WAEC SSCE - Further Mathematics - 2021
Pergunta 2 Relatório
The table shows the distribution of marks obtained by some students in a test
| Marks | 0-9 | 10-19 | 20-29 | 30-39 | 40-49 |
| Frequency | 4 | 12 | 16 | 6 | 2 |
Find the modal class mark.
Detalhes da Resposta
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).
Pergunta 3 Relatório
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
Detalhes da Resposta
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.
Pergunta 4 Relatório
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?
Detalhes da Resposta
Pergunta 5 Relatório
The table shows the distribution of marks obtained by some students in a test
| Marks | 0-9 | 10-19 | 20-29 | 30-39 | 40-49 |
| Frequency | 4 | 12 | 16 | 6 | 2 |
What is the upper class boundary of the upper quartile class?
Detalhes da Resposta
Pergunta 7 Relatório
For what range of values of x is x\(^2\) - 2x - 3 ≤ 0
Detalhes da Resposta
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.
Pergunta 8 Relatório
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.
Detalhes da Resposta
Pergunta 9 Relatório
A fair die is tossed 60 times and the results are recorded in the table
| Number of die | 1 | 2 | 3 | 4 | 5 | 6 |
| Frequency | 15 | 10 | 14 | 2 | 8 | 11 |
Find the probability of obtaining a prime number.
Detalhes da Resposta
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.
Pergunta 10 Relatório
If 2i +pj and 4i -2j are perpendicular, find the value of p.
Detalhes da Resposta
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.
Pergunta 13 Relatório
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.
Detalhes da Resposta
Pergunta 14 Relatório
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))\(^6\)
Detalhes da Resposta
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.
Pergunta 15 Relatório
If \(\frac{15 - 2x}{(x+4)(x-3)}\) = \(\frac{R}{(x+4)}\) \(\frac{9}{7(x-3)}\), find the value of R
Detalhes da Resposta
Pergunta 16 Relatório
Three forces, F\(_1\) (8N, 030°), F\(_\2) (10N, 150° ) and F\(_\3) ( KN, 240° )are in equilibrium. Find the value of N
Detalhes da Resposta
Pergunta 17 Relatório
If α and β are the roots of 3x\(^2\) - 7x + 6 = 0, find \(\frac{1}{α}\) + \(\frac{1}{β}\)
Detalhes da Resposta
Pergunta 18 Relatório
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.
Detalhes da Resposta
Pergunta 19 Relatório
Find the inverse of \(\begin{pmatrix} 4 & 2 \\ -3 & -2 \end{pmatrix}\)
Detalhes da Resposta
Pergunta 20 Relatório
In how many ways can 8 persons be seated on a bench if only three seats are available?
Detalhes da Resposta
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.
Pergunta 21 Relatório
Given that M = \(\begin{pmatrix} 3 & 2 \\ -1 & 4 \end{pmatrix}\) and N = \(\begin{pmatrix} 5 & 6 \\ -2 & -3 \end{pmatrix}\), calculate (3M - 2N)
Detalhes da Resposta
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) \(\begin{pmatrix} -1 & -6 \\ 1 & 18 \end{pmatrix}\).
Pergunta 22 Relatório
If sin x = \(\frac{12}{13}\) and sin y = \(\frac{4}{5}\), where x and y are acute angles, find cos (x + y)
Detalhes da Resposta
Pergunta 23 Relatório
For what value of k is 4x\(^2\) - 12x + k, a perfect square?
Detalhes da Resposta
Pergunta 24 Relatório
Find the equation of the normal to the curve y= 2x\(^2\) - 5x + 10 at P(1, 7).
Detalhes da Resposta
Pergunta 25 Relatório
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"?
Detalhes da Resposta
Pergunta 26 Relatório
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.
Detalhes da Resposta
Pergunta 27 Relatório
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.
Detalhes da Resposta
Pergunta 28 Relatório
If ( 1- 2x)\(^4\) = 1 + px + qx\(^2\) - 32x\(^3\) + 16\(^4\), find the value of (q - p)
Detalhes da Resposta
Pergunta 29 Relatório
If f(x) = 4x\(^3\) + px\(^2\) + 7x - 23 is divided by (2x -5), the remainder is 7. find the value of p
Detalhes da Resposta
Pergunta 30 Relatório
Find the value of the derivative of y = 3x\(^2\) (2x +1) with respect to x at the point x = 2.
