WAEC SSCE - Further Mathematics - 2021

Simplify \(\frac{1}{3}\) log8 + \(\frac{1}{3}\) log 64 - 2 log6

Bayanin Amsa

Given that f: x --> x\(^2\) - x + 1 is defined on the Set Q = { x : 0 ≤ x < 20, x is a multiple of 5}. find the set of range of F.

Bayanin Amsa

In △PQR, \(\overline{PQ}\) = 5i - 2j and \(\overline{QR}\) = 4i + 3j. Find \(\overline{RP}\).

Bayanin Amsa

Three forces, F\(_1\) (8N, 030°), F\(_\2) (10N, 150° ) and F\(_\3) ( KN, 240° )are in equilibrium. Find the value of N

Bayanin Amsa

The first term of an AP is 4 and the sum of the first three terms is 18. Find the product of the first three terms

Bayanin Amsa

Let the common difference of the AP be denoted by d. Then, the first three terms of the AP are 4, 4 + d, and 4 + 2d, respectively. The sum of the first three terms of the AP is given as 18. Therefore, we have: 4 + (4 + d) + (4 + 2d) = 18 Simplifying the above equation, we get: 3d + 12 = 18 3d = 6 d = 2 Hence, the common difference of the AP is 2. Therefore, the first three terms of the AP are: 4, 6, 8 The product of the first three terms is: 4 x 6 x 8 = 192 Therefore, the answer is 192.

Find the inverse of  \(\begin{pmatrix} 4 & 2 \\ -3 & -2 \end{pmatrix}\)

Bayanin Amsa

Given that P = { x: 0 ≤ x ≤ 36, x is a factor of 36 divisible by 3} and Q = { x: 0 ≤ x ≤ 36, x is an even number and a perfect square}, find P n Q.

Bayanin Amsa

If ( 1- 2x)\(^4\) = 1 + px + qx\(^2\) - 32x\(^3\) + 16\(^4\), find the value of (q - p)

Bayanin Amsa

Consider the following statements:

X: Benita is polite

y: Benita is neat

z: Benita is intelligent

Which of the following symbolizes the statement: "Benita is neat if and only if she is neither polite nor intelligent"?

Bayanin Amsa

Find the radius of the circle 2x\(^2\) - 4x + 2y\(^2\) - 6y -2 = 0. 

Bayanin Amsa

To find the radius of a circle from its equation in general form, we need to rewrite the equation in the standard form of a circle, which is: (x - h)\(^2\) + (y - k)\(^2\) = r\(^2\) Where (h, k) is the center of the circle, and r is its radius. To do this, we complete the square for both x and y terms in the given equation. We start by rearranging the terms as follows: 2x\(^2\) - 4x + 2y\(^2\) - 6y -2 = 0 2x\(^2\) - 4x + 2y\(^2\) - 6y = 2 Now, we need to add and subtract appropriate constants to complete the square for x and y terms separately. For the x terms, we take half of the coefficient of x (-4/2 = -2) and square it to get 4. So, we add and subtract 4 to the equation: 2x\(^2\) - 4x + 4 - 4 + 2y\(^2\) - 6y = 2 We can now group the first three terms and factor it as a perfect square: 2(x - 1)\(^2\) + 2y\(^2\) - 6y - 2 = 0 For the y terms, we take half of the coefficient of y (-6/2 = -3) and square it to get 9. So, we add and subtract 9 to the equation: 2(x - 1)\(^2\) + 2(y - 3)\(^2\) - 2 - 9 = 0 2(x - 1)\(^2\) + 2(y - 3)\(^2\) = 11 Now, we have the equation in the standard form of a circle, where the center is at (1, 3) and the radius is the square root of 11/2. We can simplify this expression to get: sqrt(11/2) = sqrt(11)/sqrt(2) = sqrt(2)*sqrt(11)/2 Therefore, the answer is option (C), 17/√2.

A bag contains 8 red, 4 blue and 2 green identical balls. Two balls are drawn randomly from the bag without replacement. Find the probability that the balls drawn are red and blue.

