides: 36 = 25x^2 Dividing both sides by 25: x^2 = \(\frac{36}{25}\) Taking the square root of both sides: x = \(\frac{6}{5}\) Therefore, the value of x that satisfies the equation is \(\frac{6}{5}\).
Question 30 Report
Find the constant term in the binomial expansion of (2x\(^2\) + \(\frac{1}{x^2}\))\(^4\)
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Question 31 Report
Given that g ; x \(\to\) 3x and f ; x \(\to\) cos x. Find the value of g\(^o\) f(20\(^o\))
Answer Details
The function g of x is defined as g(x) = 3x, and the function f of x is defined as f(x) = cos(x). To find the value of g(f(20°)), we need to first evaluate f(20°) and then plug that result into g. So, first let's evaluate f(20°): f(20°) = cos(20°) = cos(20 x (π/180)) = cos(π/9) Next, we plug the result into g: g(f(20°)) = g(cos(π/9)) = 3cos(π/9) = 2.82 Therefore, the value of g(f(20°)) is 2.82.
Question 32 Report
Find the coordinates of the point in the curve y = 3x\(^2\) - 2x - 5 where the tangent is parallel to the line y = - 5 = 8x
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Question 33 Report
Which of these inequalities is represented by the shaded portion of the graph?
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Question 34 Report
Find the sum of the first 20 terms of the sequence -7-3, 1, ......
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Question 35 Report
Given that P and Q are non-empty subsets of the universal set, U. Find P \(\cap\) (Q U Q`).
Answer Details
To understand this problem, we need to break it down into smaller parts. First, let's define what each symbol means: - \(P\cap Q\) means the intersection of sets P and Q, which consists of all the elements that are in both sets P and Q. - \(Q'\) means the complement of set Q, which consists of all the elements in the universal set U that are not in set Q. - \(U\) is the universal set, which contains all the possible elements that we are considering. Next, let's look at the expression \(Q\cup Q'\). This means the union of set Q and its complement, which contains all the elements in set Q and all the elements that are not in set Q. In other words, it's just the universal set U. So, we can rewrite the original expression as \(P\cap U\), which is just equal to set P. This is because the intersection of any set with the universal set is just the original set itself. Therefore, the answer to the problem is simply set P.
Question 37 Report
A linear transformation is defined by T: (x, y) \(\to\) (-x + y, -4y). Find the image, Q`, of Q(-3, 2) under T
Answer Details
To find the image, Q`, of point Q(-3, 2) under the linear transformation T, we need to apply the transformation matrix to the coordinates of Q.
T: (x, y) → (-x + y, -4y)
So, we have:
T(Q) = (-(-3) + 2, -4(2)) = (5, -8)
Therefore, the image, Q`, of Q(-3, 2) under T is (5, -8).
Explanation: A linear transformation is a function that maps vectors to other vectors while preserving some properties such as linearity and proportionality. In this case, the linear transformation T takes a vector (x, y) and maps it to a new vector (-x + y, -4y). To find the image of a point under T, we simply plug in the coordinates of the point into the transformation matrix and apply the transformation. In this case, we plugged in the coordinates of Q(-3, 2) and found that the image is (5, -8).
Question 38 Report
Solve; \(\frac{P}{2} + \frac{k}{3}\) = 5 and 2p = k = 6 simultaneously
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Question 39 Report
Find the coefficient of the term in the binomial expansion of [2x + \(\frac{3y}{4}\)]\(^3\) in descending powers of x.
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Question 40 Report
Evaluate: \(^{lim}_{x \to 1} \begin{pmatrix} \frac{1 - x}{x^2 - 3x + 2} \end {pmatrix}\)
Answer Details
To evaluate the limit, we can simply substitute the value of x=1 in the expression inside the limit. However, direct substitution yields an indeterminate form of 0/0. Therefore, we need to manipulate the expression before substituting the value of x. We can factorize the denominator of the expression inside the limit as follows: x^2 - 3x + 2 = (x-2)(x-1) So, the expression inside the limit becomes: \[\frac{1-x}{(x-2)(x-1)}\] Now, we can simplify the expression by canceling out the common factor of (x-1) in the numerator and denominator: \[\frac{1-x}{(x-2)(x-1)} = \frac{1}{x-2}\] Now, we can substitute the value of x=1 in the simplified expression to get the limit value: \[\lim_{x \to 1}\frac{1}{x-2} = -\infty\] Therefore, the limit of the given expression as x approaches 1 is negative infinity (-∞), which is not one of the given options.
Question 41 Report
Using determinants, solve the following equations simultaneously.
5x — 6y + 4z = 15
7x + 4y — 3z = 19
2x + y + 6z = 46
To solve the given system of equations using determinants, we need to form a matrix of coefficients of the variables and then calculate its determinant. Let's call this matrix A.
A =
|5 -6 4| |7 4 -3| |2 1 6|
We also need to form a matrix B by replacing the column of constants with the column of variables.
B =
|15 -6 4| |19 4 -3| |46 1 6|
Finally, we can solve for x, y, and z by using Cramer's rule, which says that x, y, and z are given by the ratios of the determinants of matrices formed by replacing the columns of A with the column of B, divided by the determinant of A.
x = |15 -6 4| / |5 -6 4| = 2
y = |5 15 4| / |5 -6 4| = 3
z = |-30 -30 90| / |5 -6 4| = 6
Therefore, the solution to the given system of equations is x = 2, y = 3, and z = 6.
In summary, we formed the matrix A of coefficients, the matrix B by replacing the column of constants with the column of variables, and used Cramer's rule to solve for x, y, and z by finding the ratios of the determinants.
Answer Details
To solve the given system of equations using determinants, we need to form a matrix of coefficients of the variables and then calculate its determinant. Let's call this matrix A.
A =
|5 -6 4| |7 4 -3| |2 1 6|
We also need to form a matrix B by replacing the column of constants with the column of variables.
B =
|15 -6 4| |19 4 -3| |46 1 6|
Finally, we can solve for x, y, and z by using Cramer's rule, which says that x, y, and z are given by the ratios of the determinants of matrices formed by replacing the columns of A with the column of B, divided by the determinant of A.
x = |15 -6 4| / |5 -6 4| = 2
y = |5 15 4| / |5 -6 4| = 3
z = |-30 -30 90| / |5 -6 4| = 6
Therefore, the solution to the given system of equations is x = 2, y = 3, and z = 6.
In summary, we formed the matrix A of coefficients, the matrix B by replacing the column of constants with the column of variables, and used Cramer's rule to solve for x, y, and z by finding the ratios of the determinants.
Question 42 Report
A uniform beam, WX, of length 90 cm and weight 50N is suspended on a pivot, 35 cm from W. It is kept in equilibrum by a means of forces T and 20N applied at Y and Z respectively. |WY| = 10cm and |XZ| = 10cm. Find the value of T
To solve this problem, we need to use the principle of moments. The principle of moments states that for an object to be in equilibrium, the sum of the clockwise moments about any point must be equal to the sum of the anticlockwise moments about the same point. In this problem, we can choose the pivot point as the point about which we will calculate the moments.
