WAEC SSCE - Further Mathematics - 2022
Tambaya 1 Rahoto
If f(x-1) = x3
+ 3x2
+ 4x - 5, find f(2)
Bayanin Amsa
x - 1 = 2
x = 3
f(2) = (3)3
+ 3(3)2
+ 4(3) - 5
f(2) = 27 + 27 + 12 - 5
= 61
Tambaya 2 Rahoto
The table shows the distribution of the distance (in km) covered by 40 hunters while hunting.
| Distance(km) | 3 | 4 | 5 | 6 | 7 | 8 |
| Frequency | 5 | 4 | x | 9 | 2x | 1 |
If a hunter is selected at random, find the probability that the hunter covered at least 6km.
Bayanin Amsa
5+4+x+9+2x+1 = 40
19+3x = 40
3x = 21
x = 7
| Distance(km) | 3 | 4 | 5 | 6 | 7 | 8 |
| Frequency | 5 | 4 | 7 | 9 | 14 | 1 |
The probability that the hunter covered at least 6km, means the hunter covered either 6km or 7km, or 8km.
24 hunters covered at least 6km
| 2440 | = | 35 |
Tambaya 3 Rahoto
The probability that a student will graduate from college is 0.4. If 3 students are selected from the college, what is the probability that at least one student will graduate?
Bayanin Amsa
To find the probability that at least one student will graduate, we can calculate the probability that no student will graduate and then subtract that from 1. The probability that a student will not graduate from college is 1 - 0.4 = 0.6. So, the probability that none of the three selected students will graduate is: 0.6 * 0.6 * 0.6 = 0.216 Therefore, the probability that at least one student will graduate is: 1 - 0.216 = 0.784 So, the answer is 0.78. In summary, the probability that at least one student will graduate from college given that 3 students are selected is 0.784 or 0.78, which is obtained by subtracting the probability that none of the three students will graduate from 1.
Tambaya 4 Rahoto
If Un = kn2 + pn, U1 = -1, U5 = 15, find the values of k and p.
Bayanin Amsa
Un = kn2 + pn,
U1 = -1,
U5 = 15,
when n = 1
U1 = k(1)2
+ p(1) = -1
k + p = -1 --------eqn1
when n = 5
U5
= k(5)2
+ p(5) = 15
25k + 5p = 15 --------eqn2
multiply eqn1 by 5 and eqn2 by 1
5k + 5p = -5 -------eqn3
25k + 5p = 15 -------eqn4
eqn4 - eqn3
20k = 20
k = 1
sub for k in eqn1
1 + p = -1
p = -1 -1 = -2
Tambaya 5 Rahoto
A binary operation ∆ is defined on the set of real numbers R, by x∆y = x+y−xy4−−−−−−−−−√
, where x, yER. Find the value of 4∆3
Bayanin Amsa
x∆y = x+y−xy4−−−−−−−−−√
4∆3 = 4+3−4∗34−−−−−−−−−√
= 4+3−3−−−−−−−−√
= 4–√
= 2.
Tambaya 6 Rahoto
A particle is acted upon by forces F = (10N, 060º), P = (15N, 120º) and Q = (12N, 200º). Express the force that will keep the particle in equilibrium in the form xi + yj, where x and y are scalars.
Bayanin Amsa
Converting the forces to their rectangular forms
F = (10N, 060º)
Fx = 10cos60 = 5i
fy = 10sin60 = 8.66j
F = 5i + 8.66j
P = (15N, 120º)
Px = 15cos120 = -7.5i
py = 15sin120 = 12.99j
P = -7.5i + 12.99j
Q = (12N, 200º)
Qx = 12cos200 = -11.28i
Qy = 12sin200 = -4.1j
Q = -11.28i -4.1j
The resultant force = F + P + Q
R = 5i + 8.66j + (-7.5i + 12.99j) + (-11.28i -4.1j)
R = -13.78i + 17.55j
Tambaya 7 Rahoto
The gradient of a function at any point (x,y) 2x - 6. If the function passes through (1,2), find the function.
Bayanin Amsa
dy/dx = 2x - 6
y = ∫ 2x - 6
y = 2x
2−6+c
y = x2
- 6x + c
passes through (1,2)
2 = 12
- 6(1) + c
2 = 1 - 6 + c
c = 7
y = x2
- 6x + c
y = x2 - 6x + 7
Tambaya 8 Rahoto
Find the coefficient of x2 in the binomial expansion of (x+2x2)5
Bayanin Amsa
(x+2x2)5
n = 5, r = 4, p = x and q = 2x2
5C4 x4 (2x2 )1 = 5C4 2x4x2
5C4 2x2 = 5![5−4]!4! * 2x2
5∗4!4!∗2x2 = 5 * 2x2 = 10x2
The coefficient is 10.
Tambaya 9 Rahoto
A body of mass 18kg moving with velocity 4ms-1 collides with another body of mass 6kg moving in the opposite direction with velocity 10ms-1. If they stick together after the collision, find their common velocity.
Bayanin Amsa
m1u1 + m2u2 = (m1 + m2)v
m1 = 18kg, m2 = 6kg, u1 = 4ms-1, u2 = -10m/s
18(4) + 6(-10) = (18+6)v
72 - 60 = 24v
12 = 24v
v = 12
m/s
Tambaya 10 Rahoto
In how many ways can six persons be paired?
Bayanin Amsa
6C2=6![6−2]![2!]
6∗5∗4!4!∗2!
= 6∗52
= 15
Tambaya 11 Rahoto
Simplify 9∗3n+1−3n+23n+1−3n
Bayanin Amsa
9∗3n+1−3n+23n+1−3n
= 3n∗3n∗31∗−32∗323n∗31−3n
= 3n(32∗31)3n(31−1)
= 27−93−1
= 182
= 9
Tambaya 12 Rahoto
The first, second and third terms of an exponential sequence (G.P) are (x - 4), (x + 2), and (3x + 1) respectively. Find the values of x.
Bayanin Amsa
U1 = x - 4
U2 = x + 2
U3 = 3x + 1
u2u1=u3u2
x+2x−4=3x+1x+2
(x+2)(x+2) = (x-4)(3x+1)
x2
+ 4x + 4 = 3x2 - 11x - 4
collecting like terms
2x2
- 15x - 8 =0
2x2
+ x - 16x - 8 = 0
x(2x + 1) - 8(2x + 1) = 0
(x-8)(2x+1) = 0
x = (−12,8 )
Tambaya 13 Rahoto
Find the coefficient of x3 y2 in the binomial expansion of (x-2y)5
Bayanin Amsa
x3 y2 in (x-2y)5
n = 5, r = 3, p = x, q = -2y
5C3 * x3 2
5C3 = 5![5−3]!3!
5∗4∗3!2!3! → 5∗42
5C3 = 10
: 5C3 * x3 -2y2 = 10 * x3 4y2
40x3
y2
the coefficient is 40
Tambaya 14 Rahoto
The table shows the distribution of the distance (in km) covered by 40 hunters while hunting.
