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Ajụjụ 2 Ripọtì
Find the equation of the set of points which are equidistant from the parallel lines x = 1 and x = 7
Akọwa Nkọwa
To find the set of points which are equidistant from the parallel lines x=1 and x=7, we can begin by finding the midpoint of the segment connecting the two parallel lines. The midpoint of the segment joining the two parallel lines is ((1+7)/2, 0) = (4,0). Now, let (x,y) be any point that is equidistant from the two parallel lines. Then, the distance from (x,y) to the line x=1 is |x-1|, and the distance from (x,y) to the line x=7 is |x-7|. Since the point (x,y) is equidistant from the two lines, we have: |x-1| = |x-7| Solving for x, we get: x-1 = -(x-7) or x-1 = x-7 Solving each equation for x, we get: x = 4 or x = -6 Since the distance from (x,y) to x=1 is the same as the distance from (x,y) to x=7, it follows that the set of points that are equidistant from the two parallel lines is the vertical line passing through the midpoint of the segment joining the two parallel lines, namely, the line x=4. Therefore, the equation of the set of points which are equidistant from the parallel lines x=1 and x=7 is x = 4. Thus, the correct option is x = 4.
Ajụjụ 3 Ripọtì
The ninary operation * is defined on the set of integers p and q by p*q = pq + p + q. Find 2 * (3 * 4).
Akọwa Nkọwa
Ajụjụ 4 Ripọtì
A hunter 1.6 m tall, views a bird on top of a tree at an angle of 45 degrees. if the distance between the hunter and the tree is 10.4 m, find the height of the tree.
Akọwa Nkọwa
Ajụjụ 5 Ripọtì
If dy/dx = 2x - 3 and y = 3 when x = 0, find y in terms of x.
Akọwa Nkọwa
To solve this problem, we need to integrate both sides of the equation with respect to x. dy/dx = 2x - 3 Integrating both sides with respect to x gives: y = x^2 - 3x + c where c is the constant of integration. To find the value of c, we use the fact that y = 3 when x = 0: 3 = 0^2 - 3(0) + c c = 3 Therefore, the equation of y in terms of x is: y = x^2 - 3x + 3
Ajụjụ 7 Ripọtì
In a school, 220 students offer Biology or Mathematics or both, 125 offer Biology and 110 Mathematics. How many offer Biology but not Mathematics?
Ajụjụ 8 Ripọtì
Simplify 52.4 - 5.7 - 3.45 - 1.75
Akọwa Nkọwa
To simplify the given expression, we can simply subtract the given values as follows: 52.4 - 5.7 - 3.45 - 1.75 = 41.5 Therefore, the answer is 41.5.
Ajụjụ 9 Ripọtì
If -2 is the solution of the equation 2x + 1 - 3c = 2c + 3x - 7, find the value of c.
Akọwa Nkọwa
To find the value of c, we need to substitute -2 for x in the given equation and then solve for c. Substituting -2 for x, we get: 2(-2) + 1 - 3c = 2c + 3(-2) - 7 Simplifying the equation, we get: -4 + 1 - 3c = 2c - 6 - 7 Combining like terms, we get: -3c - 3 = 2c - 13 Moving all the c terms to one side and all the constant terms to the other side, we get: -3c - 2c = -13 + 3 Simplifying, we get: -5c = -10 Dividing both sides by -5, we get: c = 2 Therefore, the value of c is 2.
Ajụjụ 11 Ripọtì
The range of the data: k+2, k-3, k+4, k-2, k, k-5, k+3, k-1, and k+6 is
Akọwa Nkọwa
To find the range of a data set, we need to subtract the smallest value from the largest value. In this case, the smallest value is k-5 and the largest value is k+6. Therefore, the range is (k+6) - (k-5) = k+6 - k+5 = 11. So the answer is 11.
Ajụjụ 13 Ripọtì
Calculate the mean deviation of the set of numbers 7, 3, 14, 9, 7, and 8.
