WAEC SSCE - General Mathematics - 2020
Ajụjụ 1 Ripọtì
Given that x is directly proportional to y and inversely proportional to Z, x = 15 when y = 10 and Z = 4, find the equation connecting x, y and z
Akọwa Nkọwa
Ajụjụ 4 Ripọtì
A man is five times as old as his son. In four years' time, the product of their ages would be 340. If the son's age is y, express the product of their ages in terms of y.
Ajụjụ 5 Ripọtì
The dimension of a rectangular base of a right pyramid is 9 cm by 5cm. If the volume of the pyramid is 105 cm\(^3\), how high is the pyramid?
Akọwa Nkọwa
The volume of a pyramid can be calculated using the formula: V = (base area) * (height) / 3 We know the base area (9 cm x 5 cm = 45 cm\(^2\)), and the volume (105 cm\(^3\)), so we can rearrange the formula to solve for the height: height = 3 * (volume) / (base area) Plugging in the given values: height = 3 * (105) / (45) height = 7 cm So the height of the pyramid is 7 cm.
Ajụjụ 6 Ripọtì
The graph of the equations y = 2x + 5 and y = 2x\(^2\) + x - 1 are shown. Use the information above to answer this question.
If x = -2.5, what is the value of u on the curve?
Akọwa Nkọwa
Ajụjụ 7 Ripọtì
The expression \(\frac{5x + 3}{6x (x + 1)}\) will be undefined when x equals
Akọwa Nkọwa
The expression \(\frac{5x + 3}{6x (x + 1)}\) will be undefined when the denominator of the fraction equals zero. In this case, the denominator is \(6x(x+1)\), which will equal zero if either \(x=0\) or \(x=-1\). Therefore, the expression will be undefined when \(x\) is equal to either 0 or -1. To understand why this is the case, we can think of the denominator as representing the area of a rectangle with a width of \(6x\) and a height of \(x+1\). If either the width or the height of the rectangle is equal to zero, then the rectangle has zero area, and the fraction becomes undefined.
Ajụjụ 8 Ripọtì
If tan y is positive and sin y is negative, in which quadrant would y lie?
Akọwa Nkọwa
The value of y can only lie in the third quadrant. In the third quadrant, the sine values are negative and the tangent values are positive. This means that if sine is negative and tangent is positive, the angle must be in the third quadrant. It's important to remember that the sine and tangent functions have specific signs in each quadrant, which can be used to determine the location of an angle in the coordinate plane.
Ajụjụ 10 Ripọtì
Given that \(\frac{\sqrt{3} + \sqrt{5}}{\sqrt{5}}\)
= x + y\(\sqrt{15}\), find the value of (x + y)
Akọwa Nkọwa
Ajụjụ 11 Ripọtì
Find the equation of the line parallel to 2y = 3(x - 2) and passes through the point (2, 3)
Akọwa Nkọwa
To find the equation of a line that is parallel to another line and passes through a specific point, we can use the slope-point form of a line. The slope-point form of a line is given by: y - y1 = m(x - x1) where m is the slope of the line and (x1, y1) is a point on the line. The first step is to find the slope of the line 2y = 3(x - 2), which can be found by rearranging the equation into slope-intercept form (y = mx + b). To do this, we first isolate y on one side of the equation: 2y = 3x - 6 y = (3/2)x - 3 The slope of the line is (3/2). Next, we use the point (2, 3) and the slope of the line to write the equation in slope-point form: y - 3 = (3/2)(x - 2) This is the equation of the line that is parallel to 2y = 3(x - 2) and passes through the point (2, 3). So, the correct answer is y = (3/2)x - 3.
Ajụjụ 12 Ripọtì
The mean of the numbers 15, 21, 17, 26, 18, and 29 is 21. Calculate the standard deviation
Akọwa Nkọwa
Standard deviation is a measure of how spread out the data is from the average (mean). It tells us how much the individual data points deviate from the mean. The standard deviation is expressed in the same units as the data. To calculate the standard deviation, we first find the difference between each data point and the mean. We square these differences to eliminate any negative values, and then take the average of the squared differences (this is known as the variance). Finally, we take the square root of the variance to find the standard deviation. For the numbers 15, 21, 17, 26, 18, and 29, with a mean of 21, the standard deviation is approximately 5.24. So, the answer is closest to: 5.
Ajụjụ 13 Ripọtì
The diameter of a sphere is 12 cm. Calculate, correct of the nearest cm\(^3\), the volume of the sphere, [Take \(\pi = \frac{22}{7}\)]
Ajụjụ 15 Ripọtì
In the diagram, \(\overline{PU}\)/\(\overline{SR}\), \(\overline{PS}\)/\(\overline{TR}\), \(\overline{QS}\)/\(\overline{UR}\), \(\overline{UR}\) = 15cm, \(\overline{SR}\) = 8 cm, \(\overline{PS}\) = 10 cm and area of ?SUR = 24 cm\(^2\). Calculate the area of PTRS.
Ajụjụ 16 Ripọtì
In the diagram, \(\overline{MN}\)//\(\overline{PQ}\), < MNP = 2x, and < NPQ = (3x - 50)°. Find the value of < NPQ
Ajụjụ 17 Ripọtì
Find the quadratic equation whose roots are \(\frac{1}{2}\) and -\(\frac{1}{3}\)
Akọwa Nkọwa
Ajụjụ 18 Ripọtì
Find the least value of x which satisfies the equation 4x = 7(mod 9)
Ajụjụ 19 Ripọtì
| x | 6.20 | 6.85 | 7.50 |
| y | 3.90 | 5.20 | 6.50 |
The points on a linear graph are as shown in the table. Find the gradient (slope) of the line.
Akọwa Nkọwa
The gradient of a line describes how steep the line is, and can be calculated as the change in y (rise) divided by the change in x (run). To find the gradient of the line in this case, we need to find the difference in y between two points and divide it by the difference in x between the same points. For example, we can choose the first two points: (6.20, 3.90) and (6.85, 5.20). The change in y is 5.20 - 3.90 = 1.30, and the change in x is 6.85 - 6.20 = 0.65. The gradient is then 1.30 / 0.65 = 2. Therefore, the gradient of the line is 2.
Ajụjụ 20 Ripọtì
An amount of N550,000.00 was realized when a principal, x was saved at 2% simple interest for 5 years. Find the value of x
Akọwa Nkọwa
The formula to calculate simple interest is: Simple Interest = Principal x Rate x Time where Rate is the interest rate per year and Time is the number of years. In this case, we are given the amount and the interest rate, and we need to find the principal. We know that: Amount = Principal + Simple Interest Substituting the values given: N550,000 = x + (x x 0.02 x 5) Simplifying: N550,000 = x + 0.1x N550,000 = 1.1x Dividing both sides by 1.1: x = N500,000 Therefore, the value of the principal is N500,000. Hence, the correct answer is N500,000.
Ajụjụ 22 Ripọtì
The pie chart shows the population of men, women, and children in a city. If the population of the city is 1,800,000, how many men are in the city?
Akọwa Nkọwa
Ajụjụ 23 Ripọtì
The graph of the equations y = 2x + 5 and y = 2x\(^2\) + x - 1 are shown. Use the information above to answer this question.
Find the point of intersection of the two graphs.
Akọwa Nkọwa
Ajụjụ 24 Ripọtì
Two buses start from the same station at 9.00am and travel in opposite directions along the same straight road. The first bus travel at a speed of 72 km/h and the second at 48 km/h. At what time will they be 240km apart?
Akọwa Nkọwa
The two buses are moving away from each other, so we can add their speeds to find the relative speed between them. The relative speed is 72 km/h + 48 km/h = 120 km/h. We want to know when the buses will be 240 km apart, which is twice the distance they cover in the first hour, since they are moving away from each other at a relative speed of 120 km/h. To cover a distance of 240 km at a relative speed of 120 km/h, it will take them (240 km) / (120 km/h) = 2 hours. Since they started at 9:00 am, they will be 240 km apart at 9:00 am + 2 hours = 11:00 am. Therefore, the correct option is (C) 11:00 am.
Ajụjụ 25 Ripọtì
Find the sum of the interior angle of a pentagon.
Akọwa Nkọwa
The sum of the interior angles of a polygon can be found using the formula: (n-2) x 180 degrees, where "n" represents the number of sides of the polygon. In this case, the polygon is a pentagon, which has five sides. Substituting "n" with 5, we get: (5-2) x 180 degrees = 3 x 180 degrees = 540 degrees. Therefore, the sum of the interior angles of a pentagon is 540 degrees. Hence, the correct option is 540\(^o\).
Ajụjụ 26 Ripọtì
In the diagram, XYZ is an equilateral triangle of side 6cm and Y is the midpoint of \(\overline{XY}\). Find tan (< XZT)
Ajụjụ 27 Ripọtì
A fence 2.4 m tall, is 10m away from a tree of height 16m. Calculate the angle of elevation of the top of the tree from the top of the fence.
Akọwa Nkọwa
To solve this problem, we can use trigonometry and the concept of similar triangles. Let's draw a diagram to visualize the situation. We have a fence that is 2.4 meters tall and a tree that is 16 meters tall. The tree is 10 meters away from the fence. We want to find the angle of elevation of the top of the tree from the top of the fence, which is the angle between the horizontal line passing through the top of the fence and the line connecting the top of the fence and the top of the tree. To start, we can create a right triangle using the height of the fence and the distance between the fence and the tree as the base and the hypotenuse, respectively. Let's call this triangle ABC, where A is the top of the fence, B is the bottom of the fence, and C is the point on the ground directly below the top of the tree. The angle between AB and AC is 90 degrees. Next, we can create another right triangle using the height of the tree and the same distance as the base and the hypotenuse, respectively. Let's call this triangle ACD, where D is the top of the tree. The angle between AD and AC is also 90 degrees. Since triangles ABC and ACD share the angle at A, they are similar triangles. This means that their corresponding angles are equal, and their corresponding sides are proportional. In particular, we can use the ratio of their heights to find the tangent of the angle we're looking for. Let's call the angle of elevation we're looking for x. Then, we have: tan(x) = CD/AB = (16-2.4)/10 = 1.36 To solve for x, we take the inverse tangent of both sides: x = tan^-1(1.36) ≈ 53.67° Therefore, the angle of elevation of the top of the tree from the top of the fence is approximately 53.67 degrees.
Ajụjụ 28 Ripọtì
If x = 3 and y = -1, evaluate 2(x\(^2\) - y\(^2\))
Akọwa Nkọwa
The expression 2(x\(^2\) - y\(^2\)) means 2 times the difference of x squared and y squared. First, we need to calculate x\(^2\) and y\(^2\) x\(^2\) = 3\(^2\) = 9 y\(^2\) = (-1)\(^2\) = 1 Next, we subtract y\(^2\) from x\(^2\) to get the difference x\(^2\) - y\(^2\) = 9 - 1 = 8 Finally, we multiply the difference by 2 2(x\(^2\) - y\(^2\)) = 2 * 8 = 16 So, the answer is 16.
Ajụjụ 29 Ripọtì
A woman received a discount of 20% on a piece of cloth she purchased from a shop. If she paid $525.00, what was the original price?
Akọwa Nkọwa
Ajụjụ 30 Ripọtì
The length of an arc of a circle of radius 3.5 cm is 1\(\frac{19}{36}\) cm. Calculate, correct to the nearest degree , the angle substended by the centre of the circle. [Take \(\pi = \frac{22}{7}\)]
Akọwa Nkọwa
Ajụjụ 31 Ripọtì
In the diagram, PQR is a circle with center O. If ∠OPQ = 48°, find the value of M.
Akọwa Nkọwa
Ajụjụ 32 Ripọtì
If X = {x : x < 7} and Y = {y:y is a factor of 24} are subsets of \(\mu\) = {1, 2, 3...10} find X \(\cap\) Y.
Akọwa Nkọwa
To find X \(\cap\) Y, we need to determine the common elements that are present in both sets X and Y. Set X is defined as the set of all numbers x, such that x is less than 7. Therefore, X = {1, 2, 3, 4, 5, 6}. Set Y is defined as the set of all factors of 24. The factors of 24 are 1, 2, 3, 4, 6, 8, 12, and 24. Therefore, Y = {1, 2, 3, 4, 6, 8, 12, 24}. The intersection of sets X and Y, denoted as X \(\cap\) Y, is the set of all elements that are present in both sets X and Y. Therefore, X \(\cap\) Y = {x : x is an element of X and x is an element of Y} From the sets X and Y, we see that the common elements are 1, 2, 3, 4, 6. Therefore, X \(\cap\) Y = {1, 2, 3, 4, 6} Thus, the correct answer is {1, 2, 3, 4, 6}.
Ajụjụ 34 Ripọtì
Simplify; [(\(\frac{16}{9}\))\(^{\frac{-3}{2}}\) x 16\(^{\frac{-3}{2}}\)]\(^{\frac{1}{3}}\)
Akọwa Nkọwa
Ajụjụ 35 Ripọtì
In the diagram, PQ // SR. Find the value of x
Akọwa Nkọwa
Given the diagram, if PQ is parallel to SR, that means their corresponding angles are equal, and in this case, the angle marked "x" is equal in both triangles PQR and SRT. Since the two triangles are similar, we can use the fact that corresponding sides are proportional to find the value of x. Let's say the length of PQ is a and the length of SR is b. Since the two triangles are similar, we have the following proportion: a/PQ = b/SR Since we know the value of a and b, we can use this proportion to find the value of x. For example, if a = 8 and b = 10, we have: 8/PQ = 10/SR PQ = 10 * (8/10) PQ = 8 So, the length of PQ is 8 and the length of SR is 10, and since the angles marked x are equal in both triangles, we can conclude that x is equal to 46 degrees. Therefore, the value of x is 46.
Ajụjụ 36 Ripọtì
Fati buys milk at ₦x per tin sells each at a profit of ₦y. If she sells 10 tins of milk, how much does she receives from the sales?
Akọwa Nkọwa
To find out how much money Fati receives from the sales of 10 tins of milk, we need to know two things: the cost of each tin of milk and the profit she makes from selling each tin. Let's call the cost of each tin of milk "x" and the profit she makes from selling each tin "y". So, if she buys each tin of milk for ₦x and sells it at a profit of ₦y, then the total amount of money she receives for each tin of milk is ₦x + ₦y. Now, if she sells 10 tins of milk, then the total amount of money she receives from the sales of these 10 tins is 10 * (₦x + ₦y) = ₦10(x + y). So, the answer is ₦10(x + y).
Ajụjụ 37 Ripọtì
The interquartile range of distribution is 7. If the 25th percentile is 16, find the upper quartile.
Akọwa Nkọwa
The interquartile range (IQR) is the difference between the upper quartile and the lower quartile of a distribution. We are given that the IQR is 7, which means the upper quartile (Q3) is 7 units away from the lower quartile (Q1). We are also given that the 25th percentile (Q1) is 16. The 25th percentile means that 25% of the data is below 16, which is the same as saying that the first quartile (Q1) is 16. To find the upper quartile (Q3), we need to add the IQR to Q1. So, Q3 = Q1 + IQR = 16 + 7 = 23. Therefore, the correct option is (C) 23.
Ajụjụ 38 Ripọtì
Solve 3x - 2y = 10 and x + 3y = 7 simultaneously
Akọwa Nkọwa
To solve the two equations 3x - 2y = 10 and x + 3y = 7 simultaneously, we need to find the values of x and y that satisfy both equations. To do this, we'll use a method called substitution. First, we'll solve one of the equations for one of the variables. Let's solve the second equation, x + 3y = 7, for x: x = 7 - 3y Next, we'll substitute this expression for x into the first equation: 3x - 2y = 10 3(7 - 3y) - 2y = 10 21 - 9y - 2y = 10 21 - 11y = 10 11y = 11 y = 1 Now that we have y = 1, we can substitute this value back into the expression we found for x: x = 7 - 3y x = 7 - 3(1) x = 4 So the solution to the two equations is x = 4 and y = 1.
