WAEC SSCE - Further Mathematics - 2019

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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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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 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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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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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 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 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 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 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).

Akọwa Nkọwa

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.

Akọwa Nkọwa

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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.

Akọwa Nkọwa

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 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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To find the angle between two vectors, we can use the dot product formula:

\(\vec{a} \cdot \vec{b} = |\vec{a}| |\vec{b}| \cos\theta\)

where \(\vec{a}\) and \(\vec{b}\) are the two vectors, \(|\vec{a}|\) and \(|\vec{b}|\) are their magnitudes, and \(\theta\) is the angle between them.

First, we need to find the magnitudes of the two vectors:

\(|\overline{OP}| = \sqrt{(-3)^2 + (-4)^2} = 5\)

\(|\overline{OQ}| = \sqrt{8^2 + (-15)^2} = 17\)

Next, we need to find the dot product of the two vectors:

\(\overline{OP} \cdot \overline{OQ} = (-3)(8) + (-4)(-15) = 9\)

Now, we can substitute the values into the dot product formula to find the angle:

\(9 = 5 \cdot 17 \cdot \cos\theta\)

\(\cos\theta = \frac{9}{85}\)

\(\theta = \cos^{-1}\left(\frac{9}{85}\right)\)

Using a calculator, we get:

\(\theta \approx 71.4^\circ\)

Therefore, the angle between \(\overline{OP}\) and \(\overline{OQ}\) is approximately \(71.4^\circ\).

Akọwa Nkọwa

To solve this problem, we need to use the principles of equilibrium. This means that the forces acting on the mass must balance out, so that the net force is zero and the mass remains stationary.

First, let's draw a free-body diagram of the mass. This diagram shows all the forces acting on the mass and their directions. We have two forces acting on the mass: the weight of the mass, which acts downwards, and the tension in the string, which acts upwards and at an angle of 45 degrees to the vertical.

Next, we can resolve the forces into their vertical and horizontal components. The weight of the mass has a vertical component of 12g (where g is the acceleration due to gravity, approximately 9.81 m/s²) acting downwards, and no horizontal component. The tension in the string has both vertical and horizontal components. The vertical component is T cos(45), where T is the tension in the string, and the horizontal component is T sin(45).

Since the mass is in equilibrium, the net force in the horizontal direction must be zero. This means that the horizontal component of the tension in the string must be equal and opposite to the horizontal force R:

T sin(45) = R

Since we know the value of the mass and the acceleration due to gravity, we can calculate the weight of the mass and the vertical component of the tension in the string. The net force in the vertical direction must also be zero, so the vertical component of the tension in the string must balance out the weight of the mass:

T cos(45) = 12g

We can now solve these two equations simultaneously to find the tension in the string and the value of R.

To do this, we can first use the second equation to find the value of T:

T = 12g / cos(45)
T = 12g / 0.707
T = 169.7 N (to two significant figures)

We can then substitute this value of T into the first equation to find the value of R:

R = T sin(45)
R = 169.7 N x 0.707
R = 120 N (to two significant figures)

Therefore, the tension in the string is 169.7 N and the value of R is 120 N.

In summary, we can solve this problem by drawing a free-body diagram, resolving the forces into their components, and using the principles of equilibrium to find the values of the tension in the string and the horizontal force R.

Akọwa Nkọwa

Differentiation is a mathematical process that allows us to find the rate of change of a function at a certain point. To differentiate the function (3x² + 2x - 1) with respect to x from first principles, we'll use the power rule of differentiation.

The power rule states that if we have a term in the form of xⁿ, then its derivative is given by nx^(n-1).

Applying this rule to the function (3x² + 2x - 1), we get:

d/dx (3x²) = 6x

d/dx (2x) = 2

The derivative of a constant is 0, so:

d/dx (-1) = 0

Putting it all together, we find that the derivative of (3x² + 2x - 1) with respect to x is:

d/dx (3x² + 2x - 1) = 6x + 2

So, the rate of change of the function (3x² + 2x - 1) at any point x is given by 6x + 2.

Akọwa Nkọwa

To find the angle between the forces 10N and 16N, we can use the fact that the particle is in equilibrium. This means that the net force acting on the particle is zero.

