When solving anproblem including money a negative answer can represent debt (how much you owe)
Answer:
6/12, 8/12, and 9/12.
Step-by-step explanation:
Let's find the least common denominator:
First, the least common multiple of 2, 3, and 4 is 12.
12 is divisible by 2, 3, and 4.
<u>Next, multiply the denominators with the numerators:</u>
Products: 6/12, 8/12, and 9/12
(yX3)+1 because product means multipy
Answer:
1+i
Step-by-step explanation:
To find the 8th roots of unity, you have to find the trigonometric form of unity.
1. Since 
then
and
This gives you 
Thus,
2. The 8th roots can be calculated using following formula:
![\sqrt[8]{z}=\{\sqrt[8]{|z|} (\cos\dfrac{\varphi+2\pi k}{8}+i\sin \dfrac{\varphi+2\pi k}{8}), k=0,\ 1,\dots,7\}.](https://tex.z-dn.net/?f=%5Csqrt%5B8%5D%7Bz%7D%3D%5C%7B%5Csqrt%5B8%5D%7B%7Cz%7C%7D%20%28%5Ccos%5Cdfrac%7B%5Cvarphi%2B2%5Cpi%20k%7D%7B8%7D%2Bi%5Csin%20%5Cdfrac%7B%5Cvarphi%2B2%5Cpi%20k%7D%7B8%7D%29%2C%20k%3D0%2C%5C%201%2C%5Cdots%2C7%5C%7D.)
Now
at k=0, ![z_0=\sqrt[8]{1} (\cos\dfrac{0+2\pi \cdot 0}{8}+i\sin \dfrac{0+2\pi \cdot 0}{8})=1\cdot (1+0\cdot i)=1;](https://tex.z-dn.net/?f=z_0%3D%5Csqrt%5B8%5D%7B1%7D%20%28%5Ccos%5Cdfrac%7B0%2B2%5Cpi%20%5Ccdot%200%7D%7B8%7D%2Bi%5Csin%20%5Cdfrac%7B0%2B2%5Cpi%20%5Ccdot%200%7D%7B8%7D%29%3D1%5Ccdot%20%281%2B0%5Ccdot%20i%29%3D1%3B)
at k=1, ![z_1=\sqrt[8]{1} (\cos\dfrac{0+2\pi \cdot 1}{8}+i\sin \dfrac{0+2\pi \cdot 1}{8})=1\cdot (\dfrac{\sqrt{2}}{2}+i\dfrac{\sqrt{2}}{2})=\dfrac{\sqrt{2}}{2}+i\dfrac{\sqrt{2}}{2};](https://tex.z-dn.net/?f=z_1%3D%5Csqrt%5B8%5D%7B1%7D%20%28%5Ccos%5Cdfrac%7B0%2B2%5Cpi%20%5Ccdot%201%7D%7B8%7D%2Bi%5Csin%20%5Cdfrac%7B0%2B2%5Cpi%20%5Ccdot%201%7D%7B8%7D%29%3D1%5Ccdot%20%28%5Cdfrac%7B%5Csqrt%7B2%7D%7D%7B2%7D%2Bi%5Cdfrac%7B%5Csqrt%7B2%7D%7D%7B2%7D%29%3D%5Cdfrac%7B%5Csqrt%7B2%7D%7D%7B2%7D%2Bi%5Cdfrac%7B%5Csqrt%7B2%7D%7D%7B2%7D%3B)
at k=2, ![z_2=\sqrt[8]{1} (\cos\dfrac{0+2\pi \cdot 2}{8}+i\sin \dfrac{0+2\pi \cdot 2}{8})=1\cdot (0+1\cdot i)=i;](https://tex.z-dn.net/?f=z_2%3D%5Csqrt%5B8%5D%7B1%7D%20%28%5Ccos%5Cdfrac%7B0%2B2%5Cpi%20%5Ccdot%202%7D%7B8%7D%2Bi%5Csin%20%5Cdfrac%7B0%2B2%5Cpi%20%5Ccdot%202%7D%7B8%7D%29%3D1%5Ccdot%20%280%2B1%5Ccdot%20i%29%3Di%3B)
at k=3, ![z_3=\sqrt[8]{1} (\cos\dfrac{0+2\pi \cdot 3}{8}+i\sin \dfrac{0+2\pi \cdot 3}{8})=1\cdot (-\dfrac{\sqrt{2}}{2}+i\dfrac{\sqrt{2}}{2})=-\dfrac{\sqrt{2}}{2}+i\dfrac{\sqrt{2}}{2};](https://tex.z-dn.net/?f=z_3%3D%5Csqrt%5B8%5D%7B1%7D%20%28%5Ccos%5Cdfrac%7B0%2B2%5Cpi%20%5Ccdot%203%7D%7B8%7D%2Bi%5Csin%20%5Cdfrac%7B0%2B2%5Cpi%20%5Ccdot%203%7D%7B8%7D%29%3D1%5Ccdot%20%28-%5Cdfrac%7B%5Csqrt%7B2%7D%7D%7B2%7D%2Bi%5Cdfrac%7B%5Csqrt%7B2%7D%7D%7B2%7D%29%3D-%5Cdfrac%7B%5Csqrt%7B2%7D%7D%7B2%7D%2Bi%5Cdfrac%7B%5Csqrt%7B2%7D%7D%7B2%7D%3B)
