WAEC SSCE - Further Mathematics - 2019

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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\(^{-1}\).

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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).

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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.

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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\(^2\).

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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) \(\frac{3}{(5 - x)^4}\).

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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.

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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, -\(\frac{3}{2}\))

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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.

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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.

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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.

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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.

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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 \(\frac{6}{5}\).

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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).

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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 -\(\frac{4}{5}\). ()

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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\(^o\).

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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.

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Histogram of Mass Distribution

Since the class intervals are unequal, the vertical axis is frequency density:

\[\text{Frequency density}=\frac{\text{frequency}}{\text{class width}}\]

Mass (kg)	Class boundaries	Class width	Frequency	Frequency density
10.5–14.4	10.45–14.45	4	2	0.50
14.5–24.4	14.45–24.45	10	6	0.60
24.5–44.4	24.45–44.45	20	18	0.90
44.5–47.4	44.45–47.45	3	2	0.67
47.5–49.4	47.45–49.45	2	1	0.50

The required histogram is:

For example, the area of the bar from \(24.45\) kg to \(44.45\) kg is \(20\times0.90=18\), equal to its frequency.

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Set distances from end \(W\). The beam is \(90\,\text{cm}\) long.

• \(Y\): \(|WY| = 10\,\text{cm}\), so \(Y\) is at \(10\,\text{cm}\).
• Pivot: \(35\,\text{cm}\) from \(W\).
• Weight \(50\,\text{N}\) acts at the centre (uniform beam): \(45\,\text{cm}\).
• \(Z\): \(|XZ| = 10\,\text{cm}\), so \(Z\) is at \(90 - 10 = 80\,\text{cm}\).

Taking moments about the pivot (which removes the pivot reaction). Distances from the pivot:

• \(T\) at \(Y\): \(35 - 10 = 25\,\text{cm}\) on the \(W\) side.
• Weight \(50\,\text{N}\): \(45 - 35 = 10\,\text{cm}\) on the \(X\) side.
• \(20\,\text{N}\) at \(Z\): \(80 - 35 = 45\,\text{cm}\) on the \(X\) side.

For equilibrium, the anticlockwise moment of \(T\) balances the clockwise moments of the weight and the \(20\,\text{N}\) force:

\(T = 56\,\text{N}\).

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The series \(-3,\ -1,\ 1,\ \dots\) is arithmetic with first term \(a=-3\) and common difference \(d=2\).

Set \(S_n = 165\):

Taking the positive value, \(n = \dfrac{30}{2} = 15\).

15 terms are required.

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(a) The matrix M can be obtained by applying M to the standard basis vectors (1,0) and (0,1):

M(1,0) = (7, -0) = (7,0)
M(0,1) = (0, -1) = (0,-1)

Therefore, the matrix M is:

M =
[7 0;
0 -1]

Similarly, the matrix N can be obtained as follows:

N(1,0) = (2, 5)
N(0,1) = (-1, 3)

Therefore, the matrix N is:

N =
[2 -1;
5 3]

(b) To find the image of P(2, -3) under the linear transformation N followed by M, we need to compute the product MNP, where P is the column vector (2,-3). That is,

MN(2,-3) = M(N(2,-3)) = M(2(-3) - (-1)(-3); 5(2) + 3(-3)) = M(-3, 11) = (77, -11)

Therefore, the image of P(2,-3) under the linear transformation N followed by M is the point (77,-11).

(c) To find the coordinates of the point Q whose image is Q(2,4) under the linear transformation N, we need to solve the equation N(x,y) = (2,4). That is,

2x - y = 2
5x + 3y = 4

Solving for x and y, we get x = 2 and y = -1. Therefore, the point Q is (2,-1).

Now we need to find the preimage of Q under the linear transformation M. That is, we need to solve the equation M(x,y) = (2,-1). That is,

7x = 2
-y = -1

Solving for x and y, we get x = 2/7 and y = 1. Therefore, the preimage of Q(2,4) under the linear transformation N followed by M is the point (2/7,1).

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(a) \(P(\text{bad})=0.3,\ P(\text{good})=0.7,\ n=12\), \(P(X=r)=\binom{12}{r}(0.3)^r(0.7)^{12-r}\) (\(X\)=number bad).

(i) exactly 6 bad: \(\binom{12}{6}(0.3)^6(0.7)^6=924(0.000729)(0.117649)=0.0793\).