A. 12/91 B. C. D.

Bayanin Amsa

A committee consists of 6 boys and 4 girls. In how many ways can a sub-committee consisting of 3 boys and 2 girls be formed if one particular boy and one particular girl must be on the sub-committee?

Bayanin Amsa

The table shows the distribution of marks obtained by some students in a test

What is the upper class boundary of the upper quartile class?

Bayanin Amsa

A binary operation * is defined on the set of real numbers, R, by

P * q = \(\frac{q^2 - p^2}{2pq}\). Find 3 * 2

Bayanin Amsa

The binary operation * is defined as: P * q = (q^2 - p^2) / (2pq). To find 3 * 2, we need to substitute the values of p and q in the equation: 3 * 2 = (2^2 - 3^2) / (2 * 3 * 2) = (-5) / (12) = -5/12 So, the answer is -5/12.

If 2i +pj and 4i -2j are perpendicular, find the value of p.

Bayanin Amsa

To find the value of p, we need to use the concept of perpendicular vectors. Two vectors are perpendicular if and only if the dot product of the two vectors is equal to zero. The dot product of two vectors can be calculated as the product of the magnitudes of the two vectors and the cosine of the angle between them. If the angle between the two vectors is 90 degrees, the cosine of the angle is zero and the dot product is also zero. Therefore, to find the value of p, we need to calculate the dot product of the two vectors, 2i + pj and 4i - 2j, and set it equal to zero. The dot product of two vectors (a, b) and (c, d) is given by: (a, b) * (c, d) = ac + bd So, the dot product of 2i + pj and 4i - 2j is: (2i + pj) * (4i - 2j) = (2 * 4) + (p * -2) = 8 - 2p = 0 Therefore, 2p = 8 and p = 4. So, the value of p is 4.

For what range of values of x is x\(^2\) - 2x - 3 ≤ 0

Bayanin Amsa

To solve the inequality x\(^2\) - 2x - 3 ≤ 0, we can use factoring or the quadratic formula. Factoring gives us (x - 3)(x + 1) ≤ 0, which means that the expression is less than or equal to zero when x is between or equal to -1 and 3, since the factors change sign at these values. Therefore, the correct answer is {x: -1 ≤ x ≤ 3}. Alternatively, we can use the quadratic formula to find the roots of the equation x\(^2\) - 2x - 3 = 0, which are x = -1 and x = 3. Since the quadratic function is a parabola that opens upward, it is negative in the interval between these two roots. Therefore, the expression x\(^2\) - 2x - 3 is less than or equal to zero when x is between or equal to -1 and 3.

Using binomial expansion of ( 1 + x)\(^6\) = 1 + 6x + 15x\(^2\) + 20x\(^3\) + 6x\(^5\) + x)\(^6\), find, correct to three decimal places, the value of (1.998))\(^6\)

Bayanin Amsa

To use the binomial expansion of (1 + x)\(^6\), we substitute x = 1.998, which gives: (1 + 1.998)\(^6\) = 1 + 6(1.998) + 15(1.998)\(^2\) + 20(1.998)\(^3\) + 6(1.998)\(^5\) + (1.998)\(^6\) We are interested in finding the value of (1.998)\(^6\), which is the last term on the right-hand side of the equation. We can solve for it by subtracting the other terms from both sides of the equation: (1.998)\(^6\) = (1 + 1.998)\(^6\) - 1 - 6(1.998) - 15(1.998)\(^2\) - 20(1.998)\(^3\) - 6(1.998)\(^5\) Using a calculator, we can evaluate the right-hand side of the equation to get: (1.998)\(^6\) ≈ 63.167 Therefore, the correct answer is option B, 63.167, rounded to three decimal places. Explanation: The binomial expansion of (1 + x)\(^6\) is a formula that allows us to expand the expression into a sum of terms involving powers of x. By substituting x = 1.998, we can use the formula to find the value of (1.998)\(^6\). We then use a calculator to evaluate the expression to obtain the final answer.

A stone is thrown vertically upward and distance, S metres after t seconds is given by S = 12t + \(\frac{5}{2t^2}\) - t\(^3\).

