WAEC SSCE - General Mathematics - 1991
Ajụjụ 1 Ripọtì
Solve the equation: 3a + 10 = a\(^2\)
Akọwa Nkọwa
To solve the given equation, we need to bring all the terms to one side and then factorize it to find the values of 'a'. We start by subtracting '3a' from both sides of the equation: 3a + 10 - 3a = a^2 - 3a 10 = a^2 - 3a Now, we can rearrange the terms and factorize the equation as: a^2 - 3a - 10 = 0 (a - 5)(a + 2) = 0 Using the zero product property, we get: a - 5 = 0 or a + 2 = 0 a = 5 or a = -2 Therefore, the values of 'a' that satisfy the given equation are a = 5 or a = -2. Hence, the answer is a = 5 or a = -2.
Ajụjụ 2 Ripọtì
Find the number whose logarithm to base 10 is 2.6025
Akọwa Nkọwa
The logarithm to base 10 of a number is the power to which 10 must be raised to give the number. So, if the logarithm to base 10 of a number is 2.6025, we can write it as: 10^2.6025 = x Using a calculator, we get: 10^2.6025 ≈ 400.401 Therefore, the number whose logarithm to base 10 is 2.6025 is approximately 400.401. Thus, the correct option is (a) 400.4.
Ajụjụ 3 Ripọtì
A box contains 2 white and 3 blue identical marbles. If two marbles are picked at random, one after the other without replacement, what is the probability of picking two marbles of different colors?
Akọwa Nkọwa
There are a total of 5 marbles in the box, 2 of which are white and 3 are blue. If two marbles are picked one after the other without replacement, there are two possible scenarios: either a white marble is picked first or a blue marble is picked first. Case 1: A white marble is picked first. In this case, there are 4 marbles left, out of which 3 are blue. Therefore, the probability of picking a blue marble after picking a white marble is 3/4. Case 2: A blue marble is picked first. In this case, there are 4 marbles left, out of which 2 are white. Therefore, the probability of picking a white marble after picking a blue marble is 2/4 or 1/2. Since there are two possible scenarios and each scenario is mutually exclusive, we can add the probabilities of the two scenarios to get the probability of picking two marbles of different colors: Probability = (Probability of picking a white marble first x Probability of picking a blue marble second) + (Probability of picking a blue marble first x Probability of picking a white marble second) Probability = (2/5 x 3/4) + (3/5 x 1/2) Probability = 3/10 + 3/10 Probability = 6/10 or 3/5 Therefore, the probability of picking two marbles of different colors is 3/5 or 0.6. Answer is the correct answer.
Ajụjụ 4 Ripọtì
A cylinder of base radius 4cm is open at one end . If the ratio of the area of its base to that of its curved surface is 1:4, calculate the height of the cylinder
Ajụjụ 5 Ripọtì
Mrs. Kofi sold an article for C7.50 instead of C12.75. Calculate her percentage of error, correct to one decimal place.
Ajụjụ 6 Ripọtì
Simplify: 1/4(2n - 2n+2)
Ajụjụ 7 Ripọtì
If events X and Y are mutually exclusive, . P(X) = 1/3 and P(Y) = 2/5, P(X∩Y) is
Ajụjụ 8 Ripọtì
Express 0.0462 in standard form
Ajụjụ 9 Ripọtì
Find the area of the enclosed region, PXROY correct to the nearest whole number
Ajụjụ 10 Ripọtì
A car is travelling at an average speed of 80km/hr. Its speed in meters per second (m/s) is
Akọwa Nkọwa
We know that 1 kilometer is equal to 1000 meters and 1 hour is equal to 3600 seconds. Therefore, to convert the car's speed from kilometers per hour (km/hr) to meters per second (m/s), we can use the following conversion factors: 1 km = 1000 m 1 hr = 3600 s We start by converting the speed from km/hr to meters per hour: 80 km/hr x (1000 m/km) = 80,000 m/hr Next, we convert from hours to seconds: 80,000 m/hr x (1 hr/3600 s) = 22.22 m/s (rounded to two decimal places) Therefore, the car's speed in meters per second (m/s) is approximately 22.22 m/s. Answer: 22.2m/s.
