<span>The
third root of the given complex number 27(cos(pi/5)+isin(pi/5)) is <span>3(cos(pi/15)+i sin(pi/15))
</span>The solution would be like this
for this specific problem:</span>
<span>2^5 =
32 so you need a 2 out front the 5th root of cos(x) + i sin(x) is
cos(x/5) + i sin(x/5). Additionally, 5 roots are located at even
intervals around the circle. They are spaced every 2 pi/5 or 6 pi/15 radians.
</span>
<span>Roots
are located at pi/15, pi/15+ 10pi/15 = 11 pi/15 and pi/15+ 20pi/15 = 21 pi/15
(or 7 pi /5 ).</span>
An yes
Step-by-step explanation:
2=2 is 5
Step-by-step explanation:
12) y = 4 + 7x
x = 3,-1
y = 4 + 7(3) = 25
13) y = x - 11
x = 72,
y = 72 - 11 = 61
14) y = 6x - 5
y = 6(-4) - 5 = -29
y = -11 (x = -1)
y = -5 (x = 0)
y = 13 (x = 3)
15) y = x + 1, each relation is a function
16) An = 22 + (n-1)11
A(2) = 22 + 1(11) = 33
A(5) = 22 + 4(11) = 66
lesson 5,1
1 ------- A
2 ------ C
3 ------ A
Answer:
kendra is 8 and tim is 4
Step-by-step explanation:
