Answer:
Step-by-step explanation:
We are given that 
is in <em>fourth</em> quadrant.
is always positive in 4th quadrant and
is always negative in 4th quadrant.
Also, we know the following identity about 
and 
:
Using \theta_1 in place of \theta:
We are given that 
is in <em>4th quadrant </em>so is negative.
So, value of
Answer:
9
Step-by-step explanation:
Answer:
i have a Similar question I’m stuck on, just remember z rule, vertically opposite etc.
The only way 3 digits can have product 24 is
1 x 3 x 8 = 241 x 4 x 6 = 242 x 2 x 6 = 242 x 3 x 4 = 24
So the digits comprises of 1,3,8 or 1,4,6, or 2,2,6, or 2,3,4
To be divisible by 3 the sum of the digits must be divisible by 3.
1+ 3+ 8=12, 1+ 4+ 6= 11, 2 +2 + 6=10, 2 +3 + 4=9Of those sums of digits, only 12 and 9 are divisible by 3.
So we have ruled out all but integers whose digits consist of1,3,8, and 2,3,4.
Meanwhile they must be odd they either must end in 1 or 3.
The only ones which can end in 1 are 381 and 831.
The others must end in 3.
They must be greater than 152 which is 225. So the
First digit cannot be 1. So the only way its digits can contain of1,3,8 and close in 3 is to be 813.
The rest must contain of the digits 2,3,4, and the only way they can end in 3 is to be 243 or 423.
So there are precisely five such three-digit integers: 381, 831, 813, 243, and 423.