We can break down the three forces into their horizontal and vertical components. Let's call the angle between the 14N force and the horizontal axis as θ1, and the angle between the 16N force and the horizontal axis as θ2.

The horizontal components of the forces are:

N cos(θ1)
14N cos(0)
16N cos(θ2)

Since the particle is in equilibrium, the net horizontal force must be zero. So we have:

N cos(θ1) + 14N cos(0) + 16N cos(θ2) = 0

Simplifying this equation, we get:

N cos(θ1) + 16N cos(θ2) = -14N

We can do the same thing for the vertical components of the forces:

N sin(θ1)
14N sin(0)
16N sin(θ2)

Since the particle is in equilibrium, the net vertical force must be zero. So we have:

N sin(θ1) + 14N sin(0) + 16N sin(θ2) = 0

Simplifying this equation, we get:

N sin(θ1) + 16N sin(θ2) = 0

We can now solve these two equations for N and sin(θ1):

N = -16N cos(θ2) / cos(θ1)
sin(θ1) = -16 sin(θ2) / cos(θ1)

Now, we can use the Pythagorean theorem to find the magnitude of the 10N force:

sqrt((10N)^2 + (N sin(θ1))^2) = sqrt((10N)^2 + (16 sin(θ2))^2)

Finally, we can use the arctangent function to find the angle between the 10N and 16N forces:

tan(θ10-16) = (N sin(θ1)) / (10N) = (-16 sin(θ2)) / (10N cos(θ2))

θ10-16 = arctan((-16 sin(θ2)) / (10N cos(θ2)))

Therefore, we can find the angle between the forces 10N and 16N by using the above formula with the values we have calculated.

Akọwa Nkọwa

To calculate the mean and standard deviation of the distribution, we need to use the formulae:

Mean = Assumed Mean + (Σfd)/n

Standard Deviation = √(Σf(x-ḿ)²)/n

Where:

• Σfd represents the sum of the products of the frequency (f) and the deviation from the assumed mean (d = x - ḿ)
• n represents the total frequency, which is the sum of all frequencies (Σf)

Using the table and the given assumed mean of 67, we can calculate the values needed for the mean and standard deviation as follows:

a) Mean:

We need to find Σfd and n to plug into the formula. To find Σfd, we need to calculate the deviation from the assumed mean for each class interval and multiply it by the frequency of that interval. For example, for the first interval (50-54), the midpoint is 52, so the deviation from the assumed mean of 67 is -15. We multiply this by the frequency of 5 to get -75. We repeat this process for all intervals and sum up the products to get Σfd = -204. The total frequency n is the sum of all frequencies, which is 5 + 15 + 20 + 28 + 12 + 9 + 7 + 4 = 100. Now we can plug these values into the formula to get:

Mean = 67 + (-204)/100 = 65.8 (correct to one decimal place)

Therefore, the mean of the distribution is 65.8.

b) Standard deviation:

We need to find Σf(x-ḿ)² and n to plug into the formula. To find Σf(x-ḿ)², we need to calculate the squared deviation from the assumed mean for each class interval, multiply it by the frequency of that interval, and sum up the products. For example, for the first interval (50-54), the midpoint is 52, so the deviation from the assumed mean of 67 is -15. We square this to get 225, then multiply by the frequency of 5 to get 1125. We repeat this process for all intervals and sum up the products to get Σf(x-ḿ)² = 8444. The total frequency n is the same as before, 100. Now we can plug these values into the formula to get:

Standard deviation = √(Σf(x-ḿ)²)/n = √(8444/100) = 9.19 (correct to one decimal place)

Therefore, the standard deviation of the distribution is 9.19.

In summary, the mean of the distribution is 65.8 and the standard deviation is 9.19.

