Final assignment

1. lim tn
n- 00 tn= 1/n lim tn = 0

n /tn The limit is 0 as n approaches infinity. The table justifys the answe
1 1
2 1/2
3 1/3
15 1/15

10 10 10
3. ∑ 5k-k^2 5∑ K - ∑ k^2 5(10/2 X 11) - (10) (11) (21) / 6 = - 110
K = 1 K=1 K = 1



5. sin 300 Turns from 0 to 300 degrees which leaves 60 degrees as reference angle. Radius is 2, X is 1, and the Y is square root of 3. So sin is y/r so the answer is = - square root of 3 (because y in the 4th quadrant is negative) / 2.


6. 27 degrees to radians. 27 degrees x Л / 180 degrees = 0.471 radians.


7. secθ = -2 sec = r/x

cos -1 = -1/2 = 120 degrees (reference angle is 60 degrees).

R cant be negative, so x is negative, only can be - in 2nd and 3rd quadrant. 180+60=240 degrees. And 180 - 60 = 120 degrees.


8. for the sequence 2, 4, 6 ....... write the recursive, explicit defintion, and find t23.

Recursive - t1=2 tn=(tn-1) + 2
Explicit - tn=2+(n-1)(2) t23= 2 + (22)(2) = 46 t23= 46



10. cot = -8/5. θ is in the 3rd qudrant. Cot = x/y. x is negative ( - 8), y is 5 and there is r missing. -8^2 + 5 ^2 = r^2 r= square root of 89

sinθ = 5 x square root of 89 / 89 tanθ= 5/-8 cosθ= -8 x square root of 89 / 89.
cscθ= square root of 89 / 5 secθ= squere root of 89 / -8


11. 1-2sinθ = 0 -1 -2 sinθ = -1 -2sinθ/-2 = -1/-2 sinθ= 1/2 θ= 30 degrees
180 degrees - 30 degrees = 150 degrees. 1st revolutionary angles in { 0, 360} degress are 30 and 150.



12. Find the sector, arc lenght if θ= 27 degrees and r = 10cm

27 degrees x Л / 180 = 0.47 radians Area sector = 0.47 (10cm)/2. A= 23.5 cm squared

S (arc lenght) 0.47 (10cm) = 4.7 cm




13. Given the terms of thsi series, use sigma notation to find the sum.

3 + 9 + 15+ ..........+117 tn+ 3 + (n-1)(6) tn= 3 + (6n-6) - 6
tn = -3 + 6n
117 = -3 + 6n +3
120/6 = 6n/6
n = 20
20
∑ -3 + 6n
n = 1