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Ajụjụ 1 Ripọtì
The coefficient of the 7th term in the binomial expansion of \((2 - \frac{x}{3})^{10}\) in ascending powers of x is
Akọwa Nkọwa
Ajụjụ 2 Ripọtì
Find the least value of n for which \(^{3n}C_{2} > 0, n \in R\).
Ajụjụ 3 Ripọtì
Two forces (2i - 5j)N and (-3i + 4j)N act on a body of mass 5kg. Find in \(ms^{-2}\), the magnitude of the acceleration of the body.
Akọwa Nkọwa
To find the magnitude of acceleration, we need to use Newton's Second Law, which states that the force acting on an object is equal to its mass multiplied by its acceleration. In this case, we have two forces: (2i - 5j)N and (-3i + 4j)N. To find the net force, we add these two vectors by adding their corresponding components: (2i - 5j)N + (-3i + 4j)N = -i - j N So the net force is (-i - j)N. Now, using Newton's Second Law, we can find the acceleration: F = ma where F is the net force, m is the mass, and a is the acceleration. In our case, the net force is (-i - j)N and the mass is 5kg, so we have: (-i - j)N = 5kg * a Solving for a, we get: a = (-i - j)N / 5kg To find the magnitude of a, we take the square root of the sum of the squares of its components: |a| = sqrt((-1)^2 + (-1)^2) N/kg = sqrt(2) N/kg Converting to \(ms^{-2}\) by dividing by 5, we get: |a| = sqrt(2)/5 \(ms^{-2}\) Therefore, the answer is \(\frac{\sqrt{2}}{5}\).
Ajụjụ 4 Ripọtì
If \(y = 2(2x + \sqrt{x})^{2}\), find \(\frac{\mathrm d y}{\mathrm d x}\).
Akọwa Nkọwa
To differentiate the given function with respect to x, we will use the chain rule and the power rule of differentiation. Let's start by using the power rule of differentiation to find the derivative of \((2x + \sqrt{x})^{2}\) with respect to x: \[\frac{\mathrm d}{\mathrm d x} (2x + \sqrt{x})^{2} = 2(2x + \sqrt{x}) \cdot \frac{\mathrm d}{\mathrm d x} (2x + \sqrt{x})\] Next, we use the chain rule to differentiate \(\frac{\mathrm d}{\mathrm d x} (2x + \sqrt{x})\) with respect to x: \[\frac{\mathrm d}{\mathrm d x} (2x + \sqrt{x}) = 2 + \frac{1}{2\sqrt{x}}\] Substituting this into the previous equation, we get: \[\frac{\mathrm d}{\mathrm d x} (2x + \sqrt{x})^{2} = 2(2x + \sqrt{x}) \cdot \left(2 + \frac{1}{2\sqrt{x}}\right)\] Expanding this expression, we get: \[\frac{\mathrm d}{\mathrm d x} (2x + \sqrt{x})^{2} = 4(2x + \sqrt{x}) + 2\sqrt{x}(2x + \sqrt{x})\] Simplifying further, we get: \[\frac{\mathrm d}{\mathrm d x} (2x + \sqrt{x})^{2} = 8x + 6\sqrt{x}\] Finally, using the constant multiple rule of differentiation, we can find the derivative of the given function: \[\frac{\mathrm d y}{\mathrm d x} = 2 \cdot \frac{\mathrm d}{\mathrm d x} (2x + \sqrt{x})^{2} = 2(8x + 6\sqrt{x}) = 16x + 12\sqrt{x}\] Therefore, the correct answer is: \(4(2x + \sqrt{x})(2 + \frac{1}{2\sqrt{x}})\).
Ajụjụ 5 Ripọtì
The roots of a quadratic equation are \((3 - \sqrt{3})\) and \((3 + \sqrt{3})\). Find its equation.