Let's start by calculating the clockwise moments. The weight of the beam, 50N, acts at the center of gravity of the beam, which is at a distance of 45 cm from the pivot point. Therefore, the clockwise moment due to the weight of the beam is:
50N x 45cm = 2250 Ncm
Next, let's calculate the anticlockwise moments. The force T acts at point Y, which is 10 cm from the pivot point. Therefore, the anticlockwise moment due to the force T is:
T x 10cm
Similarly, the force of 20N acts at point Z, which is also 10 cm from the pivot point. Therefore, the anticlockwise moment due to the force of 20N is:
20N x 10cm = 200 Ncm
Since the beam is in equilibrium, the sum of the clockwise moments must be equal to the sum of the anticlockwise moments. Therefore, we have:
2250 Ncm = T x 10cm + 200 Ncm
Simplifying this equation, we get:
T = (2250 Ncm - 200 Ncm)/10cm = 205N
Therefore, the value of T is 205N.
In summary, we used the principle of moments to calculate the value of T. We chose the pivot point as the point about which we calculated the moments. We calculated the clockwise moment due to the weight of the beam and the anticlockwise moments due to the forces at points Y and Z. We then equated the sum of the clockwise moments to the sum of the anticlockwise moments and solved for T. The answer is T = 205N.
Answer Details
To solve this problem, we need to use the principle of moments. The principle of moments states that for an object to be in equilibrium, the sum of the clockwise moments about any point must be equal to the sum of the anticlockwise moments about the same point. In this problem, we can choose the pivot point as the point about which we will calculate the moments.
Let's start by calculating the clockwise moments. The weight of the beam, 50N, acts at the center of gravity of the beam, which is at a distance of 45 cm from the pivot point. Therefore, the clockwise moment due to the weight of the beam is:
50N x 45cm = 2250 Ncm
Next, let's calculate the anticlockwise moments. The force T acts at point Y, which is 10 cm from the pivot point. Therefore, the anticlockwise moment due to the force T is:
T x 10cm
Similarly, the force of 20N acts at point Z, which is also 10 cm from the pivot point. Therefore, the anticlockwise moment due to the force of 20N is:
20N x 10cm = 200 Ncm
Since the beam is in equilibrium, the sum of the clockwise moments must be equal to the sum of the anticlockwise moments. Therefore, we have:
2250 Ncm = T x 10cm + 200 Ncm
Simplifying this equation, we get:
T = (2250 Ncm - 200 Ncm)/10cm = 205N
Therefore, the value of T is 205N.
In summary, we used the principle of moments to calculate the value of T. We chose the pivot point as the point about which we calculated the moments. We calculated the clockwise moment due to the weight of the beam and the anticlockwise moments due to the forces at points Y and Z. We then equated the sum of the clockwise moments to the sum of the anticlockwise moments and solved for T. The answer is T = 205N.
Question 43 Report
Find the angle between \(\over {OP}\) = (\(^{-3}_{-4}\)) and \(\over{OQ}\) = (\(^8_{-15}\))
To find the angle between two vectors, we can use the dot product formula:
\(\vec{a} \cdot \vec{b} = |\vec{a}| |\vec{b}| \cos\theta\)
where \(\vec{a}\) and \(\vec{b}\) are the two vectors, \(|\vec{a}|\) and \(|\vec{b}|\) are their magnitudes, and \(\theta\) is the angle between them.
First, we need to find the magnitudes of the two vectors:
\(|\overline{OP}| = \sqrt{(-3)^2 + (-4)^2} = 5\)
\(|\overline{OQ}| = \sqrt{8^2 + (-15)^2} = 17\)
Next, we need to find the dot product of the two vectors:
\(\overline{OP} \cdot \overline{OQ} = (-3)(8) + (-4)(-15) = 9\)
Now, we can substitute the values into the dot product formula to find the angle:
\(9 = 5 \cdot 17 \cdot \cos\theta\)
\(\cos\theta = \frac{9}{85}\)
\(\theta = \cos^{-1}\left(\frac{9}{85}\right)\)
Using a calculator, we get:
\(\theta \approx 71.4^\circ\)
Therefore, the angle between \(\overline{OP}\) and \(\overline{OQ}\) is approximately \(71.4^\circ\).
Answer Details
To find the angle between two vectors, we can use the dot product formula:
\(\vec{a} \cdot \vec{b} = |\vec{a}| |\vec{b}| \cos\theta\)
where \(\vec{a}\) and \(\vec{b}\) are the two vectors, \(|\vec{a}|\) and \(|\vec{b}|\) are their magnitudes, and \(\theta\) is the angle between them.
First, we need to find the magnitudes of the two vectors:
\(|\overline{OP}| = \sqrt{(-3)^2 + (-4)^2} = 5\)
\(|\overline{OQ}| = \sqrt{8^2 + (-15)^2} = 17\)
Next, we need to find the dot product of the two vectors:
\(\overline{OP} \cdot \overline{OQ} = (-3)(8) + (-4)(-15) = 9\)
Now, we can substitute the values into the dot product formula to find the angle:
\(9 = 5 \cdot 17 \cdot \cos\theta\)
\(\cos\theta = \frac{9}{85}\)
\(\theta = \cos^{-1}\left(\frac{9}{85}\right)\)
Using a calculator, we get:
\(\theta \approx 71.4^\circ\)
Therefore, the angle between \(\overline{OP}\) and \(\overline{OQ}\) is approximately \(71.4^\circ\).
Question 44 Report
(a) Given that m = i - i, n = 2i + 3j and 2m + n - r = 0, find |r|
(b) The distance, S metres of a moving particle at any time tseconds is given by
S = 3t - \(\frac{t^3}{3}\) + 9
Find the;
(i) time
(ii) distance travelled
When the particle is momentarily at rest
(a) Given that m = i - i, m = 0. Also, n = 2i + 3j.
Since 2m + n = 2(0) + n = n = 2i + 3j, the equation becomes:
2m + n - r = 2(0) + n - r = n - r = 0
Therefore, r = 2i + 3j.
The magnitude of a vector can be found using the Pythagorean theorem, which states that the magnitude of a vector (a, b) is given by √(a^2 + b^2).
So, |r| = √(2^2 + 3^2) = √(4 + 9) = √13.
(b) (i) The particle is momentarily at rest when its velocity is 0, i.e., when its derivative with respect to time is 0. The velocity of the particle is given by the derivative of its position with respect to time.
So, the velocity is given by:
v = dS/dt = 3 - t^2
Setting this equal to 0 and solving for t, we get:
0 = 3 - t^2
t^2 = 3
t = ±√3
So, the particle is momentarily at rest when t = ±√3.
(ii) To find the distance travelled when the particle is momentarily at rest, we need to substitute t = ±√3 into the expression for S:
S = 3t - t^3/3 + 9
S = 3(±√3) - (±√3)^3/3 + 9
S = 9 ± 3√3 - (3 ± 3√3) + 9
S = 18
So, the particle has travelled a distance of 18 metres when it is momentarily at rest.
Answer Details
(a) Given that m = i - i, m = 0. Also, n = 2i + 3j.
Since 2m + n = 2(0) + n = n = 2i + 3j, the equation becomes:
2m + n - r = 2(0) + n - r = n - r = 0
Therefore, r = 2i + 3j.
The magnitude of a vector can be found using the Pythagorean theorem, which states that the magnitude of a vector (a, b) is given by √(a^2 + b^2).
So, |r| = √(2^2 + 3^2) = √(4 + 9) = √13.
(b) (i) The particle is momentarily at rest when its velocity is 0, i.e., when its derivative with respect to time is 0. The velocity of the particle is given by the derivative of its position with respect to time.