Ajụjụ 11 Ripọtì
In the diagram above , |AD| = 10cm, |DC| = 8cm and |CF| = 15cm. Which of the following is correct?
Ajụjụ 12 Ripọtì
The first term a of an A.P is equal to twice the
common difference d. Find, in terms of d, the 5th
term of the A.P.
Akọwa Nkọwa
In an arithmetic progression (A.P), the difference between any two consecutive terms is constant. Let's call this constant difference "d". We're told that the first term "a" is equal to twice the common difference, so: a = 2d To find the 5th term of the A.P, we can use the formula: an = a + (n-1)d where "an" is the nth term of the A.P. Substituting the given values, we get: a5 = 2d + (5-1)d a5 = 2d + 4d a5 = 6d Therefore, the 5th term of the A.P is 6d. Answer: 6d.
Ajụjụ 14 Ripọtì
Two towns, P and Q, are on (4oN 40oW) and (4oN
20oE) respectively. What is the distance between
them, along their line of latitude? (Give your
answer in teems of π and R, the radius of the earth).
Akọwa Nkọwa
To find the distance between towns P and Q along their line of latitude, we need to calculate the length of the arc formed by the angle between the two towns and the center of the Earth. The distance between P and Q can be calculated using the formula: Distance = angle (in radians) x radius of the Earth First, we need to find the angle between P and Q. Since both towns are on the same line of latitude, the angle between them is simply the difference between their longitudes, which is: 20oE - 40oW = 60o Next, we need to convert this angle to radians by multiplying it by π/180, since there are π radians in 180 degrees. 60o x π/180 = π/3 radians Finally, we can plug this value into the formula for distance: Distance = angle (in radians) x radius of the Earth = (π/3) x R = (πR)/3 Therefore, the correct answer is:
Ajụjụ 15 Ripọtì
Solve: 6(x - 4) + 3(x + 7) = 3
Akọwa Nkọwa
We can solve the given equation by using the distributive property of multiplication and combining like terms:
6(x - 4) + 3(x + 7) = 3 6x - 24 + 3x + 21 = 3 9x - 3 = 3 9x = 6 x = 6/9
Simplifying the fraction 6/9, we get:
x = 2/3
Therefore, the solution to the given equation is x = 2/3. So, the correct answer is:
2/3
Ajụjụ 16 Ripọtì
Find the median of the distribution
Akọwa Nkọwa
To find the median of a distribution, we need to arrange the values in order from lowest to highest or highest to lowest. Then, we locate the middle value of the distribution. If the number of values is odd, the median is the middle value. In this case, we have five values, which is an odd number. Therefore, we can directly locate the middle value. If the number of values is even, the median is the average of the two middle values. In this case, we do not have an even number of values. Arranging the given values in order from lowest to highest, we have: 4, 4.5, 5, 6.5, 9 The middle value of this distribution is 5, which is the third value in the ordered set. Therefore, the median of this distribution is 5.
Ajụjụ 18 Ripọtì
Simplify: log6 + log2 - log12
Akọwa Nkọwa
We can use the logarithmic rule that states loga(b) + loga(c) = loga(bc) to simplify the expression. Applying this rule, we have: log6(6) + log6(2) - log6(12) = 1 + log6(2) - log6(2\*2\*3) = 1 + log6(2) - (log6(2) + log6(2\*3)) = 1 + log6(2) - (log6(2) + log6(6)) = 1 + log6(2) - (log6(2) + 1) = log6(2) - log6(2) - 1 + 1 = 0 Therefore, the simplified expression is 0. The correct option is (C) 0.