Akọwa Nkọwa
The roots of a quadratic equation of the form \(ax^{2} + bx + c = 0\) are given by the formula: $$x = \frac{-b \pm \sqrt{b^{2} - 4ac}}{2a}$$ If the roots are \((3 - \sqrt{3})\) and \((3 + \sqrt{3})\), then we have: $$x = 3 - \sqrt{3} \quad \text{or} \quad x = 3 + \sqrt{3}$$ Therefore, we can write the equation as: $$(x - (3 - \sqrt{3}))(x - (3 + \sqrt{3})) = 0$$ Expanding the brackets, we get: $$x^{2} - (3 - \sqrt{3} + 3 + \sqrt{3})x + (3 - \sqrt{3})(3 + \sqrt{3}) = 0$$ Simplifying, we get: $$x^{2} - 6x + 6 = 0$$ Hence, the equation of the quadratic with roots \((3 - \sqrt{3})\) and \((3 + \sqrt{3})\) is \(x^{2} - 6x + 6 = 0\). Therefore, the answer is.
Ajụjụ 6 Ripọtì
Find the value of \(\cos(60° + 45°)\) leaving your answer in surd form.
Akọwa Nkọwa
Using the sum to product formula, we have: \begin{align*} \cos(60° + 45°) &= \cos 60° \cos 45° - \sin 60° \sin 45° \\ &= \frac{1}{2} \cdot \frac{\sqrt{2}}{2} - \frac{\sqrt{3}}{2} \cdot \frac{\sqrt{2}}{2} \\ &= \frac{\sqrt{2} - \sqrt{6}}{4} \end{align*} Therefore, the correct answer is option C: \(\frac{\sqrt{2} - \sqrt{6}}{4}\).
Ajụjụ 7 Ripọtì
A bicycle wheel of diameter 70 cm covered a distance of 350 cm in 2 seconds. How many radians per second did it turn?
Akọwa Nkọwa
To find the radians per second that the bicycle wheel turns, we need to find the angle in radians that the wheel turns in 1 second. The distance covered by the bicycle wheel in 1 second can be found by dividing the distance covered in 2 seconds by 2: \[\frac{350 \, \text{cm}}{2 \, \text{s}} = 175 \, \text{cm/s}\] To convert this to radians per second, we need to know the circumference of the wheel in centimeters. The circumference of the wheel is given by: \[\pi \times \text{diameter} = \pi \times 70 \, \text{cm} = 220 \pi \, \text{cm}\] So, in one revolution, the wheel covers a distance of 220π cm. The angle in radians that the wheel turns in 1 second can be found by dividing the distance covered in 1 second by the circumference of the wheel: \[\frac{175 \, \text{cm/s}}{220 \pi \, \text{cm/rev}} = \frac{5}{2\pi} \, \text{rev/s}\] To convert revolutions per second to radians per second, we multiply by 2π: \[\frac{5}{2\pi} \times 2\pi = 5 \, \text{rad/s}\] Therefore, the bicycle wheel turns at a rate of 5 radians per second. So the answer is (A) 5.
Ajụjụ 8 Ripọtì
Two forces \(F_{1} = (7i + 8j)N\) and \(F_{2} = (3i + 4j)N\) act on a particle. Find the magnitude and direction of \(F_{1} - F_{2}\).
Akọwa Nkọwa
To find the magnitude and direction of \(F_{1} - F_{2}\), we need to subtract the components of \(F_{2}\) from \(F_{1}\). \begin{align*} F_{1} - F_{2} &= (7i + 8j) - (3i + 4j)\\ &= 4i + 4j\\ \end{align*} The magnitude of \(F_{1} - F_{2}\) is given by: \begin{align*} |F_{1} - F_{2}| &= \sqrt{(4)^2 + (4)^2}\\ &= 4\sqrt{2} N\\ \end{align*} The direction of \(F_{1} - F_{2}\) is given by: \begin{align*} \theta &= \tan^{-1}\left(\frac{4j}{4i}\right)\\ &= \tan^{-1}(1)\\ &= 45°\\ \end{align*} Therefore, the magnitude and direction of \(F_{1} - F_{2}\) are \((4\sqrt{2} N, 045°)\). Answer is correct.