(ii) at least 4 bad \(=1-\big[P(0)+P(1)+P(2)+P(3)\big]\):

(iii) no bad one: \(P(0)=(0.7)^{12}=0.0138\).

(b) 8 boys, 5 girls; committee of 7 with 4 boys and 3 girls:

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a) To find the range of values of p for which 4x\(^2\) - px + 1 = 0, we need to use the discriminant of the quadratic equation.

The discriminant is given by the expression b\(^2\) - 4ac, where a = 4, b = -p and c = 1.

So, the discriminant is:

b\(^2\) - 4ac = (-p)\(^2\) - 4(4)(1) = p\(^2\) - 16

For the quadratic equation to have real roots, the discriminant must be greater than or equal to zero.

So, we have:

p\(^2\) - 16 \(\geq\) 0

Solving for p, we get:

p \(\leq\) -4 or p \(\geq\) 4

Therefore, the range of values of p for which 4x\(^2\) - px + 1 = 0 has real roots is p \(\leq\) -4 or p \(\geq\) 4.

b)

i) To expand (1 + 3x)\(^6\) in ascending powers of x, we can use the binomial theorem. The general term in the expansion is given by:

C(n, r) a\(^r\) b\(^{n-r}\)

where C(n, r) is the binomial coefficient, n is the power of the binomial, a is the first term, and b is the second term.

In this case, we have:

n = 6, a = 1, b = 3x

So, the expansion is:

(1 + 3x)\(^6\) = C(6,0) 1\(^6\) (3x)\(^0\) + C(6,1) 1\(^5\) (3x)\(^1\) + C(6,2) 1\(^4\) (3x)\(^2\) + C(6,3) 1\(^3\) (3x)\(^3\) + C(6,4) 1\(^2\) (3x)\(^4\) + C(6,5) 1\(^1\) (3x)\(^5\) + C(6,6) 1\(^0\) (3x)\(^6\)

Simplifying and collecting like terms, we get:

(1 + 3x)\(^6\) = 1 + 18x + 135x\(^2\) + 540x\(^3\) + 1215x\(^4\) + 1458x\(^5\) + 729x\(^6\)

ii) To find the value of (1.03)\(^6\) correct to four significant figures, we can substitute x = 0.03 into the expression we obtained in part (i):

(1 +

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Measure each direction anticlockwise from the positive x-axis and resolve. For equilibrium the sum of the components in each direction is zero.

x-components:

y-components:

The y-equation gives \(\sqrt3\,P + Q = 5\), so \(Q = 5 - \sqrt3\,P\).

The x-equation gives \(P - \sqrt3\,Q = 6 - 5\sqrt3\). Substituting:

So \(P = 1.5\), and \(Q = 5 - 1.5\sqrt3 = 5 - 2.598 = 2.402\).

\(P \approx 1.5\,\text{N}, \quad Q \approx 2.4\,\text{N}\) (to 1 d.p.).

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This is a concurrent-forces (Lami's theorem) problem in which a load of 120 N is held in equilibrium by a tension \(T\) and a reaction \(R\), with the geometry giving angles of \(90^\circ\) and \(135^\circ\) opposite \(T\) and the load respectively.

(a) By Lami's theorem, \(\dfrac{T}{\sin 90^\circ}=\dfrac{120}{\sin 135^\circ}\), so

(b) Similarly \(\dfrac{R}{\sin 135^\circ}=\dfrac{120}{\sin 135^\circ}\), giving

Hence \(T\approx\textbf{169.71 N}\) and \(R=\textbf{120 N}\), confirming the stated working. (The full configuration/diagram was not supplied, so the angle assignments are taken from the given ratios.)

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Given \(\sin x = \dfrac{P-Q}{P+Q}\) with \(0^{o}\le x\le 90^{o}\).

Hence

\(\displaystyle 1 - \tan^2 x = \frac{6PQ - P^2 - Q^2}{4PQ}\).

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The 12 kg mass hangs from a string whose upper end is fixed at \(P\). A horizontal force \(R\) pulls the mass sideways so that the string makes \(45^{\circ}\) with the vertical. Three forces act at the junction (knot): the tension \(T\) along the string, the horizontal pull \(R\), and the weight \(W\) acting vertically downward.