Detalhes da Resposta
Pergunta 31 Relatório
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\(^2\) ]
Detalhes da Resposta
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.
Pergunta 32 Relatório
A binary operation * is defined on the set of real numbers, R, by
P * q = \(\frac{q^2 - p^2}{2pq}\). Find 3 * 2
Detalhes da Resposta
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.
Pergunta 33 Relatório
If √5 cosx + √15sinx = 0, for 0° < x < 360°, find the values of x.
Detalhes da Resposta
Pergunta 34 Relatório
In △PQR, \(\overline{PQ}\) = 5i - 2j and \(\overline{QR}\) = 4i + 3j. Find \(\overline{RP}\).
Detalhes da Resposta
Pergunta 35 Relatório
The gradient ofy= 3x\(^2\) + 11x + 7 at P(x.y) is -1. Find the coordinates of P.
Detalhes da Resposta
Pergunta 36 Relatório
Find the radius of the circle 2x\(^2\) - 4x + 2y\(^2\) - 6y -2 = 0.
Detalhes da Resposta
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.
Pergunta 37 Relatório
Simplify ( \(\frac{1}{2 - √3}\) + \(\frac{2}{2 + √3}\) )\(^{-1}\)
Detalhes da Resposta
Pergunta 38 Relatório
Simplify \(\frac{1}{3}\) log8 + \(\frac{1}{3}\) log 64 - 2 log6
Detalhes da Resposta
Pergunta 39 Relatório
Evaluate: \(^9{?}_1\) \(\frac{x(2x-3)}{?x}\) dx
9∫1 x(2x−3)√x dx = 2x2−3x)x1/2
= x −1/2 (2x2 - 3x)
= 2x3/2 - 3x1/2
= 2x3/23/2+1 - 3x1/21/2+1
= 2x3/25/2 - 3x1/23/2
= 9∫1 [4x3/25 - 6x1/23 ]
[ 4(9)3/25 - 6(9)1/23 ] - [ 4(9)1/25 - 6(1)1/23 ]
= [ 4(27)5 - 6(3)3 - [ 45 - 2 ]
= [ 1085- 2 ] - [ 65 ]
= 785 + 65= 845
164/5
Detalhes da Resposta
9∫1 x(2x−3)√x dx = 2x2−3x)x1/2
= x −1/2 (2x2 - 3x)
= 2x3/2 - 3x1/2
= 2x3/23/2+1 - 3x1/21/2+1
= 2x3/25/2 - 3x1/23/2
= 9∫1 [4x3/25 - 6x1/23 ]
[ 4(9)3/25 - 6(9)1/23 ] - [ 4(9)1/25 - 6(1)1/23 ]
= [ 4(27)5 - 6(3)3 - [ 45 - 2 ]
= [ 1085- 2 ] - [ 65 ]
= 785 + 65= 845
164/5
Pergunta 40 Relatório
The table shows the distribution of monthly income (in thousands of naira) of workers in a factory
| Monthly Income (N'1000) | 135-139 | 140-149 | 150-154 | 155-164 | 165-169 |
| Number of workers | 20 | 42 | 28 | 38 | 22 |
(a) Draw a histogram for the distribution.
(b) Use your graph to estimate the mode of the distribution.
(a) Draw a Histogram
Given the data on monthly income distribution, we can create a histogram. Here's a step-by-step process for drawing the histogram:
- Class Intervals: The monthly income classes are given as ranges (in thousands of naira).
- Frequency: The number of workers in each class interval represents the frequency.
Let's list the class intervals and their corresponding frequencies:
- 135-139: 20 workers
- 140-149: 42 workers
- 150-154: 28 workers
- 155-164: 38 workers
- 165-169: 22 workers
We'll plot these intervals on the x-axis and the number of workers on the y-axis.
(b) Estimate the Mode
The mode of a distribution is the value that appears most frequently. In a histogram, the mode corresponds to the class interval with the highest frequency.
Identify the class interval with the highest frequency.
Use the following formula for estimating the mode in a grouped frequency distribution:
Mode=L+((2fm??f1??f2?)(fm??f1?)?)×wwhere:
- L is the lower boundary of the modal class,
- fm? is the frequency of the modal class,
- f1? is the frequency of the class preceding the modal class,
- f2? is the frequency of the class succeeding the modal class,
- w is the width of the class intervals.