So, the velocity is given by:
v = dS/dt = 3 - t^2
Setting this equal to 0 and solving for t, we get:
0 = 3 - t^2
t^2 = 3
t = ±√3
So, the particle is momentarily at rest when t = ±√3.
(ii) To find the distance travelled when the particle is momentarily at rest, we need to substitute t = ±√3 into the expression for S:
S = 3t - t^3/3 + 9
S = 3(±√3) - (±√3)^3/3 + 9
S = 9 ± 3√3 - (3 ± 3√3) + 9
S = 18
So, the particle has travelled a distance of 18 metres when it is momentarily at rest.
Question 45 Report
The table shows the distribution of masks obtained by students in an examination.
Marks | 50 - 54 | 55 - 59 | 60 - 64 | 65 - 69 | 70 - 74 | 75 - 79 | 80 - 84 | 85 - 89 |
Frequency | 5 | 15 | 20 | 28 | 12 | 9 | 7 | 4 |
Using an assumed mean of 67, calculate, correct to one decimal place. the
a) Mean
b) Standard deviation of the distribution
To calculate the mean and standard deviation of the distribution, we need to use the formulae:
Mean = Assumed Mean + (Σfd)/n
Standard Deviation = √(Σf(x-ḿ)²)/n
Where:
Using the table and the given assumed mean of 67, we can calculate the values needed for the mean and standard deviation as follows:
a) Mean:
We need to find Σfd and n to plug into the formula. To find Σfd, we need to calculate the deviation from the assumed mean for each class interval and multiply it by the frequency of that interval. For example, for the first interval (50-54), the midpoint is 52, so the deviation from the assumed mean of 67 is -15. We multiply this by the frequency of 5 to get -75. We repeat this process for all intervals and sum up the products to get Σfd = -204. The total frequency n is the sum of all frequencies, which is 5 + 15 + 20 + 28 + 12 + 9 + 7 + 4 = 100. Now we can plug these values into the formula to get:
Mean = 67 + (-204)/100 = 65.8 (correct to one decimal place)
Therefore, the mean of the distribution is 65.8.
b) Standard deviation:
We need to find Σf(x-ḿ)² and n to plug into the formula. To find Σf(x-ḿ)², we need to calculate the squared deviation from the assumed mean for each class interval, multiply it by the frequency of that interval, and sum up the products. For example, for the first interval (50-54), the midpoint is 52, so the deviation from the assumed mean of 67 is -15. We square this to get 225, then multiply by the frequency of 5 to get 1125. We repeat this process for all intervals and sum up the products to get Σf(x-ḿ)² = 8444. The total frequency n is the same as before, 100. Now we can plug these values into the formula to get:
Standard deviation = √(Σf(x-ḿ)²)/n = √(8444/100) = 9.19 (correct to one decimal place)
Therefore, the standard deviation of the distribution is 9.19.
In summary, the mean of the distribution is 65.8 and the standard deviation is 9.19.
Answer Details
To calculate the mean and standard deviation of the distribution, we need to use the formulae:
Mean = Assumed Mean + (Σfd)/n
Standard Deviation = √(Σf(x-ḿ)²)/n
Where:
Using the table and the given assumed mean of 67, we can calculate the values needed for the mean and standard deviation as follows:
a) Mean:
We need to find Σfd and n to plug into the formula. To find Σfd, we need to calculate the deviation from the assumed mean for each class interval and multiply it by the frequency of that interval. For example, for the first interval (50-54), the midpoint is 52, so the deviation from the assumed mean of 67 is -15. We multiply this by the frequency of 5 to get -75. We repeat this process for all intervals and sum up the products to get Σfd = -204. The total frequency n is the sum of all frequencies, which is 5 + 15 + 20 + 28 + 12 + 9 + 7 + 4 = 100. Now we can plug these values into the formula to get:
Mean = 67 + (-204)/100 = 65.8 (correct to one decimal place)
Therefore, the mean of the distribution is 65.8.
b) Standard deviation:
We need to find Σf(x-ḿ)² and n to plug into the formula. To find Σf(x-ḿ)², we need to calculate the squared deviation from the assumed mean for each class interval, multiply it by the frequency of that interval, and sum up the products. For example, for the first interval (50-54), the midpoint is 52, so the deviation from the assumed mean of 67 is -15. We square this to get 225, then multiply by the frequency of 5 to get 1125. We repeat this process for all intervals and sum up the products to get Σf(x-ḿ)² = 8444. The total frequency n is the same as before, 100. Now we can plug these values into the formula to get:
Standard deviation = √(Σf(x-ḿ)²)/n = √(8444/100) = 9.19 (correct to one decimal place)
Therefore, the standard deviation of the distribution is 9.19.
In summary, the mean of the distribution is 65.8 and the standard deviation is 9.19.
Question 46 Report
Find the equation of the circle centre (2. 3) which passes through the y - intercept of the line 3x - 2y + 6 = 0
To find the equation of the circle with center (2, 3) that passes through the y-intercept of the line 3x - 2y + 6 = 0, we need to follow these steps:
To find the y-intercept, we set x = 0 and solve for y:
3(0) - 2y + 6 = 0
-2y + 6 = 0
-2y = -6
y = 3
So the y-intercept of the line is (0, 3).
The radius of the circle is the distance between the center (2, 3) and the y-intercept (0, 3):
r = sqrt((2 - 0)^2 + (3 - 3)^2) = sqrt(4) = 2
So the radius of the circle is 2.
The equation of a circle with center (h, k) and radius r is:
(x - h)^2 + (y - k)^2 = r^2
Plugging in the values we found, we get:
(x - 2)^2 + (y - 3)^2 = 2^2
Simplifying, we get:
(x - 2)^2 + (y - 3)^2 = 4
So the equation of the circle with center (2, 3) that passes through the y-intercept of the line 3x - 2y + 6 = 0 is (x - 2)^2 + (y - 3)^2 = 4.
Answer Details
To find the equation of the circle with center (2, 3) that passes through the y-intercept of the line 3x - 2y + 6 = 0, we need to follow these steps:
To find the y-intercept, we set x = 0 and solve for y:
3(0) - 2y + 6 = 0
-2y + 6 = 0
-2y = -6
y = 3
So the y-intercept of the line is (0, 3).
The radius of the circle is the distance between the center (2, 3) and the y-intercept (0, 3):
r = sqrt((2 - 0)^2 + (3 - 3)^2) = sqrt(4) = 2
So the radius of the circle is 2.
The equation of a circle with center (h, k) and radius r is:
(x - h)^2 + (y - k)^2 = r^2
Plugging in the values we found, we get:
(x - 2)^2 + (y - 3)^2 = 2^2
Simplifying, we get:
(x - 2)^2 + (y - 3)^2 = 4
So the equation of the circle with center (2, 3) that passes through the y-intercept of the line 3x - 2y + 6 = 0 is (x - 2)^2 + (y - 3)^2 = 4.