Akọwa Nkọwa
To find the mean deviation of a set of numbers, we first need to find the mean (average) of the set. Mean = (7+3+14+9+7+8) / 6 = 48/6 = 8 Next, we need to find the deviation of each number from the mean. To do this, we subtract the mean from each number: (7-8), (3-8), (14-8), (9-8), (7-8), (8-8) -1, -5, 6, 1, -1, 0 Notice that some of the deviations are negative and some are positive. To avoid canceling out the deviations that are opposite in sign, we take the absolute value of each deviation: 1, 5, 6, 1, 1, 0 Finally, we find the mean of these absolute deviations: Mean Deviation = (1+5+6+1+1+0) / 6 = 14/6 = 7/3 Therefore, the mean deviation of the given set of numbers is 7/3.
Ajụjụ 14 Ripọtì
The sum to infinity of the series: 1 + (1/3) + (1/9) + (1/27) + ... is
Akọwa Nkọwa
This is a geometric series with first term 1 and common ratio 1/3. The formula for the sum of an infinite geometric series is S = a/(1-r), where a is the first term and r is the common ratio. Plugging in the values, we get S = 1/(1 - 1/3) = 3/2. Therefore, the answer is 3/2.
Ajụjụ 15 Ripọtì
If the 9th term of an A.P is five times the 5th term, find the relationship between a and d.
Akọwa Nkọwa
Let the fifth term of the AP be 'a + 4d'. Then, the ninth term will be 'a + 8d'. The problem states that the ninth term is five times the fifth term, which can be represented mathematically as: a + 8d = 5(a + 4d) Simplifying the equation gives: a + 8d = 5a + 20d Subtracting 'a' and 20d from both sides gives: -12d = -4a Dividing both sides by -4 gives: 3d = a So the relationship between 'a' and 'd' is a = 3d. Thus, the answer is a + 3d = 0.
Ajụjụ 16 Ripọtì
| No. of days | 1 | 2 | 3 | 4 | 5 | 6 |
| No. of students | 20 | x | 50 | 40 | 2x | 60 |
The distribution above shows the number of days a group of 260 students were absents from school in a particular term. How many students were absent for at least four days in the term
Akọwa Nkọwa
20 + X + 50 + 40 + 2X + 60 = 260
3X + 170 = 260
3X = 260 - 170
3x = 90
x = 30
Absent for at least 4 days
| 4 | 5 | 6 |
| 40 | 2x | 60 |
Ajụjụ 18 Ripọtì
The venn diagram above shows the numbers of students offering music and history in a class of 80 students. If a student is picked at random from the class, what is the probability that he offers Music only?
Akọwa Nkọwa
30 - x + x + 40 - x = 80 - 20
70 - x = 60
- x = 60 - 70
- x = - 10
∴ x = 10
Music only = 30 - x
= 30 - 10
20
P(music only) = 20/80
= 1/4
= 0.25
Ajụjụ 19 Ripọtì
Simplify (√0.7 + √70)2
Akọwa Nkọwa
To simplify (√0.7 + √70)², we can use the formula (a + b)² = a² + 2ab + b², where a = √0.7 and b = √70. So, we have: (√0.7 + √70)² = (√0.7)² + 2(√0.7)(√70) + (√70)² Simplifying, we get: 0.7 + 2√(0.7 × 70) + 70 = 0.7 + 2√49 + 70 = 0.7 + 2(7) + 70 = 84.7 Therefore, the answer is 84.7.
Ajụjụ 20 Ripọtì
In the diagram, a cylinder is surmounted by a hemispherical bowl. Calculate the volume of the solid
Akọwa Nkọwa
V = πr2h+23πr3
N.B h = r
V = πx(3)2 x 20 = 23×π×(3)3
180π+18π=198cm3
Ajụjụ 21 Ripọtì
How many three-digit numbers can be formed from 32564 without repeating any of the digits?