Ajụjụ 9 Ripọtì
The distance s metres of a particle from a fixed point at time t seconds is given by \(s = 7 + pt^{3} + t^{2}\), where p is a constant. If the acceleration at t = 3 secs is \(8 ms^{-2}\), find the value of p.
Akọwa Nkọwa
Given, distance of a particle from a fixed point at time t seconds is given by, $$s = 7 + pt^{3} + t^{2}$$ We can find the acceleration of the particle by differentiating the distance equation twice with respect to time. $$\frac{d}{dt} s = \frac{d}{dt} (7 + pt^{3} + t^{2})$$ $$\Rightarrow \frac{ds}{dt} = 3pt^2 + 2t$$ $$\frac{d^{2}}{dt^{2}} s = \frac{d}{dt} (3pt^2 + 2t)$$ $$\Rightarrow \frac{d^{2}s}{dt^{2}} = 6pt + 2$$ We are given that the acceleration at t = 3 sec is \(8 ms^{-2}\), so we can substitute the values in the second derivative of the distance equation to get, $$6p(3) + 2 = 8$$ $$\Rightarrow 18p = 6$$ $$\Rightarrow p = \frac{6}{18} = \frac{1}{3}$$ Hence, the value of p is \(\frac{1}{3}\). Therefore, is correct.
Ajụjụ 10 Ripọtì
Given that \(P = {x : \text{x is a factor of 6}}\) is the domain of \(g(x) = x^{2} + 3x - 5\), find the range of x.
Ajụjụ 11 Ripọtì
Simplify \(\frac{\log_{5} 8}{\log_{5} \sqrt{8}}\).
Akọwa Nkọwa
We can use the properties of logarithms to simplify this expression. Recall that \(\log_{a} b^{c} = c\log_{a} b\) and \(\log_{a} \sqrt{b} = \frac{1}{2}\log_{a} b\). So, \[\frac{\log_{5} 8}{\log_{5} \sqrt{8}} = \frac{\log_{5} 8}{\frac{1}{2}\log_{5} 8} = \frac{2\log_{5} 8}{\log_{5} 8} = \boxed{2}.\] Therefore, the answer is 2.
Ajụjụ 13 Ripọtì
Two particles are fired together along a smooth horizontal surface with velocities 4 m/s and 5 m/s. If they move at 60° to each other, find the distance between them in 2 seconds.
Akọwa Nkọwa
Ajụjụ 14 Ripọtì
If \(16^{3x} = \frac{1}{4}(32^{x - 1})\), find the value of x.
Ajụjụ 15 Ripọtì
Find the direction cosines of the vector \(4i - 3j\).
Akọwa Nkọwa
Direction cosines are the cosines of the angles that a vector makes with the positive x, y and z axes. Let's denote the vector by \(\mathbf{v} = 4\mathbf{i} - 3\mathbf{j} + 0\mathbf{k}\). Then the direction cosines of \(\mathbf{v}\) are: \begin{align*} \cos\alpha &= \frac{4}{5} \\ \cos\beta &= \frac{-3}{5} \\ \cos\gamma &= 0 \end{align*} Therefore, the correct option is: - \(\frac{4}{5}, -\frac{3}{5}\)
Ajụjụ 16 Ripọtì
The third of geometric progression (G.P) is 10 and the sixth term is 80. Find the common ratio.
Akọwa Nkọwa
We can use the formula for the nth term of a geometric progression to solve this problem. Let a be the first term and r be the common ratio, then the third term is ar^2 and the sixth term is ar^5. We are given that ar^2 = 10 and ar^5 = 80. Dividing the second equation by the first, we get: (ar^5)/(ar^2) = 80/10 Simplifying and canceling a, we get: r^3 = 8 Taking the cube root of both sides, we get: r = 2 Therefore, the common ratio is 2, and the answer is (a) 2.
Ajụjụ 17 Ripọtì
For what values of x is \(\frac{x^{2} - 9x + 18}{x^{2} + 2x - 35}\) undefined?