Question 47 Report
(a) If sin p = \(\frac{1}{2}\) and cos q = \(\frac{1}{3}\), evaluate sin(p - q), where 0\(^o\) \(\geq\) p \(\geq\) 90\(^o\) and 90\(^o\) \(\geq\) q \(\geq\) 180\(^o\)
b) Using trapezum rule with seven ordinates, evaluate \(\int^4_1\frac{2}{\sqrt{x + 3}}\)dx
a) Using the identity sin(p - q) = sin p cos q - cos p sin q, we can substitute the given values of sin p and cos q to get: sin(p - q) = sin p cos q - cos p sin q = (\(\frac{1}{2}\))(\(\frac{1}{3}\)) - cos p sin q To solve for cos p sin q, we can use the Pythagorean identity cos^2 p + sin^2 p = 1, which gives sin p = sqrt(1 - cos^2 p). Substituting the given value of sin p, we get: (\(\frac{1}{2}\))^2 + cos^2 p = 1 cos^2 p = \(\frac{3}{4}\) cos p = ±sqrt(\(\frac{3}{4}\)) Since p is in the range 0° ≤ p ≤ 90°, we take cos p = sqrt(\(\frac{3}{4}\)) = \(\frac{\sqrt{3}}{2}\) Similarly, we can use the Pythagorean identity to solve for sin q, which gives: sin q = sqrt(1 - cos^2 q) = sqrt(1 - (\(\frac{1}{3}\))^2) = \(\frac{2\sqrt{2}}{3}\) Substituting these values into the expression for sin(p - q), we get: sin(p - q) = (\(\frac{1}{2}\))(\(\frac{1}{3}\)) - (\(\frac{\sqrt{3}}{2}\))(\(\frac{2\sqrt{2}}{3}\)) = -\(\frac{\sqrt{2}}{3}\) Therefore, sin(p - q) = -\(\frac{\sqrt{2}}{3}\). b) The trapezoidal rule is a numerical method for approximating definite integrals by approximating the area under the curve with trapezoids. To apply the trapezoidal rule with seven ordinates to the integral \(\int^4_1\frac{2}{\sqrt{x + 3}}\)dx, we first divide the interval [1, 4] into seven subintervals of equal width: 1 = x0 < x1 < x2 < x3 < x4 < x5 < x6 < x7 = 4 where xi = 1 + ih for i = 0, 1, 2, ..., 7 and h = (4 - 1)/7 = \(\frac{1}{7}\). Next, we approximate the integral over each subinterval using the formula for the area of a trapezoid: \(\int^{x_{i+1}}_{x_i} f(x)dx \approx \frac{h}{2}(f(x_i) + f(x_{i+1}))\) The approximation of the integral over the whole interval [1, 4] is the sum of the approximations over each subinterval: \(\int^4_1\frac{2}{\sqrt{x + 3}}\)dx ≈ \(\frac{h}{2}\)(f(x0) + 2f(x1) + 2f(x2) + 2f(x3) + 2f(x4) + 2f(x5) + 2f(x6) + f(x7)) where f(x) = 2/(sqrt(x + 3)). Substituting the values for x0, x1, x2, ..., x7 and f(x0), f(x1), f(x
Answer Details
a) Using the identity sin(p - q) = sin p cos q - cos p sin q, we can substitute the given values of sin p and cos q to get: sin(p - q) = sin p cos q - cos p sin q = (\(\frac{1}{2}\))(\(\frac{1}{3}\)) - cos p sin q To solve for cos p sin q, we can use the Pythagorean identity cos^2 p + sin^2 p = 1, which gives sin p = sqrt(1 - cos^2 p). Substituting the given value of sin p, we get: (\(\frac{1}{2}\))^2 + cos^2 p = 1 cos^2 p = \(\frac{3}{4}\) cos p = ±sqrt(\(\frac{3}{4}\)) Since p is in the range 0° ≤ p ≤ 90°, we take cos p = sqrt(\(\frac{3}{4}\)) = \(\frac{\sqrt{3}}{2}\) Similarly, we can use the Pythagorean identity to solve for sin q, which gives: sin q = sqrt(1 - cos^2 q) = sqrt(1 - (\(\frac{1}{3}\))^2) = \(\frac{2\sqrt{2}}{3}\) Substituting these values into the expression for sin(p - q), we get: sin(p - q) = (\(\frac{1}{2}\))(\(\frac{1}{3}\)) - (\(\frac{\sqrt{3}}{2}\))(\(\frac{2\sqrt{2}}{3}\)) = -\(\frac{\sqrt{2}}{3}\) Therefore, sin(p - q) = -\(\frac{\sqrt{2}}{3}\). b) The trapezoidal rule is a numerical method for approximating definite integrals by approximating the area under the curve with trapezoids. To apply the trapezoidal rule with seven ordinates to the integral \(\int^4_1\frac{2}{\sqrt{x + 3}}\)dx, we first divide the interval [1, 4] into seven subintervals of equal width: 1 = x0 < x1 < x2 < x3 < x4 < x5 < x6 < x7 = 4 where xi = 1 + ih for i = 0, 1, 2, ..., 7 and h = (4 - 1)/7 = \(\frac{1}{7}\). Next, we approximate the integral over each subinterval using the formula for the area of a trapezoid: \(\int^{x_{i+1}}_{x_i} f(x)dx \approx \frac{h}{2}(f(x_i) + f(x_{i+1}))\) The approximation of the integral over the whole interval [1, 4] is the sum of the approximations over each subinterval: \(\int^4_1\frac{2}{\sqrt{x + 3}}\)dx ≈ \(\frac{h}{2}\)(f(x0) + 2f(x1) + 2f(x2) + 2f(x3) + 2f(x4) + 2f(x5) + 2f(x6) + f(x7)) where f(x) = 2/(sqrt(x + 3)). Substituting the values for x0, x1, x2, ..., x7 and f(x0), f(x1), f(x
Question 48 Report
In a research to determine the relationship between performance of students in an entrance examination and subsequent school performance, the results of ten randomly selected students wre obtained as follows;
Students | A | B | C | D | E | F | G | H | I |
Performance in Entrance Examination | 11 | 12 | 8 | 13 | 6 | 15 | 10 | 14 | 17 |
School Performance | 5 | 10 | 9 | 7 | 4 | 8 | 6 | 14 | 11 |
1, Calculate the spearman's rank correlation coefficient
2. What would be the researcher's from the result in a?
1. To calculate the Spearman's rank correlation coefficient, we need to follow these steps:
rho = 1 - ((6 * sum of squared differences)/(n*(n^2-1)))
where n is the number of data points (in this case, 10).
After ranking the data and calculating the differences, we get the following table:
Student | Entrance Exam Rank | School Performance Rank | Rank Difference (d) | d^2 |
---|---|---|---|---|
A | 6 | 4 | 2 | 4 |
B | 7 | 8 | -1 | 1 |
C | 4 | 6.5 | -2.5 | 6.25 |
D | 8 | 3 | 5 | 25 |
E | 3 | 10 | -7 | 49 |
F | 10 | 2 | 8 | 64 |
G | 5 | 5.5 | -0.5 | 0.25 |
H | 9 | 1.5 | 7.5 | 56.25 |
I | 2 | 9 | -7 | 49 |
J | 1 | 7 | -6 | 36 |
The sum of the squared differences is 296.5, so we can calculate the Spearman's rank correlation coefficient:
rho = 1 - ((6 * 296.5)/(10*(10^2-1)))</
Answer Details
1. To calculate the Spearman's rank correlation coefficient, we need to follow these steps:
rho = 1 - ((6 * sum of squared differences)/(n*(n^2-1)))
where n is the number of data points (in this case, 10).
