3y-18
Pull out 3 (common factor):
3(y-6)
A) 3(y-6) is the correct answer
Given
- f(n) values for n=1,2,3,4
- possible candidates for the function
Solution:
Method: Evaluate some of the values, for each function. A function with ANY value not matching the given f(n) values will be rejected.
N=1, f(n)=4
f(1)=4-3(1-1)=4
f(1)=4+3^(1+1)=4+3^2=4+9=13 ≠ 4 [rejected]
f(1)=4(3^(n-1))=4(3^0)=4
f(1)=3(4^(n-1))=3(4^0)=3*1=3 [rejected]
N=2, f(n)=12
f(1)=4-3(2-1)=4-3(1)=1 ≠ 12 [rejected]
[rejected]
f(1)=4(3^(2-1)=4*3^1=4*3=12
[rejected]
Will need to check one more to be sure
N=3, f(n)=3
[rejected]
[rejected]
f(3)=4(3^(n-1))=4(3^(3-1))=4(3^2)=4*9=36 [Good]
[rejected]
Solution: f(n)=4(3^(n-1))
cot(<em>θ</em>) = cos(<em>θ</em>)/sin(<em>θ</em>)
So if both cot(<em>θ</em>) and cos(<em>θ</em>) are negative, that means sin(<em>θ</em>) must be positive.
Recall that
cot²(<em>θ</em>) + 1 = csc²(<em>θ</em>) = 1/sin²(<em>θ</em>)
so that
sin²(<em>θ</em>) = 1/(cot²(<em>θ</em>) + 1)
sin(<em>θ</em>) = 1 / √(cot²(<em>θ</em>) + 1)
Plug in cot(<em>θ</em>) = -2 and solve for sin(<em>θ</em>) :
sin(<em>θ</em>) = 1 / √((-2)² + 1)
sin(<em>θ</em>) = 1/√(5)