Weight of the mass (taking \(g = 10\,\text{m s}^{-2}\)):

Since the system is in equilibrium, resolve the tension into vertical and horizontal components. The string is \(45^{\circ}\) from the vertical, so the vertical component of \(T\) is \(T\cos 45^{\circ}\) and the horizontal component is \(T\sin 45^{\circ}\).

(a) Tension in the string

Resolving vertically (the vertical component of the tension supports the weight):

So the tension is about \(170\,\text{N}\).

(b) Value of \(R\)

Resolving horizontally (the horizontal force balances the horizontal component of the tension):

Hence \(R = 120\,\text{N}\).

Check: Because the string is at \(45^{\circ}\), the vertical and horizontal components of \(T\) are equal, so \(R\) equals the weight, \(120\,\text{N}\), which agrees with the result above. (Using \(g = 9.8\,\text{m s}^{-2}\) gives \(W = 117.6\,\text{N}\), \(T \approx 166.3\,\text{N}\) and \(R = 117.6\,\text{N}\).)

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(a) Velocity-time graph

The velocity increases uniformly from \(20\text{ m s}^{-1}\) to \(30\text{ m s}^{-1}\) in 4 s, remains constant for 8 s, and then decreases uniformly to zero.

(b) Calculation of \(t\)

Acceleration \(=2\frac{1}{2}=\frac{5}{2}\text{ m s}^{-2}\).

Let the retardation be \(r\text{ m s}^{-2}\). Since

\[\frac{5}{2}:r=3:4,\]

\[r=\frac{4}{3}\times\frac{5}{2}=\frac{10}{3}\text{ m s}^{-2}.\]

Velocity after the first 4 s is

\[v=20+\left(\frac{5}{2}\times4\right)=30\text{ m s}^{-1}.\]

During retardation,

\[0=30-\frac{10}{3}t.\]

\[t=\frac{30}{10/3}=9\text{ s}.\]

Thus, the body comes to rest at \(4+8+9=21\) s from the start of the journey.

(c) Total distance travelled

The total distance is the area under the velocity-time graph:

\[\begin{aligned} \text{Distance}&=\frac{1}{2}(20+30)(4)+(30\times8)+\frac{1}{2}(30)(9)\\ &=100+240+135\\ &=475\text{ m}. \end{aligned}\]

Total distance travelled = \(475\text{ m}\).

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a)

Let the point dividing the line joining (7, -5) and (-2, 7) externally in the ratio 3 : 2 be (x, y).

We can use the section formula to find the coordinates of the point:

x = (2*7 + 3*(-2))/5 = 1

y = (2*(-5) + 3*7)/5 = 1

Therefore, the coordinates of the point are (1, 1).

b)

To simplify the expression \(\frac{2}{1 + \sqrt{2}}\) - \(\frac{2}{2 + \sqrt{2}}\), we need to use the conjugate of the denominator to eliminate the radicals in the denominator.

Notice that the conjugate of \(1 + \sqrt{2}\) is \(1 - \sqrt{2}\), and the conjugate of \(2 + \sqrt{2}\) is \(2 - \sqrt{2}\).

Multiplying the first fraction by \(\frac{2 - \sqrt{2}}{2 - \sqrt{2}}\) and the second fraction by \(\frac{1 - \sqrt{2}}{1 - \sqrt{2}}\), we get:

\(\frac{2(2-\sqrt{2})}{(1+\sqrt{2})(2-\sqrt{2})}-\frac{2(1-\sqrt{2})}{(2+\sqrt{2})(1-\sqrt{2})}\)

Simplifying the numerators and denominators, we get:

\(\frac{4-2\sqrt{2}}{1}-\frac{2-4\sqrt{2}}{1} = 2\sqrt{2}-2\)

Therefore, the answer is in the form p + q\(\sqrt{n}\), where p = -2, q = 2, and n = 2.

The reasoning behind this is that we have a rational number (2) added to an irrational number (\(2\sqrt{2}\)), which gives us an expression in the form p + q\(\sqrt{n}\), where p and q are rational numbers and n is an integer. We can then identify p, q, and n by comparing the coefficients of the rational and irrational parts of the expression.

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Method. Assumed mean \(A = 13\), class width \(c = 5\), \(u = \dfrac{x - 13}{5}\) where \(x\) is the class midpoint.

Mean age.