Ajụjụ 22 Ripọtì
Without using tables, evaluate (343)1/3 x (
0.14)-1 x (25)-1/2
Akọwa Nkọwa
We can simplify the expression using the laws of exponents and basic arithmetic operations. First, let's evaluate each exponent: - (343)^(1/3) = 7, since 7^3 = 343 - (0.14)^(-1) = 1/0.14 ≈ 7.14 - 25^(-1/2) = 1/√25 = 1/5 Next, let's substitute these values back into the original expression: (343)^(1/3) x (0.14)^(-1) x 25^(-1/2) = 7 x 7.14 x 1/5 Multiplying these three numbers gives us: 7 x 7.14 x 1/5 ≈ 10 Therefore, the answer is 10.
Ajụjụ 23 Ripọtì
Find the value of a if the line 2y - ax + 4 = 0 is perpendicular to the line y + (x/4) - 7 = 0
Akọwa Nkọwa
Ajụjụ 24 Ripọtì
(0.21 x 0.072 x 0.00054) (0.006 x 1.68 x 0.063), correct to 4.s.f
Akọwa Nkọwa
To find the product of the two numbers, we can simply multiply them together: (0.21 x 0.072 x 0.00054) x (0.006 x 1.68 x 0.063) = 0.000007728 x 0.000634176 Next, we can use a calculator to evaluate this expression and round to 4 significant figures: 0.000007728 x 0.000634176 = 0.00000489479045568 Rounding to 4 significant figures, we look at the fifth digit (9) and round up because it is greater than 5. Therefore, the final answer is: 0.01286 So the answer is 0.01286, correct to 4 significant figures.
Ajụjụ 25 Ripọtì
The mean of a set of six numbers is 60. If the mean of the first five is 50, find the sixth number in the set.
Akọwa Nkọwa
The mean of a set of six numbers is 60. Therefore, the sum of the six numbers is 6 x 60 = 360. The mean of the first five is 50. Therefore, the sum of the first five numbers is 5 x 50 = 250. To find the sixth number, we can subtract the sum of the first five numbers from the sum of all six numbers: 6th number = sum of all six numbers - sum of the first five numbers 6th number = 360 - 250 6th number = 110 Therefore, the sixth number in the set is 110.
Ajụjụ 26 Ripọtì
The time taken to do a piece of work is inversely proportional to the number of men employed. if it takes 45 men to do a piece of work in 5 days, how long will it take 25 men?
Akọwa Nkọwa
The time taken to do a piece of work is inversely proportional to the number of men employed, which means that the more men there are, the less time it will take to complete the work. In this case, we are told that 45 men can do the work in 5 days. If we let x be the number of days it takes for 25 men to do the work, then we can set up a proportion: 45 men x 5 days = 25 men x (number of days) Solving for x, we get: x = (45 x 5) / 25 = 9 days Therefore, it would take 25 men 9 days to do the same amount of work that 45 men can do in 5 days. The answer is 9 days.
Ajụjụ 27 Ripọtì
If x varies directly as √n and x = 9 when n = 9, find x when n = (17/9)
Akọwa Nkọwa
If x varies directly as the square root of n, we can write the equation as: x = k√n where k is a constant of proportionality. To find k, we can use the given values of x and n: 9 = k√9 Solving for k, we get: k = 3 So the equation becomes: x = 3√n To find x when n = 17/9, we substitute n into the equation: x = 3√(17/9) = 3(√17/√9) = 3(√17/3) = √17 Therefore, x = √17.
Ajụjụ 29 Ripọtì
In the diagram, XZ is the diameter of the circle XYZW, with centre O and radius 152
cm. If XY = 12cm, find the area of the triangle XYZ
Akọwa Nkọwa
Ajụjụ 30 Ripọtì
A trader bought goats for N4000 each. He sold them for N180,000 at a loss of 25%. How many goats did he buy?