Ajụjụ 19 Ripọtì
In the diagram above, ATR is a tangent at the point T to the circle center O, if ?TOB = 145°, find ?TAO
Akọwa Nkọwa
Ajụjụ 20 Ripọtì
P = {2, 1,3, 9, 1/2}; Q = {1,21/2,3, 7} and R = {5, 4, 21/2}. Find P∩Q∩R
Akọwa Nkọwa
Ajụjụ 21 Ripọtì
For what value of y is the expression \(\frac{y + 2}{y^{2} - 3y - 10}\) undefined?
Akọwa Nkọwa
The expression \(\frac{y + 2}{y^{2} - 3y - 10}\) is undefined when the denominator is equal to zero, since division by zero is undefined. So we can set the denominator equal to zero and solve for y: y^2 - 3y - 10 = 0 We can factor this quadratic equation as: (y - 5)(y + 2) = 0 And using the zero product property, we know that this equation is only true when either (y - 5) = 0 or (y + 2) = 0. Therefore, the expression is undefined when either y = 5 or y = -2. Therefore, the answer to the question is y = 5 or y = -2.
Ajụjụ 22 Ripọtì
The shaded portion shows the outer boundary
of the half plane defined by the inequality
Akọwa Nkọwa
To understand this problem, let's first define a half-plane. A half-plane is a part of the plane that lies on one side of a straight line and extends infinitely far in that direction. In this problem, we have an inequality that defines a half-plane.
The inequality is: 4x + 3y ≥ 6
To graph this inequality, we can first plot the line 4x + 3y = 6. To plot this line, we can find two points on the line by setting x = 0 and y = 0 and solving for the other variable.
When x = 0, we get: 3y = 6, y = 2
When y = 0, we get: 4x = 6, x = 3/2
So the two points on the line are (0, 2) and (3/2, 0). We can plot these points and draw a straight line passing through them.
|
|
|
------|-------
|
|
|
Now we need to determine which side of the line represents the half-plane defined by the inequality 4x + 3y ≥ 6.
To do this, we can choose a test point not on the line, such as the origin (0,0), and substitute its coordinates into the inequality:
4(0) + 3(0) ≥ 6
0 ≥ 6
Since this is false, the point (0,0) is not in the half-plane defined by the inequality. Therefore, we shade the half-plane that does not include the origin:
|xxxxxxx
|xxxxxxx
|xxxxxxx
------|-------
xxxxx|
xxxxx|
xxxxx|
The shaded portion shows the outer boundary of the half-plane defined by the inequality 4x + 3y ≥ 6.
Answer: 4x + 3y ≥ 6.
Ajụjụ 23 Ripọtì
In a class of 80 students, every student had to study Economics or Geography or both Economics and Geography. lf 65 students studied Economics and 50 studied Geography, how many studied both subjects?
Akọwa Nkọwa
We are given that every student had to study Economics or Geography or both. Therefore, the total number of students in the class is 80.
Let's denote the number of students who studied only Economics by 'E', the number of students who studied only Geography by 'G', and the number of students who studied both subjects by 'B'. Then we can use a Venn diagram to represent the information given in the problem:
-----------------------------------------
| | |
| Economics | Geography |
| | |
-----------------------------------------
| |
| |
| |
B G
| |
| |
| |
-----------------------------------------
| | |
| Only Economics | Only Geography |
| | |
-----------------------------------------
We are given that 65 students studied Economics, which includes both those who studied only Economics and those who studied both subjects. So we can write:
E + B = 65
Similarly, we are given that 50 students studied Geography, which includes both those who studied only Geography and those who studied both subjects. So we can write:
G + B = 50
We want to find the value of B, the number of students who studied both subjects. To do this, we can add the two equations above:
E + B = 65
G + B = 50
----------
E + G + 2B = 115
We know that the total number of students in the class is 80, so we can write:
E + G + B = 80
Substituting the expression for E + G + 2B into this equation, we get:
115 - B = 80
Solving for B, we get:
B = 35
Therefore, 35 students studied both Economics and Geography.