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(a) Taking \(m=i-j\) and \(n=2i+3j\), with \(2m+n-r=0\) we get \(r=2m+n\).

\[r=2(i-j)+(2i+3j)=(2i-2j)+(2i+3j)=4i+j\]

\[|r|=\sqrt{4^{2}+1^{2}}=\sqrt{17}\approx4.12\]

(b) \(S=3t-\dfrac{t^{3}}{3}+9\). Velocity is \(v=\dfrac{dS}{dt}=3-t^{2}\).

(i) Time when momentarily at rest: \(v=0\):

\[3-t^{2}=0\ \Rightarrow\ t^{2}=3\ \Rightarrow\ t=\sqrt{3}\ \text{s}\ (\approx1.73\ \text{s})\]

(ii) Distance at that instant:

\[S=3\sqrt{3}-\frac{(\sqrt{3})^{3}}{3}+9=3\sqrt{3}-\sqrt{3}+9=2\sqrt{3}+9\approx12.46\ \text{m}\]

Akọwa Nkọwa

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:

1. Find the y-intercept of the line 3x - 2y + 6 = 0.

   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).

2. Find the radius of the circle.

   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.

3. Write the equation of the circle.

   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.

Akọwa Nkọwa

Method (assumed mean / coding). Take assumed mean \(A = 67\) and class width \(c = 5\). For each class let \(x\) be the midpoint and \(u = \dfrac{x - A}{c}\).

(a) Mean.

(b) Standard deviation.

Akọwa Nkọwa

This is a binomial situation. Let \(X\) be the number who fail. A candidate fails with probability \(p=0.4\) and passes with probability \(0.6\), with \(n=10\).

\[P(X=r)=\binom{10}{r}(0.4)^{r}(0.6)^{10-r}\]

(1) At least two failed: \(P(X\ge 2)\)

\[P(X\ge2)=1-P(0)-P(1)\]

\(P(0)=(0.6)^{10}=0.00605\) and \(P(1)=\binom{10}{1}(0.4)(0.6)^{9}=10(0.4)(0.010078)=0.04031\).

\[P(X\ge2)=1-0.00605-0.04031=0.954\]

(2) Exactly half passed means \(5\) passed and \(5\) failed, i.e. \(X=5\).

\[P(X=5)=\binom{10}{5}(0.4)^{5}(0.6)^{5}=252(0.01024)(0.07776)=0.201\]

(3) At most two failed: \(P(X\le 2)=P(0)+P(1)+P(2)\)

\(P(2)=\binom{10}{2}(0.4)^{2}(0.6)^{8}=45(0.16)(0.016796)=0.12093\).

\[P(X\le2)=0.00605+0.04031+0.12093=0.167\]

Akọwa Nkọwa

For \(3x^2+4x-5=0\) with roots \(\alpha,\beta\): sum \(\alpha+\beta=-\dfrac{4}{3}\), product \(\alpha\beta=-\dfrac{5}{3}\).

Use \((\alpha-\beta)^2=(\alpha+\beta)^2-4\alpha\beta\):

So \(\alpha-\beta=\dfrac{2\sqrt{19}}{3}\) (taking the positive surd).

Akọwa Nkọwa

Hit probabilities: \(P(X)=\tfrac13,\ P(Y)=\tfrac15,\ P(Z)=\tfrac14\); miss probabilities: \(\tfrac23,\tfrac45,\tfrac34\). "Only one hits" means exactly one succeeds while the other two miss (events independent):

Adding (LCD 30):

Akọwa Nkọwa

(a) The line \(y + 2x - 3 = 0\) gives \(y = 3 - 2x\). At intersection with \(y = 7 - \dfrac{6}{x}\):

Multiply through by \(x\):

When \(x = 1:\ y = 3 - 2 = 1\). When \(x = -3:\ y = 3 + 6 = 9\).

The points are \((1,\ 1)\) and \((-3,\ 9)\).

(b) Midpoint \(= \left(\dfrac{1 + (-3)}{2},\ \dfrac{1 + 9}{2}\right) = (-1,\ 5)\).

Gradient of the join \(= \dfrac{9 - 1}{-3 - 1} = \dfrac{8}{-4} = -2\); the perpendicular gradient is \(\dfrac{1}{2}\).

Perpendicular bisector: \(x - 2y + 11 = 0\).

Akọwa Nkọwa

Write every base as a power of \(5\): \(625=5^{4}\), \(125=5^{3}\).