Akọwa Nkọwa
Let's start by using a bit of math to understand the problem. We know that the trader bought each goat for N4000, and sold them for N180,000. We also know that he sold them at a loss of 25%. If the trader sold the goats at a loss of 25%, it means that he sold them for only 75% of their original value. We can calculate the original value of the goats by dividing the selling price by 75% or multiplying it by 4/3. So, the original value of the goats = 180,000 * (4/3) = 240,000. Since the trader bought each goat for N4000, we can find out how many goats he bought by dividing the total value of the goats (N240,000) by the price he paid for each goat (N4000). Number of goats = 240,000 / 4000 = 60. Therefore, the trader bought 60 goats. To summarize, the trader bought 60 goats and sold them for N180,000 at a loss of 25%. We found the answer by calculating the original value of the goats and dividing it by the price the trader paid for each goat.
Ajụjụ 31 Ripọtì
Find the coordinates of the mid-point of x and y intercepts of the line 2y = 4x - 8
Akọwa Nkọwa
To find the x-intercept of the line 2y = 4x - 8, we set y = 0 and solve for x. So, we have: 2(0) = 4x - 8 4x = 8 x = 2 This means the x-intercept is (2, 0). To find the y-intercept of the line, we set x = 0 and solve for y. So, we have: 2y = 4(0) - 8 2y = -8 y = -4 This means the y-intercept is (0, -4). The midpoint of two points (x1, y1) and (x2, y2) is: ((x1 + x2)/2, (y1 + y2)/2) Using this formula, we can find the midpoint of the x and y intercepts: ((2 + 0)/2, (0 + (-4))/2) = (1, -2) Therefore, the coordinates of the midpoint of the x and y intercepts of the line 2y = 4x - 8 is (1, -2). Answer option number 2.
Ajụjụ 33 Ripọtì
The acres for rice, pineapple, cassava, cocoa, and palm oil in a certain district are given respectively as 2, 5, 3, 11, and 9. What is the angle sector for cassava in a pie chart?
Akọwa Nkọwa
To find the angle sector for cassava in a pie chart, we need to calculate what percentage of the total area is occupied by cassava. First, we need to find the total area of the pie chart, which represents the total acres of all the crops in the district. The total acres can be found by adding up the acres for each crop: 2 + 5 + 3 + 11 + 9 = 30 So the total area of the pie chart is 30 acres. Next, we need to find what percentage of the total area is occupied by cassava. We can do this by dividing the acres of cassava by the total acres and then multiplying by 100: (3 / 30) x 100 = 10% Therefore, the angle sector for cassava in the pie chart is 10% of the total area of the pie chart. To find the angle in degrees, we need to multiply the percentage by 360 (the total number of degrees in a circle): 10% x 360 = 36° So the angle sector for cassava in the pie chart is 36°.
Ajụjụ 34 Ripọtì
The probability of a student passing any examination is 2/3. If the students takes three examination, what is the probability that he will not pass any of them
Akọwa Nkọwa
The probability of passing any one of the three examinations is 2/3. Therefore, the probability of not passing any one of the exams is 1 - 2/3 = 1/3. Since the student takes three examinations, the probability of not passing any of them is the probability of not passing the first exam multiplied by the probability of not passing the second exam multiplied by the probability of not passing the third exam. So, the probability of not passing any of the three exams is (1/3) x (1/3) x (1/3) = 1/27. Therefore, the answer is option (D), which is 1/27.
