To get this you would do 8x8 which equals 64
Answer:
Step-by-step explanation:
roots of a complex number is given by DeMoivre's formula.
![\sf \boxed{\bf r^{\frac{1}{n}}\left[Cos \dfrac{\theta + 2\pi k}{n}+i \ Sin \ \dfrac{\theta+2\pi k}{n}\right]}](https://tex.z-dn.net/?f=%5Csf%20%5Cboxed%7B%5Cbf%20r%5E%7B%5Cfrac%7B1%7D%7Bn%7D%7D%5Cleft%5BCos%20%5Cdfrac%7B%5Ctheta%20%2B%202%5Cpi%20k%7D%7Bn%7D%2Bi%20%5C%20Sin%20%5C%20%5Cdfrac%7B%5Ctheta%2B2%5Cpi%20k%7D%7Bn%7D%5Cright%5D%7D)
Here, k lies between 0 and (n -1) ; n is the exponent.
a = -1 and b = √3
n = 4
For k = 0,
![\sf z = \sqrt[4]{10}\left[Cos \ \dfrac{\dfrac{-\pi}{3} +0}{4}+iSin \ \dfrac{\dfrac{-\pi}{3}+0}{4}\right] \\\\\\z= \sqrt[4]{10} \left[Cos \ \dfrac{ -\pi }{12}+iSin \ \dfrac{-\pi}{12}\right]\\\\\\z = \sqrt[4]{10}\left[-Cos \ \dfrac{\pi}{12}-i \ Sin \ \dfrac{\pi}{12}\right]](https://tex.z-dn.net/?f=%5Csf%20z%20%3D%20%5Csqrt%5B4%5D%7B10%7D%5Cleft%5BCos%20%5C%20%5Cdfrac%7B%5Cdfrac%7B-%5Cpi%7D%7B3%7D%20%2B0%7D%7B4%7D%2BiSin%20%20%5C%20%5Cdfrac%7B%5Cdfrac%7B-%5Cpi%7D%7B3%7D%2B0%7D%7B4%7D%5Cright%5D%20%5C%5C%5C%5C%5C%5Cz%3D%20%5Csqrt%5B4%5D%7B10%7D%20%5Cleft%5BCos%20%5C%20%5Cdfrac%7B%20-%5Cpi%20%20%7D%7B12%7D%2BiSin%20%20%5C%20%5Cdfrac%7B-%5Cpi%7D%7B12%7D%5Cright%5D%5C%5C%5C%5C%5C%5Cz%20%3D%20%5Csqrt%5B4%5D%7B10%7D%5Cleft%5B-Cos%20%5C%20%5Cdfrac%7B%5Cpi%7D%7B12%7D-i%20%5C%20Sin%20%5C%20%5Cdfrac%7B%5Cpi%7D%7B12%7D%5Cright%5D)
For k =1,
![\sf z = \sqrt[4]{10}\left[Cos \ \dfrac{5\pi}{12}+i \ Sin \ \dfrac{5\pi}{12}\right]](https://tex.z-dn.net/?f=%5Csf%20z%20%3D%20%5Csqrt%5B4%5D%7B10%7D%5Cleft%5BCos%20%5C%20%5Cdfrac%7B5%5Cpi%7D%7B12%7D%2Bi%20%5C%20Sin%20%5C%20%5Cdfrac%7B5%5Cpi%7D%7B12%7D%5Cright%5D)
For k =2,
![z = \sqrt[4]{10}\left[Cos \ \dfrac{11\pi}{12}+i \ Sin \ \dfrac{11\pi}{12}\right]](https://tex.z-dn.net/?f=z%20%3D%20%5Csqrt%5B4%5D%7B10%7D%5Cleft%5BCos%20%5C%20%5Cdfrac%7B11%5Cpi%7D%7B12%7D%2Bi%20%5C%20Sin%20%5C%20%5Cdfrac%7B11%5Cpi%7D%7B12%7D%5Cright%5D)
For k = 3,
![\sf z = \sqrt[4]{10}\left[Cos \ \dfrac{17\pi}{12}+i \ Sin \ \dfrac{17\pi}{12}\right]](https://tex.z-dn.net/?f=%5Csf%20z%20%3D%20%5Csqrt%5B4%5D%7B10%7D%5Cleft%5BCos%20%5C%20%5Cdfrac%7B17%5Cpi%7D%7B12%7D%2Bi%20%5C%20Sin%20%5C%20%5Cdfrac%7B17%5Cpi%7D%7B12%7D%5Cright%5D)
For k = 4,
![\sf z =\sqrt[4]{10}\left[Cos \ \dfrac{23\pi}{12}+i \ Sin \ \dfrac{23\pi}{12}\right]](https://tex.z-dn.net/?f=%5Csf%20z%20%3D%5Csqrt%5B4%5D%7B10%7D%5Cleft%5BCos%20%5C%20%5Cdfrac%7B23%5Cpi%7D%7B12%7D%2Bi%20%5C%20Sin%20%5C%20%5Cdfrac%7B23%5Cpi%7D%7B12%7D%5Cright%5D)
Answer:
60cm^2
Step-by-step explanation:
We assume that is a circumscribing quadrilateral, rather than one that is circumscribed. It is also called a "tangential quadrilateral" and its area is ...
K = sr
where s is the semi-perimeter, the sum of opposite sides, and r is the radius of the incircle.
K = (12 cm) (5cm) = 60 cm²
_____
A quadrilateral can only be tangential if pairs of opposite sides add to the same length. Hence the given sum is the semiperimeter.
Answer:
7000
Step-by-step explanation:
