Use algebra for such problems
let, Angle COD = x
Angle KOD = y
Angle KPC = z
Given ,. x - y = 61° ( Equation 1 )
x - z = 53° ( Equation 2 )
Subtract 1st equation from 2nd and you'll get :
z - y = 8° ( Equation 3 )
Now since , x + y + z = 180° ( Equation 4 )
Add Equation 3 to Equation 4 and you'll get
x + 2z = 188° ( Equation 5)
From Equation 2 we know that , x -z = 53°
or, x = 53° + z ( Equation 6 )
Put this value of 'x' in Equation 5 and solve for z. you'll get :
(53° + z) + 2z = 188°
or
3z = 188 - 53 = 135°
solving for z we get
z = 45°
put this value of z in Equation 5
x + ( 2 x 45° ) = 188°
or
x = 188° - 90° = 98°
hence , Angle COD = 98°
Answer:
x= -15/2 y=-33/2
Step-by-step explanation:
i dont know how you want to solve it but if it is "system of equations" then first solve for the first variable in on e of the equations, then substitute the result into the other equation.
We have to calculate the fourth roots of this complex number:
![z=9+9\sqrt[]{3}i](https://tex.z-dn.net/?f=z%3D9%2B9%5Csqrt%5B%5D%7B3%7Di)
We start by writing this number in exponential form:
![\begin{gathered} r=\sqrt[]{9^2+(9\sqrt[]{3})^2} \\ r=\sqrt[]{81+81\cdot3} \\ r=\sqrt[]{81+243} \\ r=\sqrt[]{324} \\ r=18 \end{gathered}](https://tex.z-dn.net/?f=%5Cbegin%7Bgathered%7D%20r%3D%5Csqrt%5B%5D%7B9%5E2%2B%289%5Csqrt%5B%5D%7B3%7D%29%5E2%7D%20%5C%5C%20r%3D%5Csqrt%5B%5D%7B81%2B81%5Ccdot3%7D%20%5C%5C%20r%3D%5Csqrt%5B%5D%7B81%2B243%7D%20%5C%5C%20r%3D%5Csqrt%5B%5D%7B324%7D%20%5C%5C%20r%3D18%20%5Cend%7Bgathered%7D)
![\theta=\arctan (\frac{9\sqrt[]{3}}{9})=\arctan (\sqrt[]{3})=\frac{\pi}{3}](https://tex.z-dn.net/?f=%5Ctheta%3D%5Carctan%20%28%5Cfrac%7B9%5Csqrt%5B%5D%7B3%7D%7D%7B9%7D%29%3D%5Carctan%20%28%5Csqrt%5B%5D%7B3%7D%29%3D%5Cfrac%7B%5Cpi%7D%7B3%7D)
Then, the exponential form is:
The formula for the roots of a complex number can be written (in polar form) as:
Then, for a fourth root, we will have n = 4 and k = 0, 1, 2 and 3.
To simplify the calculations, we start by calculating the fourth root of r:
![r^{\frac{1}{4}}=18^{\frac{1}{4}}=\sqrt[4]{18}](https://tex.z-dn.net/?f=r%5E%7B%5Cfrac%7B1%7D%7B4%7D%7D%3D18%5E%7B%5Cfrac%7B1%7D%7B4%7D%7D%3D%5Csqrt%5B4%5D%7B18%7D)
<em>NOTE: It can not be simplified anymore, so we will leave it like this.</em>
Then, we calculate the arguments of the trigonometric functions:
We can now calculate for each value of k:
![\begin{gathered} k=0\colon \\ z_0=\sqrt[4]{18}\cdot(\cos (\pi(\frac{1}{8}+\frac{0}{2}))+i\cdot\sin (\pi(\frac{1}{8}+\frac{0}{2}))) \\ z_0=\sqrt[4]{18}\cdot(\cos (\frac{\pi}{8})+i\cdot\sin (\frac{\pi}{8}) \\ z_0=\sqrt[4]{18}\cdot e^{i\frac{\pi}{8}} \end{gathered}](https://tex.z-dn.net/?f=%5Cbegin%7Bgathered%7D%20k%3D0%5Ccolon%20%5C%5C%20z_0%3D%5Csqrt%5B4%5D%7B18%7D%5Ccdot%28%5Ccos%20%28%5Cpi%28%5Cfrac%7B1%7D%7B8%7D%2B%5Cfrac%7B0%7D%7B2%7D%29%29%2Bi%5Ccdot%5Csin%20%28%5Cpi%28%5Cfrac%7B1%7D%7B8%7D%2B%5Cfrac%7B0%7D%7B2%7D%29%29%29%20%5C%5C%20z_0%3D%5Csqrt%5B4%5D%7B18%7D%5Ccdot%28%5Ccos%20%28%5Cfrac%7B%5Cpi%7D%7B8%7D%29%2Bi%5Ccdot%5Csin%20%28%5Cfrac%7B%5Cpi%7D%7B8%7D%29%20%5C%5C%20z_0%3D%5Csqrt%5B4%5D%7B18%7D%5Ccdot%20e%5E%7Bi%5Cfrac%7B%5Cpi%7D%7B8%7D%7D%20%5Cend%7Bgathered%7D)