Ajụjụ 36 Ripọtì
Find the mean of the data: 7, -3, 4, -2, 5, -9, 4, 8, -6, 12
Akọwa Nkọwa
Ajụjụ 37 Ripọtì
In the diagram above , XZ is the diameter of the circle XZW, with center O and radius 15/2 cm. If XY = 12 cm, find the area of the triangle XYZ
Akọwa Nkọwa
Ajụjụ 38 Ripọtì
The triangle PQR above is
Akọwa Nkọwa
∠PQR = 180 - 128 (∠s on a straight line)
= 52∘
∠QPR + 76 + 52 = 180(extr ∠ = sum of intr. opp ∠s)
∠QPR + 128 = 180
∠QPR = 180 - 128
= 52∘
∴ΔPQR is an isosceles triangle
Ajụjụ 39 Ripọtì
Using the graph find the values of p and q if px + qy ? 4
Akọwa Nkọwa
m = y2?y1x2?x1=2?00?(4)=24=12
y2?y1x2?x1?m
y?0x+4?12
2y ? ? x + 4, -x + 2y ? 4 = px + qy ? 4
p = -1, q = 2
Ajụjụ 40 Ripọtì
A chord of a circle subtends an angle of 12o degrees at the centre of a circle of diameter 4√3 cm. Calculate the area of the major sector.
Akọwa Nkọwa
Ajụjụ 41 Ripọtì
A bucket is 12 cm in diameter at the top, 8 cm in diameterat the bottom and 4 cm deep. Calculate its volume.
Akọwa Nkọwa
To calculate the volume of the bucket, we need to use the formula for the volume of a frustum of a cone which is given by: V = (1/3) × π × h × (r₁² + r₂² + r₁ × r₂) where h is the height of the frustum, r₁ and r₂ are the radii of the top and bottom faces, respectively. In this case, the height of the frustum is 4 cm, the top radius is 6 cm (half of the diameter), and the bottom radius is 4 cm. Substituting these values into the formula, we get: V = (1/3) × π × 4 × (6² + 4² + 6 × 4) V = (1/3) × π × 4 × (36 + 16 + 24) V = (1/3) × π × 4 × 76 V = 304π/3 cubic centimeters Therefore, the volume of the bucket is 304π/3 cubic centimeters. Answer (1) is correct.
Ajụjụ 42 Ripọtì
The locus of a point P which is equidistant from two given points S and T is
Akọwa Nkọwa
The correct answer is "the perpendicular bisector of ST". The locus of a point that is equidistant from two points S and T is the set of all points that are at the same distance from S and T. This set of points is a line that is equidistant from S and T, and this line is the perpendicular bisector of the line segment ST. Therefore, the answer is "the perpendicular bisector of ST".
Ajụjụ 43 Ripọtì
In the diagram are two concentric circles of radii r and R respectively with centre O. If r = 25
R, express the area of the shaded portion in terms of π
and R
Akọwa Nkọwa
Ashaded portion = πR2−πr2 r = 25 R
Ashaded portion = πR2−π(23R)2=πR2−4πR225
= πR2(1−23)=2125πR2
Ajụjụ 44 Ripọtì
In the diagram above, PST is a straight line, PQ = QS = RS. If ∠RST = 72∘
, find x
Akọwa Nkọwa
In Δ PQS, ∠PSQ = X(base ∠s of isoc Δ PQS)
In Δ QRS, ∠RQS = ∠PSQ + X(Extr ∠ = sum of two intr. opp ∠s)
∴ ∠RQS = X + X
= 2X
Also ∠QRS = 2X(base ∠s of isoc Δ QRS in Δ PRS,
72 = ∠RPS + ∠PRS(Extr, ∠ = sum of two intr. opp ∠s)
∴ 72 = x + 2x
72 = 3x
x = 72/3
x = 24∘
Ajụjụ 45 Ripọtì
In the diagram PQ is parallel to RS. What is the value of α+β+γ
Ajụjụ 46 Ripọtì
In the diagram above, a cylinder is summounted by a hemisphere bowl. Calculate the volume of the solid.
