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Question 2 Report
If y = (2x + 1)3 find dy/dx
Answer Details
y = (2x + 1)3
dy/dx = 3(2x + 1)3-1 x 2
= 3(2x + 1)2 x 2
= 6(2x + 1)2
Question 3 Report
If x * y = x + y2, find then value of (2*3)*5
Answer Details
x * y = x + y2
2 * 3 = 2 + 32
= 2 + 9
= 11
(2 * 3) * 5 = 11 + 52
= 11 + 25
= 36
Question 4 Report
A man bought a second-hand photocopying machine for N34,000. He serviced is at a cost od N2,000 and then sold it i profit of 15%. What was the selling price?
Answer Details
To find the selling price of the photocopying machine, we need to first calculate the profit the man made. He bought the machine for N34,000 and spent an additional N2,000 on servicing it, which brings his total cost to N36,000. He then sold the machine for a profit of 15%, which means he earned 15% of his cost as profit. To calculate the profit, we can use the formula: Profit = (Profit Percentage/100) * Cost Price Substituting the given values, we get: Profit = (15/100) * 36,000 Profit = 5,400 Therefore, the man made a profit of N5,400. To find the selling price, we can add the profit to the cost price: Selling Price = Cost Price + Profit Selling Price = 36,000 + 5,400 Selling Price = 41,400 Therefore, the selling price of the photocopying machine was N41,400. Therefore, the correct option is: N41,400
Question 5 Report
Marks | 1 | 2 | 3 | 4 | 5 |
Frequency | 2 | 2 | 8 | 4 | 4 |
The table above show the marks obtained in a given test.
Find the mean mark
Answer Details
To find the mean mark, we need to calculate the sum of all the marks obtained and divide it by the total number of students. The sum of all the marks obtained can be found by multiplying each mark by its corresponding frequency and adding up the results. So, sum of all marks = (1 x 2) + (2 x 2) + (3 x 8) + (4 x 4) + (5 x 4) = 2 + 4 + 24 + 16 + 20 = 66 The total number of students can be found by adding up all the frequencies. So, total number of students = 2 + 2 + 8 + 4 + 4 = 20 Therefore, the mean mark = (sum of all marks) / (total number of students) = 66 / 20 = 3.3 Hence, the answer is 3.3.
Question 6 Report
Marks12345Frequency22844
The table above shows the marks obtained in a given test. Find the mean mark.
Answer Details
To find the mean, we need to calculate the sum of all the marks obtained, and then divide by the total number of students who took the test. The sum of all marks obtained is: 1 x 2 + 2 x 2 + 3 x 8 + 4 x 4 + 5 x 4 = 2 + 4 + 24 + 16 + 20 = 66 The total number of students who took the test is: 2 + 2 + 8 + 4 + 4 = 20 Therefore, the mean mark is: 66 / 20 = 3.3 So the answer is option (C) 3.3
Question 7 Report
Solve for x and y if x - y = 2 and x2 - y2 = 8
Answer Details
x - y = 2 ...........(1)
x2 - y2 = 8 ........... (2)
x - 2 = y ............ (3)
Put y = x -2 in (2)
x2 - (x - 2)2 = 8
x2 - (x2 - 4x + 4) = 8
x2 - x2 + 4x - 4 = 8
4x = 8 + 4 = 12
x = 124
= 3
from (3), y = 3 - 2 = 1
therefore, x = 3, y = 1
Question 8 Report
Determine the value of x for which (x2 - 1) > 0
Answer Details
We want to solve the inequality (x² - 1) > 0 for x. To do this, we can factor the left-hand side of the inequality: (x² - 1) = (x - 1)(x + 1) Now we have the inequality: (x - 1)(x + 1) > 0 The product of two factors is positive if and only if both factors are positive or both factors are negative. So we can break the inequality into two cases: Case 1: (x - 1) > 0 and (x + 1) > 0 This simplifies to x > 1, which means x is greater than 1. Case 2: (x - 1) < 0 and (x + 1) < 0 This simplifies to x < -1, which means x is less than -1. Therefore, the solution to the inequality (x² - 1) > 0 is: x < -1 or x > 1 So the answer is: x < -1 or x > 1.
Question 9 Report
In the diagram, the tangent MN makes an angle of 55o with the chord PS. IF O is the centre of the circle, find < RPS
Answer Details
Join SR
< PRS = 90?