After ranking the data and calculating the differences, we get the following table:
Student | Entrance Exam Rank | School Performance Rank | Rank Difference (d) | d^2 |
---|---|---|---|---|
A | 6 | 4 | 2 | 4 |
B | 7 | 8 | -1 | 1 |
C | 4 | 6.5 | -2.5 | 6.25 |
D | 8 | 3 | 5 | 25 |
E | 3 | 10 | -7 | 49 |
F | 10 | 2 | 8 | 64 |
G | 5 | 5.5 | -0.5 | 0.25 |
H | 9 | 1.5 | 7.5 | 56.25 |
I | 2 | 9 | -7 | 49 |
J | 1 | 7 | -6 | 36 |
The sum of the squared differences is 296.5, so we can calculate the Spearman's rank correlation coefficient:
rho = 1 - ((6 * 296.5)/(10*(10^2-1)))</
Question 49 Report
The curve y = 7 - \(\frac{6}{x}\) and the line y + 2x - 3 = 0 intersect at two point. Finf the;
(a) coordinates of the two points
(b) equation of the perpendicular bisector of the line joining the two points
(a)
To find the coordinates of the two points where the curve \(y = 7 - \frac{6}{x}\) intersects the line \(y + 2x - 3 = 0\), we can substitute \(y\) in the equation of the line with the expression for \(y\) in the equation of the curve and solve for \(x\).
\(y + 2x - 3 = 0\) becomes:
\(7 - \frac{6}{x} + 2x - 3 = 0\)
Simplifying this expression, we get:
\(2x^2 - x - 2 = 0\)
We can factorize this quadratic equation to get:
\((2x + 1)(x - 2) = 0\)
This gives us two solutions for \(x\): \(x = -\frac{1}{2}\) and \(x = 2\).
We can substitute these values of \(x\) into either equation to get the corresponding values of \(y\).
For \(x = -\frac{1}{2}\):
\(y + 2(-\frac{1}{2}) - 3 = 0\)
\(y = 4\)
So one point of intersection is \(-\frac{1}{2}, 4\).
For \(x = 2\):
\(y + 2(2) - 3 = 0\)
\(y = -1\)
So the other point of intersection is \(2, -1\).
Therefore, the coordinates of the two points of intersection are \(-\frac{1}{2}, 4\) and \(2, -1\).
(b)
To find the equation of the perpendicular bisector of the line joining the two points of intersection, we first need to find the midpoint of the line segment joining the two points.
The midpoint formula gives us:
\(\left(\frac{x_1 + x_2}{2}\), \(\frac{y_1 + y_2}{2}\right)\)
Substituting the values of the two points, we get:
\(\left(\frac{-\frac{1}{2} + 2}{2}\), \(\frac{4 - 1}{2}\right)\)
Simplifying this expression, we get the midpoint \(\left(\frac{3}{4}, \frac{3}{2}\right)\).
Next, we need to find the slope of the line joining the two points.
The slope formula gives us:
\(\frac{y_2 - y_1}{x_2 - x_1}\)
Substituting the values of the two points, we get:
\(\frac{-1 - 4}{2 - (-\frac{1}{2})}\)
Simplifying this expression, we get the slope \(-\frac{10}{9}\).
The perpendicular bisector of the line segment joining the two points will have a slope that is the negative reciprocal of this slope, which is \(\frac{9}{10}\).
We
Answer Details
(a)
To find the coordinates of the two points where the curve \(y = 7 - \frac{6}{x}\) intersects the line \(y + 2x - 3 = 0\), we can substitute \(y\) in the equation of the line with the expression for \(y\) in the equation of the curve and solve for \(x\).
\(y + 2x - 3 = 0\) becomes:
\(7 - \frac{6}{x} + 2x - 3 = 0\)
Simplifying this expression, we get:
\(2x^2 - x - 2 = 0\)
We can factorize this quadratic equation to get:
\((2x + 1)(x - 2) = 0\)
This gives us two solutions for \(x\): \(x = -\frac{1}{2}\) and \(x = 2\).
We can substitute these values of \(x\) into either equation to get the corresponding values of \(y\).
For \(x = -\frac{1}{2}\):
\(y + 2(-\frac{1}{2}) - 3 = 0\)
\(y = 4\)
So one point of intersection is \(-\frac{1}{2}, 4\).
For \(x = 2\):
\(y + 2(2) - 3 = 0\)
\(y = -1\)
So the other point of intersection is \(2, -1\).
Therefore, the coordinates of the two points of intersection are \(-\frac{1}{2}, 4\) and \(2, -1\).
(b)
To find the equation of the perpendicular bisector of the line joining the two points of intersection, we first need to find the midpoint of the line segment joining the two points.
The midpoint formula gives us:
\(\left(\frac{x_1 + x_2}{2}\), \(\frac{y_1 + y_2}{2}\right)\)
Substituting the values of the two points, we get:
\(\left(\frac{-\frac{1}{2} + 2}{2}\), \(\frac{4 - 1}{2}\right)\)
Simplifying this expression, we get the midpoint \(\left(\frac{3}{4}, \frac{3}{2}\right)\).
Next, we need to find the slope of the line joining the two points.
The slope formula gives us:
\(\frac{y_2 - y_1}{x_2 - x_1}\)
Substituting the values of the two points, we get:
\(\frac{-1 - 4}{2 - (-\frac{1}{2})}\)
Simplifying this expression, we get the slope \(-\frac{10}{9}\).
The perpendicular bisector of the line segment joining the two points will have a slope that is the negative reciprocal of this slope, which is \(\frac{9}{10}\).
We
Question 50 Report
A body, moving at 20ms\(^{-1}\) accelerates uniformly at 2\(\frac{1}{2}ms^{-2}\) for 4 seconds. It continues the journey at this speed for 8 seconds, before coming to rest seconds at tseconds after with uniform retardation. If the ratio of the acceleration to retardation is 3 : 4
(a( sketch the velocity - times graph of the journey
(b) find t
(c) find the total distance of the journey
a)
${}^{}$
${}^{}$
${}^{}$
$b)$
If we let r be the retardation, then st
$\frac{\mathit{s}}{\mathit{t}}$ : r = 3 : 4
when simplified,
r = 103 $\frac{\mathrm{\mathrm{}}}{}$
Velocity after 4 seconds
= 20 + 52 $\frac{\mathrm{\mathrm{}}}{}$ x 4 = 30ms−1 ${}^{-1}$
${\mathrm{}}^{}$
So that 52 $\frac{\mathrm{\mathrm{}}}{}$ : 30t $\frac{\mathrm{}}{}$ = 34 $\frac{\mathrm{\mathrm{}}}{}$
t = 9 seconds
Answer Details
a)
${}^{}$
${}^{}$
${}^{}$
$b)$
If we let r be the retardation, then st
$\frac{\mathit{s}}{\mathit{t}}$ : r = 3 : 4
when simplified,
r = 103 $\frac{\mathrm{\mathrm{}}}{}$
Velocity after 4 seconds
= 20 + 52 $\frac{\mathrm{\mathrm{}}}{}$ x 4 = 30ms−1 ${}^{-1}$
${\mathrm{}}^{}$
So that 52 $\frac{\mathrm{\mathrm{}}}{}$ : 30t $\frac{\mathrm{}}{}$ = 34 $\frac{\mathrm{\mathrm{}}}{}$
t = 9 seconds
Question 51 Report
If \(\alpha\) and \(\beta\) are the roots of the equation 3x\(^2\) + 4x - 5 = 0, find the value of (\(\alpha - \beta\)), leaving the answer in surd form.