Calculate the distance travelled in the third second.

Bayanin Amsa

Find the equation of the normal to the curve y= 2x\(^2\) - 5x + 10 at P(1, 7).

Bayanin Amsa

Simplify ( \(\frac{1}{2 - √3}\) + \(\frac{2}{2 + √3}\) )\(^{-1}\)

Bayanin Amsa

If \(\frac{15 - 2x}{(x+4)(x-3)}\) = \(\frac{R}{(x+4)}\) \(\frac{9}{7(x-3)}\), find the value of R

Bayanin Amsa

Given that M = \(\begin{pmatrix} 3 & 2 \\ -1 & 4 \end{pmatrix}\) and N = \(\begin{pmatrix} 5 & 6 \\ -2 & -3 \end{pmatrix}\), calculate (3M - 2N)

Bayanin Amsa

To calculate 3M - 2N, we need to multiply each matrix by its scalar factor and then subtract the results. First, we have: 3M = 3 x \(\begin{pmatrix} 3 & 2 \\ -1 & 4 \end{pmatrix}\) = \(\begin{pmatrix} 9 & 6 \\ -3 & 12 \end{pmatrix}\) Next, we have: 2N = 2 x \(\begin{pmatrix} 5 & 6 \\ -2 & -3 \end{pmatrix}\) = \(\begin{pmatrix} 10 & 12 \\ -4 & -6 \end{pmatrix}\) Now we can subtract these two matrices to get: 3M - 2N = \(\begin{pmatrix} 9 & 6 \\ -3 & 12 \end{pmatrix}\) - \(\begin{pmatrix} 10 & 12 \\ -4 & -6 \end{pmatrix}\) = \(\begin{pmatrix} -1 & -6 \\ 1 & 18 \end{pmatrix}\) Therefore, the correct answer is (B) \(\begin{pmatrix} -1 & -6 \\ 1 & 18 \end{pmatrix}\).

If √5 cosx + √15sinx = 0, for 0° < x < 360°, find the values of x.

Bayanin Amsa

A body of mass 15kg is placed on a smooth plane which is inclined at 60° to the horizontal. If the box is at rest,

calculate the normal reaction to the plane. [ Take g = 10m/s\(^2\) ]

Bayanin Amsa

To find the normal reaction to the plane, we can use the following approach: Step 1: Draw a diagram of the situation. In this case, we have a body of mass 15kg on a smooth plane inclined at 60° to the horizontal. The weight of the body acts vertically downwards and is equal to mg, where m is the mass of the body and g is the acceleration due to gravity. Step 2: Resolve the weight of the body into two components, one perpendicular to the plane and the other parallel to the plane. The component of the weight perpendicular to the plane is equal to mg cos 60°, which is equal to (15 kg) x (10 m/s\(^2\)) x cos 60° = 75 N. Step 3: The normal reaction to the plane is equal in magnitude but opposite in direction to the component of the weight perpendicular to the plane. Therefore, the normal reaction to the plane is 75 N in the upward direction. Step 4: Check the options given in the question and select the one that matches the value obtained in Step 3. In this case, the correct option is 75N. Therefore, the normal reaction to the plane is 75 N.

If α and β are the roots of 3x\(^2\) - 7x + 6 = 0, find \(\frac{1}{α}\) + \(\frac{1}{β}\)

Bayanin Amsa

If f(x) = 4x\(^3\) + px\(^2\) + 7x - 23 is divided by (2x -5), the remainder is 7. find the value of p

Bayanin Amsa

g(x) = 2x + 3 and f(x) = 3x\(^2\) - 2x + 4

Bayanin Amsa

A fair die is tossed 60 times and the results are recorded in the table

Find the probability of obtaining a prime number.