Ajụjụ 25 Ripọtì
In the diagram above, O is the center of the circle, |SQ| = |QR| and ?PQR = 68°. Calculate ?PRS
Akọwa Nkọwa
Ajụjụ 27 Ripọtì
The common ratio of a G.P. is 2. If the 5th term is
greater than the 1st term by 45, find the 5th term,
Akọwa Nkọwa
Let the first term of the G.P. be a. The formula for the nth term of a G.P. with first term a and common ratio r is given by: an = ar^(n-1) Since the common ratio is 2, the 5th term will be a x 2^(5-1) = 16a. It is given that the 5th term is greater than the 1st term by 45. Therefore, 16a - a = 45 Simplifying this equation gives: 15a = 45 Dividing both sides by 15 gives: a = 3 Therefore, the 5th term is 16a = 16 x 3 = 48. Hence, the answer is 48.
Ajụjụ 28 Ripọtì
Mrs. Jones is expecting a baby. The probability that it will be a boy is 1/2 and probability that the baby will have blue eyes is 1/4. What is the probability that she will have a blue-eyed boy?
Akọwa Nkọwa
To determine the probability of having a blue-eyed boy, we need to consider the probability of two independent events happening together: the baby being a boy and having blue eyes. The probability of having a boy is 1/2, and the probability of having blue eyes is 1/4. Since these events are independent, we can multiply their probabilities to find the probability of both events happening together: 1/2 * 1/4 = 1/8 Therefore, the probability that Mrs. Jones will have a blue-eyed boy is 1/8.
Ajụjụ 29 Ripọtì
P={2, 1,3, 9, 1/2}; Q = {1,21/2,3, 7} and R = {5, 4, 21/2}. Find P∪Q∪R
Akọwa Nkọwa
To find P∪Q∪R, we need to combine all the elements in P, Q and R, without repeating any element. The elements in P are {2, 1, 3, 9, 1/2}. The elements in Q are {1, 2^(1/2), 3, 7}. The elements in R are {5, 4, 2^(1/2)}. So, P∪Q∪R = {1/2, 1, 2, 2^(1/2), 3, 4, 5, 7, 9}. Therefore, the correct answer is (1/2, 1, 2, 21/2, 3, 4, 5, 7, 9).
Ajụjụ 30 Ripọtì
In a given regular polygon, the ratio of the exterior angle to the interior angles is 1:3. How many side has the polygon?
Akọwa Nkọwa
In any polygon, the sum of the exterior angles is always 360 degrees. Therefore, the measure of each exterior angle of a regular polygon with n sides is 360/n degrees. The ratio of the exterior angle to the interior angle is 1:3. This means that the measure of the exterior angle is three times the measure of the interior angle. Let x be the measure of the interior angle. Then, the measure of the exterior angle is 3x. We know that the sum of the interior angles of any polygon is (n-2) times 180 degrees. Therefore, the measure of each interior angle of a regular polygon with n sides is [(n-2) x 180]/n degrees. Now we can set up an equation to solve for n. 3x = 360/n x = (n-2) x 180/n Substituting 3x for the exterior angle in the first equation: 3x = 360/n x = 120/n Substituting 120/n for x in the second equation: 120/n = (n-2) x 180/n Simplifying: 120 = 180(n-2) 120 = 180n - 360 540 = 180n n = 3 However, a polygon with 3 sides is not possible since it would be a triangle. Therefore, the answer is not 3. We can try other answer options by substituting each value of n in the formula for the measure of each exterior angle (360/n) and checking if the ratio of the exterior angle to the interior angle is 1:3. For n=4, the measure of each exterior angle is 90 degrees, and the measure of each interior angle is 90 degrees. Therefore, the ratio of the exterior angle to the interior angle is 1:1, which is not 1:3. For n=5, the measure of each exterior angle is 72 degrees, and the measure of each interior angle is 108 degrees. Therefore, the ratio of the exterior angle to the interior angle is 2:3, which is not 1:3. For n=6, the measure of each exterior angle is 60 degrees, and the measure of each interior angle is 120 degrees. Therefore, the ratio of the exterior angle to the interior angle is 1:2, which is not 1:3. For n=8, the measure of each exterior angle is 45 degrees, and the measure of each interior angle is 135 degrees. Therefore, the ratio of the exterior angle to the interior angle is 1:3, which satisfies the given condition. Therefore, the polygon has 8 sides, and the answer is 8.