Akọwa Nkọwa
The expression is undefined when the denominator is equal to zero because division by zero is undefined. Therefore, we can find the values of x that make the denominator zero by setting it equal to zero and solving for x: $$x^{2} + 2x - 35 = 0$$ We can factor this quadratic equation as: $$(x + 7)(x - 5) = 0$$ So, the denominator is equal to zero when x = -7 or x = 5. Therefore, the expression is undefined for these values of x. So the correct answer is: - -7 or 5
Ajụjụ 18 Ripọtì
If \(h(x) = x^{3} - \frac{1}{x^{3}}\), evaluate \(h(a) - h(\frac{1}{a})\).
Akọwa Nkọwa
We can start by computing the values of \(h(a)\) and \(h(\frac{1}{a})\), and then find their difference. First, we have \(h(a) = a^{3} - \frac{1}{a^{3}}\). Next, we have \(h(\frac{1}{a}) = (\frac{1}{a})^{3} - \frac{1}{(\frac{1}{a})^{3}} = \frac{1}{a^{3}} - a^{3}\). Therefore, \begin{align*} h(a) - h(\frac{1}{a}) &= \left(a^{3} - \frac{1}{a^{3}}\right) - \left(\frac{1}{a^{3}} - a^{3}\right) \\ &= a^{3} - \frac{1}{a^{3}} - \frac{1}{a^{3}} + a^{3} \\ &= 2a^{3} - \frac{2}{a^{3}}. \end{align*} So the answer is \(2a^{3} - \frac{2}{a^{3}}\). Therefore, is the correct answer.
Ajụjụ 19 Ripọtì
The probabilities that a husband and wife will be alive in 15 years time are m and n respectively. Find the probability that only one of them will be alive at that time.
Akọwa Nkọwa
The probability that only one of the husband and wife will be alive in 15 years can happen in two ways: either the husband is alive and the wife is not, or the wife is alive and the husband is not. The probability of the first case is (m)(1-n) = m - mn, and the probability of the second case is (n)(1-m) = n - mn. Therefore, the total probability that only one of them will be alive is (m - mn) + (n - mn) = m + n - 2mn. Hence, the correct answer is: m + n - 2mn.
Ajụjụ 20 Ripọtì
Calculate, correct to one decimal place, the length of the line joining points X(3, 5) and Y(5, 1).
Akọwa Nkọwa
Ajụjụ 21 Ripọtì
If \(\overrightarrow{OA} = 3i + 4j\) and \(\overrightarrow{OB} = 5i - 6j \) where O is the origin and M is the midpoint of AB, find OM.
Akọwa Nkọwa
We can find the midpoint M of AB by using the midpoint formula: \[\overrightarrow{OM} = \frac{\overrightarrow{OA} + \overrightarrow{OB}}{2}\] Substituting the values we have: \[\overrightarrow{OM} = \frac{(3i + 4j) + (5i - 6j)}{2} = \frac{8i - 2j}{2} = 4i - j\] Therefore, the answer is 4i - j.
Ajụjụ 22 Ripọtì
Find the domain of \(f(x) = \frac{x}{3 - x}, x \in R\), the set of real numbers.
Akọwa Nkọwa
The function \(f(x) = \frac{x}{3-x}\) is defined for all real numbers except when the denominator is zero, i.e., when \(3-x=0\). Therefore, the domain of the function is all real numbers except \(x=3\), or in interval notation: \({x : x \in R, x \neq 3}\). So the correct option is: \({x : x \in R, x \neq 3}\).
Ajụjụ 24 Ripọtì
In a class of 50 pupils, 35 like Science and 30 like History. What is the probability of selecting a pupil who likes both Science and History?