Bayanin Amsa

A prime number is a number greater than 1 that is only divisible by 1 and itself. The prime numbers on a die are 2, 3, and 5. We know that the die is fair, which means that each number has an equal probability of being rolled. There are a total of 60 rolls, so we can use the frequency table to count the number of times each prime number appears on the die. The frequency of the number 2 is 10, the frequency of the number 3 is 14, and the frequency of the number 5 is 8. Therefore, the total number of times a prime number was rolled is: 10 + 14 + 8 = 32 The probability of rolling a prime number on a fair die is the total number of times a prime number was rolled divided by the total number of rolls: 32/60 = 8/15 Therefore, the answer is option D, 8/15.

A stone is thrown vertically upward and distance, S metres after t seconds is given by S = 12t + \(\frac{5}{2t^2}\) - t\(^3\).

Calculate the maximum height reached.

Bayanin Amsa

If sin x = \(\frac{12}{13}\) and sin y = \(\frac{4}{5}\), where x and y are acute angles, find  cos (x + y)

Bayanin Amsa

In how many ways can 8 persons be seated on a bench if only three seats are available?

Bayanin Amsa

The first person has 8 choices for a seat, the second person has 7 choices remaining, and the third person has 6 choices. However, the order in which they choose their seats does not matter, so we need to divide by the number of ways that the three people can be arranged. This is given by 3 factorial (3!), which is 3 x 2 x 1 = 6. Therefore, the total number of ways that 8 persons can be seated on a bench if only three seats are available is: 8 x 7 x 6 / 3! = 336 So the correct answer is (C) 336.

Given that F\(^1\)(x) = x\(^3\) √x, find f(x)

Bayanin Amsa

Find the value of the derivative of y = 3x\(^2\) (2x +1) with respect to x at the point x = 2. 

Bayanin Amsa

The gradient ofy= 3x\(^2\) + 11x + 7 at P(x.y) is -1. Find the coordinates of P. 

Bayanin Amsa

For what value of k is 4x\(^2\) - 12x + k, a perfect square?

Bayanin Amsa

Solve (\(\frac{1}{9}\))\(^{x + 2}\) = 243\(^{x - 2}\) 

Bayanin Amsa

The table shows the distribution of marks obtained by some students in a test

Find the modal class mark.

Bayanin Amsa

To find the modal class mark, we need to first identify the class with the highest frequency, which in this case is the class with marks 20-29 (16 students). The modal class mark is the midpoint of this class, which is calculated by adding the lower and upper limits of the class and dividing by 2: Modal class mark = (20 + 29) / 2 = 24.5 Therefore, the modal class mark is 24.5, which is option (C).

If 2y\(^2\) + 7 = 3y - xy, find \(\frac{dy}{dx}\)

Bayanin Amsa

The table shows the frequency distribution of heights (in cm) of pupils in a certain school.

 (a) (i) Construct a cumulative frequency table. (ii) Use the table to draw a cumulative frequency curve.

(b) Using the curve, estimate the: (i)median height; (ii) inter quartile range (iii) percentage of students whose heights are most 130cm.

Bayanin Amsa

(a) (i) To construct a cumulative frequency table, we need to add up the frequencies for each height interval and all the intervals before it. The cumulative frequency for each interval is shown in the table below:

(a) (ii) To draw a cumulative frequency curve, we plot the cumulative frequency for each interval against the upper limit of the interval. We then join the points with straight lines to form the curve, as shown below:

(b) (i) To estimate the median height, we locate the value on the cumulative frequency curve that corresponds to the 50th percentile. This is the point where the curve intersects the line at y = 0.5. We draw a line down to the x-axis to find the corresponding height. From the graph, we can see that the median height is around 124-125cm.

(b) (ii) To estimate the interquartile range (IQR), we need to find the heights corresponding to the 25th and 75th percentiles on the cumulative frequency curve. We draw lines down from these points to the x-axis to find the corresponding heights. The IQR is then the difference between these heights. From the graph, we can estimate the 25th percentile at around 116-117cm and the 75th percentile at around 134-135cm. Therefore, the IQR is around 18-19cm.

(b) (iii) To estimate the percentage of students whose heights are most 130cm, we locate the point on the cumulative frequency curve that corresponds to 130cm. We then draw a line across to the y-axis to find the corresponding percentile. From the graph, we can see that the percentage of students whose heights are most 130cm is around 60-65%.