Ajụjụ 31 Ripọtì
In the diagram above, PQ and XY are two concentric arc; center O, the ratio of the length of the two arc is 1:3, find the ratio of the areas of the two sectors OPQ and OXY
Ajụjụ 32 Ripọtì
If events X and Y are mutually exclusive, P(X) = 1/3 and P(Y) = 2/5, P(X∪Y) is
Akọwa Nkọwa
If events X and Y are mutually exclusive, it means they cannot occur at the same time. In other words, if X happens, Y cannot happen, and vice versa. This also means that the probability of both events happening at the same time (P(X∩Y)) is equal to zero. We can use the formula for the probability of the union of two events: P(X∪Y) = P(X) + P(Y) - P(X∩Y) Since X and Y are mutually exclusive, we know that P(X∩Y) is equal to zero: P(X∪Y) = P(X) + P(Y) - 0 We can substitute the given probabilities into this formula: P(X∪Y) = 1/3 + 2/5 To add these fractions, we need a common denominator. The smallest common multiple of 3 and 5 is 15, so we can rewrite the fractions with this denominator: P(X∪Y) = 5/15 + 6/15 P(X∪Y) = 11/15 Therefore, the answer is 11/15.
Ajụjụ 33 Ripọtì
Given that sin \(\theta\) = -0.9063, where O \(\leq\) \(\theta\) \(\leq\) 270°, find \(\theta\).
Akọwa Nkọwa
Ajụjụ 34 Ripọtì
From the top of a cliff, the angle of depression of
a boat on the sea is 60o, if the top of the cliff is
25m above the sea level, calculate the horizontal
distance from the bottom of the cliff to the boat.
Akọwa Nkọwa
Ajụjụ 35 Ripọtì
Find the mean of the distribution
Ajụjụ 39 Ripọtì
A headmaster contributes 7% of his income into a fund and his wife contributes 4% of her income. If the husband earns N5,500 per annum (p.a) and the wife earns N4,000 (P.a), find the sum of their annual contribution to the fund
Akọwa Nkọwa
To find the sum of their annual contribution to the fund, we need to calculate the amount that each of them contributes and then add them up. The headmaster contributes 7% of his income, which is 7/100 * N5,500 = N385 per annum. The wife contributes 4% of her income, which is 4/100 * N4,000 = N160 per annum. Therefore, the sum of their annual contribution to the fund is N385 + N160 = N545. So, the correct answer is N545.
Ajụjụ 40 Ripọtì
Factorize 3a\(^2\) - 11a + 6
Akọwa Nkọwa
To factorize 3a\(^2\) - 11a + 6, we need to find two binomials that, when multiplied together, result in the original expression. To do this, we can use a technique called "factoring by grouping." First, we need to identify two numbers that multiply to 3 x 6 = 18 and add up to -11. These numbers are -2 and -9. Next, we can split the middle term -11a into -2a - 9a, and then group the terms as follows: 3a\(^2\) - 2a - 9a + 6 Now, we can factor out the greatest common factor from the first two terms, and the greatest common factor from the last two terms: a(3a - 2) - 3(3a - 2) Notice that we have a common binomial factor of (3a - 2), which we can factor out: (3a - 2)(a - 3) Therefore, the correct option is (3a - 2)(a - 3).