Akọwa Nkọwa
We can use the formula: P(A and B) = P(A) + P(B) - P(A or B) where A is the event of liking Science, B is the event of liking History, P(A and B) is the probability of selecting a pupil who likes both Science and History, and P(A or B) is the probability of selecting a pupil who likes either Science or History or both. We know that there are 50 pupils in the class, 35 like Science and 30 like History. This means that the maximum number of pupils who could like both Science and History is 30 (since 30 already like History). Therefore, we have: P(A and B) <= 30/50 = 0.6 Also, we know that P(A or B) = P(A) + P(B) - P(A and B) (as shown above). Therefore: P(A and B) = P(A) + P(B) - P(A or B) = 35/50 + 30/50 - P(A or B) = 65/50 - P(A or B) We also know that P(A or B) <= 1, since it is the probability of selecting a pupil who likes either Science or History or both. Therefore: P(A and B) >= 65/50 - 1 = 15/50 = 0.3 Thus, the probability of selecting a pupil who likes both Science and History is between 0.3 and 0.6. The only option that falls within this range is 0.30, so the answer is: - 0.30
Ajụjụ 25 Ripọtì
P, Q, R, S are points in a plane such that PQ = 8i - 5j, QR = 5i + 7j, RS = 7i + 3j and PS = xi + yj. Find (x, y).
Akọwa Nkọwa
Given that PQ = 8i - 5j, QR = 5i + 7j, RS = 7i + 3j, and PS = xi + yj. Since PQ + QR + RS = PS, we have: (8i - 5j) + (5i + 7j) + (7i + 3j) = xi + yj Simplifying the left-hand side, we get: 20i + 5j = xi + yj Equating the corresponding components, we have: 20 = x and 5 = y Therefore, the coordinates of point S are (20, 5). Hence, the correct answer is (20, 5).
Ajụjụ 26 Ripọtì
Calculate, correct to one decimal place, the acute angle between the lines 3x - 4y + 5 = 0 and 2x + 3y - 1 = 0.
Akọwa Nkọwa
Ajụjụ 28 Ripọtì
A stone is thrown vertically upwards and its height at any time t seconds is \(h = 45t - 9t^{2}\). Find the maximum height reached.
Akọwa Nkọwa
The height of the stone at any time t seconds is given by \(h = 45t - 9t^{2}\). To find the maximum height reached, we need to find the vertex of the parabolic function \(h\), since the vertex represents the highest point of the parabola. The vertex of the parabola is given by the formula \(\frac{-b}{2a}\), where \(a\) and \(b\) are the coefficients of the quadratic function. In this case, \(a = -9\) and \(b = 45\). Substituting these values into the formula for the vertex, we get: \[\frac{-b}{2a} = \frac{-45}{2(-9)} = \frac{45}{18} = 2.5.\] Therefore, the stone reaches its maximum height after 2.5 seconds. To find the maximum height, we substitute \(t = 2.5\) into the equation for \(h\): \[h = 45(2.5) - 9(2.5)^2 = 56.25\text{ m}.\] Therefore, the maximum height reached by the stone is 56.25 meters. The answer is (D) 56.25 m.
Ajụjụ 30 Ripọtì
Yomi was asked to label four seats S, R, P, Q. What is the probability he labelled them in alphabetical order?
Akọwa Nkọwa
There are 4 seats to be labeled, so Yomi can label the first seat with any of the 4 letters. After labeling the first seat, there are 3 letters left to choose from for the second seat, 2 letters for the third seat, and only 1 letter left for the last seat. Thus, there are a total of 4 x 3 x 2 x 1 = 24 possible ways to label the seats. Out of these 24 possible ways, only one arrangement is in alphabetical order (i.e., S, P, Q, R). Therefore, the probability that Yomi labeled the seats in alphabetical order is 1/24. Therefore, the correct option is (a) \(\frac{1}{24}\).
Ajụjụ 32 Ripọtì
If \(\frac{5}{\sqrt{2}} - \frac{\sqrt{8}}{8} = m\sqrt{2}\), where m is a constant. Find m.
Akọwa Nkọwa
Ajụjụ 33 Ripọtì
The mean age of 15 pupils in a class is 14.2 years. One new pupil joined the class and the mean changed to 14.1 years. Calculate the age of the new pupil.