(Angle in a semicircle)
< PRS = 55?
(Angle between a chord and a tangent = Angle in the alternate segment)
< PSR + < PRS + < RSP = 180?
90v + 55?
+ < RSP = 180?
< RSP = 180?
- 145?
= 35?
Question 10 Report
Evaluate ∣∣ ∣∣205463891∣∣ ∣∣
Answer Details
∣∣ ∣∣205463891∣∣ ∣∣
= 2(6 - 27) - 0(4 - 24) + 5(36 - 48)
= 2(-21) - 0 + 5(-12)
= -42 + 5(-12)
= -42 - 60
= -102
Question 11 Report
From the cyclic quadrilateral TUVW above, find the value of x
Answer Details
TUVW is a cyclic quad
3χ + 20 + 88 = 180 (opp ∠ s of a cyclic quad are supplementary)
3χ + 108 = 180
3χ = 180 - 108
3χ = 72
χ = 72/3χ = 24∘
Question 12 Report
If two smaller sides of a right angled triangle are 4cm and 5cm, find its area
Answer Details
To find the area of a right angled triangle, we can use the formula: Area = (base x height) / 2 In a right angled triangle, the two smaller sides that form the right angle are the base and height. Therefore, we can substitute 4 cm for the base and 5 cm for the height in the formula: Area = (4 cm x 5 cm) / 2 = 10 cm^2 Therefore, the area of the right angled triangle is 10 cm^2. The answer is (A) 10 cm^2.
Question 13 Report
actorize completely x3+3x2−10x2x2−8
Answer Details
x3+3x2−10x2x2−8
= x(x2+3x−10)2(x2−4)
= x(x2+5x−2x−10)2(x+2)(x−2)
= x(x−2)(x+5)2(x+2)(x−2)
= x(x+5)2(x+2)
Question 14 Report
If Cos θ = 1213 . Find θ + cos2θ
Answer Details
Cos θ
= 1213
x2 + 122 = 132
x2 = 169- 144 = 25
x = 25
= 5
Hence, tanθ
= 512
and cosθ
= 1213
If cos2θ
= 1 + 1tan2θ
= 1 + 1(5)212
= 1 + 125144
= 1 + 14425
= 25+14425
= 16925
Question 15 Report
A cylindrical pipe 50cm long with radius 7m has one end open. What is the total surface area of the pipe?
Answer Details
To calculate the total surface area of the cylindrical pipe, we need to add the surface area of the curved part and the surface area of the two circular ends. The surface area of the curved part can be calculated by multiplying the circumference of the circle (2πr) by the length of the pipe (50cm), which gives us: 2πr x h = 2π x 7m x 50cm = 7π m^2 The surface area of one circular end can be calculated by multiplying the area of the circle (πr^2) by 1, since one end of the pipe is open and has no surface area. Thus, the total surface area of both circular ends is: 2πr^2 = 2π x 7m^2 = 14π m^2 Finally, we add the surface area of the curved part and the surface area of the two circular ends to get the total surface area of the pipe: 7π m^2 + 14π m^2 = 21π m^2 Therefore, the total surface area of the pipe is 21π square meters. The closest option to this answer is 749π, but it is not the correct answer.
Question 16 Report
An arc subtends an angle of 50∘ at the center of circle of radius 6cm. Calculate the area of the sector formed
Answer Details
Area of a sector = | θ | x πr2 |
360 |
Question 17 Report
The interior angles of a quadrilateral are (x + 15)o, (2x - 45)o and (x + 10)o. Find the value of the least interior angle.
Answer Details
(x + 15)o + (2x - 45)o + (x + 10)o = (2n - 4)90o
when n = 4
x + 15o + 2x - 45o + x - 30o + x + 10o = (2 x 4 - 4) 90o
5x - 50o = (8 - 4)90o
5x - 50o = 4 x 90o = 360o
5x = 360o + 50o
5x = 410o
x = 410o5
= 82o
Hence, the value of the least interior angle is (x - 30o)
= (82 - 30)o
= 52o
Question 18 Report
W is directly proportional to U. If W = 5 when U = 3, find U when W = 27
Answer Details
W α
U
W = ku
u = wk
; 27
x 35
= 635
Question 19 Report
If [x327] = 15, find the value of x.
