We will have the following:
![(-2+3i)+2(5+6i)=-2+3i+10+12i](https://tex.z-dn.net/?f=%28-2%2B3i%29%2B2%285%2B6i%29%3D-2%2B3i%2B10%2B12i)
![=(10-2)+(3i+12i)=8+15i](https://tex.z-dn.net/?f=%3D%2810-2%29%2B%283i%2B12i%29%3D8%2B15i)
So, the solution would be:
<u>ANSWER TO PART A</u>
The given triangle has vertices ![J(-4,1), K(-4,-2),L(-3,-1)](https://tex.z-dn.net/?f=J%28-4%2C1%29%2C%20K%28-4%2C-2%29%2CL%28-3%2C-1%29)
The mapping for rotation through
counterclockwise has the mapping
![(x,y)\rightarrow (-y,x)](https://tex.z-dn.net/?f=%28x%2Cy%29%5Crightarrow%20%28-y%2Cx%29)
Therefore
![J(-4,1)\rightarrow J'(-1,-4)](https://tex.z-dn.net/?f=J%28-4%2C1%29%5Crightarrow%20J%27%28-1%2C-4%29)
![K(-4,-2)\rightarrow K'(2,-4)](https://tex.z-dn.net/?f=K%28-4%2C-2%29%5Crightarrow%20K%27%282%2C-4%29)
![L(-3,-1)\rightarrow L'(1,-3)](https://tex.z-dn.net/?f=L%28-3%2C-1%29%5Crightarrow%20L%27%281%2C-3%29)
We plot all this point and connect them with straight lines.
ANSWER TO PART B
For a reflection across the y-axis we negate the x coordinates.
The mapping is
![(x,y)\rightarrow (-x,y)](https://tex.z-dn.net/?f=%28x%2Cy%29%5Crightarrow%20%28-x%2Cy%29)
Therefore
![J(-4,1)\rightarrow J''(4,1)](https://tex.z-dn.net/?f=J%28-4%2C1%29%5Crightarrow%20J%27%27%284%2C1%29)
![K(-4,-2)\rightarrow K''(4,-2)](https://tex.z-dn.net/?f=K%28-4%2C-2%29%5Crightarrow%20K%27%27%284%2C-2%29)
![L(-3,-1)\rightarrow L''(3,-1)](https://tex.z-dn.net/?f=L%28-3%2C-1%29%5Crightarrow%20L%27%27%283%2C-1%29)
We plot all this point and connect them with straight lines.
See graph in attachment
6.23+ -12.49 -2.6= 6.23+ -12.49+ -2.6
-6.26+-2.6= -8.86
Given f(x) = 8x + 1 and g(x) = f(x − 2), which equation represents g substitute x-2 for x in f(x). we have. g(x)=8(x-2)+1. =8x-16+1. =8x-15.
Answer:
Step-by-step explanation:
12 and 1/8, I just simplified 2/16 to 1/8