Answer:
15
19
14
Step-by-step explanation:
I´d say "d" is the distance from the eye to the wall.
Now substracting 1.2-1 you´ll get the distance of the wall of the smallest triangle = 0.2 And you do 1.5-0.2= 0.3 that´s the distance of the wall of the other triangle. Then you solve everything with Pitagoras theorem. You have 2 rectangle triangles.
B+alfa=45°
tan^-1(0.2/d)=B
tan^-1(1.3/d)=alfa
THEN:
tan^-1(0.2/d)+tan^-1(1.3/d)=45°
Now you have 3 ecs and 3 variables.
alfa,B and "d"
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)
Not all 30-story buildings are the same height. Depends on how tall the ceiling is on each floor. But usually it is around 300 feet
Answer:
Altogether they made $24 which is enough to earn $12 each
Step-by-step explanation:
Important information:
1 cup: $0.75
4 cups = 1 quart
8 quarts of lemonade altogether
If we are looking to solve for the revenue earned based on quarts, we must first solve for how much money they make on 1 quart
Since 1 cup sells for $0.75 and 4 cups makes up a quart, we must multiply
0.75 × 4 = 3
They make $3 per 1 quart
We are looking for the amount for 8 quarts though, not just 1. Again we must multiply
3 × 8 = 24
They will make $24 if they sell all 8 quarts
The questions ask if they will <em>each</em> earn $12
Since there are two of them, Miles and Lucy, we must divide revenue earned by 2, the number of people
24 ÷ 2 = 12
$12
