Rounding 345.67 to the nearest
thousand: 0
hundred: 300
ten: 340
unit: 346
tenth: 345.7
hundredth: 345.67
You can select the largest!
Did you mean ... for \(n\ge2\) rather than ...for \(x\ge2 \)?
If so:
Rewrite \(x^2-x-1=0 \) as
\(x^2=x+1\) (1)
Mutiply (1) by x
\(x^3=x^2+x\) Rewrite as
\(x^3-x^2=x \) (2)
Subtract (1) from (2)
\(x^3-x^2-x^2=x-(x+1)\)
Simplify
\(x^3-2x^2=-1\)
In polar coordinates: \(z=re^{i\theta} \text{ and } \bar{z} = re^{-i\theta}\) so \((\bar{z})^n=(re^{-i\theta})^n=r^ne^{-in\theta}\) and \(\bar{z^n}=\bar{(r^ne^{in\theta})}=r^ne^{-in\theta}\)
Here's one way of tackling this:
Do you mean 1/16 are yellow? (You don't get a whole number of balls with the figures you've specified!).
The following graph should help:
As follows:
(5-4i)-2(3+6i) = 5 - 4i - 6 - 6i
Combine all real parts together and all imaginary parts together, s0: (5-4i)-2(3+6i) = -1 -10i
Since median and mode are both forms of average, I'm assuming that where you used the word average you were referring to the mean. In which case:
+33659 