What is the mode of the distribution?
| Distance(km) | 3 | 4 | 5 | 6 | 7 | 8 |
| Frequency | 5 | 4 | x | 9 | 2x | 1 |
Bayanin Amsa
To find the mode of the distribution, we need to identify the value with the highest frequency. In the table, the frequency is given for each distance (in km) covered by the hunters. Let's first find the frequency of distance 5 km covered by the hunters. From the table, we know that the sum of all frequencies should be equal to 40, the total number of hunters. Sum of frequencies = 5 + 4 + x + 9 + 2x + 1 = 40 Simplifying the equation, we get: 3x + 19 = 40 3x = 21 x = 7 Now we know that the frequency of distance 5 km is x = 7. The frequencies for the other distances are: - Frequency of distance 3 km = 5 - Frequency of distance 4 km = 4 - Frequency of distance 6 km = 9 - Frequency of distance 7 km = 2x = 14 - Frequency of distance 8 km = 1 The highest frequency is 14, which corresponds to the distance of 7 km. Therefore, the mode of the distribution is 7 km. Hence, the answer is option (C) 7.
Tambaya 16 Rahoto
A linear transformation T is defined by T: (x,y) → (3x - y, x + 4y). Find the image of (2, -1) under T.
Bayanin Amsa
The linear transformation T is defined as T: (x, y) → (3x - y, x + 4y). To find the image of (2, -1) under T, we need to apply T to (2, -1) and see what we get. So, T(2, -1) = (3(2) - (-1), 2 + 4(-1)) = (7, -2) Therefore, the image of (2, -1) under T is (7, -2). Answer is correct.
Tambaya 17 Rahoto
Given that 8x+mx2−3x−4≡5x+1+3x−4
Bayanin Amsa
8x+mx2−3x−4≡5x+1+3x−4
8x+mx2−3x−4 ≡ 5(x−1)+3(x+4)x2−3x−4
multiplying both sides by x2-3x-4
8x+m ≡ 5(x-4)+3(x+1)
8x + m ≡ 5x - 20 + 3x + 3
8x - 5x - 3x + m = -20 + 3
m = -17
Tambaya 18 Rahoto
If g(x) = √(1-x2 ), find the domain of g(x)
Bayanin Amsa
1 - x2
≥ 0
-x2
≥ -1
x2
≤ 1
√x2
≤ 1
|x| ≤ 1
-1 ≤ x ≤ 1
Tambaya 19 Rahoto
Evaluate 4p2+4C2−4p3
Bayanin Amsa
4p2+4C2−4p3
npr=n![n−r]!andnCr=n![n−r]!r!
= 4![4−2]!+4![4−2]!2!−4![4−3]!=4!2!+4!2!2!−4!1!
= 4∗3∗2!2!+4∗3∗2!2!2!−4∗3∗2∗11!
12 + 6 - 24 = -6
Tambaya 21 Rahoto
Solve: 32x−2−28(3x−2)+3=0
Bayanin Amsa
32x−2−28(3x−2)+3=0
32x32−28.3x32+3=0
32x9−28.3x9+3=0
let p = 3x
p29−28p9+3=0
multiply through by 9
p2
- 28p + 27 = 0
p2
- p - 27p + 27 = 0
p (p - 1) - 27(p - 1) = 0
(p-1)(p-27) = 0
p = 1 or 27
when p = 1
p = 3x
3x
= 1
3x
= 30
x = 0
when p = 27
3x = 27
3x = 33
x = 3
Tambaya 24 Rahoto
A particle of mass 3kg moving along a straight line under the action of a F N, covers a line distance, d, at time, t, such that d = t2 + 3t. Find the magnitude of F at time t.
Bayanin Amsa
F = m * a
d = t2 + 3t.
a = d2ddt2
d[d]dt = 2t + 3
d2ddt2 = 2m/s2
a = 2m/s2
F = m * a
F = 3 × 2 = 6N
Tambaya 26 Rahoto
Express 4π2 radians in degrees.
Tambaya 27 Rahoto
The mean heights of three groups of students consisting of 20, 16 and 14 students each are 1.67m, 1.50m and 1.40m respectively. Find the mean height of all the students.
Bayanin Amsa
To find the mean height of all the students, you need to add up the heights of all the students and divide by the total number of students. First, you need to find the total number of students, which is the sum of 20, 16 and 14 students, which is 50 students in total. Then, you need to find the total height of all the students by multiplying the number of students in each group by their respective mean height and adding them all up. Total height of all students = (20 x 1.67) + (16 x 1.50) + (14 x 1.40) = 33.4 + 24 + 19.6 = 77 meters Finally, to find the mean height of all the students, you need to divide the total height of all the students by the total number of students: Mean height of all students = Total height of all students / Total number of students = 77 / 50 = 1.54 meters Therefore, the answer is 1.54m.
Tambaya 28 Rahoto
The length of the line joining points (x,4) and (-x,3) is 7 units. Find the value of x.
Bayanin Amsa
The problem provides us with two points, (x, 4) and (-x, 3), and tells us that the distance between them is 7 units. We need to find the value of x that satisfies this condition. To find the distance between two points, we can use the distance formula: d = sqrt((x2 - x1)^2 + (y2 - y1)^2) where (x1, y1) and (x2, y2) are the coordinates of the two points, and d is the distance between them. Using this formula, we can calculate the distance between the given points: d = sqrt((-x - x)^2 + (3 - 4)^2) d = sqrt(4x^2 + 1) We are given that this distance is equal to 7 units: sqrt(4x^2 + 1) = 7 To solve for x, we need to isolate it on one side of the equation. To do this, we will square both sides of the equation: 4x^2 + 1 = 49 Now we can solve for x: 4x^2 = 48 x^2 = 12 x = sqrt(12) = 2sqrt(3) Therefore, the value of x that satisfies the given conditions is 2sqrt(3), which is option D.
Tambaya 30 Rahoto
Bayanin Amsa
∣∣∣21−34∣∣∣∣∣∣−6k∣∣∣∣∣∣3−26∣∣∣=15
∣∣∣2[−6]1[−6]−3k+4k∣∣∣=∣∣∣3−26∣∣∣
∣∣∣−12−6−3k+4k∣∣∣=∣∣∣3−26∣∣∣
-12 - 3k = 3
-3k = 3 + 12
k = 15−3
k = -5
Tambaya 31 Rahoto
Bayanin Amsa
This is a question about vector addition. To find →PR, we need to add →PQ and →RQ, but in reverse order (→PR = →PQ + →RQ). Given →PQ = -2i + 5j and →RQ = -i - 7j, we can add them as follows: →PR = →PQ + →RQ →PR = (-2i + 5j) + (-i - 7j) →PR = -2i - i + 5j - 7j →PR = -3i - 2j Therefore, the answer is "-3i - 2j". Option (B) "-3i + 12j", (C) "-i + 12j", and (D) "i - 12j" are incorrect.