Akọwa Nkọwa

a)

${}^{b)}$

If we let r be the retardation, then st

$\frac{\mathrm{<span\; style="font-family:\; Ubuntu;\; font-size:\; 16px;">𝑠</span>}}{\mathrm{<span\; style="font-family:\; Ubuntu;\; font-size:\; 16px;">𝑡</span>}}$ : r = 3 : 4

when simplified,

r = 103 $\frac{\mathrm{<br>}}{\mathrm{<br>}}$

Velocity after 4 seconds

= 20 + 52 $\frac{\mathrm{<br>}}{\mathrm{<br>}}$ x 4 = 30ms−1 ${}^{-1}$

So that 52 $\frac{\mathrm{<br>}}{\mathrm{<br>}}$ : 30t $\frac{\mathrm{<br>}}{\mathrm{<br>}}$ = 34 $\frac{\mathrm{<br>}}{\mathrm{<br>}}$

t = 9 seconds

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To solve this problem, we need to use the principle of moments. The principle of moments states that for an object to be in equilibrium, the sum of the clockwise moments about any point must be equal to the sum of the anticlockwise moments about the same point. In this problem, we can choose the pivot point as the point about which we will calculate the moments.

Let's start by calculating the clockwise moments. The weight of the beam, 50N, acts at the center of gravity of the beam, which is at a distance of 45 cm from the pivot point. Therefore, the clockwise moment due to the weight of the beam is:

50N x 45cm = 2250 Ncm

Next, let's calculate the anticlockwise moments. The force T acts at point Y, which is 10 cm from the pivot point. Therefore, the anticlockwise moment due to the force T is:

T x 10cm

Similarly, the force of 20N acts at point Z, which is also 10 cm from the pivot point. Therefore, the anticlockwise moment due to the force of 20N is:

20N x 10cm = 200 Ncm

Since the beam is in equilibrium, the sum of the clockwise moments must be equal to the sum of the anticlockwise moments. Therefore, we have:

2250 Ncm = T x 10cm + 200 Ncm

Simplifying this equation, we get:

T = (2250 Ncm - 200 Ncm)/10cm = 205N

Therefore, the value of T is 205N.

In summary, we used the principle of moments to calculate the value of T. We chose the pivot point as the point about which we calculated the moments. We calculated the clockwise moment due to the weight of the beam and the anticlockwise moments due to the forces at points Y and Z. We then equated the sum of the clockwise moments to the sum of the anticlockwise moments and solved for T. The answer is T = 205N.

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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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(a)

To find the coordinates of the two points where the curve \(y = 7 - \frac{6}{x}\) intersects the line \(y + 2x - 3 = 0\), we can substitute \(y\) in the equation of the line with the expression for \(y\) in the equation of the curve and solve for \(x\).

\(y + 2x - 3 = 0\) becomes:

\(7 - \frac{6}{x} + 2x - 3 = 0\)

Simplifying this expression, we get:

\(2x^2 - x - 2 = 0\)

We can factorize this quadratic equation to get:

\((2x + 1)(x - 2) = 0\)

This gives us two solutions for \(x\): \(x = -\frac{1}{2}\) and \(x = 2\).

We can substitute these values of \(x\) into either equation to get the corresponding values of \(y\).

For \(x = -\frac{1}{2}\):

\(y + 2(-\frac{1}{2}) - 3 = 0\)

\(y = 4\)

So one point of intersection is \(-\frac{1}{2}, 4\).

For \(x = 2\):

\(y + 2(2) - 3 = 0\)

\(y = -1\)

So the other point of intersection is \(2, -1\).

Therefore, the coordinates of the two points of intersection are \(-\frac{1}{2}, 4\) and \(2, -1\).

(b)

To find the equation of the perpendicular bisector of the line joining the two points of intersection, we first need to find the midpoint of the line segment joining the two points.

The midpoint formula gives us:

\(\left(\frac{x_1 + x_2}{2}\), \(\frac{y_1 + y_2}{2}\right)\)

Substituting the values of the two points, we get:

\(\left(\frac{-\frac{1}{2} + 2}{2}\), \(\frac{4 - 1}{2}\right)\)

Simplifying this expression, we get the midpoint \(\left(\frac{3}{4}, \frac{3}{2}\right)\).

Next, we need to find the slope of the line joining the two points.

The slope formula gives us:

\(\frac{y_2 - y_1}{x_2 - x_1}\)

Substituting the values of the two points, we get:

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

Simplifying this expression, we get the slope \(-\frac{10}{9}\).

The perpendicular bisector of the line segment joining the two points will have a slope that is the negative reciprocal of this slope, which is \(\frac{9}{10}\).

We

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The expression given is: (^2x + 1₃) - (^2x - 1₃) - 2(^x₂)

We can simplify this expression by using the formula (^r_n) = ^nC_r, which represents the number of ways of choosing r objects from a set of n distinct objects.