![\begin{gathered} k=1\colon \\ z_1=\sqrt[4]{18}\cdot(\cos (\pi(\frac{1}{8}+\frac{1}{2}))+i\cdot\sin (\pi(\frac{1}{8}+\frac{1}{2}))) \\ z_1=\sqrt[4]{18}\cdot(\cos (\frac{5\pi}{8})+i\cdot\sin (\frac{5\pi}{8})) \\ z_1=\sqrt[4]{18}e^{i\frac{5\pi}{8}} \end{gathered}](https://tex.z-dn.net/?f=%5Cbegin%7Bgathered%7D%20k%3D1%5Ccolon%20%5C%5C%20z_1%3D%5Csqrt%5B4%5D%7B18%7D%5Ccdot%28%5Ccos%20%28%5Cpi%28%5Cfrac%7B1%7D%7B8%7D%2B%5Cfrac%7B1%7D%7B2%7D%29%29%2Bi%5Ccdot%5Csin%20%28%5Cpi%28%5Cfrac%7B1%7D%7B8%7D%2B%5Cfrac%7B1%7D%7B2%7D%29%29%29%20%5C%5C%20z_1%3D%5Csqrt%5B4%5D%7B18%7D%5Ccdot%28%5Ccos%20%28%5Cfrac%7B5%5Cpi%7D%7B8%7D%29%2Bi%5Ccdot%5Csin%20%28%5Cfrac%7B5%5Cpi%7D%7B8%7D%29%29%20%5C%5C%20z_1%3D%5Csqrt%5B4%5D%7B18%7De%5E%7Bi%5Cfrac%7B5%5Cpi%7D%7B8%7D%7D%20%5Cend%7Bgathered%7D)
![\begin{gathered} k=2\colon \\ z_2=\sqrt[4]{18}\cdot(\cos (\pi(\frac{1}{8}+\frac{2}{2}))+i\cdot\sin (\pi(\frac{1}{8}+\frac{2}{2}))) \\ z_2=\sqrt[4]{18}\cdot(\cos (\frac{9\pi}{8})+i\cdot\sin (\frac{9\pi}{8})) \\ z_2=\sqrt[4]{18}e^{i\frac{9\pi}{8}} \end{gathered}](https://tex.z-dn.net/?f=%5Cbegin%7Bgathered%7D%20k%3D2%5Ccolon%20%5C%5C%20z_2%3D%5Csqrt%5B4%5D%7B18%7D%5Ccdot%28%5Ccos%20%28%5Cpi%28%5Cfrac%7B1%7D%7B8%7D%2B%5Cfrac%7B2%7D%7B2%7D%29%29%2Bi%5Ccdot%5Csin%20%28%5Cpi%28%5Cfrac%7B1%7D%7B8%7D%2B%5Cfrac%7B2%7D%7B2%7D%29%29%29%20%5C%5C%20z_2%3D%5Csqrt%5B4%5D%7B18%7D%5Ccdot%28%5Ccos%20%28%5Cfrac%7B9%5Cpi%7D%7B8%7D%29%2Bi%5Ccdot%5Csin%20%28%5Cfrac%7B9%5Cpi%7D%7B8%7D%29%29%20%5C%5C%20z_2%3D%5Csqrt%5B4%5D%7B18%7De%5E%7Bi%5Cfrac%7B9%5Cpi%7D%7B8%7D%7D%20%5Cend%7Bgathered%7D)
![\begin{gathered} k=3\colon \\ z_3=\sqrt[4]{18}\cdot(\cos (\pi(\frac{1}{8}+\frac{3}{2}))+i\cdot\sin (\pi(\frac{1}{8}+\frac{3}{2}))) \\ z_3=\sqrt[4]{18}\cdot(\cos (\frac{13\pi}{8})+i\cdot\sin (\frac{13\pi}{8})) \\ z_3=\sqrt[4]{18}e^{i\frac{13\pi}{8}} \end{gathered}](https://tex.z-dn.net/?f=%5Cbegin%7Bgathered%7D%20k%3D3%5Ccolon%20%5C%5C%20z_3%3D%5Csqrt%5B4%5D%7B18%7D%5Ccdot%28%5Ccos%20%28%5Cpi%28%5Cfrac%7B1%7D%7B8%7D%2B%5Cfrac%7B3%7D%7B2%7D%29%29%2Bi%5Ccdot%5Csin%20%28%5Cpi%28%5Cfrac%7B1%7D%7B8%7D%2B%5Cfrac%7B3%7D%7B2%7D%29%29%29%20%5C%5C%20z_3%3D%5Csqrt%5B4%5D%7B18%7D%5Ccdot%28%5Ccos%20%28%5Cfrac%7B13%5Cpi%7D%7B8%7D%29%2Bi%5Ccdot%5Csin%20%28%5Cfrac%7B13%5Cpi%7D%7B8%7D%29%29%20%5C%5C%20z_3%3D%5Csqrt%5B4%5D%7B18%7De%5E%7Bi%5Cfrac%7B13%5Cpi%7D%7B8%7D%7D%20%5Cend%7Bgathered%7D)
Answer:
The four roots in exponential form are
z0 = 18^(1/4)*e^(i*π/8)
z1 = 18^(1/4)*e^(i*5π/8)
z2 = 18^(1/4)*e^(i*9π/8)
z3 = 18^(1/4)*e^(i*13π/8)
Answer:a) 870
b) 435
Step-by-step explanation:
a)
number of photos to be posted =n = 30
number of web pages on which they would be posted is 2
Since the order in which the photos appear on the web pages matters,
Number of ways = 30 permutation 2
=870 ways
b)
Since the order in which the photos appear on the web pages does not matter,
Number of ways = 30 combination 2
= 435 ways