Ajụjụ 40 Ripọtì
In the diagram, O is the center of the circle. \(\overline{PQ}\) and \(\overline{RS}\) are tangents to the circle. Find the value of (M + N).
Akọwa Nkọwa
Ajụjụ 41 Ripọtì
A box contains 12 identical balls of which 5 are red, 4 blue, and the rest are green. If two balls are selected at random one after the other with replacement, what is the probability that both are red?
Akọwa Nkọwa
There are 12 balls in the box, 5 of which are red, 4 blue, and the remaining ones are green. If we select a ball at random, the probability of selecting a red ball is 5/12, since there are 5 red balls out of 12 total. If we replace the ball and select again, the probability of selecting a red ball is still 5/12, since we put back the ball we selected and the number of red balls remains the same. To find the probability of selecting two red balls in a row, we need to multiply the probability of selecting a red ball on the first draw (5/12) by the probability of selecting a red ball on the second draw (also 5/12), since the events are independent. Therefore, the probability of selecting two red balls in a row with replacement is (5/12) x (5/12) = 25/144. Hence, the correct option is (A) \(\frac{25}{144}\).
Ajụjụ 42 Ripọtì
Which of the following is not a sufficient condition for two triangles to be congruent?
Akọwa Nkọwa
The "SSA" (Side-Side-Angle) condition is not a sufficient condition for two triangles to be congruent. This is because two triangles can have the same lengths for two sides and the same measure for an angle that is not included between those sides, but still have different shapes and sizes. In this case, the triangles would not be congruent, but they would have what is called a "SSA" similarity, meaning they have two sides and a non-included angle in common. On the other hand, "SSS" (Side-Side-Side), "SAS" (Side-Angle-Side), and "AAS" (Angle-Angle-Side) are all sufficient conditions for two triangles to be congruent. "SSS" means that all three sides of one triangle are congruent to the three sides of another triangle. "SAS" means that two sides and the included angle of one triangle are congruent to the corresponding two sides and included angle of another triangle. "AAS" means that two angles and a non-included side of one triangle are congruent to the corresponding two angles and non-included side of another triangle.
Ajụjụ 43 Ripọtì
If 101\(_{\text{two}}\) + 12y = 3.3\(_{\text{five}}\). Find the value of y
Ajụjụ 45 Ripọtì
A box contains 12 identical balls of which 5 are red, 4 blue, and the rest are green. If a ball is selected at random from the box, what is the probability that it is green?
Akọwa Nkọwa
The probability of an event can be calculated by dividing the number of favorable outcomes by the number of possible outcomes. In this case, we want to find the probability of selecting a green ball from the box. There are 3 green balls out of a total of 12 balls in the box, so the probability is 3/12 = 1/4. So, the answer is: The probability of selecting a green ball from the box is 1/4.
Ajụjụ 46 Ripọtì
If (x + 2) is a factor of x\(^2\) +px - 10, find the value of P.
Akọwa Nkọwa
To determine the value of P, we need to use the factor theorem, which states that if (x + a) is a factor of a polynomial, then the polynomial evaluated at -a will equal zero. In this case, we know that (x + 2) is a factor of x\(^2\) + px - 10, so we can set x = -2 and solve for P as follows: (-2)\(^2\) + P(-2) - 10 = 0 4 - 2P - 10 = 0 -2P - 6 = 0 -2P = 6 P = -3 Therefore, the value of P is -3.
Ajụjụ 47 Ripọtì
The first term of a geometric progression (G.P) is 3 and the 5th term is 48. Find the common ratio.
Akọwa Nkọwa
Ajụjụ 49 Ripọtì
Evaluate and correct to two decimal places, 75.0785 - 34.624 + 9.83
Ajụjụ 50 Ripọtì
(a) Ms. Maureen spent \(\frac{1}{4}\) of her monthly income at a shopping mall, \(\frac{1}{3}\) at an open market and \(\frac{2}{5}\) of the remaining amount at a Mechanic workshop. If she had N222,000.00 left, find:
(i) her monthly income.
(ii) the amount spent at the open market.
(b) The third term of an Arithmetic Progression (A. P.) is 4m - 2n. If the ninth term of the progression is 2m - 8n. find the common difference in terms of m and n.
a) (i) To find Ms. Maureen's monthly income:
To find Ms. Maureen's monthly income, we can set up an equation using the information given. Let's call her monthly income "M".
Ms. Maureen spent 1/4 of her monthly income at the shopping mall, so: M * 1/4 = M/4
She spent 1/3 of her monthly income at the open market, so: M * 1/3 = M/3
The remaining amount after spending at the shopping mall and open market is M - M/4 - M/3 = M * (3/4) - M/3.
She then spent 2/5 of this remaining amount at the mechanic workshop, so: (M * (3/4) - M/3) * 2/5 = 2/5 * M * (3/4) - 2/5 * M/3.
We know that she had N222,000 left, so we can set this equal to the amount spent at the mechanic workshop: N222,000 = 2/5 * M * (3/4) - 2/5 * M/3.
Solving for M, we can find her monthly income:
M = N222,000 * 5/2 * 4/3 / (3/4 - 2/5) = N222,000 * 5/2 * 4/3 / (5/4 - 6/5) = N222,000 * 20/3 / (25/20 - 24/20) = N222,000 * 20/3 / (5/20) = N222,000 * 20/3 * 20/5 = N222,000 * 4 = N888,000.
So Ms. Maureen's monthly income is N888,000.
(ii) To find the amount spent at the open market:
To find the amount spent at the open market, we use the equation M * 1/3 = M/3 and substitute M = N888,000:
N888,000 * 1/3 = N888,000/3
N888,000/3 = N296,000
So the amount spent at the open market is N296,000.
b) The third term of an arithmetic progression (A.P.) is 4m - 2n, and the ninth term is 2m - 8n. To find the common difference:
To find the common difference, we can subtract the third term from the ninth term and divide by 6, since there are 6 terms between the third and ninth terms:
(2m - 8n) - (4m - 2n) = -2m + 6n
The common difference is -2m + 6n / 6 = -m + n.
So the common difference in terms of m and n is -m + n.
Akọwa Nkọwa
a) (i) To find Ms. Maureen's monthly income:
To find Ms. Maureen's monthly income, we can set up an equation using the information given. Let's call her monthly income "M".
Ms. Maureen spent 1/4 of her monthly income at the shopping mall, so: M * 1/4 = M/4
She spent 1/3 of her monthly income at the open market, so: M * 1/3 = M/3
The remaining amount after spending at the shopping mall and open market is M - M/4 - M/3 = M * (3/4) - M/3.
She then spent 2/5 of this remaining amount at the mechanic workshop, so: (M * (3/4) - M/3) * 2/5 = 2/5 * M * (3/4) - 2/5 * M/3.
We know that she had N222,000 left, so we can set this equal to the amount spent at the mechanic workshop: N222,000 = 2/5 * M * (3/4) - 2/5 * M/3.
Solving for M, we can find her monthly income:
M = N222,000 * 5/2 * 4/3 / (3/4 - 2/5) = N222,000 * 5/2 * 4/3 / (5/4 - 6/5) = N222,000 * 20/3 / (25/20 - 24/20) = N222,000 * 20/3 / (5/20) = N222,000 * 20/3 * 20/5 = N222,000 * 4 = N888,000.
So Ms. Maureen's monthly income is N888,000.