Ajụjụ 41 Ripọtì
The angles of a pentagon are x°, 2x°, (x + 60)°, (x + 10)°, (x -10)°. Find the value of x.
Akọwa Nkọwa
In a pentagon, the sum of the angles is equal to (5 - 2) times 180 degrees, which is 540 degrees. So we can set up an equation to solve for x:
x + 2x + (x + 60) + (x + 10) + (x - 10) = 540
Simplifying this equation, we get:
6x + 60 = 540
Subtracting 60 from both sides, we get:
6x = 480
Dividing both sides by 6, we get:
x = 80
Therefore, the value of x is 80. So, the correct answer is:
80
Ajụjụ 42 Ripọtì
The population of a village is 5846. Express this number to three significant figures
Akọwa Nkọwa
When expressing a number to three significant figures, we only consider the first three digits of the number and round it up or down depending on the value of the fourth digit. In this case, the number is already given as 5846. Since the fourth digit is 6, which is greater than or equal to 5, we round up the last significant digit (the third digit) by adding 1 to it. Therefore, the answer rounded to three significant figures is 5850. So the correct option is: 5850.
Ajụjụ 45 Ripọtì
The angle of elevation of the top of a tree 39m
away from a point on the ground is 30o. Find the height of the tree
Akọwa Nkọwa
Ajụjụ 47 Ripọtì
The mean of 30 observations recorded in an experiment is 5. lf the observed largest value of 34 is deleted, find the mean of the remaining observations
Akọwa Nkọwa
The mean of the 30 observations is given as 5. If we remove the largest observation of 34, we will be left with 29 observations. To find the mean of the remaining observations, we need to sum them up and divide by the number of observations. The sum of the 30 observations is 30 x 5 = 150. If we remove the observation of 34, the sum of the remaining 29 observations will be 150 - 34 = 116. Therefore, the mean of the remaining 29 observations is 116/29 = 4. Hence, the answer is 4.
Ajụjụ 48 Ripọtì
Which of the following is not a measure of dispersion?
Akọwa Nkọwa
The measure of dispersion gives information on how spread out or clustered a set of data is. It quantifies the variability or diversity of the data values. Out of the options provided, the measure that is not a measure of dispersion is the mode. The mode represents the value that occurs most frequently in a dataset. It is not a measure of dispersion because it does not provide any information about how much the data values deviate or spread out from the mode. The other options provided are all measures of dispersion. - Mean deviation: measures the average distance between each data point and the mean of the dataset. - Range: measures the difference between the highest and lowest value in the dataset. - Interquartile range: measures the difference between the upper and lower quartiles of the dataset. - Standard deviation: measures how much the data values deviate from the mean of the dataset. Therefore, the mode is not a measure of dispersion, while the others are.
Ajụjụ 51 Ripọtì
A man has 9 identical balls in a bag. Out of these, 3 are black, 2 are blue and the remaining are red.
(a) If a ball is drawn at random, what is the probability that it is (i) not blue? (ii) not red?
(b) If 2 balls are drawn at random, one after the other, what is the probability that both of them will be (i) black, if there is no replacement? (ii) blue, if there is a replacement?
Akọwa Nkọwa
None
Ajụjụ 52 Ripọtì
A man bought 5 reams of duplicating paper, each of which are supposed to contain 480 sheets. The actual number of sheets in the packets were : 435, 420, 405, 415 and 440.
(a) Calculate, correct to the nearest whole number, the percentage error for the packets of paper;
(b) If the agreed price for a full ream was N35.00, find, correct to the nearest naira, the amount by which the buyer was cheated.
Akọwa Nkọwa
None
Ajụjụ 53 Ripọtì
From a horizontal distance of 8.5 km, a pilot observes that the angles of depression of the top and the base of a control tower are 30° and 33° respectively. Calculate, correct to three significant figures :
(a) the shortest distance between the pilot and the base of the control tower;
(b) the height of the control tower.