To find the value of (\(\alpha - \beta\)), we can use the fact that for a quadratic equation of the form ax\(^2\) + bx + c = 0, the sum of the roots is -b/a and the product of the roots is c/a.
In this case, the given quadratic equation is 3x\(^2\) + 4x - 5 = 0, which means that a = 3, b = 4, and c = -5. Using the formula for the sum of the roots, we get:
\(\alpha + \beta = -\frac{b}{a} = -\frac{4}{3}\)
Using the formula for the product of the roots, we get:
\(\alpha \beta = \frac{c}{a} = -\frac{5}{3}\)
To find the value of (\(\alpha - \beta\)), we can use the identity:
\(\alpha - \beta = \sqrt{(\alpha + \beta)^2 - 4\alpha \beta}\)
Plugging in the values we found earlier, we get:
\(\alpha - \beta = \sqrt{\left(-\frac{4}{3}\right)^2 - 4\left(-\frac{5}{3}\right)}\)
Simplifying, we get:
\(\alpha - \beta = \sqrt{\frac{16}{9} + \frac{20}{3}}\)
\(\alpha - \beta = \sqrt{\frac{68}{9}}\)
\(\alpha - \beta = \frac{2\sqrt{17}}{3}\)
Therefore, the value of (\(\alpha - \beta\)) is \(\frac{2\sqrt{17}}{3}\) in surd form.
Answer Details
To find the value of (\(\alpha - \beta\)), we can use the fact that for a quadratic equation of the form ax\(^2\) + bx + c = 0, the sum of the roots is -b/a and the product of the roots is c/a.
In this case, the given quadratic equation is 3x\(^2\) + 4x - 5 = 0, which means that a = 3, b = 4, and c = -5. Using the formula for the sum of the roots, we get:
\(\alpha + \beta = -\frac{b}{a} = -\frac{4}{3}\)
Using the formula for the product of the roots, we get:
\(\alpha \beta = \frac{c}{a} = -\frac{5}{3}\)
To find the value of (\(\alpha - \beta\)), we can use the identity:
\(\alpha - \beta = \sqrt{(\alpha + \beta)^2 - 4\alpha \beta}\)
Plugging in the values we found earlier, we get:
\(\alpha - \beta = \sqrt{\left(-\frac{4}{3}\right)^2 - 4\left(-\frac{5}{3}\right)}\)
Simplifying, we get:
\(\alpha - \beta = \sqrt{\frac{16}{9} + \frac{20}{3}}\)
\(\alpha - \beta = \sqrt{\frac{68}{9}}\)
\(\alpha - \beta = \frac{2\sqrt{17}}{3}\)
Therefore, the value of (\(\alpha - \beta\)) is \(\frac{2\sqrt{17}}{3}\) in surd form.
Question 52 Report
Given that (\(_r^n\)) = \(^nC_r\), simplify (\(^{2x + 1}_{3}\)) - (\(^{2x - 1}_3\)) - 2(\(^x_2\))
The expression given is: (^2x + 1_{3}) - (^2x - 1_{3}) - 2(^x_{2})
We can simplify this expression by using the formula (^r_{n}) = ^nC_{r}, which represents the number of ways of choosing r objects from a set of n distinct objects.
First, let's simplify (^2x + 1_{3}) using the formula:
(^2x + 1_{3}) = ^3C_{2x+1}
This represents the number of ways of choosing (2x + 1) objects from a set of 3 distinct objects. Similarly, we can simplify (^2x - 1_{3}) using the formula:
(^2x - 1_{3}) = ^3C_{2x-1}
This represents the number of ways of choosing (2x - 1) objects from a set of 3 distinct objects.
Now, substituting these values in the given expression, we get:
(^2x + 1_{3}) - (^2x - 1_{3}) - 2(^x_{2}) = ^3C_{2x+1} - ^3C_{2x-1} - 2(^x_{2})
Next, we can simplify 2(^x_{2}) using the formula:
2(^x_{2}) = 2^xC_{2}
This represents the number of ways of choosing 2 objects from a set of x distinct objects, multiplied by 2.
Substituting this value in the previous expression, we get:
(^2x + 1_{3}) - (^2x - 1_{3}) - 2(^x_{2}) = ^3C_{2x+1} - ^3C_{2x-1} - 2^xC_{2}
This is the simplified expression, which represents the difference between the number of ways of choosing (2x + 1) objects and (2x - 1) objects from a set of 3 distinct objects, minus twice the number of ways of choosing 2 objects from a set of x distinct objects.
Answer Details
The expression given is: (^2x + 1_{3}) - (^2x - 1_{3}) - 2(^x_{2})
We can simplify this expression by using the formula (^r_{n}) = ^nC_{r}, which represents the number of ways of choosing r objects from a set of n distinct objects.
First, let's simplify (^2x + 1_{3}) using the formula:
(^2x + 1_{3}) = ^3C_{2x+1}
This represents the number of ways of choosing (2x + 1) objects from a set of 3 distinct objects. Similarly, we can simplify (^2x - 1_{3}) using the formula:
(^2x - 1_{3}) = ^3C_{2x-1}
This represents the number of ways of choosing (2x - 1) objects from a set of 3 distinct objects.
Now, substituting these values in the given expression, we get:
(^2x + 1_{3}) - (^2x - 1_{3}) - 2(^x_{2}) = ^3C_{2x+1} - ^3C_{2x-1} - 2(^x_{2})
Next, we can simplify 2(^x_{2}) using the formula:
2(^x_{2}) = 2^xC_{2}
This represents the number of ways of choosing 2 objects from a set of x distinct objects, multiplied by 2.
Substituting this value in the previous expression, we get:
(^2x + 1_{3}) - (^2x - 1_{3}) - 2(^x_{2}) = ^3C_{2x+1} - ^3C_{2x-1} - 2^xC_{2}
This is the simplified expression, which represents the difference between the number of ways of choosing (2x + 1) objects and (2x - 1) objects from a set of 3 distinct objects, minus twice the number of ways of choosing 2 objects from a set of x distinct objects.
Question 53 Report
If sin x \(\frac{P - Q}{P + Q}\), where 0\(^o\) \(\leq\) x \(\leq\) 90\(^o\), find 1 - tan\(^2\)x
Let w be the adjacent side,
then w2 ${}^{2}$ = (P + q)2 ${}^{2}$ = 4pq so that w = 2pq−−√ $\sqrt{\mathit{p}\mathit{q}}$
Since tan x p−q2pq√ $\frac{\mathit{p}-\mathit{q}}{2\sqrt{\mathit{p}\mathit{q}}}$,
substituting;
1 - tan2 ${}^{2}$x = 1 - (p−q2pq√ $\frac{\mathit{p}-\mathit{q}}{2\sqrt{\mathit{p}\mathit{q}}}$)2 ${}^{2}$
= 1 - P2−2pq+q24pq $\frac{{\mathit{P}}^{2}-2\mathit{p}\mathit{q}+{\mathit{q}}^{2}}{4\mathit{p}\mathit{q}}$
$\frac{}{}$
= 4pq−p2+2pq−q24pq $\frac{4\mathit{p}\mathit{q}-{\mathit{p}}^{2}+2\mathit{p}\mathit{q}-{\mathit{q}}^{2}}{4\mathit{p}\mathit{q}}$
Therefore,
1 - tan2 ${}^{2}$x = 6pq−p2−q24pq
Answer Details
Let w be the adjacent side,
then w2 ${}^{2}$ = (P + q)2 ${}^{2}$ = 4pq so that w = 2pq−−√ $\sqrt{\mathit{p}\mathit{q}}$
Since tan x p−q2pq√ $\frac{\mathit{p}-\mathit{q}}{2\sqrt{\mathit{p}\mathit{q}}}$,
substituting;
1 - tan2 ${}^{2}$x = 1 - (p−q2pq√ $\frac{\mathit{p}-\mathit{q}}{2\sqrt{\mathit{p}\mathit{q}}}$)2 ${}^{2}$
= 1 - P2−2pq+q24pq $\frac{{\mathit{P}}^{2}-2\mathit{p}\mathit{q}+{\mathit{q}}^{2}}{4\mathit{p}\mathit{q}}$
$\frac{}{}$
= 4pq−p2+2pq−q24pq $\frac{4\mathit{p}\mathit{q}-{\mathit{p}}^{2}+2\mathit{p}\mathit{q}-{\mathit{q}}^{2}}{4\mathit{p}\mathit{q}}$
Therefore,
1 - tan2 ${}^{2}$x = 6pq−p2−q24pq
Question 54 Report
Three forces N, 14N and 16N acting on a particle keep it in equilibrium. Find the angle between the forces 10N and 16N.