at k=4, ![z_4=\sqrt[8]{1} (\cos\dfrac{0+2\pi \cdot 4}{8}+i\sin \dfrac{0+2\pi \cdot 4}{8})=1\cdot (-1+0\cdot i)=-1;](https://tex.z-dn.net/?f=z_4%3D%5Csqrt%5B8%5D%7B1%7D%20%28%5Ccos%5Cdfrac%7B0%2B2%5Cpi%20%5Ccdot%204%7D%7B8%7D%2Bi%5Csin%20%5Cdfrac%7B0%2B2%5Cpi%20%5Ccdot%204%7D%7B8%7D%29%3D1%5Ccdot%20%28-1%2B0%5Ccdot%20i%29%3D-1%3B)
at k=5, ![z_5=\sqrt[8]{1} (\cos\dfrac{0+2\pi \cdot 5}{8}+i\sin \dfrac{0+2\pi \cdot 5}{8})=1\cdot (-\dfrac{\sqrt{2}}{2}-i\dfrac{\sqrt{2}}{2})=-\dfrac{\sqrt{2}}{2}-i\dfrac{\sqrt{2}}{2};](https://tex.z-dn.net/?f=z_5%3D%5Csqrt%5B8%5D%7B1%7D%20%28%5Ccos%5Cdfrac%7B0%2B2%5Cpi%20%5Ccdot%205%7D%7B8%7D%2Bi%5Csin%20%5Cdfrac%7B0%2B2%5Cpi%20%5Ccdot%205%7D%7B8%7D%29%3D1%5Ccdot%20%28-%5Cdfrac%7B%5Csqrt%7B2%7D%7D%7B2%7D-i%5Cdfrac%7B%5Csqrt%7B2%7D%7D%7B2%7D%29%3D-%5Cdfrac%7B%5Csqrt%7B2%7D%7D%7B2%7D-i%5Cdfrac%7B%5Csqrt%7B2%7D%7D%7B2%7D%3B)
at k=6, ![z_6=\sqrt[8]{1} (\cos\dfrac{0+2\pi \cdot 6}{8}+i\sin \dfrac{0+2\pi \cdot 6}{8})=1\cdot (0-1\cdot i)=-i;](https://tex.z-dn.net/?f=z_6%3D%5Csqrt%5B8%5D%7B1%7D%20%28%5Ccos%5Cdfrac%7B0%2B2%5Cpi%20%5Ccdot%206%7D%7B8%7D%2Bi%5Csin%20%5Cdfrac%7B0%2B2%5Cpi%20%5Ccdot%206%7D%7B8%7D%29%3D1%5Ccdot%20%280-1%5Ccdot%20i%29%3D-i%3B)
at k=7, ![z_7=\sqrt[8]{1} (\cos\dfrac{0+2\pi \cdot 7}{8}+i\sin \dfrac{0+2\pi \cdot 7}{8})=1\cdot (\dfrac{\sqrt{2}}{2}-i\dfrac{\sqrt{2}}{2})=\dfrac{\sqrt{2}}{2}-i\dfrac{\sqrt{2}}{2};](https://tex.z-dn.net/?f=z_7%3D%5Csqrt%5B8%5D%7B1%7D%20%28%5Ccos%5Cdfrac%7B0%2B2%5Cpi%20%5Ccdot%207%7D%7B8%7D%2Bi%5Csin%20%5Cdfrac%7B0%2B2%5Cpi%20%5Ccdot%207%7D%7B8%7D%29%3D1%5Ccdot%20%28%5Cdfrac%7B%5Csqrt%7B2%7D%7D%7B2%7D-i%5Cdfrac%7B%5Csqrt%7B2%7D%7D%7B2%7D%29%3D%5Cdfrac%7B%5Csqrt%7B2%7D%7D%7B2%7D-i%5Cdfrac%7B%5Csqrt%7B2%7D%7D%7B2%7D%3B)
The 8th roots are
Option C is icncorrect.
In order to form triangle PQT and quadrilateral TQRS, point T must lie on line PS which is 16 cm. long.
If the ratio of PT to TS is 5:3 and the total length of PS is 16, then PT must be 10 and TS must be 6 (10 + 6 =16) and 10:6 is the same ratio as 5:3. Another way to think about it is 5/3 = 10/6.
Now you have all the lengths that you need to find the areas of the quadrilateral and the triangle.
Make sure you draw a diagram of it!!