First, let's simplify (^2x + 1₃) using the formula:

(^2x + 1₃) = ^3C_2x+1

This represents the number of ways of choosing (2x + 1) objects from a set of 3 distinct objects. Similarly, we can simplify (^2x - 1₃) using the formula:

(^2x - 1₃) = ^3C_2x-1

This represents the number of ways of choosing (2x - 1) objects from a set of 3 distinct objects.

Now, substituting these values in the given expression, we get:

(^2x + 1₃) - (^2x - 1₃) - 2(^x₂) = ^3C_2x+1 - ^3C_2x-1 - 2(^x₂)

Next, we can simplify 2(^x₂) using the formula:

2(^x₂) = 2^xC₂

This represents the number of ways of choosing 2 objects from a set of x distinct objects, multiplied by 2.

Substituting this value in the previous expression, we get:

(^2x + 1₃) - (^2x - 1₃) - 2(^x₂) = ^3C_2x+1 - ^3C_2x-1 - 2^xC₂

This is the simplified expression, which represents the difference between the number of ways of choosing (2x + 1) objects and (2x - 1) objects from a set of 3 distinct objects, minus twice the number of ways of choosing 2 objects from a set of x distinct objects.

Akọwa Nkọwa

The series can be rewritten as follows:

-3 + (-1 + 1) + (-3 + 1) + (-1 + 1) + ...

Simplifying, we get:

-3 + 0 + -2 + 0 + -3 + 0 + -2 + 0 + ...

The terms in the series alternate between -3 and 0. Therefore, we can group the terms as follows:

(-3 + 0) + (-2 + 0) + (-3 + 0) + (-2 + 0) + ...

Each pair of terms adds up to -3, and there are n/2 pairs of terms in the series, where n is the number of terms. Therefore, we can write the following equation:

-3(n/2) = 165

Simplifying, we get:

n = -2(165)/3

n = -110

Since the number of terms must be a positive integer, the answer is none of the given options. This means that there is no solution to the problem, as the series cannot add up to 165.

Akọwa Nkọwa

(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).

Akọwa Nkọwa

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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

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(a) Given that m = i - i, m = 0. Also, n = 2i + 3j.

Since 2m + n = 2(0) + n = n = 2i + 3j, the equation becomes:

2m + n - r = 2(0) + n - r = n - r = 0

Therefore, r = 2i + 3j.

The magnitude of a vector can be found using the Pythagorean theorem, which states that the magnitude of a vector (a, b) is given by √(a^2 + b^2).

So, |r| = √(2^2 + 3^2) = √(4 + 9) = √13.

(b) (i) The particle is momentarily at rest when its velocity is 0, i.e., when its derivative with respect to time is 0. The velocity of the particle is given by the derivative of its position with respect to time.

So, the velocity is given by:

v = dS/dt = 3 - t^2

Setting this equal to 0 and solving for t, we get:

0 = 3 - t^2

t^2 = 3

t = ±√3

So, the particle is momentarily at rest when t = ±√3.

(ii) To find the distance travelled when the particle is momentarily at rest, we need to substitute t = ±√3 into the expression for S:

S = 3t - t^3/3 + 9

S = 3(±√3) - (±√3)^3/3 + 9

S = 9 ± 3√3 - (3 ± 3√3) + 9

S = 18

So, the particle has travelled a distance of 18 metres when it is momentarily at rest.

Akọwa Nkọwa

The expression can be simplified as follows:

First, we simplify the expression inside the first set of parentheses:

(3x/4 - 1) = (x - 4/3)

Next, we simplify the expression inside the second set of parentheses:

(125^(x - 1)) = (125^x / 125) = (5^x)

Now, we substitute these simplified expressions back into the original expression:

(625(x - 4/3) + 5^x) / (5^(3x - 2))

Finally, we simplify the denominator:

(625(x - 4/3) + 5^x) / (125^(x - 2/3))

This is the simplified form of the expression. Note that this expression can be further simplified, but it depends on the specific values of x.

Akọwa Nkọwa

(a) To find the value of x and y, we can use the fact that opposite sides of a parallelogram are parallel and have equal length.

