Let's raise i to various powers starting with 0,1,2,3...
i^0 = 1
i^1 = i
i^2 = ( sqrt(-1) )^2 = -1
i^3 = i^2*i = -1*i = -i
i^4 = (i^2)^2 = (-1)^2 = 1
i^5 = i^4*i = 1*i = i
i^6 = i^5*i = i*i = i^2 = -1
We see that the pattern repeats itself after 4 iterations. The four items to memorize are
i^0 = 1
i^1 = i
i^2 = -1
i^3 = -i
It bounces back and forth between 1 and i, alternating in sign as well. This could be one way to memorize the pattern.
To figure out something like i^25, we simply divide the exponent 25 over 4 to get the remainder. In this case, the remainder of 25/4 is 1 since 24/4 = 6, and 25 is one higher than 24.
This means i^25 = i^1 = i
Likewise,
i^5689 = i^1 = i
because 5689/4 = 1422 remainder 1. The quotient doesn't play a role at all so you can ignore it entirely
Let Q be the point (a, b). Since the midpoint of P (2, 3) and Q (a, b) is (4, 6), we have
(2 + a)/2 = 4
(3 + b)/2 = 6
Solve for a and b :
(2 + a)/2 = 4
2 + a = 8
a = 6
(3 + b)/2 = 6
3 + b = 12
b = 9
So the coordinates of Q are (6, 9).
Answer:
1.9099
Step-by-step explanation:
V = (PI) * r^2 * (h/3)
32 = (PI) * (16) * (h/3)
6/PI = h
1.9099 = h
The answer is A. u do the othe angle and add it to the fisrt one
Answer:
24.08
Step-by-step explanation:
We use the Pythagorean theorem and we get (16)^2 + (18)^2 = (x)^2
256 + 324 = x^2
x^2 = 580
x is about 24.08.