To find the angle between the forces 10N and 16N, we can use the fact that the particle is in equilibrium. This means that the net force acting on the particle is zero.
We can break down the three forces into their horizontal and vertical components. Let's call the angle between the 14N force and the horizontal axis as θ1, and the angle between the 16N force and the horizontal axis as θ2.
The horizontal components of the forces are:
N cos(θ1)
14N cos(0)
16N cos(θ2)
Since the particle is in equilibrium, the net horizontal force must be zero. So we have:
N cos(θ1) + 14N cos(0) + 16N cos(θ2) = 0
Simplifying this equation, we get:
N cos(θ1) + 16N cos(θ2) = -14N
We can do the same thing for the vertical components of the forces:
N sin(θ1)
14N sin(0)
16N sin(θ2)
Since the particle is in equilibrium, the net vertical force must be zero. So we have:
N sin(θ1) + 14N sin(0) + 16N sin(θ2) = 0
Simplifying this equation, we get:
N sin(θ1) + 16N sin(θ2) = 0
We can now solve these two equations for N and sin(θ1):
N = -16N cos(θ2) / cos(θ1)
sin(θ1) = -16 sin(θ2) / cos(θ1)
Now, we can use the Pythagorean theorem to find the magnitude of the 10N force:
sqrt((10N)^2 + (N sin(θ1))^2) = sqrt((10N)^2 + (16 sin(θ2))^2)
Finally, we can use the arctangent function to find the angle between the 10N and 16N forces:
tan(θ10-16) = (N sin(θ1)) / (10N) = (-16 sin(θ2)) / (10N cos(θ2))
θ10-16 = arctan((-16 sin(θ2)) / (10N cos(θ2)))
Therefore, we can find the angle between the forces 10N and 16N by using the above formula with the values we have calculated.
Answer Details
To find the angle between the forces 10N and 16N, we can use the fact that the particle is in equilibrium. This means that the net force acting on the particle is zero.
We can break down the three forces into their horizontal and vertical components. Let's call the angle between the 14N force and the horizontal axis as θ1, and the angle between the 16N force and the horizontal axis as θ2.
The horizontal components of the forces are:
N cos(θ1)
14N cos(0)
16N cos(θ2)
Since the particle is in equilibrium, the net horizontal force must be zero. So we have:
N cos(θ1) + 14N cos(0) + 16N cos(θ2) = 0
Simplifying this equation, we get:
N cos(θ1) + 16N cos(θ2) = -14N
We can do the same thing for the vertical components of the forces:
N sin(θ1)
14N sin(0)
16N sin(θ2)
Since the particle is in equilibrium, the net vertical force must be zero. So we have:
N sin(θ1) + 14N sin(0) + 16N sin(θ2) = 0
Simplifying this equation, we get:
N sin(θ1) + 16N sin(θ2) = 0
We can now solve these two equations for N and sin(θ1):
N = -16N cos(θ2) / cos(θ1)
sin(θ1) = -16 sin(θ2) / cos(θ1)
Now, we can use the Pythagorean theorem to find the magnitude of the 10N force:
sqrt((10N)^2 + (N sin(θ1))^2) = sqrt((10N)^2 + (16 sin(θ2))^2)
Finally, we can use the arctangent function to find the angle between the 10N and 16N forces:
tan(θ10-16) = (N sin(θ1)) / (10N) = (-16 sin(θ2)) / (10N cos(θ2))
θ10-16 = arctan((-16 sin(θ2)) / (10N cos(θ2)))
Therefore, we can find the angle between the forces 10N and 16N by using the above formula with the values we have calculated.
Question 55 Report
Forces(5N, 030\(^o\)), (PN, 060\(^o\)), (QN, 150\(^o\)), (3N, 180\(^o\)) and (5N, 270\(^o\)) act on a body . If the system is in quilibrium, find, correct to one decimal place, the values of P and Q
We can solve this problem by applying the conditions for equilibrium:
Step 1: Resolve forces into components.
Step 2: Apply equilibrium conditions.
ΣFx = 0:
4.3 + P/2 - QN/2 - 3 = 0
ΣFy = 0:
2.5 + P√3/2 - QN√3/2 - 5 = 0
Step 3: Solve the system of equations.
This is a system of two equations with two unknowns (P and Q). You can solve it using various methods like elimination or substitution. Here, we'll use elimination:
Multiply the top equation by -2√3: -8.6√3 - P√3 + Q√3 + 6√3 = 0
Add this equation to the bottom equation:
-6.1√3 + Q√3 = 0
Divide both sides by √3:
Q ≈ 6.1
Substitute this value of Q back into the top equation (we can choose either equation):
4.3 + P/2 - (6.1)/2 - 3 = 0
Solve for P:
P ≈ 8.8
Step 4: Round the answers to one decimal place.
Therefore, the values of P and Q are:
Answer Details
We can solve this problem by applying the conditions for equilibrium:
Step 1: Resolve forces into components.
Step 2: Apply equilibrium conditions.
ΣFx = 0:
4.3 + P/2 - QN/2 - 3 = 0
ΣFy = 0:
2.5 + P√3/2 - QN√3/2 - 5 = 0
Step 3: Solve the system of equations.
This is a system of two equations with two unknowns (P and Q). You can solve it using various methods like elimination or substitution. Here, we'll use elimination:
Multiply the top equation by -2√3: -8.6√3 - P√3 + Q√3 + 6√3 = 0
Add this equation to the bottom equation:
-6.1√3 + Q√3 = 0
Divide both sides by √3:
Q ≈ 6.1
Substitute this value of Q back into the top equation (we can choose either equation):
4.3 + P/2 - (6.1)/2 - 3 = 0
Solve for P:
P ≈ 8.8
Step 4: Round the answers to one decimal place.
Therefore, the values of P and Q are:
Question 56 Report
How many terms of the series -3 -1 + 1 +..... add up to 165?
The series can be rewritten as follows:
-3 + (-1 + 1) + (-3 + 1) + (-1 + 1) + ...
Simplifying, we get:
-3 + 0 + -2 + 0 + -3 + 0 + -2 + 0 + ...