Answer Details
[x327]
= 15
x x 7 - 2 x 3 = 15
7x - 6 = 15
7x = 15+6
7x = 21
x = 21/7x = 3
Question 20 Report
If y = (2x + 1)3, find dydx
Answer Details
If y = (2x + 1)3, then
Let u = 2x + 1 so that, y = u3
dydu
= 3u2 and dydx
= 2
Hence by the chain rule,
dydx
= dydu
x dudx
= 3u2 x 2
= 6u2
= 6(2x + 1)2
Question 21 Report
Evaluate ∫31(X2−1)dx
Answer Details
Question 22 Report
Marks2345678No. of students3152423
From the table above, if the pass mark is 5, how many students failed the test?
Answer Details
To determine how many students failed the test, we need to add up the frequencies of the students who obtained marks less than 5, since the pass mark is 5. Looking at the table, the marks less than 5 are 2, 3, and 4. Adding up the corresponding frequencies, we get: 3 + 1 + 5 = 9 Therefore, 9 students failed the test. The answer is (C) 9.
Question 23 Report
Marks | 2 | 3 | 4 | 5 | 6 | 7 | 8 |
No. of students | 3 | 1 | 5 | 2 | 4 | 2 | 3 |
From the table above, if the pass mark is 5, how many students failed the test?
Answer Details
Marks | Number of students |
2 | 3 |
3 | 1 |
4 | 5 |
Total | 9 |
Question 24 Report
If y varies directly as the square root of x and y = 3 when x = 16. Calculate y when x = 64
Answer Details
If y varies directly as the square root of x, this means that the ratio between y and the square root of x remains constant. We can write this as: y/sqrt(x) = k where k is the constant of proportionality. We are given that y = 3 when x = 16. Using this information and the equation above, we can solve for k: 3/sqrt(16) = k 3/4 = k Now that we know k, we can use the same equation to find y when x = 64: y/sqrt(64) = 3/4 y/8 = 3/4 y = (3/4) * 8 y = 6 Therefore, when x = 64, y = 6. The answer is option B.
Question 25 Report
For what range of values of x is 12 x + 14 > 13 x + 12 ?
Answer Details
12
x + 14
> 13
x + 12
Multiply through by through by the LCM of 2, 3 and 4
12 x 12
x + 12 x 14
> 12 x 13
x + 12 x 12
6x + 3 > 4x + 6
6x - 4x > 6 - 3
2x > 3
2x2
> 32
x > 32
Question 26 Report
Marks | 1 | 2 | 3 | 4 | 5 |
Frequency | 2 | 2 | 8 | 4 | 4 |
The table above show the marks obtained in a given test.
How many student too the test
Answer Details
To find the total number of students who took the test, we need to add up the frequency of all the marks. 2 + 2 + 8 + 4 + 4 = 20 Therefore, 20 students took the test. The answer is (C) 20.
Question 27 Report
At what value of x does the function y= -3 – 2x +x2 attain a minimum value?
Answer Details
To find the minimum value of the function y = -3 - 2x + x^2, we need to determine the value of x that corresponds to the vertex of the parabolic graph. The vertex of a parabolic graph with equation y = ax^2 + bx + c is located at x = -b/2a. In this case, a = 1, b = -2, and c = -3. Therefore, x = -(-2)/(2*1) = 1. So the answer is (E) 1, and that's the value of x at which the function y attains its minimum value.
Question 28 Report
Find the sum to infinity of the following series. 0.5 + 0.05 + 0.005 + 0.0005 + .....
Answer Details
Using S∞
= a1−r
r = 0.050.5
= 110
S∞
= 0.5110
= 0.5(910)
= 0.5×109
= 59
Question 29 Report
If y = x sin x, Find d2yd2x
Answer Details
To find the second derivative of the given function, we need to differentiate it twice with respect to x. First, we differentiate y with respect to x using the product rule: y = x sin x y' = x cos x + sin x Then, we differentiate y' with respect to x using the product rule again: y' = x cos x + sin x y'' = cos x - x sin x + cos x Simplifying the expression: y'' = 2cos x - x sin x Therefore, the second derivative of y = x sin x is y'' = 2cos x - x sin x.
Question 30 Report
If P = (2?311) what is P−1
Answer Details
P = (2?311)
|P| = 2 - 1 x -3 = 5
P-1 = 15
(13?12)
= (1535?1525)