How was C2?!

[Wearewhoweare]

It was okay for me .. some parts were hard but some were really easy:S

 

[CellarCY]

It was an alright exam forgot some things but overall think it is an A

 

[shamawi]

Super Easy

 

[Shoggaine]

Alhamduellah Its was more dan awesome <3
I was jst confused for 1 part.....where we had to show da sum of Nth terms........

 

[z3r0thunder]

It was very easy.

 

[Ahmed93]

it was easy

what was the perimeter of that sector?

 

[Ahmed93]

and the minimum Area
the value of x was 2.12 ?

 

[z3r0thunder]

Value of X for Cylinder is 2.122
Minimum Area = 84.84 = 85 (Nearest Integer)

Perimeter = 5+5+(5*1.855)=19.275

Sum of nth Terms is the Proof:
1. Sn=a+ar+ar^2+...+ar^(n-1)
2. rSn=ar+ar^2+...+ar^n
1. - 2. = a-ar^n
Sn(1-r) = a(1-r^n)
Sn = (a(1-r^n))/(a-r)

 

[fofo]

what were the angles for this question >>> .... (1- 5 cos2x ) sin 2x = 0 ???

 

[fofo]

Value of X for Cylinder is 2.122
Minimum Area = 84.84 = 85 (Nearest Integer)

Perimeter = 5+5+(5*1.855)=19.275

Sum of nth Terms is the Proof:
1. Sn=a+ar+ar^2+...+ar^(n-1)
2. rSn=ar+ar^2+...+ar^n
1. - 2. = a-ar^n
Sn(1-r) = a(1-r^n)
Sn = (a(1-r^n))/(a-r)

YESSSSSSS AHAAHHAAAAAAAAY GNNA GET AN A INSHALLAH !!! =P

 

[z3r0thunder]

For that question I found 5 Answers for X
Modify Range since it is for 2x => 0≥2x≤360
Sin2x = 0
Cos2x = 1/5

2x=0°, 180°,360°
2x = 78.5°, 281.5°

Final Solutions of X=0°, 39.3°, 90°, 140.8°, 180°
All must be in 1 Decimal Place

 

[Shoggaine]

For that question I found 5 Answers for X
Modify Range since it is for 2x => 0≥2x≤360
Sin2x = 0
Cos2x = 1/5

2x=0°, 180°,360°
2x = 78.5°, 281.5°

Final Solutions of X=0°, 39.3°, 90°, 140.8°, 180°
All must be in 1 Decimal Place

Hey! My solutions were only 39.3 nd 140 :O

 

[z3r0thunder]

Hey! My solutions were only 39.3 nd 140 :O

That means you used a different method for Question 1 and did not use the given result (which you should have found in Question 1) and did not get all the Solutions.

 

[Shoggaine]

Do u still remember the question?? Its tan2x=5sin2x ryt???

 

[Shoggaine]

Do u still remember the question?? Its tan2x=5sin2x ryt???

I jst wanna knw da correct answer...

 

[z3r0thunder]

Yes:

Tan2x = 5Sin2x
Sin2x/Cos2x = 5Sin2x
Sin2x = 5Sin2xCos2x
Sin2x - 5Sin2xCos2x = 0
Sin2x(1-5Cos2x) = 0 <= This was Given

 

[CellarCY]

isnt it 1-5cos2x so it will be the inverse of Cos of NEGATIVE 1/5?

 

[z3r0thunder]

isnt it 1-5cos2x so it will be the inverse of Cos of NEGATIVE 1/5?

No, because (1-5Cos2x) to be = 0
Let Cos2x = Y
(1-5Y)=0
1=5Y
Y=1/5
Cos2x = 1/5

 

[shamawi]

For that question I found 5 Answers for X
Modify Range since it is for 2x => 0≥2x≤360
Sin2x = 0
Cos2x = 1/5

2x=0°, 180°,360°
2x = 78.5°, 281.5°

Final Solutions of X=0°, 39.3°, 90°, 140.8°, 180°
All must be in 1 Decimal Place

 

[wonderland]

z3r0thunder -- For the volume of cylinder - I am just wondering what was the formula shown in the paper - was it 2πx² + 120/4π or 2πx² + 60/4π.