Tambaya 32 Rahoto
If α and β are roots of x2 + mx - n = 0, where m and n are constants, form the
| equation | whose | roots | are | 1α | and | 1β | . |
Bayanin Amsa
x2 + mx - n = 0
a = 1, b = m, c = -n
α + β = −ba = −m1 = -m
αβ = ca = −n1 = -n
the roots are = 1α and 1β
sum of the roots = 1α + 1β
1α + 1β α+βαβ
α + β = -m
αβ = -n
α+βαβ
product of the roots = 1α * 1β
1α
+ 1β
= 1αβ
→
x2
- (sum of roots)x + (product of roots)
x2
- ( m/n )x + ( 1/-n ) = 0
multiply through by n
nx2
- mx - 1 = 0
Tambaya 33 Rahoto
(3√6√5+√543√5 )−1
Bayanin Amsa
(3√6√5+√543√5 )−1
= √5(3√5)3√6+3√6
= 3∗56√6=52√6
= 5∗2√62√6+2√6=10√64∗6
= 5√612
Tambaya 34 Rahoto
Which of the following is the semi-interquartile range of a distribution?
Bayanin Amsa
The semi-interquartile range is a measure of variability in a dataset. To calculate it, we first need to find the median of the dataset, which is the middle value when the data is arranged in order from lowest to highest. Then, we need to find the quartiles, which are the values that divide the dataset into four equal parts. The semi-interquartile range is half of the difference between the upper quartile and the lower quartile. The upper quartile is the value that separates the highest 25% of the data from the lowest 75%, while the lower quartile is the value that separates the lowest 25% of the data from the highest 75%. Looking at the given options, the formula that corresponds to the semi-interquartile range is: 1/2 (Upper Quartile - Lower Quartile) Therefore, the correct answer is option d) "1/2 (Upper Quartile - Lower Quartile)".
Tambaya 35 Rahoto
If 36, p,94 and q are consecutive terms of an exponential sequence (G.P), find the sum of p and q.
Bayanin Amsa
GP : 36, P, q4 , q, ... p + q = ?
| Recall, | common | ratio, | r | = | TnTn-1 | = | T2T1 | = | T3T2 | = | T4T3 |
| ∴ | P36 | = | 94 | ÷ | p | ; | p2 | = | 94 | x | 36 | ; | p2 | = | 81 |
| p | = | 9 | ∴ | r | = | T2T1 | = | 936 | = | 14 |
| Also | r | = | T4T3 | = | q | ÷ | 94 |
∴ 14
= q ÷ 94
;
94 = 4q
| 16q | = | 9 | , | q | = | 916 | ∴ | p | + | q | = | 9 | + | 916 | = | 9 | 916 |
Tambaya 36 Rahoto
Given that P = (-4, -5) and Q = (2,3), express →PQ in the form (k,θ). where k is the magnitude and θ the bearing.
Bayanin Amsa
To find →PQ, we need to subtract the coordinates of point P from those of point Q, giving us: →PQ = Q - P = (2-(-4), 3-(-5)) = (6, 8) The magnitude, or length, of →PQ is found using the distance formula: k = √(6² + 8²) = √100 = 10 To find the bearing θ, we use trigonometry. The tangent of θ is the ratio of the opposite side (the change in y-coordinates) to the adjacent side (the change in x-coordinates): tan θ = 8/6 = 4/3 Using a calculator, we can find that θ is approximately 53.13º. However, we need to adjust this value depending on which quadrant →PQ lies in. Since both x and y are positive, →PQ lies in the first quadrant, so we don't need to make any adjustments. Therefore: θ = 53.13º So the vector →PQ can be expressed in the form (k,θ) as: (10 units, 53.13º) Therefore, the correct answer is (a) (10 units, 053º).
Tambaya 37 Rahoto
Consider the following statement:
x: All wrestlers are strong
y: Some wresters are not weightlifters.
Which of the following is a valid conclusion?
Bayanin Amsa
Tambaya 38 Rahoto
Given that P = {x: x is a multiple of 5}, Q = {x: x is a multiple of 3} and R = {x: x is an odd number} are subsets of μ = {x: 20 ≤ x ≤ 35}, (P⋃Q)∩R.
Bayanin Amsa
P = { 20, 25, 30, 35}, Q = {21, 24, 27, 30, 33}, R = {21, 23, 25, 27, 29, 31, 33, 35}
(P⋃Q)∩R = {20, 21, 24, 25, 27, 30, 33, 35} ∩ {21, 23, 25, 27, 29, 31, 33, 35}
= {21, 25, 27, 33, 35}
Tambaya 39 Rahoto
The equation of a circle is given as 2x2 + 2y2 - x - 3y - 41 = 0. Find the coordinates of its centre.
Bayanin Amsa
2x2 + 2y2 - x - 3y - 41
standard equation of circle
(x-a)2
+ (x-b)2
= r2
General form of equation of a circle.
x2
+ y2
+ 2gx + 2fy + c = 0
a = -g, b = -f., r2 = g2 + f2 - c
the centre of the circle is (a,b)
comparing the equation with the general form of equation of circle.
2x2
+ 2y2
- x - 3y - 41
= x2
+ y2
+ 2gx + 2fy + c
2x2
+ 2y2
- x - 3y - 41 = 0
divide through by 2
g = −14 ; 2g = −12
f = −34 ; 2f = −32
a = -g → - −14 ; = 14
b = -f → - (\frac{-3}{4}\) = (\frac{3}{4}\)
therefore the centre is (14 , 34 )
Tambaya 40 Rahoto
Evaluate1∫0x2(x3+2)3
Bayanin Amsa
1∫0x2(x3+2)3 dx
let u=x3+2,du=3x2dx
when x = 1, u = 3
when x = 0, u = 2
dx = du3x2
3∫2 x2[u]33x2
3∫2 u33 du
= u43∗4 2 3
112[u4] 3
112[34−24]
112[81−16]
6512
Tambaya 41 Rahoto
The functions f:x → 2x2 + 3x -7 and g:x →5x2 + 7x - 6 are defined on the set of real numbers, R. Find the values of x for which 3f(x) = g(x).
Bayanin Amsa
To find the values of x for which 3f(x) = g(x), we need to substitute the given functions f(x) and g(x) and then solve for x. Substituting the function f(x) = 2x^2 + 3x - 7, we get: 3f(x) = 3(2x^2 + 3x - 7) = 6x^2 + 9x - 21 Substituting the function g(x) = 5x^2 + 7x - 6, we get: g(x) = 5x^2 + 7x - 6 Now we can set these two equations equal to each other and solve for x: 6x^2 + 9x - 21 = 5x^2 + 7x - 6 Subtracting 5x^2 and 7x from both sides, we get: x^2 + 2x - 15 = 0 Factoring this quadratic equation, we get: (x + 5)(x - 3) = 0 Therefore, the values of x that satisfy the equation 3f(x) = g(x) are x = -5 and x = 3. Hence, the answer is: x = -5 or 3.
