Ana ebu...
|
Tẹ & Di mu lati Gbe Yika |
|||
|
Pịa Ebe a ka Imechi |
|||
Ajụjụ 2 Ripọtì
Given that \(\frac{6x+m}{2x^{2}+7x-15} \equiv \frac{4}{x+5} - \frac{2}{2x-3}\), find the value of m.
Akọwa Nkọwa
To solve for m, we need to first create a common denominator on the right-hand side of the equation: \[\frac{4}{x+5} - \frac{2}{2x-3} = \frac{4(2x-3) - 2(x+5)}{(x+5)(2x-3)} = \frac{5}{2x-3} - \frac{3}{x+5}\] Now we have: \[\frac{6x+m}{2x^{2}+7x-15} = \frac{5}{2x-3} - \frac{3}{x+5}\] Multiplying both sides by the common denominator of the right-hand side, we get: \[(6x+m)(x+5) = 5(2x^2+7x-15) - 3(2x-3)(2x+5)\] Expanding and simplifying, we get: \[6x^2 + 41x + 5m + 15 = 4x^2 + 7x - 30\] \[2x^2 + 34x + 45 + m = 0\] Now, to find the value of m, we substitute the given roots of the equation (which are -5/2 and 9/2) into the equation and solve for m: \[m = -2x^2 - 34x - 45\] When x = -5/2, \[m = -2\left(\frac{-5}{2}\right)^2 - 34\left(\frac{-5}{2}\right) - 45 = -22\] Therefore, the value of m is -22. Option (d) is correct.
Ajụjụ 3 Ripọtì
If \(x^{2} - kx + 9 = 0\) has equal roots, find the values of k.
Akọwa Nkọwa
If the equation \(x^{2} - kx + 9 = 0\) has equal roots, it means that the discriminant is zero. The discriminant is the expression inside the square root in the quadratic formula: x = (-b ± sqrt(b^2 - 4ac)) / 2a For the equation \(x^{2} - kx + 9 = 0\), the discriminant is: b^2 - 4ac = k^2 - 4(1)(9) = k^2 - 36 Since the roots are equal, the discriminant is zero: k^2 - 36 = 0 k^2 = 36 k = ±6 Therefore, the values of k that make the equation have equal roots are \(\pm6\).
Ajụjụ 4 Ripọtì
Find the coordinates of the centre of the circle \(3x^{2}+3y^{2} - 4x + 8y -2=0\)
Akọwa Nkọwa
Ajụjụ 5 Ripọtì
| Face | 1 | 2 | 3 | 4 | 5 | 6 |
| Frequency | 12 | 18 | y | 30 | 2y | 45 |
Given the table above as the result of tossing a fair die 150 times, find the mode.
Akọwa Nkọwa
The mode is the value that occurs most frequently in the data set. From the table, we can see that the frequencies for faces 1, 2, 4, 5, and 6 are all different. However, the frequency for face 3 is "y", which is an unknown value. To find the mode, we need to determine the value of y. We know that the die was tossed 150 times, so the sum of all the frequencies should be 150. 12 + 18 + y + 30 + 2y + 45 = 150 Simplifying the equation, we get: 3y + 105 = 150 3y = 45 y = 15 Now we can see that the frequency for face 3 is 15. To find the mode, we need to determine which face has the highest frequency. We can see that face 6 has the highest frequency with 45 occurrences. Therefore, the mode is 6.
Ajụjụ 6 Ripọtì
Given that \(f(x) = 2x^{2} - 3\) and \(g(x) = x + 1\) where \(x \in R\). Find g o f(x).
Akọwa Nkọwa
Ajụjụ 7 Ripọtì
A circle with centre (4,5) passes through the y-intercept of the line 5x - 2y + 6 = 0. Find its equation.