From the given information, we know that PQ and SR are parallel, and PQ = SR since they are opposite sides of the parallelogram. Therefore, we can use the distance formula to find the length of PQ and SR:

PQ = sqrt((2 - (-1))^2 + (3 - 4)^2) = sqrt(3^2 + (-1)^2) = sqrt(10)

SR = sqrt((-2 - x)^2 + (3 - y)^2)

Since PQ = SR, we have:

sqrt(10) = sqrt((-2 - x)^2 + (3 - y)^2)

Squaring both sides, we get:

10 = (2 + x)^2 + (3 - y)^2

Expanding the squares, we get:

10 = 4 + 4x + x^2 + 9 - 6y + y^2

Simplifying, we get:

x^2 + y^2 + 4x - 6y - 3 = 0

We can rewrite this equation as:

(x + 2)^2 + (y - 3)^2 = 13

This is the equation of a circle with center (-2, 3) and radius sqrt(13). Therefore, any point (x, y) that lies on this circle will form a parallelogram with P(-1, 4), Q(2, 3), and S(-2, 3).

(b)

(i) We know that the initial velocity of the particle is zero, and that its final velocity is 20ms^-1 after travelling a distance of 8 metres. We can use the formula:

v^2 = u^2 + 2as

where v is the final velocity, u is the initial velocity, a is the acceleration, and s is the distance travelled.

Plugging in the given values, we get:

(20)^2 = 0^2 + 2a(8)

Simplifying, we get:

a = 100/8 = 12.5ms^-2

Therefore, the acceleration of the particle is 12.5ms^-2.

(ii) We can use the formula:

s = ut + 0.5at^2

where s is the distance travelled, u is the initial velocity, a is the acceleration, and t is the time taken. Since the initial velocity is zero, the formula simplifies to:

s = 0.5at^2

Plugging in the given values, we get:

40 = 0.5(12.5)t^2

Simplifying, we get:

t^2 = 6.4

Taking the square root of both sides, we get:

t = 2.53 seconds (rounded to two decimal places)

Therefore, the time taken

Akọwa Nkọwa

To find the mean age of the group of people, we need to calculate the sum of the product of each age range and the corresponding number of people, and divide by the total number of people in the group.

So, using the assume mean of 13 years, we can assume that the midpoint of the age range 1-5 is 3 years, the midpoint of 6-10 is 8 years, and so on. Then, we can calculate the sum of the products:

(3 x 18) + (8 x 12) + (13 x 25) + (18 x 15) + (23 x 20) + (28 x 10) = 1299

Next, we need to divide by the total number of people in the group:

18 + 12 + 25 + 15 + 20 + 10 = 100

So the mean age of the group is:

1299 / 100 = 12.99 years (rounded to two decimal places)

Therefore, the mean age of the group of people is approximately 12.99 years.

Akọwa Nkọwa

Let w be the adjacent side,

then w2 ${}^2$ = (P + q)2 ${}^2$ = 4pq so that w = 2pq−−√ $\sqrt{𝑝𝑞}$ 

Since tan x p−q2pq√ $\frac{𝑝-𝑞}{2\sqrt{𝑝𝑞}}$,

substituting;

1 - tan2 ${}^2$x = 1 - (p−q2pq√ $\frac{𝑝-𝑞}{2\sqrt{𝑝𝑞}}$)2 ${}^2$ 

= 1 - P2−2pq+q24pq $\frac{{𝑃}^2-2𝑝𝑞+{𝑞}^2}{4𝑝𝑞}$

= 4pq−p2+2pq−q24pq $\frac{4𝑝𝑞-{𝑝}^2+2𝑝𝑞-{𝑞}^2}{4𝑝𝑞}$

Therefore,

1 - tan2 ${}^2$x = 6pq−p2−q24pq

Akọwa Nkọwa

The given problem involves two parts: finding the tension (T) in a rope and the force (R) acting on an object at a certain angle.

In part (a), we are given the weight of an object as 120 N and the angle at which the rope is being pulled as 135 degrees. We use the formula: force = weight / sin(angle) to calculate the tension in the rope. We substitute the given values and calculate T as 169.71 N.

In part (b), we are given the weight of the same object as 120 N and the angle at which the force is acting as 135 degrees. We use the same formula: force = weight / sin(angle) to calculate the force (R) acting on the object. We substitute the given values and calculate R as 120 N.