Akọwa Nkọwa
To find the volume of the solid, we need to find the volumes of the cylinder and the hemisphere bowl separately, and then add them together. The volume of the cylinder is given by the formula V_cylinder = πr^2h, where r is the radius of the base of the cylinder and h is the height of the cylinder. Looking at the diagram, we can see that the height of the cylinder is 12 cm and the radius of the base is also 12 cm (since the diameter of the cylinder is 24 cm). Therefore, the volume of the cylinder is: V_cylinder = π(12)^2(12) = 5,408π cm^3 The volume of the hemisphere bowl is given by the formula V_hemisphere = (2/3)πr^3, where r is the radius of the hemisphere. Again looking at the diagram, we can see that the radius of the hemisphere is also 12 cm (since it is the same as the radius of the base of the cylinder). Therefore, the volume of the hemisphere bowl is: V_hemisphere = (2/3)π(12)^3 = 9,216π/3 = 3,072π cm^3 To find the volume of the solid, we add the volume of the cylinder and the volume of the hemisphere bowl together: V_solid = V_cylinder + V_hemisphere = 5,408π + 3,072π = 8,480π cm^3 Therefore, the volume of the solid is 8,480π cm^3. : 198πcm^3 is not correct.
Ajụjụ 47 Ripọtì
Solve for x in the equation x3 - 5x2 - x + 5 = 0
Akọwa Nkọwa
To solve this equation, we need to use algebraic methods to isolate the variable x on one side of the equation. First, we can combine like terms on the left-hand side of the equation: x^3 - 5x^2 - x + 5 = 0 Next, we can try to factor the equation, looking for factors that will make the equation equal to zero. By trial and error, we can see that (x-1) is a factor of the equation: (x-1)(x^2 - 4x - 5) = 0 Now we can use the zero product property, which tells us that if the product of two factors is zero, then at least one of the factors must be zero. So we can set each factor equal to zero and solve for x: x-1 = 0 or x^2 - 4x - 5 = 0 Solving for x-1, we get x = 1. Solving for x in the quadratic equation, we can use the quadratic formula: x = (-b ± sqrt(b^2 - 4ac)) / 2a In this case, a = 1, b = -4, and c = -5. Plugging these values into the quadratic formula, we get: x = (-(-4) ± sqrt((-4)^2 - 4(1)(-5))) / 2(1) Simplifying, we get: x = (4 ± sqrt(36)) / 2 x = (4 ± 6) / 2 x = 5 or x = -1 Therefore, the solutions for x are 1, 5, and -1.
Ajụjụ 48 Ripọtì
In the diagram above are two concentric circles of radii r and R respectively with center O. If r = 2/3R, express the area of the shaded portion in terms of π and R
Akọwa Nkọwa
The area of the shaded region is equal to the difference between the areas of the two circles. The area of a circle is given by the formula A = πr^2, where r is the radius of the circle. Let's find the radius of the smaller circle: r = 2/3R Now we can express the area of the shaded region in terms of π and R: Area of shaded region = Area of larger circle - Area of smaller circle A = πR^2 - π(2/3R)^2 A = πR^2 - 4/9πR^2 A = (9/9πR^2 - 4/9πR^2) A = 5/9πR^2 Therefore, the answer is 5/9πR^2.
Ajụjụ 49 Ripọtì
The sum of the interior angles of a polygon is 20 right angles. How many sides does the polygon have?
Akọwa Nkọwa
The sum of the interior angles of a polygon is given by the formula (n-2) x 180 degrees, where n is the number of sides of the polygon. In this problem, the sum of the interior angles is given as 20 right angles. One right angle is equal to 90 degrees, so 20 right angles would be 20 x 90 = 1800 degrees. Therefore, we can set up the equation (n-2) x 180 = 1800 and solve for n: (n-2) x 180 = 1800 n-2 = 10 n = 12 So the polygon has 12 sides. To summarize, the sum of the interior angles of a polygon can be calculated using (n-2) x 180 degrees, where n is the number of sides. Using this formula, we can solve for n when the sum of the interior angles is given. In this case, the polygon has 12 sides.
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