I use it as a short way to say "okay" but I don't know how others may use it.
![z_1=\sqrt{3}\left(cos\:\frac{\pi }{4}+i\:sin\:\frac{\pi }{4}\right)](https://tex.z-dn.net/?f=z_1%3D%5Csqrt%7B3%7D%5Cleft%28cos%5C%3A%5Cfrac%7B%5Cpi%20%7D%7B4%7D%2Bi%5C%3Asin%5C%3A%5Cfrac%7B%5Cpi%20%7D%7B4%7D%5Cright%29)
![z_2=\sqrt{6}\left(cos\:\frac{3\pi \:}{4}+i\:sin\:\frac{3\pi \:}{4}\right)](https://tex.z-dn.net/?f=z_2%3D%5Csqrt%7B6%7D%5Cleft%28cos%5C%3A%5Cfrac%7B3%5Cpi%20%5C%3A%7D%7B4%7D%2Bi%5C%3Asin%5C%3A%5Cfrac%7B3%5Cpi%20%5C%3A%7D%7B4%7D%5Cright%29)
![\cos \left(\frac{\pi }{4}\right)=\frac{\sqrt{2}}{2}](https://tex.z-dn.net/?f=%5Ccos%20%5Cleft%28%5Cfrac%7B%5Cpi%20%7D%7B4%7D%5Cright%29%3D%5Cfrac%7B%5Csqrt%7B2%7D%7D%7B2%7D)
![\sin \left(\frac{\pi }{4}\right)=\frac{\sqrt{2}}{2}](https://tex.z-dn.net/?f=%5Csin%20%5Cleft%28%5Cfrac%7B%5Cpi%20%7D%7B4%7D%5Cright%29%3D%5Cfrac%7B%5Csqrt%7B2%7D%7D%7B2%7D)
![\cos \left(\frac{3\pi }{4}\right)=-\frac{\sqrt{2}}{2}](https://tex.z-dn.net/?f=%5Ccos%20%5Cleft%28%5Cfrac%7B3%5Cpi%20%7D%7B4%7D%5Cright%29%3D-%5Cfrac%7B%5Csqrt%7B2%7D%7D%7B2%7D)
![\sin \left(\frac{3\pi }{4}\right)=\frac{\sqrt{2}}{2}\\](https://tex.z-dn.net/?f=%5Csin%20%5Cleft%28%5Cfrac%7B3%5Cpi%20%7D%7B4%7D%5Cright%29%3D%5Cfrac%7B%5Csqrt%7B2%7D%7D%7B2%7D%5C%5C)
![Z_1*Z_2=\sqrt{3}\left(cos\:\frac{\pi }{4}+i\:sin\:\frac{\pi }{4}\right)\cdot \sqrt{6}\left(cos\:\frac{3\pi \:}{4}+i\:sin\:\frac{3\pi \:}{4}\right)](https://tex.z-dn.net/?f=Z_1%2AZ_2%3D%5Csqrt%7B3%7D%5Cleft%28cos%5C%3A%5Cfrac%7B%5Cpi%20%7D%7B4%7D%2Bi%5C%3Asin%5C%3A%5Cfrac%7B%5Cpi%20%7D%7B4%7D%5Cright%29%5Ccdot%20%5Csqrt%7B6%7D%5Cleft%28cos%5C%3A%5Cfrac%7B3%5Cpi%20%5C%3A%7D%7B4%7D%2Bi%5C%3Asin%5C%3A%5Cfrac%7B3%5Cpi%20%5C%3A%7D%7B4%7D%5Cright%29)
![=\sqrt{3}\left(\frac{\sqrt{2}}{2}+\:i\frac{\sqrt{2}}{2}\right)\cdot \sqrt{6}\left(\frac{-\sqrt{2}}{2}+\:i\frac{\sqrt{2}}{2}\right)](https://tex.z-dn.net/?f=%3D%5Csqrt%7B3%7D%5Cleft%28%5Cfrac%7B%5Csqrt%7B2%7D%7D%7B2%7D%2B%5C%3Ai%5Cfrac%7B%5Csqrt%7B2%7D%7D%7B2%7D%5Cright%29%5Ccdot%20%5Csqrt%7B6%7D%5Cleft%28%5Cfrac%7B-%5Csqrt%7B2%7D%7D%7B2%7D%2B%5C%3Ai%5Cfrac%7B%5Csqrt%7B2%7D%7D%7B2%7D%5Cright%29)
On simplifying, we get
![Z_1* Z_2 =-3\sqrt{2}](https://tex.z-dn.net/?f=Z_1%2A%20Z_2%20%3D-3%5Csqrt%7B2%7D)
<h2>Therefore, correct option is 1st option.</h2>
Let's find the slope for line t first.
We can use the given points 2,6 and 10,1 to find the slope using the slope formula.
![\frac{y.2 - y.1}{x.2 - x.1} = m](https://tex.z-dn.net/?f=%5Cfrac%7By.2%20-%20y.1%7D%7Bx.2%20-%20x.1%7D%20%3D%20m)
1 - 6 / 10 - 2
-5/8
The slope is -5/8.
Because we know the slope of this line, we can find the slope of the next line instantly, as they are perpendicular.
When a slope is perpendicular to another, it is equal to the negative reciprocal.
Negative reciprocal of -5/8 = 8/5
<h3>The slope of line u is 8/5</h3>
James should define that the rays must be collinear. the two rays coming from the same point will form collinear rays which will form an angle at end point.
The statement of james should be precise in order to clearly explain the angle formation.
The two non collinear rays will form an angle at end point where they meet with each other.
The defined vertex will form the appropriate angle. there should be defined term for the non collinear rays.
Learn more at brainly.com/question/24373158
Answer:
Step-by-step explanation:
Given is a triangle RST and another triangle R'S'T' tranformed from RST
Vertices of RST are (0, 0), (negative 2, 3), (negative 3, 1).
Vertices of R'S'T' are (2, 0), (0, negative 3), (negative 1, negative 1).
Comparing the corresponding vertices we find that x coordinate increased by 2 while y coordinate got the different sign.
This indicates that there is both reflection and transformation horizontally to the right by 2 units
So first shifted right by 2 units so that vertices became
(2,0) (0,3) (-1,1)
Now reflected on the line y=0 i.e. x axis
New vertices are
(2,0) (0,-3) (-1,-1)