Let's plot the histogram and estimate the mode.
Drawing the Histogram and Estimating the Mode
(a) Histogram
The histogram above displays the distribution of monthly income (in thousands of naira) among workers in a factory. The x-axis represents the monthly income ranges, while the y-axis shows the number of workers in each range.
(b) Estimation of the Mode
Using the histogram and the data provided, the mode can be estimated with the following steps:
Identify the Modal Class:
- The class interval with the highest frequency is 140?149 with 42 workers.
Apply the Mode Formula:
- L=140 (lower boundary of the modal class)
- fm?=42 (frequency of the modal class)
- f1?=20 (frequency of the class preceding the modal class, which is 135?139)
- f2?=28 (frequency of the class succeeding the modal class, which is 150?154)
- w=10 (width of the class intervals)
Using the formula for the mode:
Mode=140+((2×42?20?28)(42?20)?)×10Simplifying this:
Mode=140+(84?4822?)×10 Mode=140+(3622?)×10 Mode?140+6.111 Mode?146.11
Therefore, the estimated mode of the distribution is approximately 146.11 thousand naira.
Detalhes da Resposta
(a) Draw a Histogram
Given the data on monthly income distribution, we can create a histogram. Here's a step-by-step process for drawing the histogram:
- Class Intervals: The monthly income classes are given as ranges (in thousands of naira).
- Frequency: The number of workers in each class interval represents the frequency.
Let's list the class intervals and their corresponding frequencies:
- 135-139: 20 workers
- 140-149: 42 workers
- 150-154: 28 workers
- 155-164: 38 workers
- 165-169: 22 workers
We'll plot these intervals on the x-axis and the number of workers on the y-axis.
(b) Estimate the Mode
The mode of a distribution is the value that appears most frequently. In a histogram, the mode corresponds to the class interval with the highest frequency.
Identify the class interval with the highest frequency.
Use the following formula for estimating the mode in a grouped frequency distribution:
Mode=L+((2fm??f1??f2?)(fm??f1?)?)×wwhere:
- L is the lower boundary of the modal class,
- fm? is the frequency of the modal class,
- f1? is the frequency of the class preceding the modal class,
- f2? is the frequency of the class succeeding the modal class,
- w is the width of the class intervals.
Let's plot the histogram and estimate the mode.
Drawing the Histogram and Estimating the Mode
(a) Histogram
The histogram above displays the distribution of monthly income (in thousands of naira) among workers in a factory. The x-axis represents the monthly income ranges, while the y-axis shows the number of workers in each range.
(b) Estimation of the Mode
Using the histogram and the data provided, the mode can be estimated with the following steps:
Identify the Modal Class:
- The class interval with the highest frequency is 140?149 with 42 workers.
Apply the Mode Formula:
- L=140 (lower boundary of the modal class)
- fm?=42 (frequency of the modal class)
- f1?=20 (frequency of the class preceding the modal class, which is 135?139)
- f2?=28 (frequency of the class succeeding the modal class, which is 150?154)
- w=10 (width of the class intervals)
Using the formula for the mode:
Mode=140+((2×42?20?28)(42?20)?)×10Simplifying this:
Mode=140+(84?4822?)×10 Mode=140+(3622?)×10 Mode?140+6.111 Mode?146.11
Therefore, the estimated mode of the distribution is approximately 146.11 thousand naira.
Pergunta 41 Relatório
A box contains 5 red, 7 blue and 4 green identical bulbs. Two bulbs are picked at random from the box without replacement.
Calculate the probability of picking:
(a) same color of bulbs; (6) different color of bulbs (c) at least one red bulb.