Akọwa Nkọwa
To find the equation of the circle, we need to know its center and radius. We are given that the center is (4,5), so the equation will have the form \((x-4)^2 + (y-5)^2 = r^2\), where \(r\) is the radius. To find the radius, we need to first find the y-intercept of the line. To do this, we can set x = 0 and solve for y: 5x - 2y + 6 = 0 -2y + 6 = 0 -2y = -6 y = 3 So the y-intercept of the line is (0,3). Now we can use the distance formula to find the distance between this point and the center of the circle: \(\sqrt{(0-4)^2 + (3-5)^2} = \sqrt{16 + 4} = \sqrt{20} = 2\sqrt{5}\) So the radius of the circle is \(2\sqrt{5}\). Now we can substitute the center and radius into the equation of the circle: \((x-4)^2 + (y-5)^2 = (2\sqrt{5})^2\) Simplifying: \((x-4)^2 + (y-5)^2 = 20\) Expanding and rearranging, we get: \(x^2 + y^2 - 8x - 10y + 21 = 0\) Therefore, the equation of the circle is \(x^2 + y^2 - 8x - 10y + 21 = 0\). The answer is (3).
Ajụjụ 8 Ripọtì
\(\alpha\) and \(\beta\) are the roots of the equation \(2x^{2} - 3x + 4 = 0\). Find \(\frac{\alpha}{\beta} + \frac{\beta}{\alpha}\)
Akọwa Nkọwa
Ajụjụ 9 Ripọtì
How many ways can 6 students be seated around a circular table?
Akọwa Nkọwa
To find the number of ways to seat 6 students around a circular table, we can use the following formula: N = (n-1)! where N is the number of ways to seat n people around a circular table, and ! means factorial (the product of all positive integers up to a given number). So, for 6 people, we have: N = (6-1)! = 5! Using the definition of factorial, we have: 5! = 5 x 4 x 3 x 2 x 1 = 120 Therefore, there are 120 ways to seat 6 students around a circular table.
Ajụjụ 10 Ripọtì
\(\alpha\) and \(\beta\) are the roots of the equation \(2x^{2} - 3x + 4 = 0\). Find \(\alpha + \beta\).
Akọwa Nkọwa
The sum of the roots of a quadratic equation of the form \(ax^{2} + bx + c = 0\) is given by \(-\frac{b}{a}\). In this case, the given quadratic equation is \(2x^{2} - 3x + 4 = 0\). Therefore, the sum of the roots, \(\alpha\) and \(\beta\), is given by \(-\frac{-3}{2} = \frac{3}{2}\). Hence, the answer is \(\frac{3}{2}\), which represents the sum of the roots of the given quadratic equation.
Ajụjụ 11 Ripọtì
Given that \(f(x) = 5x^{2} - 4x + 3\), find the coordinates of the point where the gradient is 6.
Akọwa Nkọwa
Ajụjụ 13 Ripọtì
If \(y = \frac{1+x}{1-x}\), find \(\frac{dy}{dx}\).
Akọwa Nkọwa
To find \(\frac{dy}{dx}\) when \(y = \frac{1+x}{1-x}\), we need to use the quotient rule of differentiation, which states that if \(y = \frac{u}{v}\), then $$ \frac{dy}{dx} = \frac{v\frac{du}{dx} - u\frac{dv}{dx}}{v^2}. $$ In this case, we have $$ u = 1+x \qquad \text{and} \qquad v = 1-x. $$ Taking the derivatives of these functions, we get $$ \frac{du}{dx} = 1 \qquad \text{and} \qquad \frac{dv}{dx} = -1. $$ Substituting into the quotient rule formula, we get \begin{align*} \frac{dy}{dx} &= \frac{(1-x)(1) - (1+x)(-1)}{(1-x)^2} \\ &= \frac{1-x+1+x}{(1-x)^2} \\ &= \frac{2}{(1-x)^2}. \end{align*} Therefore, the correct answer is \(\frac{2}{(1-x)^2}\). Option (A) is the correct answer.
Ajụjụ 14 Ripọtì
If \(log_{y}\frac{1}{8}\) = 3, find the value of y.