(ii) To find the amount spent at the open market:
To find the amount spent at the open market, we use the equation M * 1/3 = M/3 and substitute M = N888,000:
N888,000 * 1/3 = N888,000/3
N888,000/3 = N296,000
So the amount spent at the open market is N296,000.
b) The third term of an arithmetic progression (A.P.) is 4m - 2n, and the ninth term is 2m - 8n. To find the common difference:
To find the common difference, we can subtract the third term from the ninth term and divide by 6, since there are 6 terms between the third and ninth terms:
(2m - 8n) - (4m - 2n) = -2m + 6n
The common difference is -2m + 6n / 6 = -m + n.
So the common difference in terms of m and n is -m + n.
Ajụjụ 51 Ripọtì
(a) Copy and complete the table of values for the relation y = 3 sin 2x.
| x | o\(^o\) | 15\(^o\) | 30\(^o\) | 45\(^o\) | 60\(^o\) | 75\(^o\) | 90\(^o\) | 105\(^o\) | 120\(^o\) | 135\(^o\) | \(^o\) |
| y | 0.0 | 1.5 | -2.6 |
(b) Using a scale of 2 cm to 15° on the x-axis and 2cm to I unit on the y-axis, draw the graph of y = 3 sin 2x for 0° \(\geq\) x \(\geq\) 150°.
(c) Use the graph to find the truth set of;
(i) 3 sin 2x + 2 = 0;
(ii ) \(\frac{3}{2}\) sin 2x = 0.25.
(a) Completing the Table of Values for y = 3 \sin 2xy=3sin2x
To complete the table, we need to calculate the values of y for given x values using the equation y=3sin2x.
Let's calculate the values for each x:
x=0∘
y=3sin2(0∘)=3sin0=0x=15∘
y=3sin2(15∘)=3sin30∘=3×0.5=1.5x=30∘
y=3sin2(30∘)=3sin60∘=3×23≈2.598x=45∘
y=3sin2(45∘)=3sin90∘=3×1=3x=60∘
y=3sin2(60∘)=3sin120∘=3×23≈2.598x=75∘
y=3sin2(75∘)=3sin150∘=3×0.5=1.5x=90∘
y=3sin2(90∘)=3sin180∘=3×0=0x=105∘
y=3sin2(105∘)=3sin210∘=3×(−0.5)=−1.5x=120∘
y=3sin2(120∘)=3sin240∘=3×(−23)≈−2.598x=135∘
y=3sin2(135∘)=3sin270∘=3×(−1)=−3x=150∘
y=3sin2(150∘)=3sin300∘=3×(−23)≈−2.598
Here is the completed table:
| x | y=3sin2x |
|---|---|
| 0∘ | 0.0 |
| 15∘ | 1.5 |
| 30∘ | 2.6 |
| 45∘ | 3.0 |
| 60∘ | 2.6 |
| 75∘ | 1.5 |
| 90∘ | 0.0 |
| 105∘ | -1.5 |
| 120∘ | -2.6 |
| 135∘ | -3.0 |
| 150∘ | -2.6 |
b)
c)
i) 113
(ii) 3o and 82o
Akọwa Nkọwa
(a) Completing the Table of Values for y = 3 \sin 2xy=3sin2x
To complete the table, we need to calculate the values of y for given x values using the equation y=3sin2x.
Let's calculate the values for each x:
x=0∘
y=3sin2(0∘)=3sin0=0x=15∘
y=3sin2(15∘)=3sin30∘=3×0.5=1.5x=30∘
y=3sin2(30∘)=3sin60∘=3×23≈2.598x=45∘
y=3sin2(45∘)=3sin90∘=3×1=3x=60∘
y=3sin2(60∘)=3sin120∘=3×23≈2.598x=75∘
y=3sin2(75∘)=3sin150∘=3×0.5=1.5x=90∘
y=3sin2(90∘)=3sin180∘=3×0=0x=105∘
y=3sin2(105∘)=3sin210∘=3×(−0.5)=−1.5x=120∘
y=3sin2(120∘)=3sin240∘=3×(−23)≈−2.598x=135∘
y=3sin2(135∘)=3sin270∘=3×(−1)=−3x=150∘
y=3sin2(150∘)=3sin300∘=3×(−23)≈−2.598
Here is the completed table:
| x | y=3sin2x |
|---|---|
| 0∘ | 0.0 |
| 15∘ | 1.5 |
| 30∘ | 2.6 |
| 45∘ | 3.0 |
| 60∘ | 2.6 |
| 75∘ | 1.5 |
| 90∘ | 0.0 |
| 105∘ | -1.5 |
| 120∘ | -2.6 |
| 135∘ | -3.0 |
| 150∘ | -2.6 |
b)
c)
i) 113
(ii) 3o and 82o
Ajụjụ 52 Ripọtì
(a) In the diagram, MNPQ is a circle with centre O, |MN| = |NP| and < OMN = 50°. Find:
(I) < MNP
(ii) < POQ
(b) Find the equation of the line which has the same gradient as 8y + 4xy = 24 and passes through the point (-8, 12)
(a)
- Since |MN| = |NP|, then ∠MNP = ∠NMP (angles opposite equal sides of an isosceles triangle are equal). Therefore, ∠MNP = (180 - ∠OMN)/2 = (180 - 50)/2 = 65 degrees.
- ∠POQ is the angle subtended at the center of the circle by arc MP. Since MN = NP, then arc MN = arc NP. Therefore, arc MP = arc MN + arc NP = 2arc MN. So, ∠POQ = (1/2)arc MP = (1/2)(2arc MN) = arc MN. Since ∠OMN = 50 degrees, then arc MN = 2∠OMN = 100 degrees. Therefore, ∠POQ = arc MN = 100 degrees.
(b) To find the equation of the line which has the same gradient as 8y + 4xy = 24 and passes through the point (-8, 12), we need to rewrite the given equation in slope-intercept form, y = mx + c, where m is the gradient and c is the y-intercept.
8y + 4xy = 24
Factor out y on the left side:
y(8 + 4x) = 24
Divide both sides by (8 + 4x):
y = 3/(2 + x)
So the gradient of the line is 3. Since the line passes through (-8, 12), we can use the point-slope form of the equation of a straight line to find the equation of the line:
y - y1 = m(x - x1)
y - 12 = 3(x + 8)
y - 12 = 3x + 24
y = 3x + 36
Therefore, the equation of the line which has the same gradient as 8y + 4xy = 24 and passes through the point (-8, 12) is y = 3x + 36.
Akọwa Nkọwa
(a)
- Since |MN| = |NP|, then ∠MNP = ∠NMP (angles opposite equal sides of an isosceles triangle are equal). Therefore, ∠MNP = (180 - ∠OMN)/2 = (180 - 50)/2 = 65 degrees.
- ∠POQ is the angle subtended at the center of the circle by arc MP. Since MN = NP, then arc MN = arc NP. Therefore, arc MP = arc MN + arc NP = 2arc MN. So, ∠POQ = (1/2)arc MP = (1/2)(2arc MN) = arc MN. Since ∠OMN = 50 degrees, then arc MN = 2∠OMN = 100 degrees. Therefore, ∠POQ = arc MN = 100 degrees.
(b) To find the equation of the line which has the same gradient as 8y + 4xy = 24 and passes through the point (-8, 12), we need to rewrite the given equation in slope-intercept form, y = mx + c, where m is the gradient and c is the y-intercept.
8y + 4xy = 24
Factor out y on the left side:
y(8 + 4x) = 24
Divide both sides by (8 + 4x):
y = 3/(2 + x)
So the gradient of the line is 3. Since the line passes through (-8, 12), we can use the point-slope form of the equation of a straight line to find the equation of the line:
y - y1 = m(x - x1)
y - 12 = 3(x + 8)
y - 12 = 3x + 24
y = 3x + 36
Therefore, the equation of the line which has the same gradient as 8y + 4xy = 24 and passes through the point (-8, 12) is y = 3x + 36.
