Answer:
Step-by-step explanation:
If 5! is equal to 5 × 4 × 3 × 2 × 1 and 6! is equal to 6 × 5 × 4 × 3 × 2 × 1, then 4! is equal to 4 × 3 × 2 × 1. Thus, 4! = 4 × 3 × 2 × 1, which can simplify to 24. 4! = 24.
is basically 4 × 4 × 4 × 4, which can simplify to 256.
So, 
= 
. can simplify to . Therefore, = .
You can use the distance formula: d = sqrt((second x value - first x value)^2 + (second y value - first y value)^2):
answer : d = sqrt((3+2)^2+(6+6)^2) = 11.95826074 units
I need help my self lol XD
<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>.