Akọwa Nkọwa
We can use the definition of logarithms to solve for y. Recall that logarithms are the inverse function of exponential functions. That is, if we have: y = b^x Then the logarithm base b of y is: log_b y = x Using this definition, we can write: log_y (1/8) = 3 As an exponential function: y^3 = 1/8 Simplifying the right side: y^3 = (1/2)^3 Taking the cube root of both sides: y = 1/2 Therefore, the answer is \(\frac{1}{2}\).
Ajụjụ 15 Ripọtì
If \(\begin{pmatrix} 2 & 1 \\ 4 & 3 \end{pmatrix}\)\(\begin{pmatrix} 5 \\ 4 \end{pmatrix}\) = k\(\begin{pmatrix} 17.5 \\ 40.0 \end{pmatrix}\), find the value of k.
Akọwa Nkọwa
To find the value of \(k\), we need to calculate the product of the given matrices and compare it with the given result. Multiplying the first matrix by the second matrix, we get: \(\begin{pmatrix} 2 & 1 \\ 4 & 3 \end{pmatrix}\)\(\begin{pmatrix} 5 \\ 4 \end{pmatrix}\) = \(\begin{pmatrix} 2(5)+1(4) \\ 4(5)+3(4) \end{pmatrix}\) = \(\begin{pmatrix} 14 \\ 32 \end{pmatrix}\) Now, we can compare this result with the given result: k\(\begin{pmatrix} 17.5 \\ 40.0 \end{pmatrix}\) We see that the first element of the calculated result, 14, is less than the first element of the given result, 17.5. This means that k must be less than 1. Dividing both sides of the given equation by \(\begin{pmatrix} 17.5 \\ 40.0 \end{pmatrix}\), we get: \(\begin{pmatrix} 2 & 1 \\ 4 & 3 \end{pmatrix}\)\(\begin{pmatrix} 5 \\ 4 \end{pmatrix}\) / \(\begin{pmatrix} 17.5 \\ 40.0 \end{pmatrix}\) = k Calculating the left-hand side, we get: \(\begin{pmatrix} 2 & 1 \\ 4 & 3 \end{pmatrix}\)\(\begin{pmatrix} 5 \\ 4 \end{pmatrix}\) / \(\begin{pmatrix} 17.5 \\ 40.0 \end{pmatrix}\) = \(\begin{pmatrix} 14/35 \\ 32/40 \end{pmatrix}\) = \(\begin{pmatrix} 0.4 \\ 0.8 \end{pmatrix}\) Therefore, the value of \(k\) is equal to the first element of this result, which is 0.4. Hence, the correct answer is option (C) 0.8.
Ajụjụ 16 Ripọtì
Given that \(\sin x = \frac{5}{13}\) and \(\sin y = \frac{8}{17}\), where x and y are acute, find \(\cos(x+y)\).
Akọwa Nkọwa
To solve the problem, we need to use the sum formula for cosine: \[\cos(x+y) = \cos x \cos y - \sin x \sin y\] We are given the values of \(\sin x\) and \(\sin y\), so we need to find \(\cos x\) and \(\cos y\). To do this, we use the fact that \(\sin^2 x + \cos^2 x = 1\) and \(\sin^2 y + \cos^2 y = 1\) for any angle x and y. From \(\sin x = \frac{5}{13}\), we can find \(\cos x\) by using the Pythagorean identity: \[\cos x = \sqrt{1 - \sin^2 x} = \sqrt{1 - \left(\frac{5}{13}\right)^2} = \frac{12}{13}\] Similarly, from \(\sin y = \frac{8}{17}\), we can find \(\cos y\): \[\cos y = \sqrt{1 - \sin^2 y} = \sqrt{1 - \left(\frac{8}{17}\right)^2} = \frac{15}{17}\] Now we can substitute into the formula for \(\cos(x+y)\): \[\cos(x+y) = \cos x \cos y - \sin x \sin y = \frac{12}{13} \cdot \frac{15}{17} - \frac{5}{13} \cdot \frac{8}{17} = \frac{140}{221}\] Therefore, the answer is \(\frac{140}{221}\).