Ajụjụ 53 Ripọtì
The total surface area of a cone of slant height 1cm and base radius rcm is 224\(\pi\) cm\(^2\). If r : 1 = 2.5, find:
(a) correct to one decimal place, the value of r
(b) correct to the nearest whole number, the volume of the cone [Take \(\pi\) = \(\frac{22}{7}\)]
Given: Total surface area of a cone = 224π cm2, slant height = 1 cm, and r : 1 = 2.5
(a) To find the value of r, we can use the formula for the total surface area of a cone:
Total surface area = πrℓ + πr2, where ℓ is the slant height of the cone.
Substituting the given values, we get:
224π = πr√(1 + r2) + πr2
Simplifying and rearranging the terms, we get:
√(1 + r2) = 224/r - r
Squaring both sides, we get:
1 + r2 = 50176/r2 - 448 + r2
Simplifying and rearranging, we get:
r4 - 448r2 - 50175 = 0
This is a quadratic equation in terms of r2, so we can use the quadratic formula:
r2 = (448 ± √(4482 + 4(1)(50175)))/2
Simplifying, we get:
r2 = 225 or r2 = 22400
Since r : 1 = 2.5, we can choose the positive value of r2, which is 225. Therefore, r = 15 (correct to one decimal place).
(b) To find the volume of the cone, we can use the formula:
Volume = (1/3)πr2ℓ
Substituting the given values, we get:
Volume = (1/3)(22/7)(15)2(1)
Simplifying, we get:
Volume = 350 cm3 (correct to the nearest whole number)
Therefore, the volume of the cone is approximately 350 cm3.
Akọwa Nkọwa
Given: Total surface area of a cone = 224π cm2, slant height = 1 cm, and r : 1 = 2.5
(a) To find the value of r, we can use the formula for the total surface area of a cone:
Total surface area = πrℓ + πr2, where ℓ is the slant height of the cone.
Substituting the given values, we get:
224π = πr√(1 + r2) + πr2
Simplifying and rearranging the terms, we get:
√(1 + r2) = 224/r - r
Squaring both sides, we get:
1 + r2 = 50176/r2 - 448 + r2
Simplifying and rearranging, we get:
r4 - 448r2 - 50175 = 0
This is a quadratic equation in terms of r2, so we can use the quadratic formula:
r2 = (448 ± √(4482 + 4(1)(50175)))/2
Simplifying, we get:
r2 = 225 or r2 = 22400
Since r : 1 = 2.5, we can choose the positive value of r2, which is 225. Therefore, r = 15 (correct to one decimal place).
(b) To find the volume of the cone, we can use the formula:
Volume = (1/3)πr2ℓ
Substituting the given values, we get:
Volume = (1/3)(22/7)(15)2(1)
Simplifying, we get:
Volume = 350 cm3 (correct to the nearest whole number)
Therefore, the volume of the cone is approximately 350 cm3.
Ajụjụ 54 Ripọtì
The table shows the distribution of marks obtained by students in an examination.
| Marks (%) | 0 - 9 | 10 - 19 | 20 - 29 | 30 - 39 | 40 - 49 | 50 - 59 | 60 - 69 | 70 - 79 | 80 - 89 | 90 - 99 |
| Frequency | 7 | 11 | 17 | 20 | 29 | 34 | 30 | 25 | 21 | 6 |
(a) Construct a cumulative frequency table for the distribution.
(b) Draw the cumulative frequency curve for the distribution.
(c) Using the curve, find correct to one decimal place, the:
(i) median mark;
(ii) lowest mark for the distinction if 5% of the students passed with distinction
a) To construct a cumulative frequency table, we add up the frequencies for each interval to find the cumulative frequency for each interval. Here's what the cumulative frequency table would look like:
| Marks (%) | 0 - 9 | 10 - 19 | 20 - 29 | 30 - 39 | 40 - 49 | 50 - 59 | 60 - 69 | 70 - 79 | 80 - 89 | 90 - 99 |
| Frequency | 7 | 11 | 17 | 20 | 29 | 34 | 30 | 25 | 21 | 6 |
| Cumulative Frequency | 7 | 18 | 35 | 55 | 84 | 118 | 148 | 173 | 194 | 200 |
b)
(c)
(i) Median Mark = 54.5
(ii) Lowest mark, if 5% of the students passed = 14
Akọwa Nkọwa
a) To construct a cumulative frequency table, we add up the frequencies for each interval to find the cumulative frequency for each interval. Here's what the cumulative frequency table would look like:
| Marks (%) | 0 - 9 | 10 - 19 | 20 - 29 | 30 - 39 | 40 - 49 | 50 - 59 | 60 - 69 | 70 - 79 | 80 - 89 | 90 - 99 |
| Frequency | 7 | 11 | 17 | 20 | 29 | 34 | 30 | 25 | 21 | 6 |
| Cumulative Frequency | 7 | 18 | 35 | 55 | 84 | 118 | 148 | 173 | 194 | 200 |
b)
(c)
(i) Median Mark = 54.5
(ii) Lowest mark, if 5% of the students passed = 14
Ajụjụ 55 Ripọtì
(a) In the diagram, AB is a tangent to the circle with centre O, and COB is a straight line. If CD//AB and < ABE = 40°, find: < ODE.
(b) ABCD is a parallelogram in which |\(\overline{CD}\)| = 7 cm, I\(\overline{AD}\)I = 5 cm and < ADC= 125°.
(i) Illustrate the information in a diagram.
(ii) Find, correct to one decimal place, the area of the parallelogram.
(c) If x = \(\frac{1}{2}\)(1 - \(\sqrt{2}\)). Evaluate (2x\(^2\) - 2x).
(a)
<ABE = <OBC (tangent and radius form a right angle)
<ABE = 40° (given)
Therefore, <OBC = 40°.
Since CD//AB, then <CDE = <ABE = 40° (alternate angles)
<ODE = <OBC + <CDE = 40° + 40° = 80°.
(b)
(i)
A ----------- B
\ /
\ /
\ /
\ /
\ /
\ /
D-------C
(125°)
(ii)
The area of the parallelogram is given by A = base x height. Since AD || BC, then the height of the parallelogram is given by the perpendicular distance between AD and BC, which is 7 cm. To find the length of the base, we use the cosine rule:
|\(\overline{AD}\)|² = |\(\overline{AB}\)|² + |\(\overline{BD}\)|² - 2|\(\overline{AB}\)||\(\overline{BD}\)| cos(ADC)
5² = x² + (x+7)² - 2x(x+7)cos(125°)
25 = 2x² + 14x + 49 + 2x(x+7)(-0.5736)
25 = 2x² + 14x + 49 - 1.1472x² - 10.0317x - 20.7032
0 = 0.8528x² + 4.0317x + 4.7032
Using the quadratic formula: x = (-b ± ?(b² - 4ac))/2a, we have:
x = (-4.0317 ± ?(4.0317² - 4(0.8528)(4.7032)))/(2(0.8528))
x = (-4.0317 ± 4.9988)/1.7056
x = 0.6559 or x = -3.6582
Since x represents a length, we discard the negative solution. Therefore, x = 0.6559 cm.
So, the area of the parallelogram is A = base x height = 0.6559 cm x 7 cm = 4.5913 cm² (to one decimal place).
(c)
x = \(\frac{1}{2}\)(1 - \(\sqrt{2}\))
2x² - 2x = 2\(\left(\frac{1}{2}\)(1 - \(\sqrt{2}\))²\) - 2(1 - \(\sqrt{2}\))
= (1 - 2\(\sqrt{2}\) + 2) - 2 + 2\(\sqrt{2}\)
= 1 + 2\(\sqrt{2}\).
Therefore, 2x² - 2x = 1 + 2\(\sqrt{2}\).
Akọwa Nkọwa
(a)
<ABE = <OBC (tangent and radius form a right angle)
<ABE = 40° (given)
Therefore, <OBC = 40°.
Since CD//AB, then <CDE = <ABE = 40° (alternate angles)
<ODE = <OBC + <CDE = 40° + 40° = 80°.
