How many intergers between 1 and 500 inclusive are divisible by both 6 and 9 but not 4

I think the answer is 14.
the smallest common multiple of 6 and 9 is 18. all multiples of 18 is divisible by 6 and 9. 
500 ÷18=27.777777777
there are 27  numbers between 1 and 500 that are divisible by both 6 and 9, but when the quotient is an even number, the multiple of 18 is divisible by 4 as well, so only the multiples of 18 that produce an odd quotient satisfy the requirement.
from 1 to 27, there are 14 odd numbers. 

cos(θ) = -0.95

Step-by-step explanation:

Remember the relation:

sin(θ)^2 + cos(θ)^2 = 1

So if we have:

sin(θ) = 0.3

we can replace that in the above equation to get:

0.3^2 + cos(θ)^2 = 1

now we can solve this for cos(θ)

cos(θ)^2 = 1 - 0.3^2 = 0.91

cos(θ) = ±√0.91

cos(θ) = ± 0.95

Now, yo can see that there are two solutions, which one is the correct one?

Well, you can see that the endpoint of the segment that defines θ is on the second quadrant.

cos(x) is negative if the endpoint of the segment that defines the angle is on the second or third quadrant.

Then we can conclude that in this case, the correct solution is the negative one.

cos(θ) = -0.95