Tambaya 42 Rahoto
Solve: 4sin2 θ + 1 = 2, where 0º < θ < 180º
Bayanin Amsa
4sin2
2
θ + 1 = 2
4sin2 θ = 2 - 1
4sin2 θ = 1
s√in2θ
= 14−−√
sinθ = 12
θ = sin−112
θ = 30º 0r 150º
Tambaya 43 Rahoto
Find correct to the nearest degree, the acute angle formed by the lines y = 2x + 5 and 2y = x - 6
Bayanin Amsa
| tanθ | = | m1 - m21 + m1m2 |
y = 2x + 5
m1 = 2
2y = x - 6
| y | = | 12 | x | - | 3 |
| m2 | = | 12 |
| tanθ | = | 2 - 12 1+2(12 ) |
tanθ = 32 ÷ (1+1)
tanθ = 32 ÷ 2
| tanθ | = | 34 |
θ = tan−1(34)
θ = 36.87º
θ = 37º
Tambaya 44 Rahoto
A straight line makes intercepts of -3 and 2 on the x and y axes respectively. Find the equation of the line.
Bayanin Amsa
Recall:
| xa | + | yb | =1 |
Where 'a' and 'b' are the x and y intercept respectively.
| x-3 | + | y2 | =1 |
2x-3y = -6
2x - 3y + 6 = 0 -------(1)
multiply through by -1
-2x + 3y - 6 = 0
Tambaya 45 Rahoto
Given that p = (8N,030º) and q = (9N, 150º), find, in component from, the unit vector along(p - q).
p = 8cos30i + 8sin30j
q = 9cos150i + 9sin150j
p = 6.9282i + 4j
q = -7.7942i + 4.5j
pq = (-7.7942 - 6.9282)i + (4.5 - 4)j
pq = -14.7224i + 0.5j
|pq| = √((-14.7224)2 + (0.5)2)
|pq| = √(216.7491 + 0.25)
|pq| = √216.9991
|pq| = 14.7309
the unit vector in the direction of p - q = −14.7224i+0.5j14.7309
-0.9994i + 0.0339j ≌ -i + 0.0339j
Bayanin Amsa
p = 8cos30i + 8sin30j
q = 9cos150i + 9sin150j
p = 6.9282i + 4j
q = -7.7942i + 4.5j
pq = (-7.7942 - 6.9282)i + (4.5 - 4)j
pq = -14.7224i + 0.5j
|pq| = √((-14.7224)2 + (0.5)2)
|pq| = √(216.7491 + 0.25)
|pq| = √216.9991
|pq| = 14.7309
the unit vector in the direction of p - q = −14.7224i+0.5j14.7309
-0.9994i + 0.0339j ≌ -i + 0.0339j
Tambaya 46 Rahoto
Evaluate: limx→−2
x3+8x+2.
x3+8x+2
x3
+ 8 = x3
+ 23
recall that
x3
+ y3
= (x+y)(x2
- xy + y2
)
x = x, y = 2
x3
+ 8 = (x+2)(x3
- 2(x) + 22)
x3+8x+2=(x+2)(x3−2(x)+22)x+2 - 2x + 4
x = -2
(-2)2
- 2(2) + 4 = 4 - 4 + 4
= 4
Bayanin Amsa
x3+8x+2
x3
+ 8 = x3
+ 23
recall that
x3
+ y3
= (x+y)(x2
- xy + y2
)
x = x, y = 2
x3
+ 8 = (x+2)(x3
- 2(x) + 22)
x3+8x+2=(x+2)(x3−2(x)+22)x+2 - 2x + 4
x = -2
(-2)2
- 2(2) + 4 = 4 - 4 + 4
= 4
Tambaya 47 Rahoto
A solid rectangular block has a base that measures 3x cm by 2x cm. The height of the block is ycm and its volume is 72cm3 .
i. Express y in terms of x.
ii. An expression for the total surface area of the block in terms of x only;
iii. the value of x for which the total surface area has a stationary value.
The volume of a solid rectangular block is given by the formula V = lwh, where l, w, and h are the length, width, and height of the block, respectively. In this problem, we are given that the base of the block has dimensions 3x cm by 2x cm, so we have l = 3x cm and w = 2x cm. The height of the block is y cm, so h = y cm. We are also given that the volume of the block is 72 cm3, so we have:
V = lwh
72 = (3x)(2x)(y)
72 = 6x^2y
Solving for y, we get:
y = 72/6x^2
y = 12/x^2
Therefore, the height of the block is 12/x^2 cm.
b.
To find the total surface area of the solid rectangular block, we need to consider the six faces of the block: the top face, bottom face, front face, back face, left face, and right face.
Given:
Base length = 3x cm
Base width = 2x cm
Height = y cm
Volume = 72 cm^3
The volume of a rectangular block is given by the formula:
Volume = Base Area * Height
Therefore, we can write the equation:
72 cm^3 = (3x cm * 2x cm) * y cm
Simplifying this equation, we have:
72 = 6x^2 * y
Now, let's express the total surface area of the block in terms of x only.
The total surface area of the block can be calculated by adding the areas of all six faces:
Total Surface Area = 2 * (Base Area) + (Front Face Area) + (Back Face Area) + (Left Face Area) + (Right Face Area)
The base area is given by:
Base Area = Length * Width = (3x cm) * (2x cm) = 6x^2 cm^2
The front face and back face both have the same dimensions, so their areas are equal:
Front Face Area = Back Face Area = Length * Height = (3x cm) * (y cm) = 3xy cm^2
Similarly, the left face and right face both have the same dimensions, so their areas are equal:
Left Face Area = Right Face Area = Width * Height = (2x cm) * (y cm) = 2xy cm^2
Now, let's substitute these values into the equation for the total surface area:
Total Surface Area = 2 * (6x^2 cm^2) + 2 * (3xy cm^2) + 2 * (2xy cm^2)
Simplifying further, we have:
Total Surface Area = 12x^2 cm^2 + 6xy cm^2 + 4xy cm^2
Finally, we can express the total surface area of the block in terms of x only as:
Total Surface Area = 12x^2 cm^2 + 10xy cm^2
Bayanin Amsa
The volume of a solid rectangular block is given by the formula V = lwh, where l, w, and h are the length, width, and height of the block, respectively. In this problem, we are given that the base of the block has dimensions 3x cm by 2x cm, so we have l = 3x cm and w = 2x cm. The height of the block is y cm, so h = y cm. We are also given that the volume of the block is 72 cm3, so we have:
V = lwh
72 = (3x)(2x)(y)
72 = 6x^2y
Solving for y, we get:
y = 72/6x^2
y = 12/x^2
Therefore, the height of the block is 12/x^2 cm.
b.
To find the total surface area of the solid rectangular block, we need to consider the six faces of the block: the top face, bottom face, front face, back face, left face, and right face.
Given:
Base length = 3x cm
Base width = 2x cm
Height = y cm
Volume = 72 cm^3
The volume of a rectangular block is given by the formula:
Volume = Base Area * Height
Therefore, we can write the equation:
72 cm^3 = (3x cm * 2x cm) * y cm
Simplifying this equation, we have:
72 = 6x^2 * y
Now, let's express the total surface area of the block in terms of x only.