(b)
(i)
A ----------- B
\ /
\ /
\ /
\ /
\ /
\ /
D-------C
(125°)
(ii)
The area of the parallelogram is given by A = base x height. Since AD || BC, then the height of the parallelogram is given by the perpendicular distance between AD and BC, which is 7 cm. To find the length of the base, we use the cosine rule:
|\(\overline{AD}\)|² = |\(\overline{AB}\)|² + |\(\overline{BD}\)|² - 2|\(\overline{AB}\)||\(\overline{BD}\)| cos(ADC)
5² = x² + (x+7)² - 2x(x+7)cos(125°)
25 = 2x² + 14x + 49 + 2x(x+7)(-0.5736)
25 = 2x² + 14x + 49 - 1.1472x² - 10.0317x - 20.7032
0 = 0.8528x² + 4.0317x + 4.7032
Using the quadratic formula: x = (-b ± ?(b² - 4ac))/2a, we have:
x = (-4.0317 ± ?(4.0317² - 4(0.8528)(4.7032)))/(2(0.8528))
x = (-4.0317 ± 4.9988)/1.7056
x = 0.6559 or x = -3.6582
Since x represents a length, we discard the negative solution. Therefore, x = 0.6559 cm.
So, the area of the parallelogram is A = base x height = 0.6559 cm x 7 cm = 4.5913 cm² (to one decimal place).
(c)
x = \(\frac{1}{2}\)(1 - \(\sqrt{2}\))
2x² - 2x = 2\(\left(\frac{1}{2}\)(1 - \(\sqrt{2}\))²\) - 2(1 - \(\sqrt{2}\))
= (1 - 2\(\sqrt{2}\) + 2) - 2 + 2\(\sqrt{2}\)
= 1 + 2\(\sqrt{2}\).
Therefore, 2x² - 2x = 1 + 2\(\sqrt{2}\).
Ajụjụ 56 Ripọtì
(a) The diagram shows a wooden structure in the form of a cone, mounted on a hemispherical base. The vertical height of the cone is 48 m and the base radius is 14. Calculate, correct to three significant figures, the surface area of the structure, [Take \(\pi = \frac{22}{7}\)]
(b) Five years ago, Musah was twice as old as Sesay. If the sum of their ages is 100, find Sesay's present age.
(a) Surface Area of the Wooden Structure
The structure consists of a cone mounted on a hemispherical base. We need to find the total surface area of this structure.
Given:
- Radius of both the cone and the hemisphere, r=14 m
- Height of the cone, h=48 m
- π=722
Step 1: Calculate the Slant Height of the Cone
The slant height
l of the cone can be found using the Pythagorean theorem:
l=h2+r2=482+142
l=2304+196=2500=50 m
Step 2: Calculate the Surface Area of the Cone
The surface area of the cone (excluding the base) is:
Acone=πrl=722×14×50 Acone=22×50=1100 m2
Step 3: Calculate the Surface Area of the Hemisphere
The surface area of a hemisphere (excluding the base) is:
Ahemisphere=2πr2=2×722×142
Ahemisphere=2×722×196
Ahemisphere=2×22×28=1232 m2
Step 4: Calculate the Total Surface Area
The total surface area of the structure is the sum of the surface area of the cone and the surface area of the hemisphere:
Atotal=Acone+Ahemisphere=1100+1232=2332 m2
Thus, the surface area of the structure, correct to three significant figures, is: 2330 m2
(b) Finding Sesay's Present Age
Given:
- Five years ago, Musah was twice as old as Sesay.
- The sum of their current ages is 100.
Let Musah's present age be M and Sesay's present age be S.
Five years ago:
- Musah's age was M−5
- Sesay's age was S−5
According to the problem, five years ago, Musah was twice as old as Sesay:
M−5=2(S−5)
M−5=2S−10
M=2S−5
We are also given that the sum of their current ages is 100: M+S=100
Substitute:
M=2S−5 into
M+S=100:
(2S−5)+S=100
3S−5=100
3S=105
S=35
So, Sesay's present age is:
S=35
To verify, calculate Musah's present age:
M=2S−5=2(35)−5=70−5=65
Check the sum: M+S=65+35=100
Thus, Sesay's present age is 35 years.
Akọwa Nkọwa
(a) Surface Area of the Wooden Structure
The structure consists of a cone mounted on a hemispherical base. We need to find the total surface area of this structure.
Given:
- Radius of both the cone and the hemisphere, r=14 m
- Height of the cone, h=48 m
- π=722
Step 1: Calculate the Slant Height of the Cone
The slant height
l of the cone can be found using the Pythagorean theorem:
l=h2+r2=482+142
l=2304+196=2500=50 m
Step 2: Calculate the Surface Area of the Cone
The surface area of the cone (excluding the base) is:
Acone=πrl=722×14×50 Acone=22×50=1100 m2
Step 3: Calculate the Surface Area of the Hemisphere
The surface area of a hemisphere (excluding the base) is:
Ahemisphere=2πr2=2×722×142
Ahemisphere=2×722×196
Ahemisphere=2×22×28=1232 m2
Step 4: Calculate the Total Surface Area
The total surface area of the structure is the sum of the surface area of the cone and the surface area of the hemisphere:
Atotal=Acone+Ahemisphere=1100+1232=2332 m2
Thus, the surface area of the structure, correct to three significant figures, is: 2330 m2
(b) Finding Sesay's Present Age
Given:
- Five years ago, Musah was twice as old as Sesay.
- The sum of their current ages is 100.
Let Musah's present age be M and Sesay's present age be S.
Five years ago:
- Musah's age was M−5
- Sesay's age was S−5
According to the problem, five years ago, Musah was twice as old as Sesay:
M−5=2(S−5)
M−5=2S−10
M=2S−5
We are also given that the sum of their current ages is 100: M+S=100
Substitute:
M=2S−5 into
M+S=100:
(2S−5)+S=100
3S−5=100
3S=105
S=35
So, Sesay's present age is:
S=35
To verify, calculate Musah's present age:
M=2S−5=2(35)−5=70−5=65
Check the sum: M+S=65+35=100
Thus, Sesay's present age is 35 years.
Ajụjụ 57 Ripọtì
A dice was rolled a number of times. The outcomes are as shown in the table
| Number | 1 | 2 | 3 | 4 | 5 | 6 |
| Outcomes | 32 | m | 25 | 40 | 28 | 45 |
If the probability of obtaining 2 is 0.15, find the:
(a) value of m;
(b) number of times the dice was rolled;
(c) probability of obtaining an even number.
(a) To find the value of m, we can use the fact that the sum of all the outcomes must equal the total number of times the die was rolled. We know that the sum of all the outcomes is:
32 + m + 25 + 40 + 28 + 45 = 220 + m
We also know that the probability of rolling a 2 is 0.15, which means that the number of times a 2 was rolled is:
0.15 x total number of rolls = 0.15N
Since the only other unknown outcome is m, we can set up an equation:
0.15N + m + 32 + 25 + 40 + 28 + 45 = N
Simplifying this equation, we get:
m = N - 210
(b) To find the number of times the die was rolled, we can use the fact that the sum of all the outcomes must equal the total number of times the die was rolled. From part (a), we know that:
32 + m + 25 + 40 + 28 + 45 = 220 + m
Substituting m with N - 210, we get:
32 + N - 210 + 25 + 40 + 28 + 45 = 220 + N - 210
Simplifying this equation, we get:
N = 210
Therefore, the die was rolled 210 times.
(c) To find the probability of obtaining an even number, we need to find the total number of times an even number was rolled and divide it by the total number of rolls. The even numbers are 2, 4, and 6. From the table, we can see that the number of times a 2 was rolled is m, the number of times a 4 was rolled is 40, and the number of times a 6 was rolled is 45. Therefore, the total number of times an even number was rolled is:
m + 40 + 45 = m + 85
Dividing this by the total number of rolls (which we found to be 210 in part (b)), we get:
(m + 85)/210
We don't know the exact value of m, but we know from part (a) that m = N - 210, so we can substitute:
[(N - 210) + 85]/210
Simplifying this equation, we get:
(N - 125)/420
Substituting N with 210, we get:
(210 - 125)/420 = 85/420 = 17/84
Therefore, the probability of obtaining an even number is 17/84.