The terms in the series alternate between -3 and 0. Therefore, we can group the terms as follows:
(-3 + 0) + (-2 + 0) + (-3 + 0) + (-2 + 0) + ...
Each pair of terms adds up to -3, and there are n/2 pairs of terms in the series, where n is the number of terms. Therefore, we can write the following equation:
-3(n/2) = 165
Simplifying, we get:
n = -2(165)/3
n = -110
Since the number of terms must be a positive integer, the answer is none of the given options. This means that there is no solution to the problem, as the series cannot add up to 165.
Answer Details
The series can be rewritten as follows:
-3 + (-1 + 1) + (-3 + 1) + (-1 + 1) + ...
Simplifying, we get:
-3 + 0 + -2 + 0 + -3 + 0 + -2 + 0 + ...
The terms in the series alternate between -3 and 0. Therefore, we can group the terms as follows:
(-3 + 0) + (-2 + 0) + (-3 + 0) + (-2 + 0) + ...
Each pair of terms adds up to -3, and there are n/2 pairs of terms in the series, where n is the number of terms. Therefore, we can write the following equation:
-3(n/2) = 165
Simplifying, we get:
n = -2(165)/3
n = -110
Since the number of terms must be a positive integer, the answer is none of the given options. This means that there is no solution to the problem, as the series cannot add up to 165.
Question 57 Report
(a) In a bakery, 30% of loaves of bread produced are of bad quality. If twelve loaves are selected at random from the bakery, calculate, correct to four decimal places. the probabshty of getting
(i) exactly 6 bad ones:
(ii) at least 4 bad ones;
(ii) no bad one.
(b) A group consists of 8 boys and 5 girls. A committee of 7 members is chosen from the group. Find the probability that the committee is made up of 4 boys and 3 girls.
(a) In this problem, we are dealing with a binomial distribution since we are selecting loaves of bread with replacement and there are only two possible outcomes: bad or good.
(b) In this problem, we are dealing with a hypergeometric distribution since we are selecting a committee without replacement from a group with a fixed number of boys and girls.
The number of ways to choose 4 boys out of 8 is (8 choose 4), and the number of ways to choose 3 girls out of 5 is (5 choose 3). The total number of ways to choose 7 members out of 13 is (13 choose 7).
Therefore, the probability of choosing a committee with 4 boys and 3 girls is:
(8 choose 4) * (5 choose 3) / (13 choose 7) ≈ 0.2593
So the probability of choosing a committee with 4 boys and 3 girls is approximately 0.2593.
Answer Details
(a) In this problem, we are dealing with a binomial distribution since we are selecting loaves of bread with replacement and there are only two possible outcomes: bad or good.
(b) In this problem, we are dealing with a hypergeometric distribution since we are selecting a committee without replacement from a group with a fixed number of boys and girls.
The number of ways to choose 4 boys out of 8 is (8 choose 4), and the number of ways to choose 3 girls out of 5 is (5 choose 3). The total number of ways to choose 7 members out of 13 is (13 choose 7).
Therefore, the probability of choosing a committee with 4 boys and 3 girls is:
(8 choose 4) * (5 choose 3) / (13 choose 7) ≈ 0.2593
So the probability of choosing a committee with 4 boys and 3 girls is approximately 0.2593.
Question 58 Report
(a). \(\frac{T}{\sin 90^o}\) = \(\frac{120}{sin 135^o}\) and found T = 169.71N
(b) \(\frac{R}{\sin 135^o}\) = \(\frac{120}{\sin 135^o}\)
R = 120N
The given problem involves two parts: finding the tension (T) in a rope and the force (R) acting on an object at a certain angle.
In part (a), we are given the weight of an object as 120 N and the angle at which the rope is being pulled as 135 degrees. We use the formula: force = weight / sin(angle) to calculate the tension in the rope. We substitute the given values and calculate T as 169.71 N.
In part (b), we are given the weight of the same object as 120 N and the angle at which the force is acting as 135 degrees. We use the same formula: force = weight / sin(angle) to calculate the force (R) acting on the object. We substitute the given values and calculate R as 120 N.
Therefore, we can conclude that in part (a), we found the tension in the rope when the object was being pulled at an angle of 135 degrees, and in part (b), we found the force acting on the object when the force was being applied at an angle of 135 degrees.
Answer Details
The given problem involves two parts: finding the tension (T) in a rope and the force (R) acting on an object at a certain angle.
In part (a), we are given the weight of an object as 120 N and the angle at which the rope is being pulled as 135 degrees. We use the formula: force = weight / sin(angle) to calculate the tension in the rope. We substitute the given values and calculate T as 169.71 N.
In part (b), we are given the weight of the same object as 120 N and the angle at which the force is acting as 135 degrees. We use the same formula: force = weight / sin(angle) to calculate the force (R) acting on the object. We substitute the given values and calculate R as 120 N.
Therefore, we can conclude that in part (a), we found the tension in the rope when the object was being pulled at an angle of 135 degrees, and in part (b), we found the force acting on the object when the force was being applied at an angle of 135 degrees.
Question 59 Report
Differentiate from first principles, with respect to x, (3x\(^2\) + 2x - 1)
Differentiation is a mathematical process that allows us to find the rate of change of a function at a certain point. To differentiate the function (3x^{2} + 2x - 1) with respect to x from first principles, we'll use the power rule of differentiation.
The power rule states that if we have a term in the form of x^{n}, then its derivative is given by nx^{(n-1)}.
Applying this rule to the function (3x^{2} + 2x - 1), we get:
d/dx (3x^{2}) = 6x
d/dx (2x) = 2
The derivative of a constant is 0, so:
d/dx (-1) = 0
Putting it all together, we find that the derivative of (3x^{2} + 2x - 1) with respect to x is:
d/dx (3x^{2} + 2x - 1) = 6x + 2
So, the rate of change of the function (3x^{2} + 2x - 1) at any point x is given by 6x + 2.
Answer Details
Differentiation is a mathematical process that allows us to find the rate of change of a function at a certain point. To differentiate the function (3x^{2} + 2x - 1) with respect to x from first principles, we'll use the power rule of differentiation.
The power rule states that if we have a term in the form of x^{n}, then its derivative is given by nx^{(n-1)}.
Applying this rule to the function (3x^{2} + 2x - 1), we get:
d/dx (3x^{2}) = 6x
d/dx (2x) = 2
The derivative of a constant is 0, so:
d/dx (-1) = 0
Putting it all together, we find that the derivative of (3x^{2} + 2x - 1) with respect to x is:
d/dx (3x^{2} + 2x - 1) = 6x + 2
So, the rate of change of the function (3x^{2} + 2x - 1) at any point x is given by 6x + 2.
Question 60 Report
Two fair dice are thrown together two times. Find the probability of obtaining a sum of seven in the first throw and a sum of four in the second throw.
+ | 1 | 2 | 3 | 4 | 5 | 6 |
1 | 2 | 3 | 4 | 5 | 6 | 7 |
2 | 3 | 4 | 5 | 6 | 7 | 8 |
3 | 4 | 5 | 6 | 7 | 8 | 9 |
4 | 5 | 6 | 7 | 8 | 9 | 10 |
5 | 6 | 7 | 8 | 9 | 10 | 11 |
6 | 7 | 8 | 9 | 10 | 11 | 12 |
P(a sum of 7)
= 636 $\frac{6}{36}$
and P(a sum of 4)
= 336 $\frac{3}{36}$
P(a sum of 4)
= 3