The total surface area of the block can be calculated by adding the areas of all six faces:
Total Surface Area = 2 * (Base Area) + (Front Face Area) + (Back Face Area) + (Left Face Area) + (Right Face Area)
The base area is given by:
Base Area = Length * Width = (3x cm) * (2x cm) = 6x^2 cm^2
The front face and back face both have the same dimensions, so their areas are equal:
Front Face Area = Back Face Area = Length * Height = (3x cm) * (y cm) = 3xy cm^2
Similarly, the left face and right face both have the same dimensions, so their areas are equal:
Left Face Area = Right Face Area = Width * Height = (2x cm) * (y cm) = 2xy cm^2
Now, let's substitute these values into the equation for the total surface area:
Total Surface Area = 2 * (6x^2 cm^2) + 2 * (3xy cm^2) + 2 * (2xy cm^2)
Simplifying further, we have:
Total Surface Area = 12x^2 cm^2 + 6xy cm^2 + 4xy cm^2
Finally, we can express the total surface area of the block in terms of x only as:
Total Surface Area = 12x^2 cm^2 + 10xy cm^2
Tambaya 48 Rahoto
Solve 2(2y+1)−5(2y)+2 = 0
2(2y+1)−5(2y)+2 = 0
Let p = 2y
22y(21)−5(2y)
+ 2 = 0
2p2
- 5p + 2 = 0
2p2
- p - 4p + 2 = 0
p (2p - 1) - 2(2p - 1) = 0
(p - 2)(2p - 1) = 0
p = 2 or 12
p = 2y
when p = 2
2y
= 2
y = 1
when p = 12
2y = 12
2y = 2−1
y = -1
y = -1 or 1
Bayanin Amsa
2(2y+1)−5(2y)+2 = 0
Let p = 2y
22y(21)−5(2y)
+ 2 = 0
2p2
- 5p + 2 = 0
2p2
- p - 4p + 2 = 0
p (2p - 1) - 2(2p - 1) = 0
(p - 2)(2p - 1) = 0
p = 2 or 12
p = 2y
when p = 2
2y
= 2
y = 1
when p = 12
2y = 12
2y = 2−1
y = -1
y = -1 or 1
Tambaya 49 Rahoto
The table shows the corresponding values of two variables X and Y.
| X | 14 | 16 | 17 | 18 | 22 | 24 | 27 | 28 | 31 | 33 |
| Y | 22 | 19 | 15 | 13 | 10 | 12 | 3 | 5 | 3 | 2 |
a. plot a scatter diagram to represent the data
b i. Calculate:x?, the mean of X and ?, the mean of Y;
ii. Caculate:
x?1, the mean of X values below x? and ?1, the mean of the corresponding Y values below x?
c. Draw the line of best fit through (x?,?) and (x?1,?1).
d. From the graph, determine the relationship between X and Y;
ii. From the graph, determine the value of Y when X is 20.
a. To plot a scatter diagram, we need to plot the given X and Y values as individual points on a graph, where X values are taken as horizontal axis and Y values are taken as vertical axis. The scatter diagram shows the relationship between two variables.
b i. To calculate x?, the mean of X, we add up all the X values and divide by the number of values, i.e., x? = (14+16+17+18+22+24+27+28+31+33) / 10 = 23.
Similarly, to calculate ?, the mean of Y, we add up all the Y values and divide by the number of values, i.e., ? = (22+19+15+13+10+12+3+5+3+2) / 10 = 11.
b ii. To calculate x?1, we need to find the X values that are below x? and then calculate their mean. From the X values given, 14, 16, 17, 18, and 22 are below x? = 23. So, x?1 = (14+16+17+18+22) / 5 = 17.4.
Similarly, we need to find the corresponding Y values that are below x? and then calculate their mean. From the Y values given, corresponding Y values for the X values 14, 16, 17, 18, and 22 are 22, 19, 15, 13, and 10 respectively. So, ?1 = (22+19+15+13+10) / 5 = 15.8.
c. To draw the line of best fit, we need to plot the points (x?, ?) and (x?1, ?1) on the scatter diagram and then draw a straight line passing through these two points. The line should be such that it has an equal number of points above and below it.
d i. From the scatter diagram, we can see that there is a negative relationship between X and Y. As X increases, Y decreases.
d ii. To determine the value of Y when X is 20, we can draw a vertical line from the point X = 20 on the X-axis to the line of best fit. From the point where the line intersects the Y-axis, we can read off the value of Y, which is approximately 12.
Bayanin Amsa
a. To plot a scatter diagram, we need to plot the given X and Y values as individual points on a graph, where X values are taken as horizontal axis and Y values are taken as vertical axis. The scatter diagram shows the relationship between two variables.
b i. To calculate x?, the mean of X, we add up all the X values and divide by the number of values, i.e., x? = (14+16+17+18+22+24+27+28+31+33) / 10 = 23.
Similarly, to calculate ?, the mean of Y, we add up all the Y values and divide by the number of values, i.e., ? = (22+19+15+13+10+12+3+5+3+2) / 10 = 11.
b ii. To calculate x?1, we need to find the X values that are below x? and then calculate their mean. From the X values given, 14, 16, 17, 18, and 22 are below x? = 23. So, x?1 = (14+16+17+18+22) / 5 = 17.4.
Similarly, we need to find the corresponding Y values that are below x? and then calculate their mean. From the Y values given, corresponding Y values for the X values 14, 16, 17, 18, and 22 are 22, 19, 15, 13, and 10 respectively. So, ?1 = (22+19+15+13+10) / 5 = 15.8.
c. To draw the line of best fit, we need to plot the points (x?, ?) and (x?1, ?1) on the scatter diagram and then draw a straight line passing through these two points. The line should be such that it has an equal number of points above and below it.
d i. From the scatter diagram, we can see that there is a negative relationship between X and Y. As X increases, Y decreases.
d ii. To determine the value of Y when X is 20, we can draw a vertical line from the point X = 20 on the X-axis to the line of best fit. From the point where the line intersects the Y-axis, we can read off the value of Y, which is approximately 12.
Tambaya 50 Rahoto
The probability that Abiola will be late to the office on a given day is 2/5. In a given working week of six days, find, correct to four significant figures, the probability that he will:
(a) only be late for 3 days.