Akọwa Nkọwa
None
Ajụjụ 54 Ripọtì
(a) Using a ruler and a pair of compasses only, construct a parallelogram PQRS with diagonals |PR| = 9cm and |QS| = 6cm, intersecting at K and < QKR = 60°.
(b) Construct a rectangle PABS which is equal in area to PQRS in (a) above and on the same side of PS as PQRS. Measure |PA|.
Ajụjụ 55 Ripọtì
The table below shows the distribution of the waiting times for some customers in a certain petrol station.
| Waiting time (in mins) | No of customers |
| 1.5 - 1.9 | 3 |
| 2.0 - 2.4 | 10 |
| 2.5 - 2.9 | 18 |
| 3.0 - 3.4 | 10 |
| 3.5 - 3.9 | 7 |
| 4.0 - 4.4 | 2 |
(a) Write down the class boundaries of the distribution.
(b) Construct a cumulative frequency curve for the data;
(c) Using your graph, estimate: (i) the interquartile range of the distribution ; (ii) the proportion of customers who could have waited for more than 3 minutes.
None
Akọwa Nkọwa
None
Ajụjụ 56 Ripọtì
Simplify :
(i) \(2\frac{2}{3} - (2\frac{1}{2} - 1\frac{4}{5})\)
(ii) \(\frac{3.25 - 1.64}{2.47 - 2.01}\)
Akọwa Nkọwa
None
Ajụjụ 57 Ripọtì
(a) Prove that the angle which an arc of a circle subtends at the centre is twice that which it subtends at any point on the remaining part of the circumference.
(b) 
In the diagram, O is the centre of the circle, < OQR = 32° and < MPQ = 15°. Calculate (i) < QPR ; (ii) < MQO.
Ajụjụ 58 Ripọtì
ABCDE is a regular pentagon and a rectangle AXYE is drawn on the side AE such that the vertices X and Y lie on the sides BC and CD respectively. Calculate the size of
(i) an interior angle of the pentagon ;
(ii) < BXA.
Akọwa Nkọwa
None
Ajụjụ 59 Ripọtì
(a) Using logarithm table, evaluate \(\frac{\sqrt[3]{1.376}}{\sqrt[4]{0.007}}\) correct to three significant figure.
(b) Without using Mathematical tables, find the value of \(\frac{\log 81}{\log \frac{1}{3}}\).
Ajụjụ 60 Ripọtì
Using a scale of 2cm to 1 unit on the x- axis and 1cm to 1 unit on the y- axis, draw on the same axes the graphs of \(y = 3 + 2x - x^{2}; y = 2x - 3\) for \(-3 \leq x \leq 4\). Using your graph:
(i) solve the equation \(6 - x^{2} = 0\);
(ii) find the maximum value of \(3 + 2x - x^{2}\);
(iii) find the range of x for which \(3 + 2x - x^{2} \leq 1\), expressing all your answers correct to one decimal place.
Akọwa Nkọwa
None
Ajụjụ 61 Ripọtì
(a) Solve the equation, correct to two decimal places \(2x^{2} + 7x - 11 = 0\)
(b) Using the substitution \(P = \frac{1}{x}; Q = \frac{1}{y}\), solve the simultaneous equations : \(\frac{2}{x} + \frac{1}{y} = 3 ; \frac{1}{x} - \frac{5}{y} = 7\)
Akọwa Nkọwa
None
Ajụjụ 62 Ripọtì
In a certain class, 22 pupils take one or more of Chemistry, Economics and Government. 12 take Economics (E), 8 take Government (G) and 7 take Chemistry (C). Nobody takes Economics and Chemistry and 4 pupils take Economics and Government.
(a)(i) Using set notation and the letters indicated above, write down the two statements in the last sentence; (ii) Draw a Venn diagram to illustrate the information.
(b) How many pupils take (i) both Chemistry and Government ? (ii) Government only?
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Akọwa Nkọwa
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