Ajụjụ 17 Ripọtì
The function f: x \(\to \sqrt{4 - 2x}\) is defined on the set of real numbers R. Find the domain of f.
Akọwa Nkọwa
The domain of a function is the set of all possible values of x for which the function is defined. In the given function, we have a square root with an expression inside it. For the square root to be defined, the expression inside it must be non-negative. Therefore, we need to solve the inequality: \(4 - 2x \geq 0\) Simplifying this inequality, we get: \(2x \leq 4\) \(x \leq 2\) So the domain of the function f is all real numbers such that \(x \leq 2\). Therefore, the correct answer is.
Ajụjụ 19 Ripọtì
Simplify \(\frac{1}{(1-\sqrt{3})^{2}}\)
Akọwa Nkọwa
We can simplify the given expression as follows: \begin{align*} \frac{1}{(1-\sqrt{3})^2} &= \frac{1}{1-2\sqrt{3}+3} \\ &= \frac{1}{4-2\sqrt{3}} \\ &= \frac{4+2\sqrt{3}}{(4-2\sqrt{3})(4+2\sqrt{3})} \\ &= \frac{4+2\sqrt{3}}{16-12} \\ &= \frac{4+2\sqrt{3}}{4} \\ &= 1 + \frac{1}{2}\sqrt{3} \end{align*} Therefore, the answer is (B) \(1+ \frac{1}{2}\sqrt{3}\).
Ajụjụ 20 Ripọtì
| Face | 1 | 2 | 3 | 4 | 5 | 6 |
| Frequency | 12 | 18 | y | 30 | 2y | 45 |
Given the table above as the results of tossing a fair die 150 times. Find the probability of obtaining a 5.
Akọwa Nkọwa
The table shows the frequency of each face when a fair die is tossed 150 times. To find the probability of obtaining a 5, we need to look at the frequency of the face with the value 5, which is given as 2y. We are not given the value of y, but we know that the die is fair, which means that each face has an equal probability of appearing on any given toss. Since there are six faces on a die, the probability of obtaining a 5 on any given toss is 1/6. Therefore, the probability of obtaining a 5 in this experiment is the frequency of the face with value 5 (2y) divided by the total number of tosses (150), multiplied by the probability of obtaining a 5 on any given toss (1/6): P(5) = (2y/150) x (1/6) = y/225 We don't know the value of y, but we can simplify the expression by noting that the probabilities of all six faces must add up to 1: P(1) + P(2) + P(3) + P(4) + P(5) + P(6) = 1 We can substitute the expressions for the probabilities in terms of y and simplify: 12/150 + 18/150 + y/150 + 30/150 + 2y/150 + 45/150 = 1 105/150 + 3y/150 = 1 3y/150 = 45/150 y = 15 Substituting y = 15 into the expression for P(5), we get: P(5) = 15/225 = 1/15 Therefore, the probability of obtaining a 5 is 1/15, which corresponds to option C.
Ajụjụ 21 Ripọtì
A force (10i + 4j)N acts on a body of mass 2kg which is at rest. Find the velocity after 3 seconds.
Akọwa Nkọwa
Ajụjụ 22 Ripọtì
Given that \( a = \begin{pmatrix} 2 \\ 3 \end{pmatrix}\) and \(b = \begin{pmatrix} -1 \\ 4 \end{pmatrix}\), evaluate \((2a - \frac{1}{4}b)\).