(b) not be late in the week:
(c) be late throughout the six days.
he will be late to office = 2/5
he will not be late to office is = 1 - 2/5 = 3/5
If he will be late for 3 days only, he will also not be late for 3 days
25 * 25 * 25 * 35 * 35 * 35
p = 0.0138
(b) not be late to office is = 1 - 2/5 = 3/5
35 * 35 * 35 * 35 * 35 * 35
p = 0.0467
(c) be late to office = 2/5
25 * 25 * 25 * 25 * 25 * 25
p = 0.0041
Bayanin Amsa
he will be late to office = 2/5
he will not be late to office is = 1 - 2/5 = 3/5
If he will be late for 3 days only, he will also not be late for 3 days
25 * 25 * 25 * 35 * 35 * 35
p = 0.0138
(b) not be late to office is = 1 - 2/5 = 3/5
35 * 35 * 35 * 35 * 35 * 35
p = 0.0467
(c) be late to office = 2/5
25 * 25 * 25 * 25 * 25 * 25
p = 0.0041
Tambaya 51 Rahoto
The table shows the scores obtained by a group of artistes in Vocal (X) and the instrument (Y) musical competition.
| Vocal (X) | 63 | 69 | 72 | 59 | 82 | 91 | 95 | 68 |
| Instrument (Y) | 58 | 61 | 67 | 51 | 53 | 79 | 92 | 57 |
Calculate the spearman's rank correlation coefficient between the scores.
p = 1−6Σ(di)2n(n2−1
p = Spearman's rank correlation coefficient
di = difference between the two ranks of each observation
n = number of observation
| X | Rx | Y | Ry | DI (Rx - Ry ) | Di 2 |
| 63 | 2 | 58 | 4 | -2 | 4 |
| 69 | 4 | 61 | 5 | -1 | 1 |
| 72 | 5 | 67 | 6 | -1 | 1 |
| 59 | 1 | 51 | 1 | 0 | 0 |
| 82 | 6 | 53 | 2 | 4 | 16 |
| 91 | 7 | 79 | 7 | 0 | 0 |
| 95 | 8 | 92 | 8 | 0 | 0 |
| 68 | 3 | 57 | 3 | 0 | 0 |
P = 1−6(4+1+1+0+16+0+0+0)8(82−1)
p = 1−6(22)8(63)
p = 1−132504
p = 1−1142
p = 1142 or 0.7381
Bayanin Amsa
p = 1−6Σ(di)2n(n2−1
p = Spearman's rank correlation coefficient
di = difference between the two ranks of each observation
n = number of observation
| X | Rx | Y | Ry | DI (Rx - Ry ) | Di 2 |
| 63 | 2 | 58 | 4 | -2 | 4 |
| 69 | 4 | 61 | 5 | -1 | 1 |
| 72 | 5 | 67 | 6 | -1 | 1 |
| 59 | 1 | 51 | 1 | 0 | 0 |
| 82 | 6 | 53 | 2 | 4 | 16 |
| 91 | 7 | 79 | 7 | 0 | 0 |
| 95 | 8 | 92 | 8 | 0 | 0 |
| 68 | 3 | 57 | 3 | 0 | 0 |
P = 1−6(4+1+1+0+16+0+0+0)8(82−1)
p = 1−6(22)8(63)
p = 1−132504
p = 1−1142
p = 1142 or 0.7381
Tambaya 52 Rahoto
Given that nC4 , nC5 and nC6 are the terms of a linear sequence (A.P), find the :
i. value of n
ii. common differences of the sequence.
nC4 = n![n−4]!4! n[n−1][n−2][n−3][n−4]![n−4]!4!
nC5 = n![n−5]!5! n[n−1][n−2][n−3][n−4][n−5]![n−5]!5!]
and nC6 = n![n−6]!6! → n[n−1][n−2][n−3][n−4][n−5][n−6]![n−6]!6!]
d = U2 - U1 = U3 - U2
nC5 - nC4 = n(n−1)(n−2)(n−3)(n−9)5!
nC6 - nC5 = n(n−1)(n−2)(n−3)(n−4)(n−11)5!6
nC5 - nC4 = nC6 - nC5
n(n−1)(n−2)(n−3)(n−9)5! = n(n−1)(n−2)(n−3)(n−4)(n−11)5!6
divide both sides by n(n-1)(n-2)(n-3)
n−95!
= [n−4][n−11]5!6
multiply both sides by 5! * 6
6(n-9) = (n-4)(n-11)
6n - 54 = n2
-11n - 4n + 44
6n - 54 = n2
- 15n + 44
n2
- 21n + 98 = 0
n2
- 7n - 14n + 98 = 0
n(n - 7) -14(n - 7) = 0
(n-7)(n-14) = 0
n = 7 or 14
ii.
d = U2 - U1
when n = 7
d = 7C5
- 7C4
d = 7∗6∗5!2!∗5! - 7∗6∗5∗4!3!∗4!
d = 21 - 35
d = -14
when n = 14
d = 14C5 - 14C4
d = 14∗13∗12∗11∗10∗9!9!∗5! - 14∗13∗12∗11∗10!10!∗4!
d = 2002 - 1001
d = 1001
Bayanin Amsa
nC4 = n![n−4]!4! n[n−1][n−2][n−3][n−4]![n−4]!4!
nC5 = n![n−5]!5! n[n−1][n−2][n−3][n−4][n−5]![n−5]!5!]
and nC6 = n![n−6]!6! → n[n−1][n−2][n−3][n−4][n−5][n−6]![n−6]!6!]
d = U2 - U1 = U3 - U2
nC5 - nC4 = n(n−1)(n−2)(n−3)(n−9)5!
nC6 - nC5 = n(n−1)(n−2)(n−3)(n−4)(n−11)5!6
nC5 - nC4 = nC6 - nC5
n(n−1)(n−2)(n−3)(n−9)5! = n(n−1)(n−2)(n−3)(n−4)(n−11)5!6
divide both sides by n(n-1)(n-2)(n-3)
n−95!
= [n−4][n−11]5!6
multiply both sides by 5! * 6
6(n-9) = (n-4)(n-11)
6n - 54 = n2
-11n - 4n + 44
6n - 54 = n2
- 15n + 44
n2
- 21n + 98 = 0
n2
- 7n - 14n + 98 = 0
n(n - 7) -14(n - 7) = 0
(n-7)(n-14) = 0
n = 7 or 14
ii.
d = U2 - U1
when n = 7
d = 7C5
- 7C4
d = 7∗6∗5!2!∗5! - 7∗6∗5∗4!3!∗4!
d = 21 - 35
d = -14
when n = 14
d = 14C5 - 14C4
d = 14∗13∗12∗11∗10∗9!9!∗5! - 14∗13∗12∗11∗10!10!∗4!
d = 2002 - 1001
d = 1001
Tambaya 53 Rahoto
A binary operation * is defined on the set T = {-2,-1,1,2} by p*q = p2
+ 2pq - q2
, where p,q ∊ T.
