| No. of Days | 1 | 2 | 3 | 4 | 5 | 6 |
| No. of students | 20 | 2x | 60 | 40 | x | 50 |
The distribution above shows the number of days a group of 260 students were absent from school in a particular term. How many students were absent for at least four days in the term?
180
120
110
40
Correct answer is B
20 + 2x + 60 + 40 + x + 50 = 260
170 + 3x = 260
3x = 260 - 170
3x = 90
x = 90/3 = 30
At least 4 days
4 days = 40
5 days = 30
6 days = 50
Total = 120
sec θ tan θ + k
tan θ + k
2sec θ + k
sec θ + k
Correct answer is B
∫sec2θ dθ = ∫ 1/cos2 dθ
∫(cos)-2 dθ,
let u = cos θ
∴∫u-2 = 1/u + c
∫cos θ = sin θ + c
∫sec-2θ = 1/u sin θ + c
= (sinθ / cosθ) + c
= Tan θ + c
2.3 cm
4.0 cm
5.2 cm
6.0 cm
Correct answer is B
S = t\(^3\) - t\(^2\) - t + 5
ds/dt = 3t\(^2\) - 2t - 1
As ds = 0
3t\(^2\) - 2t - 1 = 0
(3t+1)(t-1) = 0
∴ t = 1 or -1/3
At min pt t = 1
S = t\(^3\) - t\(^2\) - t + 5
put t = 1
= 1\(^3\) - \1(^2\) - 1 + 5
= 1 - 1 - 1 + 5
= 4
If s = (2 + 3t)(5t - 4), find ds/dt when t = 4/5 secs
0 units per sec
15 units per sec
22 unit per sec
26 units per sec
Correct answer is C
x = (2+3t)(5t-4)
Let u = 2+3t ∴du/dt = 3
and v = 5t-4 ∴dv/dt = 5
dx/dt = Vdu/dt + Udv/dt
= (5t-4)3 + (2+3t)5
= 15t - 12 + 10 + 15t
= 30t - 2
= 30x4/5 - 2
= 24 - 2
= 22
100m
75√3 m
100√3m
200√3m
Correct answer is C
| Tan 60 = | 300 |
| x |
| x = | 300 |
| Tan 60 |
| x = | 300 |
| √3 |
x = (300/√3) x (√3/√3)
| x = | 300√3 |
| .....3 |
x = 100√3