Akọwa Nkọwa
To evaluate \((2a - \frac{1}{4}b)\), we first need to perform scalar multiplication on vectors \(a\) and \(b\) and then subtract the resulting vectors: \begin{align*} 2a - \frac{1}{4}b &= 2\begin{pmatrix} 2 \\ 3 \end{pmatrix} - \frac{1}{4}\begin{pmatrix} -1 \\ 4 \end{pmatrix}\\ &= \begin{pmatrix} 4 \\ 6 \end{pmatrix} - \begin{pmatrix} -\frac{1}{4} \\ 1 \end{pmatrix}\\ &= \begin{pmatrix} \frac{17}{4} \\ 5 \end{pmatrix} \end{align*} Therefore, the answer is \(\begin{pmatrix} \frac{17}{4} \\ 5 \end{pmatrix}\).
Ajụjụ 23 Ripọtì
In a class of 10 boys and 15 girls, the average score in a Biology test is 90. If the average score for the girls is x, find the average score for the boys in terms of x.
Akọwa Nkọwa
Let the average score for the boys be y. Then the total score for the class is: 10y (boys' total score) + 15x (girls' total score) = 25(90) (total class score) Simplifying, we get: 10y + 15x = 2250 Dividing by 10, we get: y + \(\frac{3}{2}\)x = 225 Subtracting \(\frac{3}{2}\)x from both sides, we get: y = 225 - \(\frac{3}{2}\)x Therefore, the average score for the boys in terms of x is \(225 - \frac{3}{2}x\). So the correct answer is.
Ajụjụ 24 Ripọtì
A curve is given by \(y = 5 - x - 2x^{2}\). Find the equation of its line of symmetry.
Akọwa Nkọwa
The line of symmetry of a parabola is a vertical line that passes through the vertex of the parabola. The equation of the line of symmetry for a parabola of the form \(y = a(x - h)^2 + k\) is simply the vertical line \(x = h\). The given curve is a quadratic function of the form \(y = -2x^2 - x + 5\), which is a downward-facing parabola. The line of symmetry for this parabola is the vertical line that passes through its vertex. To find the vertex of the parabola, we can first find the x-coordinate of the vertex using the formula \(x = \frac{-b}{2a}\). In this case, \(a = -2\) and \(b = -1\), so we have: $$x = \frac{-b}{2a} = \frac{-(-1)}{2(-2)} = \frac{1}{4}$$ To find the corresponding y-coordinate, we can substitute this value of x into the equation for the curve: $$y = -2x^2 - x + 5 = -2\left(\frac{1}{4}\right)^2 - \frac{1}{4} + 5 = \frac{41}{8}$$ Therefore, the vertex of the parabola is \(\left(\frac{1}{4}, \frac{41}{8}\right)\), and the equation of its line of symmetry is simply: $$x = \frac{1}{4}$$ So the correct option is \(\boxed{\text{(c) } x = \frac{1}{4}}\).
Ajụjụ 25 Ripọtì
There are 7 boys in a class of 20. Find the number of ways of selecting 3 girls and 2 boys
Akọwa Nkọwa
To solve this problem, we need to use the combination formula: nCr = n! / r!(n-r)! where n is the total number of objects, r is the number of objects we want to select, and ! represents factorial, which means the product of all positive integers up to and including that number. In this problem, we want to select 3 girls and 2 boys out of 20 students, where 13 are girls and 7 are boys. The number of ways to select 3 girls out of 13 is: 13C3 = 13! / 3!(13-3)! = 13! / 3!10! = (13x12x11) / (3x2x1) = 286 Similarly, the number of ways to select 2 boys out of 7 is: 7C2 = 7! / 2!(7-2)! = 7! / 2!5! = (7x6) / (2x1) = 21 To find the total number of ways to select 3 girls and 2 boys out of 20, we need to multiply the number of ways to select 3 girls and 2 boys: 286 x 21 = 6006 Therefore, there are 6006 ways to select 3 girls and 2 boys from a class of 20. Hence, the correct answer is option (C) 6006.