Numerator:

\[625^{\left(\frac{3x}{4}-1\right)}=5^{4\left(\frac{3x}{4}-1\right)}=5^{3x-4},\qquad 125^{(x-1)}=5^{3(x-1)}=5^{3x-3}\]

So the expression is

\[\frac{5^{3x-4}+5^{3x-3}}{5^{3x-2}}=\frac{5^{3x-4}}{5^{3x-2}}+\frac{5^{3x-3}}{5^{3x-2}}=5^{-2}+5^{-1}\]

\[=\frac{1}{25}+\frac{1}{5}=\frac{1}{25}+\frac{5}{25}=\frac{6}{25}\]

Akọwa Nkọwa

Write each term using \(\binom{n}{r}=\dfrac{n!}{r!(n-r)!}\).

Their difference:

And \(2\binom{x}{2}=2\cdot\dfrac{x(x-1)}{2}=x^2-x\). Therefore

Akọwa Nkọwa

System: \(5x-6y+4z=15,\ 7x+4y-3z=19,\ 2x+y+6z=46\). Use Cramer's rule.

So \(x=3,\ y=4,\ z=6\). Check in equation 2: \(7(3)+4(4)-3(6)=21+16-18=19\).

Akọwa Nkọwa

\(\vec{OP}=\begin{pmatrix}-3\\-4\end{pmatrix},\ \vec{OQ}=\begin{pmatrix}8\\-15\end{pmatrix}\).

Dot product: \(\vec{OP}\cdot\vec{OQ}=(-3)(8)+(-4)(-15)=-24+60=36\).

Magnitudes: \(|\vec{OP}|=\sqrt{9+16}=5\), \(|\vec{OQ}|=\sqrt{64+225}=\sqrt{289}=17\).

The angle between the vectors is about \(65^\circ\).

Akọwa Nkọwa

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

Akọwa Nkọwa

1. To calculate the Spearman's rank correlation coefficient, we need to follow these steps:

• Rank the data for both the entrance examination performance and school performance separately.
• Calculate the difference between the ranks for each student.
• Square the differences and add them up.
• Use the formula to calculate the Spearman's rank correlation coefficient:

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:

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)))</

Akọwa Nkọwa

With two fair dice there are \(36\) equally likely outcomes.

Sum of 7: \((1,6),(2,5),(3,4),(4,3),(5,2),(6,1)\) gives \(6\) outcomes, so \(P(7) = \dfrac{6}{36} = \dfrac{1}{6}\).

Sum of 4: \((1,3),(2,2),(3,1)\) gives \(3\) outcomes, so \(P(4) = \dfrac{3}{36} = \dfrac{1}{12}\).

The two throws are independent, so

Probability \(= \dfrac{1}{72}\).

Akọwa Nkọwa

Let \(f(x)=3x^2+2x-1\). By first principles \(f'(x)=\displaystyle\lim_{h\to 0}\frac{f(x+h)-f(x)}{h}\).

So \(\dfrac{d}{dx}(3x^2+2x-1)=6x+2\).

Akọwa Nkọwa

The three forces are \(10\,\text{N},\ 14\,\text{N}\) and \(16\,\text{N}\) in equilibrium, so the resultant of the \(10\,\text{N}\) and \(16\,\text{N}\) forces must be equal in magnitude to \(14\,\text{N}\) (and opposite to it).

Let \(\theta\) be the angle between the \(10\,\text{N}\) and \(16\,\text{N}\) forces. Using the resultant (parallelogram) formula:

The angle between the \(10\,\text{N}\) and \(16\,\text{N}\) forces is \(120^{o}\).

Akọwa Nkọwa

(a) For parallelogram \(PQRS\) the diagonals bisect each other, so midpoint of \(PR\) = midpoint of \(QS\).

So \(R(1,2)\), i.e. \(x=1,\ y=2\). (Check: \(\vec{PQ}=(3,-1)=\vec{SR}\).)

(b) From rest, \(u=0\), reaches \(v=20\ \text{m/s}\) after \(s=8\ \text{m}\).

(i) Acceleration: \(v^2=u^2+2as\Rightarrow 20^2=2a(8)\Rightarrow a=\dfrac{400}{16}=25\ \text{m/s}^2\).

(ii) Time to travel 40 m: \(s=ut+\tfrac12 at^2\Rightarrow 40=\tfrac12(25)t^2=12.5t^2\).