Given that (p + 1/2√3)(1 - √3)\(^2\) = 3- √3,

 find x the value of p.

Bayanin Amsa

To solve for p in the given equation, we can start by simplifying the left-hand side of the equation using the identity (a + b)(a - b) = a² - b²:

(p + 1/2√3)(1 - √3)² = (p + 1/2√3)(1 - 2√3 + 3)
          = (p + 1/2√3)(4 - 2√3)
          = 4p - 2p√3 + √3/2 - 1/2
          = (4p - 1/2) - (2p√3 - √3/2)

Now we can set this expression equal to the right-hand side of the equation and solve for p:

4p - 1/2 - (2p√3 - √3/2) = 3 - √3
4p - 2p√3 = 7/2 - √3/2
2p(2 - √3) = 7/2 - √3/2
p = (7/4 - √3/4)/(2 - √3)

To simplify this expression, we can multiply the numerator and denominator by the conjugate of the denominator, which is 2 + √3:

p = [(7/4 - √3/4)(2 + √3)]/[(2 - √3)(2 + √3)]
p = (7/2 + √3/2 - 2√3/4 - √3/4)/(4 - 3)
p = (7/2 - 3√3/4)/1
p = 14/4 - 3√3/4
p = (7 - √3)/2

Therefore, the value of p that satisfies the given equation is (7 - √3)/2.

P and Q are two linear transformations in the X-Y plane defined by

P: (x, y) → (-3x + 6y, 4x + y) and

Q: (x, y) → (2x-3y, -4x - 6y).

(a) Write down the matrices of P and Q. (b) What is the image of (-2,-3) under the transformation Q?

(c) Obtain a single transformation representing the transformation Q followed by P.

(d) Find the image of (1,4) when transformed by Q followed by P.

(e) Find the image P\(^1\) of the point (-√2,2√2) under an anticlockwise rotation of 225° about the origin.

Bayanin Amsa

(a) The matrix representation of a linear transformation can be obtained by writing the image of the standard basis vectors (1,0) and (0,1). The matrix of P is given by: P = [[-3, 6], [4, 1]] Similarly, the matrix of Q is given by: Q = [[2, -3], [-4, -6]] (b) To find the image of a point (-2,-3) under the transformation Q, we first write the point as a column vector: (-2,-3) = [-2, -3]^T Next, we multiply the matrix Q with the vector: Q * [-2, -3]^T = [2, -3] * [-2, -3]^T = [-12, 18]^T So the image of (-2,-3) under the transformation Q is (-12, 18). (c) To find the single transformation representing the transformation Q followed by P, we multiply the matrices Q and P: Q * P = [[2, -3], [-4, -6]] * [[-3, 6], [4, 1]] = [[36, -18], [-24, -30]] So the single transformation representing the transformation Q followed by P is given by the matrix [[36, -18], [-24, -30]]. (d) To find the image of (1,4) when transformed by Q followed by P, we first write the point as a column vector: (1, 4) = [1, 4]^T Next, we multiply the matrix representing Q followed by P with the vector: [[36, -18], [-24, -30]] * [1, 4]^T = [36, -18] * 1 + [-24, -30] * 4 = [36, -18] + [-96, -120] = [-60, -138]^T So the image of (1,4) when transformed by Q followed by P is (-60, -138). (e) To find the image P^1 of the point (-√2,√2) under an anticlockwise rotation of 225° about the origin, we can first rotate the point by 225° in the counterclockwise direction and then apply the transformation P. The counterclockwise rotation of 225° can be represented by the matrix: R = [[cos(225), -sin(225)], [sin(225), cos(225)]] = [[-√2/2, √2/2], [-√2/2, -√2/2]] Next, we multiply the matrix R with the vector representing the point (-√2,√2): R * [-√2, √2]^T = [[-√2/2, √2/2], [-√2/2, -√2/2]] * [-√2, √2]^T = [-√2/2 * -√2 + √2/2 * √2, -√2/2 * -√2 - √2/2 * √2]^T = [√2, -√2]^T So the image of the point (-√2,√2) under the counterclockwise rotation of 225° is (√2, -√2). Finally, to find

The position vectors of P, Q and R with respect to the origin are (4i-5j), (i+3j) and (-5i+2j) respectively. If PQRM is a parallelogram, find:

(a) the coordinates of M;

(b) the acute angle between \(\overline{PM}\) and \(\overline{PQ}\), correct to the nearest degree.

