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**Question 2**
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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 3**
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Evaluate tan 75\(^o\); leaving the answer in surd form (radicals)

**Question 4**
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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 5**
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Find the constant term in the binomial expansion of (2x\(^2\) + \(\frac{1}{x^2}\))\(^4\)

**Question 6**
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Find the coefficient of the term in the binomial expansion of [2x + \(\frac{3y}{4}\)]\(^3\) in descending powers of x.

**Question 8**
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Solve; \(\frac{P}{2} + \frac{k}{3}\) = 5 and 2p = k = 6 simultaneously

**Question 9**
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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 10**
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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 11**
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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

**Question 12**
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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.

**Question 13**
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If the mean of 2, 5, (x + 1), (x + 2), 7 and 9 is 6, find the median.

**Question 14**
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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.

**Question 15**
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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 17**
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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 18**
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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?

**Question 19**
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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"?

**Question 20**
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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 21**
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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 22**
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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 24**
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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 25**
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Which of the following vectors is perpendicular to \(\begin{pmatrix} -1 & 3 \end{pmatrix}\)?

**Question 26**
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**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 27**
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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.

**Question 28**
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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 29**
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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 31**
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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 32**
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Find the sum of the first 20 terms of the sequence -7-3, 1, ......

**Question 33**
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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 34**
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Solve, correct to three significant figures, (0.3)\(^x\) = (0,5)\(^8\)

**Question 35**
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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 36**
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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

**Question 37**
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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.

**Question 38**
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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.

**Question 39**
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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

**Question 40**
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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 41**
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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

None

**Answer Details**

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**Question 42**
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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.

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**Question 43**
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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

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**Question 44**
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Using determinants, solve the following equations simultaneously.

5x — 6y + 4z = 15

7x + 4y — 3z = 19

2x + y + 6z = 46

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**Question 45**
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Differentiate from first principles, with respect to x, (3x\(^2\) + 2x - 1)

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**Question 46**
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The table shows the age distribution in years of a group of people

Age(in years) | 1 - 5 | 6 - 10 | 11 - 15 | 16 - 20 | 21 - 25 | 26 - 30 |

Number of people | 18 | 12 | 25 | 15 | 20 | 10 |

Using an assume mean of 13 years, find the mean age of the people.

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**Answer Details**

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**Question 47**
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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

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**Question 48**
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(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

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**Question 49**
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(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

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**Question 50**
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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.

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**Question 51**
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Find the angle between \(\over {OP}\) = (\(^{-3}_{-4}\)) and \(\over{OQ}\) = (\(^8_{-15}\))

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**Question 52**
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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

None

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**Question 53**
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(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

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**Question 54**
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Three soldies, X, Y and Z have probabilities \(\frac{1}{3}, \frac{1}{5}\) and \(\frac{1}{4}\) respectively of hitting a target. If each of them fires once, find, correct to two decimal places, the probability that only one of them hits the target

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**Question 55**
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Three forces N, 14N and 16N acting on a particle keep it in equilibrium. Find the angle between the forces 10N and 16N.

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**Question 56**
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(a) P(-1, 4), Q(2, 3), R(x, y) and S(-2, 3) are the verticles of a parallelogram. Find the value of x and y.

(b) A particle starts from rest and moves in a straight line. It attains a velocity of 20ms\(^{-1}\) after travelling a distance of 8 metres. Calculate;

(ii) Iis acceleration

(ii) the time taken to travel 40 metres

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**Answer Details**

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**Question 57**
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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

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**Question 58**
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If sin x \(\frac{P - Q}{P + Q}\), where 0\(^o\) \(\leq\) x \(\leq\) 90\(^o\), find 1 - tan\(^2\)x

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**Question 59**
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In an examination, 60% of the candidates passed. If 10 candidates are selected at random, find the probability that;

(1) at least two of the, failed

(2) exactly half of them passed

(3) at most two of them failed

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**Question 60**
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How many terms of the series -3 -1 + 1 +..... add up to 165?

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**Question 61**
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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?

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**Question 62**
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In the diagram, a mass of 12kg hanging from a light inextensible string is pulled aside by a horizontal force, R, such that the string is inclined at 45\(^o\) to the vertical. If the system is in equilibrium, calculate the;

(a) tension in the string;

(b) value of R

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**Question 63**
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Given that (\(_r^n\)) = \(^nC_r\), simplify (\(^{2x + 1}_{3}\)) - (\(^{2x - 1}_3\)) - 2(\(^x_2\))

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**Question 64**
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The distribution of the masses of a group of persons is shown in the following table

Mass/kg | 10.5 - 14.4 | 14.5 - 24.4 | 24.5 - 44.4 | 44.5 - 47.4 | 47.5 - 49.4 |

Number of Persons | 2 | 6 | 18 | 2 | 1 |

Draw a histogram for the distribution

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**Question 65**
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Find the equation of the circle centre (2. 3) which passes through the y - intercept of the line 3x - 2y + 6 = 0

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**Question 66**
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Simplify \(\frac{ 625(\frac{3x}{4} - 1) + 125^{(x - 1)} }{5^{(3x - 2)}}\)

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**Question 67**
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(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.

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**Question 68**
**Report**

Given that M : (x, y) \(\to\) (7x, 3x - y) and N : (x, y) \(\to\) (2x - y; 5x + 3y)

(a) write down matrices M and N of the linear transformation

(b) find the image of P(2, -3) under the linear transformation N followed by M;

(c) find the coordinates of the point Q whose image is Q(2, 4) under the linear transformation N

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**Answer Details**

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**Question 69**
**Report**

(a) Find the coordinates of the point which divides the line joining (7, -5) and (-2, 7) externally in the ration 3 : 2.

(b) Without using calculators or mathematical tables, evaluate \(\frac{2}{1 + \sqrt{2}}\) - \(\frac{2}{2 + \sqrt{2}}\), leaving the answer in the form p + q\(\sqrt{n}\), where p, q and n are integers.

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**Answer Details**

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**Question 70**
**Report**

(a) Find the range of value of p for which 4x\(^2\) - px + 1 = 0

(b)(i) Expand (1 + 3x)\(^6\) in ascending powers of x

(ii) Using the expression in 10

(ii) find, correct to four significant figures, the value of (1.03)\(^6\)

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**Answer Details**

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