WAEC SSCE - Further Mathematics - 2022 (Essay)
Question 1 Rapport
Evaluate: \( \lim_{x \to -2} \frac{x^3 + 8}{x + 2} \).
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
Détails de la réponse
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
Question 2 Rapport
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
Détails de la réponse
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
Question 3 Rapport
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
Détails de la réponse
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
Question 4 Rapport
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
Détails de la réponse
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
Question 5 Rapport
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.
Détails de la réponse
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.
Question 6 Rapport
A binary operation * is defined on the set \(T = \{-2,-1,1,2\}\) by \(p*q = p^2 + 2pq - q^2\), where \(p,q \in T\).
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 |
Détails de la réponse
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 |
Question 7 Rapport
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.
Détails de la réponse
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.
Question 8 Rapport
Given that \(nC_4\), \(nC_5\) and \(nC_6\) 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
Détails de la réponse
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
Question 9 Rapport
Solve \(2^{(2y+1)} - 5(2^y) + 2 = 0\)
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
Détails de la réponse
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
Question 10 Rapport
Two functions f and g are defined on the set of real numbers, R, by
\(f:x\to x^2+2\) and \(g:x\to \frac{1}{x+2}\).Find the domain of \((g\circ f)^{-1}\)
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−−−−√
Détails de la réponse
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−−−−√
Question 11 Rapport
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
Détails de la réponse
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
Question 12 Rapport
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
Détails de la réponse
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
Question 13 Rapport
A particle initially at rest moves in a straight line with an acceleration of \( (10t - 4t^2)\text{ m/s}^2 \). 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 \( g = 10\text{ m/s}^2 \)]
ii. with an acceleration of \( 3\text{ m/s}^2 \). [Take \( g = 10\text{ m/s}^2 \)]
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
Détails de la réponse
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
Question 14 Rapport
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
Détails de la réponse
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
Question 15 Rapport
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 72cm\(^3\).
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
Détails de la réponse
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
