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Sveta_85 [38]
3 years ago
7

Parallel to

-4 " alt="y= \frac{3}{4} x-4 " align="absmiddle" class="latex-formula">; through (-4,1)
Mathematics
1 answer:
KIM [24]3 years ago
7 0
Answer:  " y = \frac{3}{4} x + 4 " .
_________________________________________________
Explanation:
_________________________________________________
We are given an equation in "slope intercept form" ; that is; in the form of :

"y = mx + b" ;  in which "y" in isolated on the left-hand side of the equation; with "no-coefficient" (except for the "implied coefficient" of "1");  

                      in which: "m" is the slope of the line; and the coefficient of "x";  and "b" is the "y-intercept" (or the value of the "y-coordinate" of the graph when "x = 0" ;  
______________________________________________
  We are given:  " y = \frac{3}{4}x − 4 " ;  
 
in which the slope; "m", is "\frac{3}{4}" .

Since we want to write the equation, in slope-intercept form, for the line PARALLEL to the given line; we known that the "line" that is "parallel" will have the same slope".

So we can write:  " y = \frac{3}{4} x + b" .  

Note that we are instructed to find the "parallel line" that passed through:

 "(-4, 1)" ; 
______________________________________________________
So, in the aformentioned equation, we substitute "-4" for "x" ; and "1" for "y"; to solve for "b" ;
______________________________________
   y = \frac{3}{4} x + b ;

   1  =  \frac{3}{4} * -4  + b ;

→  1 = -3 + b ;

↔ b + (-3) = 1 ;    

↔ b <span>− 3 = 1 ;  

Add "3" to each side of the equation:
</span>
b − 3 + 3 = 1 + 3 ;

→ b = 4  .  
______________________________
Now, since we now that "b" is "positive 4" ; we can write the equation of the parallel line:

" y = \frac{3}{4} x + 4 " .
__________________________________________________
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Answer:

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Step-by-step explanation:

Given the differential equation

y'' - y' - 12y = 0.

We are required to verify that the functions e^(-3x) and e^(4x) form a fundamental set of solutions of the differential equation on the interval (−[infinity], [infinity]). They form a fundamental set if they are linearly independent, and they are linearly independent if their wronkian is not zero, otherwise, they are linearly dependent.

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