Total bulbs:
n(Red)=5, n(B) = 7, n(G) = 4
(5+7+4) = 16
p(R)= 5/16, p(B) = 7/16, p(G) = 4/16
(a) p(All same colour of bulbs)
= p(RR) Or p(BB) or p(GG)
516
* 415
+ 716
* 615
+ 416
* 315
= 112 + 740 + 120
= 10+21+6120 = 37120
(b) All different colors = 1-p (AIl the same colour) =
1 - 37120 = 83120
(c) p(At least one red) = p(RR) + p(RB) + p(RG)
516 * 415 + 516 * 715 + 516 * 415
= 112 + 748 + 112
8+748 = 1548
= 516
Detalhes da Resposta
Total bulbs:
n(Red)=5, n(B) = 7, n(G) = 4
(5+7+4) = 16
p(R)= 5/16, p(B) = 7/16, p(G) = 4/16
(a) p(All same colour of bulbs)
= p(RR) Or p(BB) or p(GG)
516
* 415
+ 716
* 615
+ 416
* 315
= 112 + 740 + 120
= 10+21+6120 = 37120
(b) All different colors = 1-p (AIl the same colour) =
1 - 37120 = 83120
(c) p(At least one red) = p(RR) + p(RB) + p(RG)
516 * 415 + 516 * 715 + 516 * 415
= 112 + 748 + 112
8+748 = 1548
= 516
Pergunta 42 Relatório
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.
(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
Detalhes da Resposta
(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
Pergunta 43 Relatório
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.
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%.
Detalhes da Resposta
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%.
Pergunta 44 Relatório
(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\(^2\)]
(a)
Time taken to reach the ground
Using S = ut + 1/2gt2
(u= 0 for a body falling from rest)
50 = 0 x t + 1/2 x 10t2
5t2 = 50;
t2 =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 = T1sin90 = T2sin56
= T1 20.sin90sin34 = T2 20.sin56sin34
T1 = 35.79N and T2 = 29.65N
Detalhes da Resposta
(a)
Time taken to reach the ground
Using S = ut + 1/2gt2
(u= 0 for a body falling from rest)
50 = 0 x t + 1/2 x 10t2
5t2 = 50;
t2 =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 = T1sin90 = T2sin56
= T1 20.sin90sin34 = T2 20.sin56sin34
T1 = 35.79N and T2 = 29.65N
Pergunta 45 Relatório
(a) A jogger is training for 15km charity race. He starts with a run of 500 metres, then he increases the distance he runs daily by 250 metres.
(i) How many days will it take the jogger to reach a distance of 15km in training?
(ii) Calculate the total distance he would have run in the training.
(b) The second term of a Geometric Progression (GP) is -3. If its sum to infinity is 25/2, find its common ratios.
(a)
(i) To reach a distance of 15km, the jogger needs to cover 15000 meters.
Let's call the number of days it takes to reach this distance "n."
On the first day, he runs 500 meters. On the second day, he runs 500 + 250 = 750 meters. On the third day, he runs 750 + 250 = 1000 meters.
In general, on the nth day, he runs 500 + 250(n-1) meters.
So we can set up an equation:
500 + 750 + 1000 + ... + (500 + 250(n-1)) = 15000
Simplifying, we get:
250n^2 + 250n - 15000 = 0
Dividing both sides by 250:
n^2 + n - 60 = 0
This equation can be factored as:
(n + 6)(n - 10) = 0
Since we're looking for a positive value for n, the answer is n = 10.
So it will take the jogger 10 days to reach a distance of 15km in training.
(ii) We can use the formula for the sum of a geometric series:
S = a(1 - r^n)/(1 - r)
where S is the sum of the series, a is the first term, r is the common ratio, and n is the number of terms.
In this case, we know that the second term is -3, so a = 500 * (-3) = -1500.
We also know that the sum to infinity is 25/2, so S = 25/2.
Plugging in these values and solving for r:
25/2 = (-1500)(1 - r^10)/(1 - r)
r = 1/2 or -1 (discarded since a negative ratio would make the terms alternate in sign, and the second term is negative)
So the common ratio is 1/2.
The total distance the jogger would have run in the training is the sum of the terms of the geometric series:
S = a/(1 - r) = (-1500)/(1 - 1/2) = 3000 meters.
(b)
We know that the second term of the GP is -3, so we can write the first two terms as:
a, ar
where ar = -3.
We also know that the sum to infinity is 25/2, so we can use the formula:
S = a/(1 - r)
25/2 = a/(1 - r)
a = 25/2 - 25r/2
Substituting this value of a into the equation ar = -3:
(25/2 - 25r/2)r = -3
Simplifying:
25r^2 - 50r + 6 = 0
We can solve for r using the quadratic formula:
r = (50
Detalhes da Resposta
(a)
(i) To reach a distance of 15km, the jogger needs to cover 15000 meters.