Akọwa Nkọwa
Let's use the formula for the mean (average) of a set of numbers: \[\text{mean} = \frac{\text{sum of all numbers}}{\text{number of numbers}}\] Let the age of the new pupil be x. Before the new pupil joined the class, the sum of ages of the 15 pupils would be: \[15 \times 14.2 = 213\] After the new pupil joined the class, the sum of ages of the 16 pupils would be: \[15 \times 14.2 + x = 213 + x\] We know that the new mean is 14.1, so we can set up an equation: \[\frac{15 \times 14.2 + x}{16} = 14.1\] Multiplying both sides by 16, we get: \[15 \times 14.2 + x = 16 \times 14.1\] Simplifying the left side, we get: \[213 + x = 225.6\] Subtracting 213 from both sides, we get: \[x = 12.6\] Therefore, the age of the new pupil is 12.6 years. Answer: (b) 12.6 years.
Ajụjụ 34 Ripọtì
The initial velocity of an object is \(u = \begin{pmatrix} -5 \\ 3 \end{pmatrix} ms^{-1}\). If the acceleration of the object is \(a = \begin{pmatrix} 3 \\ -4 \end{pmatrix} ms^{-2}\) and it moved for 3 seconds, find the final velocity.
Akọwa Nkọwa
The final velocity of an object can be found using the formula: v = u + at where v is the final velocity, u is the initial velocity, a is the acceleration, and t is the time. Substituting the given values, we have: v = (-5, 3) + 3(3, -4) v = (-5, 3) + (9, -12) v = (4, -9) Therefore, the final velocity of the object is \(\begin{pmatrix} 4 \\ -9 \end{pmatrix} ms^{-1}\). The correct option is (C).
Ajụjụ 35 Ripọtì
If \(\begin{vmatrix} 3 & x \\ 2 & x - 2 \end{vmatrix} = -2\), find the value of x.
Ajụjụ 36 Ripọtì
Find the axis of symmetry of the curve \(y = x^{2} - 4x - 12\).
Akọwa Nkọwa
To find the axis of symmetry of a quadratic equation in the form of \(y = ax^2 + bx + c\), we need to use the formula: $$x = \frac{-b}{2a}$$ Comparing the given equation \(y = x^2 - 4x - 12\) with the standard form of the quadratic equation, we have: $$a = 1, b = -4, c = -12$$ Substituting the values in the formula, we get: $$x = \frac{-(-4)}{2(1)} = 2$$ Therefore, the axis of symmetry of the curve is the vertical line passing through the point \((2,0)\) and the equation of this line is: $$x = 2$$ Hence, the correct answer is "x = 2".
Ajụjụ 37 Ripọtì
Find the maximum value of \(2 + \sin (\theta + 25)\).
Akọwa Nkọwa
The value of \(\sin(\theta + 25)\) varies between -1 and 1. Therefore, the maximum value of \(2 + \sin(\theta + 25)\) occurs when \(\sin(\theta + 25) = 1\). This occurs when \(\theta + 25 = \frac{\pi}{2} + 2n\pi\), where \(n\) is an integer. Thus, \(\theta = \frac{\pi}{2} - 25 + 2n\pi\). Substituting this value of \(\theta\) into the expression for \(2 + \sin(\theta + 25)\), we get: \[2 + \sin\left(\frac{\pi}{2} + 2n\pi\right) = 2 + 1 = 3\] Therefore, the maximum value of \(2 + \sin(\theta + 25)\) is 3.
Ajụjụ 38 Ripọtì
If \(\sin x = -\sin 70°, 0° < x < 360°\), determine the two possible values of x.
Akọwa Nkọwa
Ajụjụ 39 Ripọtì
If (x - 3) is a factor of \(2x^{2} - 2x + p\), find the value of constant p.