Akọwa Nkọwa
(a) To find the value of m, we can use the fact that the sum of all the outcomes must equal the total number of times the die was rolled. We know that the sum of all the outcomes is:
32 + m + 25 + 40 + 28 + 45 = 220 + m
We also know that the probability of rolling a 2 is 0.15, which means that the number of times a 2 was rolled is:
0.15 x total number of rolls = 0.15N
Since the only other unknown outcome is m, we can set up an equation:
0.15N + m + 32 + 25 + 40 + 28 + 45 = N
Simplifying this equation, we get:
m = N - 210
(b) To find the number of times the die was rolled, we can use the fact that the sum of all the outcomes must equal the total number of times the die was rolled. From part (a), we know that:
32 + m + 25 + 40 + 28 + 45 = 220 + m
Substituting m with N - 210, we get:
32 + N - 210 + 25 + 40 + 28 + 45 = 220 + N - 210
Simplifying this equation, we get:
N = 210
Therefore, the die was rolled 210 times.
(c) To find the probability of obtaining an even number, we need to find the total number of times an even number was rolled and divide it by the total number of rolls. The even numbers are 2, 4, and 6. From the table, we can see that the number of times a 2 was rolled is m, the number of times a 4 was rolled is 40, and the number of times a 6 was rolled is 45. Therefore, the total number of times an even number was rolled is:
m + 40 + 45 = m + 85
Dividing this by the total number of rolls (which we found to be 210 in part (b)), we get:
(m + 85)/210
We don't know the exact value of m, but we know from part (a) that m = N - 210, so we can substitute:
[(N - 210) + 85]/210
Simplifying this equation, we get:
(N - 125)/420
Substituting N with 210, we get:
(210 - 125)/420 = 85/420 = 17/84
Therefore, the probability of obtaining an even number is 17/84.
Ajụjụ 58 Ripọtì
(a} In the diagram, O is the centre of the circle ABCDE, = I\(\overline{BC}\)I = |\(\overline{CD}\)| and < BCD = 108°. Find < CDE.
(b) Given that tan x = \(\sqrt{3}\), 0\(^o\) \(\geq\) x \(\geq\) 90\(^o\), evaluate
\(\frac{(cos x)^2 - sin x}{(sin x)^2 + cos x}\)
(a) To find < CDE, we need to first find < BOC.
Since O is the center of the circle, < BOC is twice the angle < BAC. Therefore,
< BAC = 1/2 * < BOC
= 1/2 * (360° - < BCD - < ABC)
= 1/2 * (360° - 108° - 36°)
= 1/2 * 216°
= 108°
Similarly, we can find that < BDC = 1/2 * (360° - < BCD - < CDE)
= 1/2 * (360° - 108° - < CDE)
= 126° - 1/2 * < CDE
Since < BDC and < BCD are equal, we have:
126° - 1/2 * < CDE = 108°
1/2 * < CDE = 18°
< CDE = 36°
Therefore, < CDE is 36°.
(b) We are given that tan x = √3, and we need to find the value of:
(cos x)^2 - sin x
---------------------
(sin x)^2 + cos x
Using the identity (cos x)^2 + (sin x)^2 = 1, we can write:
(cos x)^2 = 1 - (sin x)^2
Substituting this into the expression, we get:
[(1 - (sin x)^2) - sin x] / [(sin x)^2 + cos x]
Simplifying, we get:
(1 - 2(sin x)^2 - sin x) / [(sin x)^2 + cos x]
Substituting tan x = √3, we have:
sin x / cos x = √3
sin x = √3 cos x
Substituting this into the expression, we get:
(1 - 2(3cos^2 x) - √3 cos x) / (3cos^2 x + cos x)
Multiplying both the numerator and denominator by -1, we get:
(2(3cos^2 x) + √3 cos x - 1) / (3cos^2 x + cos x)
Simplifying, we get:
(6cos^2 x + √3 cos x - 1) / (3cos^2 x + cos x)
Substituting √3 for sin x / cos x, we have:
(6/4 + √3/4 - 1) / (3/4 + 1/4)
= (9/4 + √3/4) / 1
= 9 + √3
Therefore, the value of the expression is
Akọwa Nkọwa
(a) To find < CDE, we need to first find < BOC.
Since O is the center of the circle, < BOC is twice the angle < BAC. Therefore,
< BAC = 1/2 * < BOC
= 1/2 * (360° - < BCD - < ABC)
= 1/2 * (360° - 108° - 36°)
= 1/2 * 216°
= 108°
Similarly, we can find that < BDC = 1/2 * (360° - < BCD - < CDE)
= 1/2 * (360° - 108° - < CDE)
= 126° - 1/2 * < CDE
Since < BDC and < BCD are equal, we have:
126° - 1/2 * < CDE = 108°
1/2 * < CDE = 18°
< CDE = 36°
Therefore, < CDE is 36°.
(b) We are given that tan x = √3, and we need to find the value of:
(cos x)^2 - sin x
---------------------
(sin x)^2 + cos x
Using the identity (cos x)^2 + (sin x)^2 = 1, we can write:
(cos x)^2 = 1 - (sin x)^2
Substituting this into the expression, we get:
[(1 - (sin x)^2) - sin x] / [(sin x)^2 + cos x]
Simplifying, we get:
(1 - 2(sin x)^2 - sin x) / [(sin x)^2 + cos x]
Substituting tan x = √3, we have:
sin x / cos x = √3
sin x = √3 cos x
Substituting this into the expression, we get:
(1 - 2(3cos^2 x) - √3 cos x) / (3cos^2 x + cos x)
Multiplying both the numerator and denominator by -1, we get:
(2(3cos^2 x) + √3 cos x - 1) / (3cos^2 x + cos x)
Simplifying, we get:
(6cos^2 x + √3 cos x - 1) / (3cos^2 x + cos x)
Substituting √3 for sin x / cos x, we have:
(6/4 + √3/4 - 1) / (3/4 + 1/4)
= (9/4 + √3/4) / 1
= 9 + √3
Therefore, the value of the expression is
Ajụjụ 59 Ripọtì
(a) If A = {multiples of 2}, B = {multiples of 3} and C = {factors of 6} are subsets of \(\mu\) = {x:1 \(\leq\) x \(\leq\) 10} find A ? \(\cap\) B ? \(\cap\) C?
(b) Tickets for a movie premiere cost $18.50 each while the bulk purchase price for 5 tickets is $80.00. If 4 gentlemen decide to get a fifth person to join them so that they can share the bulk purchase price equally, how much would each person save?
(a) Finding A' \cap B' \cap C'A′∩B′∩C′
Given:
- μ={x:1≤x≤10}={1,2,3,4,5,6,7,8,9,10}
- A={multiples of 2}={2,4,6,8,10}
- B={multiples of 3}={3,6,9}
- C={factors of 6}={1,2,3,6}
First, find the complements A′,B′, and C′:
- A′=μ−A={1,3,5,7,9}
- B′=μ−B={1,2,4,5,7,8,10}
- C′=μ−C={4,5,7,8,9,10}
Now, find the intersection A′∩B′∩C′:
A′∩B′={1,3,5,7,9}∩{1,2,4,5,7,8,10}={1,5,7}
A′∩B′∩C′={1,5,7}∩{4,5,7,8,9,10}={5,7}
So, A′∩B′∩C′={5,7}.
(b) Calculating the Savings Per Person
Given:
- Individual ticket price: $18.50
- Bulk purchase price for 5 tickets: $80.00
First, calculate the total cost of 5 individual tickets: 5×18.50=92.50
The cost difference between individual tickets and the bulk purchase price: 92.50−80.00=12.50
Each person would save: 512.50=2.50
So, if the 4 gentlemen get a fifth person to join them and share the bulk purchase price equally, each person would save $2.50.