Let's call the number of days it takes to reach this distance "n."
On the first day, he runs 500 meters. On the second day, he runs 500 + 250 = 750 meters. On the third day, he runs 750 + 250 = 1000 meters.
In general, on the nth day, he runs 500 + 250(n-1) meters.
So we can set up an equation:
500 + 750 + 1000 + ... + (500 + 250(n-1)) = 15000
Simplifying, we get:
250n^2 + 250n - 15000 = 0
Dividing both sides by 250:
n^2 + n - 60 = 0
This equation can be factored as:
(n + 6)(n - 10) = 0
Since we're looking for a positive value for n, the answer is n = 10.
So it will take the jogger 10 days to reach a distance of 15km in training.
(ii) We can use the formula for the sum of a geometric series:
S = a(1 - r^n)/(1 - r)
where S is the sum of the series, a is the first term, r is the common ratio, and n is the number of terms.
In this case, we know that the second term is -3, so a = 500 * (-3) = -1500.
We also know that the sum to infinity is 25/2, so S = 25/2.
Plugging in these values and solving for r:
25/2 = (-1500)(1 - r^10)/(1 - r)
r = 1/2 or -1 (discarded since a negative ratio would make the terms alternate in sign, and the second term is negative)
So the common ratio is 1/2.
The total distance the jogger would have run in the training is the sum of the terms of the geometric series:
S = a/(1 - r) = (-1500)/(1 - 1/2) = 3000 meters.
(b)
We know that the second term of the GP is -3, so we can write the first two terms as:
a, ar
where ar = -3.
We also know that the sum to infinity is 25/2, so we can use the formula:
S = a/(1 - r)
25/2 = a/(1 - r)
a = 25/2 - 25r/2
Substituting this value of a into the equation ar = -3:
(25/2 - 25r/2)r = -3
Simplifying:
25r^2 - 50r + 6 = 0
We can solve for r using the quadratic formula:
r = (50
Pergunta 46 Relatório
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.
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.
Detalhes da Resposta
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.
Pergunta 47 Relatório
\(^{5y}{C}_2\) = 190, find the value of y
We can use the formula for combinations to solve for y:
5yC2 = (5y)! / (2!(5y-2)!) = 190
Expanding the factorials:
(5y)(5y-1) / 2 = 190
Solving for y:
(5y)(5y-1) = 380
25y^2 - 5y = 380
25y^2 - 5y - 380 = 0
Using the quadratic formula:
y = (-b ± √(b^2 - 4ac))/(2a)
y = (-(−5) ± √((-5)^2 - 4(25)(-380)))/(2(25))
y = (5 ± √(25 + 90000))/(50)
y = (5 ± √(90025))/(50)
Taking the positive root:
y = (5 + 300)/(50)
y = 305/50
y = 6.1
So the value of y is approximately 6.1.
Detalhes da Resposta
We can use the formula for combinations to solve for y:
5yC2 = (5y)! / (2!(5y-2)!) = 190
Expanding the factorials:
(5y)(5y-1) / 2 = 190
Solving for y:
(5y)(5y-1) = 380
25y^2 - 5y = 380
25y^2 - 5y - 380 = 0
Using the quadratic formula:
y = (-b ± √(b^2 - 4ac))/(2a)
y = (-(−5) ± √((-5)^2 - 4(25)(-380)))/(2(25))
y = (5 ± √(25 + 90000))/(50)
y = (5 ± √(90025))/(50)
Taking the positive root:
y = (5 + 300)/(50)
y = 305/50
y = 6.1
So the value of y is approximately 6.1.
Pergunta 48 Relatório
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.
