The line integral along the given positively oriented curve  is  mathematically given as
![=3\left[e^{4}-1\right]](https://tex.z-dn.net/?f=%3D3%5Cleft%5Be%5E%7B4%7D-1%5Cright%5D) 
 
This is further explained below.
<h3>What is the line integral along the given positively oriented curve.?</h3>
Generally, 

M=y e^{x}
Therefore
x -->0 to 4 
y --> 0 to 3
![\end{aligned}\\&=\int_{0}^{4} \int_{0}^{3}\left(2 e^{x}-e^{x}\right) d y d x\\&=\int_{0}^{4} \int_{0}^{3} e^{x} d y d x\\&=3 \int_{0}^{4} e^{x} d x\\&=3\left[e^{x}\right]_{0}^{4}\\&=3\left[e^{4}-e^{0}\right]\\&=3\left[e^{4}-1\right]\end{aligned}](https://tex.z-dn.net/?f=%5Cend%7Baligned%7D%5C%5C%26%3D%5Cint_%7B0%7D%5E%7B4%7D%20%5Cint_%7B0%7D%5E%7B3%7D%5Cleft%282%20e%5E%7Bx%7D-e%5E%7Bx%7D%5Cright%29%20d%20y%20d%20x%5C%5C%26%3D%5Cint_%7B0%7D%5E%7B4%7D%20%5Cint_%7B0%7D%5E%7B3%7D%20e%5E%7Bx%7D%20d%20y%20d%20x%5C%5C%26%3D3%20%5Cint_%7B0%7D%5E%7B4%7D%20e%5E%7Bx%7D%20d%20x%5C%5C%26%3D3%5Cleft%5Be%5E%7Bx%7D%5Cright%5D_%7B0%7D%5E%7B4%7D%5C%5C%26%3D3%5Cleft%5Be%5E%7B4%7D-e%5E%7B0%7D%5Cright%5D%5C%5C%26%3D3%5Cleft%5Be%5E%7B4%7D-1%5Cright%5D%5Cend%7Baligned%7D) 
  
In conclusion, the line integral along the given positively oriented curve.
![=3\left[e^{4}-1\right]](https://tex.z-dn.net/?f=%3D3%5Cleft%5Be%5E%7B4%7D-1%5Cright%5D)
Read more about line integral
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Answer:
-516
Step-by-step explanation:
 
        
             
        
        
        
Image: 
S= -10,7
R: -14,4
T: -5,4
Pre-image 
R: -10,-2
S: -6,1
T: -1,-2
The vertices of the image move 4 units to the rigth and 6 units down.
        
             
        
        
        
Answer:
The time was actually showing 15:01 so it takes them 20 minutes i think?