Copy and complete the table.
| * | -2 | -1 | 1 | 2 |
| -2 | 7 | -8 | ||
| -1 | 2 | -2 | ||
| 1 | -7 | 1 | ||
| 2 | -1 | |
p*q = p2 + 2pq - q2
when p = -2 and q = -2
p*q = -22
+ 2(-2)(-2) - (-2)2
p*q = 4 + 8 - 4 = 8
when p = -2 and q = -1
p*q = -22
+ 2(-2)(-1) - (-1)2
p*q = 4 + 4 - 1 = 7
when p = -2 and q = 1
p*q = -22
+ 2(-2)(1) - (1)2
p*q = 4 - 4 -1 = -1
when p = -1 and q = -2
p*q = -12
+ 2(-1)(-2) - (-2)2
p*q = 1 + 4 - 4 = 1
when p = -1 and q = -1
p*q = -12
+ 2(-1)(-1) - (-1)2
p*q = 1 + 2 -1 = 2
when p = -1 and q = 2
p*q = -12
+ 2(-1)(2) - (2)2
p*q = 1 - 4 - 4 = -7
when p = 1 and q = -2
p*q = 12
+ 2(1)(-2) - (-2)2
p*q = 1 - 4 - 4 = -7
when p = 1 and q = -1
p*q = 12
+ 2(1)(-1) - (-1)2
p*q = 1 - 2 - 1 = -2
when p = 1 and q = 1
p*q = 12
+ 2(1)(1) - (1)2
p*q = 1 + 2 - 1 = 2
when p = 1 and q = 2
p*q = 12
+ 2(1)(2) - (2)2
p*q = 1 + 4 - 4 = 1
when p = 2 and q = -2
p*q = 22
+ 2(2)(-2) - (-2)2
p*q = 4 - 8 - 4 = -8
when p = 2 and q = -1
p*q = 22
+ 2(2)(-1) - (-1)2
p*q = 4 - 4 - 1 = -1
when p = 2 and q = 1
p*q = 22
+ 2(2)(1) - (1)2
p*q = 4 + 4 - 1 = 7
when p = 2 and q = 2
p*q = 22
+ 2(2)(2) - (2)2
p*q = 4 + 8 - 4 = 8
| * | -2 | -1 | 1 | 2 |
| -2 | 8 | 7 | -1 | -8 |
| -1 | 1 | 2 | -2 | -7 |
| 1 | -7 | -2 | 2 | 1 |
| 2 | -8 | -1 | 7 | 8 |
Bayanin Amsa
p*q = p2 + 2pq - q2
when p = -2 and q = -2
p*q = -22
+ 2(-2)(-2) - (-2)2
p*q = 4 + 8 - 4 = 8
when p = -2 and q = -1
p*q = -22
+ 2(-2)(-1) - (-1)2
p*q = 4 + 4 - 1 = 7
when p = -2 and q = 1
p*q = -22
+ 2(-2)(1) - (1)2
p*q = 4 - 4 -1 = -1
when p = -1 and q = -2
p*q = -12
+ 2(-1)(-2) - (-2)2
p*q = 1 + 4 - 4 = 1
when p = -1 and q = -1
p*q = -12
+ 2(-1)(-1) - (-1)2
p*q = 1 + 2 -1 = 2
when p = -1 and q = 2
p*q = -12
+ 2(-1)(2) - (2)2
p*q = 1 - 4 - 4 = -7
when p = 1 and q = -2
p*q = 12
+ 2(1)(-2) - (-2)2
p*q = 1 - 4 - 4 = -7
when p = 1 and q = -1
p*q = 12
+ 2(1)(-1) - (-1)2
p*q = 1 - 2 - 1 = -2
when p = 1 and q = 1
p*q = 12
+ 2(1)(1) - (1)2
p*q = 1 + 2 - 1 = 2
when p = 1 and q = 2
p*q = 12
+ 2(1)(2) - (2)2
p*q = 1 + 4 - 4 = 1
when p = 2 and q = -2
p*q = 22
+ 2(2)(-2) - (-2)2
p*q = 4 - 8 - 4 = -8
when p = 2 and q = -1
p*q = 22
+ 2(2)(-1) - (-1)2
p*q = 4 - 4 - 1 = -1
when p = 2 and q = 1
p*q = 22
+ 2(2)(1) - (1)2
p*q = 4 + 4 - 1 = 7
when p = 2 and q = 2
p*q = 22
+ 2(2)(2) - (2)2
p*q = 4 + 8 - 4 = 8
| * | -2 | -1 | 1 | 2 |
| -2 | 8 | 7 | -1 | -8 |
| -1 | 1 | 2 | -2 | -7 |
| 1 | -7 | -2 | 2 | 1 |
| 2 | -8 | -1 | 7 | 8 |
Tambaya 54 Rahoto
Solve 3cos2x - sinx = 0 for 0º≤x≤360º
3cos2x - sinx = 0
cos 2A = 1 – 2 sin2
A
3(1 - 2sin2
x) - sinx = 0
3 - 6sin2
x -sinx = 0
let sinx = p
3 - 6p2
-p = 0
-6p2 - p +3 = 0
factors of p in the eqn:
p = 0.7953 or 0.6287
when p = 0.7953
p = sinx
x = sin−1
p
x = sin−1
(0.7953)
x = 52.68º
when p = 0.6287
p = sinx
x = sin−1
p
x = sin−1
(0.6287)
x = 38.95º
Bayanin Amsa
3cos2x - sinx = 0
cos 2A = 1 – 2 sin2
A
3(1 - 2sin2
x) - sinx = 0
3 - 6sin2
x -sinx = 0
let sinx = p
3 - 6p2
-p = 0
-6p2 - p +3 = 0
factors of p in the eqn:
p = 0.7953 or 0.6287
when p = 0.7953
p = sinx
x = sin−1
p
x = sin−1
(0.7953)
x = 52.68º
when p = 0.6287
p = sinx
x = sin−1
p
x = sin−1
(0.6287)
x = 38.95º
Tambaya 55 Rahoto
Two functions f and g are defined on the set of real numbers, R, by
f:x → x2 + 2 and g:x → 1x+2.Find the domain of (g∘f)−1
f:x → x2 + 2 and g:x → 1x+2.
(g∘f)−1
(g∘f)x2+2
gx2+2
= 1(x2+2)+2
(g∘f) = 1(x2+4
let y = 1(x2+4
y((x2+4) = 1
yx2+4y=1
x2 = 1−4yy
x = 1−4yy−−−−√
(g∘f)−1 = 1−4xx−−−−√
Bayanin Amsa
f:x → x2 + 2 and g:x → 1x+2.
(g∘f)−1
(g∘f)x2+2
gx2+2
= 1(x2+2)+2
(g∘f) = 1(x2+4
let y = 1(x2+4
y((x2+4) = 1
yx2+4y=1
x2 = 1−4yy
x = 1−4yy−−−−√
(g∘f)−1 = 1−4xx−−−−√
Tambaya 56 Rahoto
A basket contains 12 fruits: orange, apple and avocado pear, all of the same size. The number of oranges, apples and avocado pear forms three consecutive integers.