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:
¯¯¯¯¯¯¯¯ = (13) - (4−5) = (−38)
MR¯¯¯¯¯¯¯¯¯ = (−52) - (xy) = (−52−x−y)
(−38) = (−52−x−y)
⇒ -5 - x = -3; x = -2
⇒ 2 - y = 8; y = -6
The coordinate of M (-2,-6)
(b) PM¯¯¯¯¯¯¯¯¯ = OM¯¯¯¯¯¯¯¯¯ - OP¯¯¯¯¯¯¯¯
= (−2−6) - (4−5) = (−6−1)
PQ¯¯¯¯¯¯¯¯ = OQ¯¯¯¯¯¯¯¯ - OP¯¯¯¯¯¯¯¯
= (13) - (4−5) = (−38)
|PM¯¯¯¯¯¯¯¯¯ | = √(-62 + [-12 ]) = √37
|PQ¯¯¯¯¯¯¯¯ | = √(-32 + 82 ) = √73
cosØ = PM.PQ|PM¯¯¯¯¯¯¯¯¯¯||PQ¯¯¯¯¯¯¯¯|
cosØ = [−6i−j].[−3i+8j]√37.√73 = 10√2710
⇒ Ø = cos−1 10√2710
: Ø = 78.91° ≈ 79°
Detalhes da Resposta
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:
¯¯¯¯¯¯¯¯ = (13) - (4−5) = (−38)
MR¯¯¯¯¯¯¯¯¯ = (−52) - (xy) = (−52−x−y)
(−38) = (−52−x−y)
⇒ -5 - x = -3; x = -2
⇒ 2 - y = 8; y = -6
The coordinate of M (-2,-6)
(b) PM¯¯¯¯¯¯¯¯¯ = OM¯¯¯¯¯¯¯¯¯ - OP¯¯¯¯¯¯¯¯
= (−2−6) - (4−5) = (−6−1)
PQ¯¯¯¯¯¯¯¯ = OQ¯¯¯¯¯¯¯¯ - OP¯¯¯¯¯¯¯¯
= (13) - (4−5) = (−38)
|PM¯¯¯¯¯¯¯¯¯ | = √(-62 + [-12 ]) = √37
|PQ¯¯¯¯¯¯¯¯ | = √(-32 + 82 ) = √73
cosØ = PM.PQ|PM¯¯¯¯¯¯¯¯¯¯||PQ¯¯¯¯¯¯¯¯|
cosØ = [−6i−j].[−3i+8j]√37.√73 = 10√2710
⇒ Ø = cos−1 10√2710
: Ø = 78.91° ≈ 79°
Pergunta 49 Relatório
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
x = (−43) and y= (−915)
Changing x and y to the form xi+yj
we have:
x= -4i + 3j and y = -9i - 15j
using:
cosØ = 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 +32 ) =√(16 +9) = √25 = 5
|y| = √(-9)2 + (-15)2 = √(81 +225) = √306
cosØ =−815√306
= −9√306170
17.49∗9170
Ø = cos−1 (-0.1029);
Ø = 95.91°
Ø = 96°
Detalhes da Resposta
x = (−43) and y= (−915)
Changing x and y to the form xi+yj
we have:
x= -4i + 3j and y = -9i - 15j
using:
cosØ = 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 +32 ) =√(16 +9) = √25 = 5
|y| = √(-9)2 + (-15)2 = √(81 +225) = √306
cosØ =−815√306
= −9√306170
17.49∗9170
Ø = cos−1 (-0.1029);
Ø = 95.91°
Ø = 96°
Pergunta 50 Relatório
The table shows the frequency distribution of heights (in cm) of pupils in a certain school.
| Heights |
100-109 | 110-119 | 120-129 | 130-139 | 140-149 | 150-159 | 160-169 |
| Frequency |
27 | 58 | 130 | 105 | 50 | 25 | 5 |
(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.
(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:
Heights |
100-109 | 110-119 | 120-129 | 130-139 | 140-149 | 150-159 | 160-169 |
Frequency |
27 | 58 | 130 | 105 | 50 | 25 | 5 |
Cumulative Frequency |
27 | 85 | 215 | 320 | 370 | 395 | 400 |
(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%.
Detalhes da Resposta
(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:
Heights |
100-109 | 110-119 | 120-129 | 130-139 | 140-149 | 150-159 | 160-169 |
Frequency |
27 | 58 | 130 | 105 | 50 | 25 | 5 |
Cumulative Frequency |
27 | 85 | 215 | 320 | 370 | 395 | 400 |
(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%.
Pergunta 51 Relatório
Given that (p + 1/2√3)(1 - √3)\(^2\) = 3- √3,
find x the value of p.
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.
Detalhes da Resposta
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.
Pergunta 52 Relatório
(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\(^2\)]
(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.
Detalhes da Resposta
(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.