Ajụjụ 26 Ripọtì
Simplify \(\frac{\sqrt{128}}{\sqrt{32} - 2\sqrt{2}}\)
Akọwa Nkọwa
We can start by simplifying the numerator and the denominator separately. The square root of 128 can be written as the square root of 64 times 2, which simplifies to 8 times the square root of 2. The denominator can be simplified using the difference of squares formula: \begin{align*} \sqrt{32} - 2\sqrt{2} &= \sqrt{16 \times 2} - 2\sqrt{2} \\ &= 4\sqrt{2} - 2\sqrt{2} \\ &= 2\sqrt{2} \end{align*} Now we can substitute these simplifications back into the original expression: \begin{align*} \frac{\sqrt{128}}{\sqrt{32} - 2\sqrt{2}} &= \frac{8\sqrt{2}}{2\sqrt{2}} \\ &= 4 \end{align*} Therefore, the simplified expression is 4.
Ajụjụ 27 Ripọtì
If P = \({n^{2} + 1: n = 0,2,3}\) and Q = \({n + 1: n = 2,3,5}\), find P\(\cap\) Q.
Ajụjụ 28 Ripọtì
Express cos150° in surd form.
Akọwa Nkọwa
To express cos150° in surd form, we need to use the trigonometric identity cos(180° - θ) = -cos(θ). Thus, cos150° = cos(180° - 30°) = -cos30°. Using the value of cos30°, which is \(\frac{\sqrt{3}}{2}\), we get: cos150° = -cos30° = -\(\frac{\sqrt{3}}{2}\) Therefore, the correct answer is option B, \(-\frac{\sqrt{3}}{2}\).
Ajụjụ 29 Ripọtì
The 3rd and 7th term of a Geometric Progression (GP) are 81 and 16. Find the 5th term.
Akọwa Nkọwa
Ajụjụ 30 Ripọtì
A binary operation \(\Delta\) is defined on the set of real numbers, R, by \(a \Delta b = \frac{a+b}{\sqrt{ab}}\), where a\(\neq\) 0, b\(\neq\) 0. Evaluate \(-3 \Delta -1\).
Akọwa Nkọwa
To evaluate \(-3 \Delta -1\), we simply need to substitute a = -3 and b = -1 in the definition of the binary operation: \(-3 \Delta -1 = \frac{-3+(-1)}{\sqrt{(-3)(-1)}} = \frac{-4}{\sqrt{3}}\) To simplify this expression, we rationalize the denominator by multiplying both the numerator and denominator by \(\sqrt{3}\): \(\frac{-4}{\sqrt{3}} = \frac{-4}{\sqrt{3}} \cdot \frac{\sqrt{3}}{\sqrt{3}} = \frac{-4\sqrt{3}}{3}\) Therefore, the correct answer is.
Ajụjụ 31 Ripọtì
The velocity, V, of a particle after t seconds, is \(V = 3t^{2} + 2t - 1\). Find the acceleration of the particle after 2 seconds.
Akọwa Nkọwa
The acceleration, A, of a particle is the rate of change of its velocity, V. So, we need to differentiate the given equation of velocity with respect to time, t to get the equation of acceleration. \[\frac{dV}{dt} = \frac{d}{dt}(3t^2 + 2t - 1)\] \[\frac{dV}{dt} = 6t + 2\] Therefore, the acceleration of the particle after 2 seconds is given by substituting t = 2 into the equation of acceleration as follows: \[A = 6t + 2\] \[A = 6(2) + 2\] \[A = 14 ms^{-2}\] Hence, the correct option is (c) 14\(ms^{-2}\).
Ajụjụ 32 Ripọtì
Given that a = 5i + 4j and b = 3i + 7j, evaluate (3a - 8b).
Akọwa Nkọwa
To evaluate (3a - 8b), we need to multiply each component of the vectors by their respective scalar, and then add the corresponding components. So, 3a = 3(5i + 4j) = 15i + 12j 8b = 8(3i + 7j) = 24i + 56j Therefore, (3a - 8b) = (15i + 12j) - (24i + 56j) = (-9i - 44j) Hence, the correct answer is option (C) -9i - 44j.