Therefore, we can conclude that in part (a), we found the tension in the rope when the object was being pulled at an angle of 135 degrees, and in part (b), we found the force acting on the object when the force was being applied at an angle of 135 degrees.

Akọwa Nkọwa

To draw a histogram for this distribution, we need to first identify the range of each class interval and its corresponding frequency.

In this case, the class intervals are:

• 10.5 kg to 14.4 kg
• 14.5 kg to 24.4 kg
• 24.5 kg to 44.4 kg
• 44.5 kg to 47.4 kg
• 47.5 kg to 49.4 kg

And their corresponding frequencies are:

• 2
• 6
• 18
• 2
• 1

To draw the histogram, we need to represent the class intervals on the x-axis and the frequencies on the y-axis. The width of each bar should correspond to the range of the class interval, and the height should correspond to the frequency.

So, we can draw the histogram as follows:

       Frequency
        |     x
        |     x                 x
        |     x        x        x
        |     x                 x
        |     x           x
        |____________________________
           10.5-14.4  14.5-24.4  24.5-44.4  44.5-47.4  47.5-49.4

In this histogram, we can see that the class intervals are represented on the x-axis, and the frequencies are represented on the y-axis. The width of each bar corresponds to the range of the class interval, and the height corresponds to the frequency.

The bars are separated by a small gap to indicate that the classes are distinct. The histogram gives us a visual representation of the distribution of the masses of the group of persons. We can see that the majority of the group falls within the range of 24.5 kg to 44.4 kg.

Akọwa Nkọwa

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 +

Akọwa Nkọwa

P(a sum of 7)

= 636 $\frac{\mathrm{<span\; style="font-size:\; 16px;\; font-family:\; Ubuntu;">6</span>}}{\mathrm{<span\; style="font-size:\; 16px;\; font-family:\; Ubuntu;">36</span>}}$ 

and P(a sum of 4)

= 336    $\frac{\mathrm{<span\; style="font-size:\; 16px;"> 3</span>}}{\mathrm{<span\; style="font-size:\; 16px;">36</span>}}$

P(a sum of 4)

= 336 $\frac{3}{36}$ 

P(a sum of 7 and a sum of 4)

= 636×336 $\frac{6}{36}\times \frac{3}{36}$ = 172 $\frac{1}{72}$ 

Akọwa Nkọwa

(a) In this problem, we are dealing with a binomial distribution since we are selecting loaves of bread with replacement and there are only two possible outcomes: bad or good.

1. (i) The probability of getting exactly 6 bad loaves can be calculated using the binomial probability formula:
   P(X = 6) = (12 choose 6) * (0.3)^6 * (0.7)^6
   where "12 choose 6" is the number of ways to choose 6 loaves out of 12. Using a calculator, we get:
   P(X = 6) ≈ 0.2120
2. (ii) The probability of getting at least 4 bad loaves can be calculated by adding the probabilities of getting 4, 5, 6, ..., or 12 bad loaves:
   P(X ≥ 4) = P(X = 4) + P(X = 5) + ... + P(X = 12)
   Using the binomial probability formula for each term and a calculator, we get:
   P(X ≥ 4) ≈ 0.7094
3. (iii) The probability of getting no bad loaves can be calculated using the binomial probability formula for X = 0:
   P(X = 0) = (12 choose 0) * (0.3)^0 * (0.7)^12
   Using a calculator, we get:
   P(X = 0) ≈ 0.0138

(b) In this problem, we are dealing with a hypergeometric distribution since we are selecting a committee without replacement from a group with a fixed number of boys and girls.

The number of ways to choose 4 boys out of 8 is (8 choose 4), and the number of ways to choose 3 girls out of 5 is (5 choose 3). The total number of ways to choose 7 members out of 13 is (13 choose 7).

Therefore, the probability of choosing a committee with 4 boys and 3 girls is:

(8 choose 4) * (5 choose 3) / (13 choose 7) ≈ 0.2593

So the probability of choosing a committee with 4 boys and 3 girls is approximately 0.2593.

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

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

To find the value of (\(\alpha - \beta\)), we can use the fact that for a quadratic equation of the form ax\(^2\) + bx + c = 0, the sum of the roots is -b/a and the product of the roots is c/a.

