Answer:
a = 94
b = -66√3
Step-by-step explanation:
Given z = 

![(-1-i\sqrt{3})^6=[(-1-i\sqrt{3})^3]^2](https://tex.z-dn.net/?f=%28-1-i%5Csqrt%7B3%7D%29%5E6%3D%5B%28-1-i%5Csqrt%7B3%7D%29%5E3%5D%5E2)
![=[(1+i\sqrt{3})^3]^2](https://tex.z-dn.net/?f=%3D%5B%281%2Bi%5Csqrt%7B3%7D%29%5E3%5D%5E2)
[(a + b)³ = a³ + b³ + 3a²b + 3ab²]
= [1 + 3i² + 3i√3 + 9i²]²
= [1 - 3 + 3i√3 - 9]²
= [-11 + 3i√3]²
= (11)² -2(3i√3)(11) + (3i√3)²
= 121 - 66i√3 + 27i²
= 121 - 66i√3 - 27
= (94 - 66i√3)
Comparing it with (a + bi),
a = 94 and b = -66√3
Answer:
$135
Step-by-step explanation:
P is the principal amount, $600.00.
r is the interest rate, 4.5% per year, or in decimal form, 4.5/100=0.045.
t is the time involved, 5....year(s) time periods.
So, t is 5....year time periods.
To find the simple interest, we multiply 600 × 0.045 × 5 to get that:
:) hope this helps
Answer:
-9
Step-by-step explanation:
y = 8
x = -3
- 2(2-y)-x = ⇒ replace x and y with their values
- 2(2 - 8) - (-3) = ⇒ solve parenthesis
- 2(-6) + 3 = ⇒ multiply
- -12 + 3= ⇒ add
- - 9 ⇒ answer
Answer:
3x^2 -2x + 1 =3(x^2-2/3x+1/3)=3(x-1/3)^2+2/9*3= 3(x-1/3)^2+2/3
(x-1/3)^2 is greater or equal to zero
3(x-1/3)^2 is greater or equal to zero
and 2/3 is greater than zero
So there sum is greater than zero
Proved
Step-by-step explanation:
3x^2 -2x + 1 =3(x^2-2/3x+1/3)
Consider x^2-2/3x+1/3
Remember that (a-b)^2 =a^2-2ab+b^2
x^2=a^2
a=x
-2/3x= -2*x*b
b=1/3
S0 (x-1/3)^2= x^2-2/3x+1/9
x^2-2/3x+1/3= x^2-2/3x+1/9+1/3-1/9= (x-1/3)^2+2/9
3x^2 -2x + 1 =3(x^2-2/3x+1/3)=3(x-1/3)^2+2/9*3= 3(x-1/3)^2+2/3
(x-1/3)^2 is greater or equal to zero
3(x-1/3)^2 is greater or equal to zero
and 2/3 is greater than zero
So there sum is greater than zero
Proved
Answer: It is only 5 points, if it were 20 points, I would answer.
My username is Gorillatactics.
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