Remember you can add like terms exg, 3y+4y=7y 3y^2+4y^2=7y^2, but cannot add 3y^2 and 4y, or 3 and 4y so 
3a+2w-5a-9w
group like trems
3a-5a+2w-9w
subtract
-2a-7w
3y^2+2w^2-y^2-w^2
group like terms (think of w^2 and y^2 as 1w^2 and 1y^2)
3y^2-y^2+2w^2-w^2
subtract
2y^2+w^2
2x+2x=4x
        
                    
             
        
        
        
Answer:
Question 4: Which equation is parallel to the above equation and passes through the point (35, 30)
 
  is the correct answer, I found this by inputting the x and y value of the coordinate (35, 30) onto the equation and solving for y-intercept since the slope of all equations is the same (since it's traveling parallel)
  is the correct answer, I found this by inputting the x and y value of the coordinate (35, 30) onto the equation and solving for y-intercept since the slope of all equations is the same (since it's traveling parallel) 

so the equation would be  
  
 
Question 5: Which equation is perpendicular to the above equation and passes through the point (35, 30)
 is the correct answer, I found this using the same method as before, input coordinate values into the equation and solve for the y-intercept (The only thing changed from the last answer is the opposite reciprocal slope).
 is the correct answer, I found this using the same method as before, input coordinate values into the equation and solve for the y-intercept (The only thing changed from the last answer is the opposite reciprocal slope).

so the equation would be 
 
        
             
        
        
        
Answer:
7
Step-by-step explanation:
 
        
             
        
        
        
Option B
i(x) = 2x - 4 is the equation of the new function
<em><u>Solution:</u></em>
Given that the function h(x) = 2x-9 is translated up 5 united to become a new function, i(x)
To find: Equation of new function
The graph of a function can be moved up, down, left, or right by adding to or subtracting from the output or the input.
Adding to the output of a function moves the graph up
Therefore,
Given function is:
h(x) = 2x - 9
Translated up by 5 units
<em><u>Therefore, new function is:</u></em>
i(x) = h(x) + 5
i(x) = 2x - 9 + 5
i(x) = 2x - 4
Thus Option B is correct
 
        
             
        
        
        
Probably the intended ellipse is the one with equation

We can parameterize  as a piece of this curve by
 as a piece of this curve by

with  . Then
. Then

etc