:
No.
Half of 48 would be 24, not 21
<3
<em>z</em> = 3<em>i</em> / (-1 - <em>i</em> )
<em>z</em> = 3<em>i</em> / (-1 - <em>i</em> ) × (-1 + <em>i</em> ) / (-1 + <em>i</em> )
<em>z</em> = (3<em>i</em> × (-1 + <em>i</em> )) / ((-1)² - <em>i</em> ²)
<em>z</em> = (-3<em>i</em> + 3<em>i</em> ²) / ((-1)² - <em>i</em> ²)
<em>z</em> = (-3 - 3<em>i </em>) / (1 - (-1))
<em>z</em> = (-3 - 3<em>i </em>) / 2
Note that this number lies in the third quadrant of the complex plane, where both Re(<em>z</em>) and Im(<em>z</em>) are negative. But arctan only returns angles between -<em>π</em>/2 and <em>π</em>/2. So we have
arg(<em>z</em>) = arctan((-3/2)/(-3/2)) - <em>π</em>
arg(<em>z</em>) = arctan(1) - <em>π</em>
arg(<em>z</em>) = <em>π</em>/4 - <em>π</em>
arg(<em>z</em>) = -3<em>π</em>/4
where I'm taking arg(<em>z</em>) to have a range of -<em>π</em> < arg(<em>z</em>) ≤ <em>π</em>.
So if 240 is 5%, double it to get 480 than multiply it by 10 to get 4800.
Answer:
The number of possible combinations is 180
Step-by-step explanation:
The numbers to select from are 3-8
These are 6 numbers
Now, the restriction we have is that the 3 characters cannot be the same
For the first character, we have 6 choices;
For the second character, we have another 6 choices
For the last character, since it cannot be the same, we can only have 5 choices
So the possible number of choices will be;
6 * 6 * 5 = 189 combinations
Substitute z with 3.
3(4)^3
Cube 4 with 3
3(64)
Multiply 3 and 64
=192