Bayanin Amsa

To solve this problem, we first need to understand what a parallelogram is. A parallelogram is a four-sided shape with opposite sides that are parallel to each other. This means that if we draw lines from opposite corners of the parallelogram, these lines will intersect at the midpoint of the two sides they are drawn from.

Now let's look at the given position vectors of P, Q, and R. We can use these to find the position vector of M, which is opposite to P in the parallelogram. To do this, we need to use the fact that the position vector of the midpoint of a line segment is the average of the position vectors of the two endpoints. So, the position vector of M is:

¯¯¯¯¯¯¯¯ $\overline{𝑃𝑄}$ =  (13) $\left(\begin{array}{c}1\\ 3\end{array}\right)$ - (4−5) $\left(\begin{array}{c}4\\ -5\end{array}\right)$ = (−38) $\left(\begin{array}{c}-3\\ 8\end{array}\right)$

MR¯¯¯¯¯¯¯¯¯ $\overline{𝑀𝑅}$ =  (−52) $\left(\begin{array}{c}-5\\ 2\end{array}\right)$ - (xy) $\left(\begin{array}{c}𝑥\\ 𝑦\end{array}\right)$ = (−52−x−y) $\left(\begin{array}{cc}-5& -𝑥\\ 2& -𝑦\end{array}\right)$

 (−38) $\left(\begin{array}{c}-3\\ 8\end{array}\right)$ = (−52−x−y) $\left(\begin{array}{cc}-5& -𝑥\\ 2& -𝑦\end{array}\right)$

⇒ -5 - x = -3; x = -2

⇒ 2 - y = 8; y = -6
The coordinate of M (-2,-6)

(b) PM¯¯¯¯¯¯¯¯¯ $\overline{𝑃𝑀}$ = OM¯¯¯¯¯¯¯¯¯ $\overline{𝑂𝑀}$ - OP¯¯¯¯¯¯¯¯ $\overline{𝑂𝑃}$

=   (−2−6) $\left(\begin{array}{c}-2\\ -6\end{array}\right)$ -  (4−5) $\left(\begin{array}{c}4\\ -5\end{array}\right)$ =  (−6−1) $\left(\begin{array}{c}-6\\ -1\end{array}\right)$

    PQ¯¯¯¯¯¯¯¯ $\overline{𝑃𝑄}$ = OQ¯¯¯¯¯¯¯¯ $\overline{𝑂𝑄}$ - OP¯¯¯¯¯¯¯¯ $\overline{𝑂𝑃}$

=  (13) $\left(\begin{array}{c}1\\ 3\end{array}\right)$ -  (4−5) $\left(\begin{array}{c}4\\ -5\end{array}\right)$ =  (−38) $\left(\begin{array}{c}-3\\ 8\end{array}\right)$

|PM¯¯¯¯¯¯¯¯¯ $\overline{𝑃𝑀}$| = √(-62 ${}^2$ + [-12 ${}^2$]) = √37

|PQ¯¯¯¯¯¯¯¯ $\overline{𝑃𝑄}$| = √(-32 ${}^2$ + 82 ${}^2$) = √73

cosØ = PM.PQ|PM¯¯¯¯¯¯¯¯¯¯||PQ¯¯¯¯¯¯¯¯| $\frac{𝑃𝑀.𝑃𝑄}{|\overline{𝑃𝑀}||\overline{𝑃𝑄}|}$

cosØ = [−6i−j].[−3i+8j]√37.√73 $\frac{[-6𝑖-𝑗].[-3𝑖+8𝑗]}{\surd 37.\surd 73}$ = 10√2710 $\frac{10}{\surd 2710}$

⇒ Ø = cos−1 ${}^{-1}$ 10√2710 $\frac{10}{\surd 2710}$

: Ø = 78.91° ≈ 79°

(a) A girl threw a stone horizontally with a velocity of 30m/s from the top of a cliff 50m high. How far from the foot of the cliff does the stone strike the ground? [Take g= 10m/s\(^2\) 

(b) A body A, of mass 2kg is held in equilibrium by means of two strings AP and AR. AP is inclined at 56° to the upward vertical and AR is horizontal.