In this case, the given quadratic equation is 3x\(^2\) + 4x - 5 = 0, which means that a = 3, b = 4, and c = -5. Using the formula for the sum of the roots, we get:

\(\alpha + \beta = -\frac{b}{a} = -\frac{4}{3}\)

Using the formula for the product of the roots, we get:

\(\alpha \beta = \frac{c}{a} = -\frac{5}{3}\)

To find the value of (\(\alpha - \beta\)), we can use the identity:

\(\alpha - \beta = \sqrt{(\alpha + \beta)^2 - 4\alpha \beta}\)

Plugging in the values we found earlier, we get:

\(\alpha - \beta = \sqrt{\left(-\frac{4}{3}\right)^2 - 4\left(-\frac{5}{3}\right)}\)

Simplifying, we get:

\(\alpha - \beta = \sqrt{\frac{16}{9} + \frac{20}{3}}\)

\(\alpha - \beta = \sqrt{\frac{68}{9}}\)

\(\alpha - \beta = \frac{2\sqrt{17}}{3}\)

Therefore, the value of (\(\alpha - \beta\)) is \(\frac{2\sqrt{17}}{3}\) in surd form.

Akọwa Nkọwa

(1) To find the probability of at least two of the 10 candidates failing, we can use the complementary probability approach: find the probability of none or only one candidate failing, and subtract that from 1. The probability of a single candidate failing is 40% (because 60% passed), and the probability of none failing is (60%)^10 = 0.604. Using the formula for combinations, we can find the number of ways to choose 0 or 1 failing candidates out of 10: C(10,0) + C(10,1) = 1 + 10 = 11. So, the probability of 0 or 1 failing is (0.604 + 0.4 * 0.6^9) = 0.604 + 0.266 = 0.87. Finally, the probability of at least two failing is 1 - 0.87 = 0.13.

(2) To find the probability of exactly half of the candidates passing, we need to find the number of ways to choose 5 passing and 5 failing candidates, and divide that by the total number of possible combinations of 10 candidates. Using combinations again, we find that there are C(10,5) = 252 ways to choose 5 passing candidates out of 10. So, the probability is 252 * (0.6)^5 * (0.4)^5 = 0.067.

(3) To find the probability of at most two of the candidates failing, we can find the probability of 0 or 1 or 2 failing and subtract that from 1. The probability of 0 failing is (0.6)^10 = 0.604. The probability of 1 failing is 10 * 0.4 * (0.6)^9 = 0.266. The probability of 2 failing is C(10,2) * (0.4)^2 * (0.6)^8 = 0.171. So, the probability of at most two failing is 0.604 + 0.266 + 0.171 = 1.04. Finally, the probability of more than two failing is 1 - 1.04 = -0.04, which is not a valid probability (probabilities must be between 0 and 1). This means that there is some error in the calculation; perhaps you meant to say "at most two of them passing," in which case the answer would be 1 - 0.604 = 0.396.

Akọwa Nkọwa

The probability that only one of the soldiers hits the target can be calculated by finding the sum of the probability that each soldier hits the target and the others miss.

Starting with soldier X, the probability that he hits the target and the others miss is: \(\frac{1}{3} \times \frac{2}{5} \times \frac{3}{4} = \frac{1}{15}\)

Similarly, the probability that soldier Y hits the target and the others miss is: \(\frac{2}{5} \times \frac{2}{3} \times \frac{3}{4} = \frac{4}{75}\)

And the probability that soldier Z hits the target and the others miss is: \(\frac{2}{3} \times \frac{4}{5} \times \frac{1}{4} = \frac{2}{45}\)

So the total probability that only one of the soldiers hits the target is: \(\frac{1}{15} + \frac{4}{75} + \frac{2}{45} = \frac{11}{45}\)

Therefore, the probability that only one of the soldiers hits the target is 0.24 or 24%.

Akọwa Nkọwa

To solve the given system of equations using determinants, we need to form a matrix of coefficients of the variables and then calculate its determinant. Let's call this matrix A.

A =

|5 -6 4|
|7 4 -3|
|2 1 6|

We also need to form a matrix B by replacing the column of constants with the column of variables.