Akọwa Nkọwa
If (x-3) is a factor of the quadratic expression, then the expression can be factored into the form: $$(x - 3)(ax + b)$$ where a and b are constants. Multiplying out the brackets gives: $$2x^{2} - 2x + p = (x - 3)(ax + b) = ax^{2} + (b - 3a)x - 3b$$ Since the coefficients of the quadratic expression are equal to those of the factored expression, we can equate the corresponding coefficients to get a system of equations: $$a = 2$$ $$b - 3a = -2$$ $$-3b = p$$ Solving these equations simultaneously, we get: $$a = 2, b = 3a - 2 = 4, p = -3b = -12$$ Therefore, the constant p is -12.
Ajụjụ 40 Ripọtì
A company took delivery of 12 vehicles made up of 7 buses and 5 saloon cars for two of its departments; Personnel and General Administration. If the Personnel department is to have at least 3 saloon cars, in how many ways can these vehicles be distributed equally between the departments?
Ajụjụ 41 Ripọtì
(a) Copy and complete the table for the relation: \(y = 2\cos x + 3\sin x\) for \(0° \leq x \leq 360°\).
| x | 0° | 30° | 60° | 90° | 120° | 150° | 180° | 210° |
| y | 2.00 | 3.23 | 1.60 | -3.23 |
(b) Using a scale of 2 cm to 60° on the x- axis and 2 cm to one unit on the y- axis, draw the graph of \(y = 2\cos x + 3\sin x\) for \(0° \leq x \leq 360°\).
(c) From the graph, find the : (i) maximum value of y, correct to two decimal places ; (ii) solution of the equation \(\frac{2}{3} \cos x + \sin x = \frac{5}{6}\).
Ajụjụ 42 Ripọtì
The images of (3, 2) and (-1, 4) under a linear transformation T are (-1, 4) and (7, 11) respectively. P is another transformation where \(P : (x, y) \to (x + y, x + 2y)\).
(a) Find the matrices T and P of the linear transformations T and P;
(b) Calculate TP.
(c) Find the image of the point X(4, 3) under TP.
Ajụjụ 43 Ripọtì
(a) Two ships M and N, moving with constant velocities, have position vectors (3i + 7j) and (4i + 5j) respectively. If the velocities of M and N are (5i + 6j) and (2i + 3j) and the distance covered by the ships after t seconds are in metres, find (i) MN ; (ii) |MN|, when t = 3 seconds.
(b) A particle is acted upon by forces \(F_{1} = 5i + pj ; F_{2} = qi + j ; F_{3} = -2pi + 3j\) and \(F_{4} = -4i + qj\), where p and q are constants. If the particle remains in equilibrium under the action of these forces, find the values of p and q.
Ajụjụ 44 Ripọtì
(a) Express \(\frac{2\sqrt{2}}{\sqrt{48} - \sqrt{8} - \sqrt{27}}\) in the form \(p + q\sqrt{r}\), where p, q and r are rational numbers.
(b) If \(V = A\log_{10} (M + N)\), express N in terms of M, V and A.
Akọwa Nkọwa
None
Ajụjụ 45 Ripọtì
(a) Eight coins are tossed at once. Find, correct to three decimal places, the probability of obtaining :
(i) exactly 8 heads ; (ii) at least 5 heads ; (iii) at most 1 head.
(b) In how many ways can four letters from the word SHEEP be arranged (i) without any restriction ; (ii) with only one E.
Akọwa Nkọwa
None
Ajụjụ 46 Ripọtì
The position vector of a particle of mass 3 kg moving along a space curve is given by \(r = (4t^{3} - t^{2})i - (2t^{2} - t)j\) at any time t seconds. Find the force acting on it at t = 2 seconds.
Ajụjụ 47 Ripọtì
A uniform plank PQ of length 8m and mass 10kg is supported horizontally at the end P and at point R, 3 metres from Q. A boy of mass 20 kg walks along the plank starting from P. If the plank is in equilibrium, calculate the
(a) reactions at P and R when he walked 1.5 metres;
(b) distance he had walked when the two reactions are equal;
(c) distance he walked before the plank tips over.