Ajụjụ 35 Ripọtì
If \((2x^{2} - x - 3)\) is a factor of \(f(x) = 2x^{3} - 5x^{2} - x + 6\), find the other factor
Akọwa Nkọwa
Ajụjụ 37 Ripọtì
A straight line 2x+3y=6, passes through the point (-1,2). Find the equation of the line.
Akọwa Nkọwa
To find the equation of a straight line, we need to know either two points that the line passes through or a point and the slope of the line. In this case, we are given one point that the line passes through, (-1,2), and the equation of the line, 2x+3y=6. We can use the point-slope form of the equation of a line to find the equation of the line passing through the given point. The point-slope form is: y - y1 = m(x - x1) where (x1, y1) is the given point and m is the slope of the line. To find the slope of the given line, we can rearrange the equation 2x+3y=6 into the slope-intercept form: y = (-2/3)x + 2 where the slope is -2/3. Substituting the values of the given point and slope into the point-slope form, we get: y - 2 = (-2/3)(x - (-1)) Simplifying this equation, we get: y - 2 = (-2/3)x - (2/3) Multiplying both sides by 3 to eliminate the fraction, we get: 3y - 6 = -2x - 2 Adding 2x to both sides, we get: 2x + 3y = 4 Therefore, the equation of the line passing through the given point is 2x+3y=4. Hence, the correct answer is option (D) 2x+3y=4.
Ajụjụ 38 Ripọtì
A fair die is tossed twice. What is its smple size?
Akọwa Nkọwa
When a fair die is tossed twice, there are 6 possible outcomes for the first toss and 6 possible outcomes for the second toss, giving a total of 6 x 6 = 36 possible outcomes for the two tosses combined. Each possible outcome can be represented by an ordered pair (a,b), where a is the outcome of the first toss and b is the outcome of the second toss. Therefore, the sample space of this experiment has 36 possible outcomes, making the correct answer.
Ajụjụ 39 Ripọtì
Find the 21st term of the Arithmetic Progression (A.P.): -4, -1.5, 1, 3.5,...
Akọwa Nkọwa
We are given the first four terms of an Arithmetic Progression (A.P.) and we are required to find the 21st term. We can notice that the common difference between any two successive terms is \(d = -1.5 - (-4) = 2.5\). Using the formula for the nth term of an A.P., which is given by: \(a_{n} = a_{1} + (n - 1) d\) where \(a_{n}\) is the nth term, \(a_{1}\) is the first term, \(n\) is the number of terms and \(d\) is the common difference. We can now find the 21st term as follows: \(a_{21} = a_{1} + (21 - 1) d\) Substituting the given values, we get: \(a_{21} = -4 + (21 - 1) \times 2.5 = 46\) Hence, the 21st term of the given A.P. is 46. Therefore, the correct option is (B) 46.
Ajụjụ 41 Ripọtì
If (x + 1) and (x - 2) are factors of the polynomial \(g(x) = x^{4} + ax^{3} + bx^{2} - 16x - 12\), find the values of a and b.
Ajụjụ 42 Ripọtì
(a) In a school, the ratio of those who passed to those who failed in a History test is 4 : 1. If 7 students are selected at random from the school, find, correct to two decimal places, the probability that :
(i) at least 3 passed the test ; (ii) between 3 and 6 students failed the test.
(b) A fair die is thrown five times; find the probability of obtaining a six three times.
Ajụjụ 43 Ripọtì
A parallelogram MNQR has vertices M(4, -6), N(10, 2), Q(8, 16) and R(x, y). Find the coordinates of R.
Ajụjụ 44 Ripọtì
Forces \(F_{1} (18N, 330°), F_{2} (10N, 090°)\) and \(F_{3} (25N, 180°)\) act on a body at rest. Find, correct to one decimal place, the magnitude and direction of the resultant force.