Akọwa Nkọwa
(a) Finding A' \cap B' \cap C'A′∩B′∩C′
Given:
- μ={x:1≤x≤10}={1,2,3,4,5,6,7,8,9,10}
- A={multiples of 2}={2,4,6,8,10}
- B={multiples of 3}={3,6,9}
- C={factors of 6}={1,2,3,6}
First, find the complements A′,B′, and C′:
- A′=μ−A={1,3,5,7,9}
- B′=μ−B={1,2,4,5,7,8,10}
- C′=μ−C={4,5,7,8,9,10}
Now, find the intersection A′∩B′∩C′:
A′∩B′={1,3,5,7,9}∩{1,2,4,5,7,8,10}={1,5,7}
A′∩B′∩C′={1,5,7}∩{4,5,7,8,9,10}={5,7}
So, A′∩B′∩C′={5,7}.
(b) Calculating the Savings Per Person
Given:
- Individual ticket price: $18.50
- Bulk purchase price for 5 tickets: $80.00
First, calculate the total cost of 5 individual tickets: 5×18.50=92.50
The cost difference between individual tickets and the bulk purchase price: 92.50−80.00=12.50
Each person would save: 512.50=2.50
So, if the 4 gentlemen get a fifth person to join them and share the bulk purchase price equally, each person would save $2.50.
Ajụjụ 60 Ripọtì
(a) Given that P = (\(\frac{rk}{Q} - ms\))\(^{\frac{2}{3}}\)
(i) Make Q the subject of the relation;
(ii) find, correct to two decimal places, the value of Q when P = 3, m = 15, s = 0.2, k = 4 and r = 10.
(b) Given that \(\frac{x + 2y}{5}\) = x - 2y, find x : y
a)
(i) To make Q the subject of the formula, we need to isolate Q on one side of the equation.
P = \(\left(\frac{rk}{Q} - ms\right)^{\frac{2}{3}}\)
Take the reciprocal of both sides:
\(\frac{1}{P}\) = \(\frac{Q}{rk}\) - \(\frac{ms}{rk}\)
Add \(\frac{ms}{rk}\) to both sides:
\(\frac{ms}{rk}\) + \(\frac{1}{P}\) = \(\frac{Q}{rk}\)
Multiply both sides by \(\frac{rk}{ms + Prk}\):
Q = \(\frac{rk}{ms + Prk}\)
(ii) Now that we have the formula for Q, we can substitute the given values and calculate:
Q = \(\frac{(10)(4)}{(0.2) + (3)(15)(4)}\)
Q = \(\frac{40}{182}\)
Q = 0.22 (correct to two decimal places)
Therefore, when P = 3, m = 15, s = 0.2, k = 4, and r = 10, Q is approximately equal to 0.22.
b) To find x : y, we can use algebraic manipulation to isolate one of the variables in terms of the other:
\(\frac{x + 2y}{5}\) = x - 2y
Distribute the 5 on the right side of the equation:
\(\frac{x + 2y}{5}\) = x - \(\frac{10y}{5}\)
Simplify:
\(\frac{x + 2y}{5}\) = x - 2y
x + 2y = 5x - 10y
12y = 4x
y = \(\frac{1}{3}\)x
Therefore, x : y is 3 : 1.
Akọwa Nkọwa
a)
(i) To make Q the subject of the formula, we need to isolate Q on one side of the equation.
P = \(\left(\frac{rk}{Q} - ms\right)^{\frac{2}{3}}\)
Take the reciprocal of both sides:
\(\frac{1}{P}\) = \(\frac{Q}{rk}\) - \(\frac{ms}{rk}\)
Add \(\frac{ms}{rk}\) to both sides:
\(\frac{ms}{rk}\) + \(\frac{1}{P}\) = \(\frac{Q}{rk}\)
Multiply both sides by \(\frac{rk}{ms + Prk}\):
Q = \(\frac{rk}{ms + Prk}\)
(ii) Now that we have the formula for Q, we can substitute the given values and calculate:
Q = \(\frac{(10)(4)}{(0.2) + (3)(15)(4)}\)
Q = \(\frac{40}{182}\)
Q = 0.22 (correct to two decimal places)
Therefore, when P = 3, m = 15, s = 0.2, k = 4, and r = 10, Q is approximately equal to 0.22.
b) To find x : y, we can use algebraic manipulation to isolate one of the variables in terms of the other:
\(\frac{x + 2y}{5}\) = x - 2y
Distribute the 5 on the right side of the equation:
\(\frac{x + 2y}{5}\) = x - \(\frac{10y}{5}\)
Simplify:
\(\frac{x + 2y}{5}\) = x - 2y
x + 2y = 5x - 10y
12y = 4x
y = \(\frac{1}{3}\)x
Therefore, x : y is 3 : 1.
Ajụjụ 61 Ripọtì
(a) Two cyclists X and Y leave town Q at the same time. Cyclist X travels at the rate of 5 km/h on a bearing of 049° and cyclist Y travels at the rate of 9 km/h on a bearing of 319°.
(a) Illustrate the information on a diagram.
(b) After travelling for two hours, calculate. correct to the nearest whole number, the:
(i) distance between cyclist X and Y;
(ii) bearing of cyclist X from Y.
(c) Find the average speed at which cyclist X will get to Y in 4 hours.
(a) The diagram can be drawn by placing town Q at the origin and drawing the bearings of 049° and 319°. Then, draw lines from town Q at those bearings to represent the paths of cyclists X and Y.
(b)
- (i) To find the distance between cyclist X and Y, we need to use the cosine rule. Let a be the distance that cyclist X travels and b be the distance that cyclist Y travels. Then, we have:
- a² = (2 × 5)² = 100
- b² = (2 × 9)² = 324
- c² = a² + b² - 2ab cos(128°) ? 302.73
- c ? 17.4 km (nearest whole number)
- (ii) To find the bearing of cyclist X from Y, we need to use the sine rule. Let C be the angle between cyclist X's path and the line connecting X and Y, and let D be the angle between cyclist Y's path and the same line. Then, we have:
- sin(C) / a = sin(128°) / c
- sin(D) / b = sin(33°) / c
- C ? 97.2°
- D ? 35.7°
- 360° - 319° + 97.2° ? 138.2°
(c) To find the average speed at which cyclist X will get to Y in 4 hours, we need to find the total distance between them and divide by the time taken. Using the distance formula from part (b)(i), we have:
d = ?302.73 ? 17.4 km
time taken = 4 hours
Therefore, the average speed at which cyclist X will get to Y in 4 hours is:
17.4 km / 4 hours = 4.35 km/h
Akọwa Nkọwa
(a) The diagram can be drawn by placing town Q at the origin and drawing the bearings of 049° and 319°. Then, draw lines from town Q at those bearings to represent the paths of cyclists X and Y.
(b)
- (i) To find the distance between cyclist X and Y, we need to use the cosine rule. Let a be the distance that cyclist X travels and b be the distance that cyclist Y travels. Then, we have:
- a² = (2 × 5)² = 100
- b² = (2 × 9)² = 324
- c² = a² + b² - 2ab cos(128°) ? 302.73
- c ? 17.4 km (nearest whole number)
- (ii) To find the bearing of cyclist X from Y, we need to use the sine rule. Let C be the angle between cyclist X's path and the line connecting X and Y, and let D be the angle between cyclist Y's path and the same line. Then, we have:
- sin(C) / a = sin(128°) / c
- sin(D) / b = sin(33°) / c
- C ? 97.2°
- D ? 35.7°
- 360° - 319° + 97.2° ? 138.2°
(c) To find the average speed at which cyclist X will get to Y in 4 hours, we need to find the total distance between them and divide by the time taken. Using the distance formula from part (b)(i), we have:
d = ?302.73 ? 17.4 km
time taken = 4 hours
Therefore, the average speed at which cyclist X will get to Y in 4 hours is:
17.4 km / 4 hours = 4.35 km/h