Find the tensions T\(_1\), and T\(_2\), in the strings [Take g= 10ms\(^2\)]

Bayanin Amsa

(a)

Time taken to reach the ground

Using S = ut + 1/2gt2 ${}^2$

 (u= 0 for a body falling from rest)

50 = 0 x t + 1/2 x 10t2 ${}^2$

5t2 ${}^2$ = 50;

t2 ${}^2$ =50/5 = 10

 t = √10 =3.16

Distance when it strikes the ground This journey is independent of gravity

d =vt =30 x 3.16 = 94.8m

(b)

Using Lami's rule,

20sin34 $\frac{\mathrm{<br>}}{\mathrm{<br>}}$ = T1sin90 $\frac{{\mathrm{<br>\; <br>}}_{}}{\mathrm{<br>}}$ = T2sin56 $\frac{{\mathrm{<br>\; <br>}}_{}}{}$

= T1 ${}_1$ 20.sin90sin34 $\frac{\mathrm{<br>}\mathrm{<br>}}{\mathrm{<br>}}$ = T2 ${}_2$ 20.sin56sin34 $\frac{\mathrm{<br>}\mathrm{<br>}}{\mathrm{<br>}}$ 

T1 ${}_1$ = 35.79N and T2 ${}_2$ = 29.65N

Given that x =  \(\begin{pmatrix} -4 \\ 3 \end{pmatrix}\) and y= \(\begin{pmatrix} -9 \\ 15 \end{pmatrix}\) calculate, correct to the nearest degree, the angle between the vectors

Bayanin Amsa

x =  (−43) $\left(\begin{array}{c}-4\\ 3\end{array}\right)$ and y= (−915) $\left(\begin{array}{c}-9\\ 15\end{array}\right)$

Changing x and y to the form xi+yj

we have:

x= -4i + 3j and y = -9i - 15j

using:

cosØ = xy|x|ly| $\frac{𝑥𝑦}{|𝑥|𝑙𝑦|}$

where Ø is the angle between x and y

xy = (-4i+3j) (-9i - 15j)

-36 + 60 x 0 - 27 x 0- 45 = -81

|x| = √(42 ${}^2$ +32 ${}^2$) =√(16 +9) = √25 = 5

|y| = √(-9)2 ${}^2$ + (-15)2 ${}^2$ = √(81 +225) = √306

cosØ =−815√306 $\frac{-81}{5\surd 306}$

= −9√306170 $\frac{-9\surd 306}{170}$

17.49∗9170 $\frac{17.49\ast 9}{170}$

Ø = cos−1 ${}^{-1}$(-0.1029);

Ø = 95.91°

Ø = 96°

The polynomial f(x) =2x\(^3\) + px+ qx - 5 has  (x-1) as a factor and a remainder of 27 when divided by (x + 2), where p and q are constants. Find the values of p and q.

2x^3 + px + qx - 5 
          ÷ x + 2
2          1    p/2    (q - 5)/2
-2x^2 + (p - 4)x + (q - 5)

-2x^2 + (p - 4)x + (q - 5) = 27

p - 4 = 0
p = 4
q - 5 = 27
q = 32

Bayanin Amsa

To find the values of p and q, we can use the Remainder Theorem and synthetic division. The Remainder Theorem states that if a polynomial f(x) is divided by x - a, then the remainder is equal to f(a).

So, in this case, when we divide f(x) by (x - 1), the remainder is 27, meaning that f(1) = 27. We can use this information to find one of the constants, p.

Next, we can use synthetic division to divide f(x) by (x + 2) and find the remainder. The process of synthetic division involves dividing each term of the polynomial by the leading coefficient of the divisor, and then using the coefficients to find the remainder.