Ajụjụ 45 Ripọtì
The table shows the frequency distribution of the ages of patients in a clinic.
| Ages (years) | 17 - 19 | 20 - 22 | 23 - 28 | 29 - 34 | 35 - 43 |
| No. of patients | 6 | 9 | 12 | 18 | 18 |
(a) Draw a histogram for the distribution
(b) Find, correct to two decimal places, the mean age of the patients.
Ajụjụ 46 Ripọtì
(a) If \(f(x) = \frac{2x - 3}{(x^{2} - 1)(x + 2)}\)
(i) find the values of x for which f(x) is undefined.
(ii) express f(x) in partial fractions.
(b) A circle with centre (-3, 1) passes through the point (3, 1). Find its equation.
Akọwa Nkọwa
None
Ajụjụ 47 Ripọtì
(a) Given that \(m = (6i + 8j)\) and \(n = (-8i + \frac{7}{3}j)\), find the :
(i) magnitudes and direction of m and n ; (ii) angle between m and n.
(b) The position vectors of points P, Q, R and S are \(\begin{pmatrix} -2 \\ 3 \end{pmatrix}, \begin{pmatrix} 10 \\ 4 \end{pmatrix}, \begin{pmatrix} 3 \\ 12 \end{pmatrix}\) and \(\begin{pmatrix} 4 \\ 0 \end{pmatrix}\) respectively. Show that \(\overrightarrow{PQ}\) is perpendicular to \(\overrightarrow{RS}\).
Ajụjụ 48 Ripọtì
The table shows the heights in cm of some seedlings in a certain garden.
| Height (cm) | 36-40 | 41-45 | 46-50 | 51-55 | 56-60 |
| Frequency | 3 | 9 | 21 | 12 | 5 |
(a) Draw the cumulative frequency curve for the distribution.
(b) Using the curve in (a), find thesemi-interquartile range.
Ajụjụ 49 Ripọtì
(a) If \(f(x) = \int (4x - x^{2}) \mathrm {d} x\) and f(3) = 21, find f(x).
(b) The second, fourth and eigth terms of an Arithmetic Progression (A.P) form the first three consecutive terms of a Geometric Progression (G.P). The sum of the third and fifth terms of the A.P is 20, find the :
(i) first four terms of the A.P
(ii) sum of the first ten terms of the A.P
Akọwa Nkọwa
None
Ajụjụ 50 Ripọtì
(a) A body P of mass q kg is suspended by two light inextensible strings AB and DB attached to a horizontal table. The strings are inclined at 30° and 60° respectively to the horizontal and the tension in AB is 48N. If the system is in equilibrium :
(i) sketch a diagram to represent the information ; (ii) calculate the tension in DB ;
Ajụjụ 52 Ripọtì
Bottles of the same sizes produced in a factory are packed in boxes. Each box contains 10 bottles. If 8% of the bottles are defective, find, correct to two decimal places, the probability that box chosen at random contains at least 3 defective bottles.
Ajụjụ 53 Ripọtì
(a) Given that \(\log_{10} p = a, \log_{10} q = b\) and \(\log_{10} s = c\), express \(\log_{10} (\frac{p^{\frac{1}{3}}q^{4}}{s^{2}}\) in terms of a, b and c.
(b) The radius of a circle is 6cm. If the area is increasing at the rate of 20\(cm^{2}s^{-1}\), find, leaving the answer in terms of \(\pi\), the rate at which the radius is increasing.
Ajụjụ 54 Ripọtì
(a) If \(f(x) = \frac{4 - 5x}{2}\), and \(g(x) = x + 6, x \in R\), find \(f \circ g^{-1}\).
(b) P(x, y) divides the line joining (7, -5) and (-2, 7) internally in 5 : 4. Find the coordinates of P.
Ajụjụ 55 Ripọtì
Evaluate : \(\int_{1}^{3} (\frac{x - 1}{(x + 1)^{2}}) \mathrm {d} x\).
Ị ga-achọ ịga n'ihu na omume a?