Pergunta 53 Relatório
(a) Find the equation of the normal to the curve y = (x\(^2\) - x + 1)(x - 2) at the point where the curve cuts the X - axis.
(b) The coordinates of the pints P, Q and R are (-1, 2), (5, 1) and (3, -4) respectively. Find the equation of the line joining Q and the midpoint of \(\overline{PR}\).
(a) To find the equation of the normal to the curve y = (x2 - x + 1)(x - 2) at the point where the curve cuts the X-axis, we first need to find the x-coordinate of the point where the curve intersects the X-axis. This is where y = 0, so we can solve the equation (x2 - x + 1)(x - 2) = 0 to find the roots of the equation. The roots are x = 1 ± i√3 and x = 2, but since we want the point where the curve intersects the X-axis, we take x = 2.
Next, we need to find the gradient of the curve at this point. We can do this by differentiating the equation y = (x2 - x + 1)(x - 2) with respect to x, giving:
dy/dx = 3x2 - 8x + 1
Substituting x = 2, we get:
dy/dx = 3(2)2 - 8(2) + 1 = -7
Therefore, the gradient of the curve at the point where it intersects the X-axis is -7.
Since the normal to the curve is perpendicular to the tangent at the point of intersection, we know that the gradient of the normal is the negative reciprocal of the gradient of the tangent. So the gradient of the normal is 1/7.
Finally, we can find the equation of the normal using the point-slope form of the equation of a straight line:
y - 0 = (1/7)(x - 2)
Simplifying this equation gives:
y = (1/7)x - 2/7
So the equation of the normal to the curve y = (x2 - x + 1)(x - 2) at the point where the curve cuts the X-axis is y = (1/7)x - 2/7.
(b) To find the equation of the line joining Q and the midpoint of PR, we first need to find the coordinates of the midpoint of PR. The midpoint of a line segment with endpoints (x1, y1) and (x2, y2) is ((x1 + x2)/2, (y1 + y2)/2). So the midpoint of PR is ((-1 + 3)/2, (2 - 4)/2), which simplifies to (1, -1).
Next, we need to find the gradient of the line joining Q and the midpoint of PR. We can use the gradient formula, which gives:
m = (y2 - y1)/(x2 - x1)
Substituting the coordinates of Q and the midpoint of PR gives:
m = (1 - (-1))/(5 - 1) = 1/2
Detalhes da Resposta
(a) To find the equation of the normal to the curve y = (x2 - x + 1)(x - 2) at the point where the curve cuts the X-axis, we first need to find the x-coordinate of the point where the curve intersects the X-axis. This is where y = 0, so we can solve the equation (x2 - x + 1)(x - 2) = 0 to find the roots of the equation. The roots are x = 1 ± i√3 and x = 2, but since we want the point where the curve intersects the X-axis, we take x = 2.
Next, we need to find the gradient of the curve at this point. We can do this by differentiating the equation y = (x2 - x + 1)(x - 2) with respect to x, giving:
dy/dx = 3x2 - 8x + 1
Substituting x = 2, we get:
dy/dx = 3(2)2 - 8(2) + 1 = -7
Therefore, the gradient of the curve at the point where it intersects the X-axis is -7.
Since the normal to the curve is perpendicular to the tangent at the point of intersection, we know that the gradient of the normal is the negative reciprocal of the gradient of the tangent. So the gradient of the normal is 1/7.
Finally, we can find the equation of the normal using the point-slope form of the equation of a straight line:
y - 0 = (1/7)(x - 2)
Simplifying this equation gives:
y = (1/7)x - 2/7
So the equation of the normal to the curve y = (x2 - x + 1)(x - 2) at the point where the curve cuts the X-axis is y = (1/7)x - 2/7.
(b) To find the equation of the line joining Q and the midpoint of PR, we first need to find the coordinates of the midpoint of PR. The midpoint of a line segment with endpoints (x1, y1) and (x2, y2) is ((x1 + x2)/2, (y1 + y2)/2). So the midpoint of PR is ((-1 + 3)/2, (2 - 4)/2), which simplifies to (1, -1).
Next, we need to find the gradient of the line joining Q and the midpoint of PR. We can use the gradient formula, which gives:
m = (y2 - y1)/(x2 - x1)
Substituting the coordinates of Q and the midpoint of PR gives:
m = (1 - (-1))/(5 - 1) = 1/2