In this case, the polynomial f(x) is divided by (x + 2), giving us:

2x^3 + px + qx - 5 
          ÷ x + 2
2          1    p/2    (q - 5)/2
-2x^2 + (p - 4)x + (q - 5)

Since the remainder is 27, we have:

-2x^2 + (p - 4)x + (q - 5) = 27

Solving for p and q, we have:

p - 4 = 0
p = 4
q - 5 = 27
q = 32

So, the values of p and q are p = 4 and q = 32.

(a) The speed of a moving bus reduced from 45m/s to 5m/s with a uniform retardation of 10m/s\(^2\). Calculate the distance covered.

(b) A bucket full of water with mass 16kg is pulled out of a well with a light inextensible rope. Find its acceleration when the tension in the rope is 240N. [Take g= 10m/s\(^2\)]

Bayanin Amsa

(a) The initial velocity of the bus is 45 m/s and the final velocity is 5 m/s, and the retardation is 10 m/s². We can use the equation of motion to find the distance covered. The equation is:

v² = u² + 2as

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

Substituting the given values, we get:

5² = 45² + 2(-10)s

25 = 2025 - 20s

20s = 2000

s = 100 meters

Therefore, the distance covered by the bus is 100 meters.

(b) The weight of the bucket is given by W = mg, where m is the mass of the bucket and g is the acceleration due to gravity. The tension in the rope is equal to the weight of the bucket, so we have:

T = W = mg

The net force acting on the bucket is given by F = T - mg, where F is the force accelerating the bucket.

Using Newton's second law of motion, F = ma, where a is the acceleration of the bucket. Substituting the values, we get:

ma = T - mg

a = (T - mg) / m

Substituting the given values, we get:

a = (240 - 16 x 10) / 16

a = 10 m/s²

Therefore, the acceleration of the bucket is 10 m/s².

A bag contains 24 mangoes out of which six are bad. If 6 mangoes are selected randomly from the bag with replacement, find the probability that not more than 3 are bad.

Bayanin Amsa

The probability of selecting a bad mango from the bag is 6/24 = 1/4. This means that the probability of selecting a good mango is 3/4.

To find the probability that not more than 3 mangoes are bad, we need to consider four cases: selecting 0 bad mangoes, selecting 1 bad mango, selecting 2 bad mangoes, or selecting 3 bad mangoes.

Case 1: Selecting 0 bad mangoes

The probability of selecting a good mango on the first draw is 3/4. This probability remains the same for each of the 6 mangoes that are selected. Therefore, the probability of selecting 0 bad mangoes is (3/4)^6 = 0.1771.

Case 2: Selecting 1 bad mango

There are 6 ways to select 1 bad mango out of 6 mangoes, and 5 ways to select 5 good mangoes out of the remaining 18 mangoes. Therefore, the probability of selecting 1 bad mango is (6/24) * (18/24)^5 * 6 = 0.3934.

Case 3: Selecting 2 bad mangoes

There are 15 ways to select 2 bad mangoes out of 6 mangoes, and 4 ways to select 4 good mangoes out of the remaining 18 mangoes. Therefore, the probability of selecting 2 bad mangoes is (15/24)^2 * (9/24)^4 * 15 = 0.3147.

Case 4: Selecting 3 bad mangoes

There are 20 ways to select 3 bad mangoes out of 6 mangoes, and 3 ways to select 3 good mangoes out of the remaining 18 mangoes. Therefore, the probability of selecting 3 bad mangoes is (6/24)^3 * (18/24)^3 * 20 = 0.0983.

The total probability of not selecting more than 3 bad mangoes is the sum of the probabilities of these four cases: 0.1771 + 0.3934 + 0.3147 + 0.0983 = 0.9835.

Therefore, the probability of not more than 3 bad mangoes being selected is 0.9835 or approximately 98.35%.

The table shows the distribution of monthly income (in thousands of naira) of workers in a factory

(a) Draw a histogram for the distribution.

(b) Use your graph to estimate the mode of the distribution.