Two fruits are drawn one after the other without replacement. Calculate the probability that:
i. the first is an orange and the second is an avocado pear.
ii.both are of same fruit;
iii. at least one is an apple
They form 3 consecutive integers
let the number of orange be x, apple = x+1 and avocado pear = x+2
x + x+1 + x+2 = 12
3x + 3 = 12
3x = 9
x = 3
apple = 4
avocado pear = 5
probability that orange is drawn = 312
= 14
probability that apple is drawn = 412 = 13
probability that avocado pear is drawn = 512
probability that the first is orange and the second is an avocado pear = 312 * 511 = 544
ii.
both are of same fruit, they are both Orange or apple or Avocado pear
p(O∩O) + p(A∩A) + p(P∩P) = 14∗211+13∗311+512∗411
= 122+111+533
= 1966
iii.
at least one is an apple = p(A∩O) or p(O∩A) or p(A∩P) or p(P∩A) or p(A∩A)
p(A∩O) + p(O∩A) + p(A∩P) + p(P∩A) + p(A∩A) =
= 412∗311+311∗412+412∗511+512∗411+412∗311
= 12132 + 12132 + 20132 + 20132 + 12132
= 76132
= 1933
Bayanin Amsa
They form 3 consecutive integers
let the number of orange be x, apple = x+1 and avocado pear = x+2
x + x+1 + x+2 = 12
3x + 3 = 12
3x = 9
x = 3
apple = 4
avocado pear = 5
probability that orange is drawn = 312
= 14
probability that apple is drawn = 412 = 13
probability that avocado pear is drawn = 512
probability that the first is orange and the second is an avocado pear = 312 * 511 = 544
ii.
both are of same fruit, they are both Orange or apple or Avocado pear
p(O∩O) + p(A∩A) + p(P∩P) = 14∗211+13∗311+512∗411
= 122+111+533
= 1966
iii.
at least one is an apple = p(A∩O) or p(O∩A) or p(A∩P) or p(P∩A) or p(A∩A)
p(A∩O) + p(O∩A) + p(A∩P) + p(P∩A) + p(A∩A) =
= 412∗311+311∗412+412∗511+512∗411+412∗311
= 12132 + 12132 + 20132 + 20132 + 12132
= 76132
= 1933
Tambaya 57 Rahoto
A particle initially at rest moves in a straight line with an acceleration of (10t - 4t2
)m/s2
Find the:
a. velocity of the particle after t seconds;
ii. average acceleration of the particle during the 4th second.
b. A load of mass 120kg is placed on a lift. Calculate the reaction between the floor of the lift and the load when the lift moves upwards at a constant velocity. [Take g = 10m/s2 ]
ii. with an acceleration of 3m/s2 . [Take g = 10m/s2 ]
a. a = (10t - 4t2
)m/s2
, t = t seconds, u = 0
from first equation of motion,
v = u + at
v = 0 + (10t - 4t2
)t
v = (10t2
- 4t3
)m/s2
ii. average acceleration of the particle during the 4th second = the average acceleration between the t = 3 and t = 4
Δa4 = 10(t4-t3) - 4(t4-t3)2
Δa4 = 10(4-1) - 4(4-3)2
Δa4 = 10(1) - 4(12)
Δa4 = (10-4)m/s2
Δa4 = 6ms-2
b. R = m(g+a)
When the lift moves with a constant velocity, a = 0
R = mg
R = 120 × 10
R = 1200N
ii. when the lift moves upward with an acceleration of 3m/s2
R = m(g+a)
R = 120(10+3)
R = 120 × 13
R = 1560N
Bayanin Amsa
a. a = (10t - 4t2
)m/s2
, t = t seconds, u = 0
from first equation of motion,
v = u + at
v = 0 + (10t - 4t2
)t
v = (10t2
- 4t3
)m/s2
ii. average acceleration of the particle during the 4th second = the average acceleration between the t = 3 and t = 4
Δa4 = 10(t4-t3) - 4(t4-t3)2
Δa4 = 10(4-1) - 4(4-3)2
Δa4 = 10(1) - 4(12)
Δa4 = (10-4)m/s2
Δa4 = 6ms-2
b. R = m(g+a)
When the lift moves with a constant velocity, a = 0
R = mg
R = 120 × 10
R = 1200N
ii. when the lift moves upward with an acceleration of 3m/s2
R = m(g+a)
R = 120(10+3)
R = 120 × 13
R = 1560N
Tambaya 58 Rahoto
The vectors 6i + 8j and 8i - 6j are parallel to ?OP and ?OQ respectively. If the magnitude of ?OP and ?OQ are 80 units and 120 units respectively, express: ?OP and ?OQ in terms of i and j;
ii. |?PQ|, in the form c?k, where c and k are constants.
The question is asking to express vectors →OP and →OQ in terms of i and j, where 6i + 8j is parallel to →OP and 8i - 6j is parallel to →OQ. We are also given the magnitudes of →OP and →OQ, which are 80 units and 120 units respectively.
To express →OP and →OQ in terms of i and j, we can simply multiply the scalar component of each vector by its respective unit vector. Thus, →OP = 80/10 (6i + 8j) = 48i + 64j, and →OQ = 120/10 (8i - 6j) = 96i - 72j.
To find the magnitude of →PQ, we can use the formula |→PQ| = |→OQ - →OP|. Substituting the values we obtained earlier, we get |→PQ| = |(96i - 72j) - (48i + 64j)| = |48i - 136j|. To simplify this, we can use the Pythagorean theorem, which states that for a right triangle with sides a and b and hypotenuse c, a² + b² = c². Thus, |→PQ| = √(48² + (-136)²) = √20800 = 40√13. Therefore, the magnitude of →PQ is 80√29 units, which is in the form c√k where c = 40 and k = 13.
Bayanin Amsa
The question is asking to express vectors →OP and →OQ in terms of i and j, where 6i + 8j is parallel to →OP and 8i - 6j is parallel to →OQ. We are also given the magnitudes of →OP and →OQ, which are 80 units and 120 units respectively.
To express →OP and →OQ in terms of i and j, we can simply multiply the scalar component of each vector by its respective unit vector. Thus, →OP = 80/10 (6i + 8j) = 48i + 64j, and →OQ = 120/10 (8i - 6j) = 96i - 72j.
To find the magnitude of →PQ, we can use the formula |→PQ| = |→OQ - →OP|. Substituting the values we obtained earlier, we get |→PQ| = |(96i - 72j) - (48i + 64j)| = |48i - 136j|. To simplify this, we can use the Pythagorean theorem, which states that for a right triangle with sides a and b and hypotenuse c, a² + b² = c². Thus, |→PQ| = √(48² + (-136)²) = √20800 = 40√13. Therefore, the magnitude of →PQ is 80√29 units, which is in the form c√k where c = 40 and k = 13.
Tambaya 59 Rahoto
A body of mass of 18kg is suspended by an inextensible string from a rigid support and is pulled by a horizontal force F until the angle of inclination of the string to the vertical is 35º. If the system is in equilibrium, calculate the:
i. value of F
ii. tension in the string
If the system is in equilibrium, then the weight of the body = the force pulling it
and the force = the vertical component of force F
since Force F is inclined to vertical then the vertical component = Fcos35º
the weight of the body = mg
Fcos35º = 18(10)
0.8192F = 180
F = 1800.8192
F = 219.7N
tension in the string
T = mg
T = 18kg × 10
T = 180N
Bayanin Amsa
If the system is in equilibrium, then the weight of the body = the force pulling it
and the force = the vertical component of force F
since Force F is inclined to vertical then the vertical component = Fcos35º
the weight of the body = mg
Fcos35º = 18(10)
0.8192F = 180
F = 1800.8192
F = 219.7N
tension in the string
T = mg
T = 18kg × 10
T = 180N
