3_Practice Quiz in Mathematics_College Algebra_Trigonometry_Probability and Statistics_Calculus with Answer and Solution

Refresher - MATHEMATICS Quiz 3
PROBLEM 1:


Find the value of x if 1 + 2 + 3 + 4 + …. x = 36
n(n + 1)
n=
Using 1 + 2 + 3 ….. 2


Solution:
n(n + 1)
= 36
2
n2 + n - 72 = 0
(n - 8)(n + 9) = 0
n=8
x=8

, Refresher - MATHEMATICS Quiz 3
PROBLEM 2:

In Boston, the number of hours of daylight D(t) at a particular time of the 1year may be approximated by
sin 210˚ = -
⎡ 2π ⎤ 2
D(t) = 3 sin ⎢ (t - 79) ⎥ + 12 with t in days and t = 0 corresponding to January 1. How many days of
⎣ 365 ⎦ 1
sin 210˚ = -
the year have more than 10.5 hours of daylight? 2
2π 7π
Solution: (t - 79) =
365 6
⎡ 2π ⎤
D(t) = 3 sin ⎢ (t - 79) ⎥ + 12 7π/6
⎣ 365 ⎦ t - 79 =
2π / 365
⎡ π ⎤
10.5 = 3 sin ⎢ (t - 79) ⎥ + 12 365(7)
⎣ 365 ⎦ t - 79 =
12
⎡ 2π ⎤
3 sin ⎢ (t - 79) ⎥ = - 15 t = 213 + 79
⎣ 365 ⎦
⎡ 2π ⎤ t = 292
sin ⎢ (t - 79) ⎥ = - 0.5
⎣ 365 ⎦ 2π 11 π
(t- 79) =
1 365 6
sin θ = -
2 π/6
7π t - 79 =
θ= or 210˚ 2π / 365
6
11(365)
11π t - 79 =
θ= θ 330˚ 12
6
1 t = 414 > 365
sin 210˚ = -
2 t = 414 - 365
1 t = 49
sin 210˚ = -
2
2π 7π There will be at least 10.5 hours of daylight from
(t - 79) = t = 49 to t = 292 days or 292˚ - 49˚ = 243 days of the
365 6 year.
7π/6
t - 79 =
2π / 365
365(7)
t - 79 =
12
t = 213 + 79
t = 292
2π 11 π
(t- 79) =
365 6
π/6
t - 79 =
2π / 365

11(365)