Ajụjụ 48 Ripọtì
If \(3x^{2} + 2y^{2} + xy + x - 7 = 0\), find \(\frac{\mathrm d y}{\mathrm d x}\) at the point (-2, 1).
Ajụjụ 50 Ripọtì
Five students are to be selected from a large population. If 60% of them are boys and the rest are girls, find the probability that :
(a) exactly 3 of them are boys;
(b) at least 3 of them are girls.
Ajụjụ 51 Ripọtì
The table shows the marks obtained by a group of students in a class test.
| Marks | 40 - 44 | 45 - 49 | 50 - 54 | 55 - 59 | 60 - 64 | 65 - 69 |
| No of students |
4 | 9 | 18 | 23 | 10 | 6 |
(a) Draw a histogram for the distribution ;
(b) Use your histogram to estimate the median of the distribution.
Ajụjụ 52 Ripọtì
The table gives the distribution of marks of 60 candidates in a test.
| Marks | 23-25 | 26-28 | 29-31 | 32-34 | 35-37 | 38-40 |
| Frequency | 3 | 7 | 15 | 21 | 10 | 4 |
(a) Draw a cumulative frequency curve of the distribution.
(b) From your curve, estimate the : (i) 80th percentile ; (ii) median ; (iii) semi-interquartile range.
Ajụjụ 53 Ripọtì
(a) A manufacturer produces light bulbs which are tested in the following way. A batch is accepted in either of the following cases:
(i) a first sample of 5 bulbs contains no faulty bulbs ; (ii) a first sample of 5 bulbs contains at least one faulty bulb but a second sample of size 5 has no faulty bulb. If 10% of the bulbs are faulty, what is the probability that the batch is accepted?
(b) A bag contains 15 identical marbles of which 3 are black, Keshi picks a marble at random from the bag and replaces it. If this is repeated 10 times; what is the probability that he :
(i) did not pick a black ball? (ii) picked a black ball at most three times?
Akọwa Nkọwa
None
Ajụjụ 54 Ripọtì
(a) The nth term of a sequence is given by \(T_{n} = 4T_{n - 1} - 3\). If twice the third term is five times the second term, find the first three terms of the sequence.
(b) Given that \(\begin{pmatrix} 2 & 0 & 1 \\ 5 & -3 & 1 \\ 0 & 4 & 6 \end{pmatrix} \begin{pmatrix} 1 \\ m \\ r \end{pmatrix} = \begin{pmatrix} k \\ 2 \\ 26 \end{pmatrix}\), find the values of the constants k, m and r.
Akọwa Nkọwa
None
Ajụjụ 55 Ripọtì
The equation of a curve is \(y = x(3 - x^{2})\). Find the equation of its normal of the point where x = 2.
Ajụjụ 56 Ripọtì
If the quadratic equation \((2x - 1) - p(x^{2} + 2) = 0\), where p is a constant, has real roots :
(a) show that \(2p^{2} + p - 1 < 0\);
(b) find the values of p.
Ajụjụ 57 Ripọtì
A particle moves from point O along a straight line such that its acceleration at any time, t seconds is \(a = (4 - 2t) ms^{-2}\). At t = 0, its distance from O is 18 metres while its velocity is \(5 ms^{-1}\).
(a) At what time will the velocity be greatest?
(b) Calculate the : (i) time ; (ii) distance of the particle from O when the particle is momentarily at rest.
Ajụjụ 58 Ripọtì
(a) The position vectors of the points X and Y are \(x = (-2i + 5j)\) and \(y = (i - 7j)\) respectively. Find :
(i) (3x + 2y) ; (ii) \(|(y - 2x)|\) ; (iii) the angle between x and y ; (iv) the unit vector in the direction of \((x + y)\).
(b) A bullet of mass 0.084kg is fired horizontally into a 20 kg block of wood at rest on a smooth floor. If they both move at a velocity of \(0.24 ms^{-1}\) after impact; Calculate, correct to two decimal places, the initial velocity of the bullet.
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