![\bf -7x-2y=4\implies -2y=7x+4\implies y=\cfrac{7x+4}{-2}\implies y=\cfrac{7x}{-2}+\cfrac{4}{-2} \\\\\\ y=\stackrel{\stackrel{m}{\downarrow }}{-\cfrac{7}{2}} x-2\qquad \impliedby \begin{array}{|c|ll} \cline{1-1} slope-intercept~form\\ \cline{1-1} \\ y=\underset{y-intercept}{\stackrel{slope\qquad }{\stackrel{\downarrow }{m}x+\underset{\uparrow }{b}}} \\\\ \cline{1-1} \end{array} \\\\[-0.35em] ~\dotfill](https://tex.z-dn.net/?f=%5Cbf%20-7x-2y%3D4%5Cimplies%20-2y%3D7x%2B4%5Cimplies%20y%3D%5Ccfrac%7B7x%2B4%7D%7B-2%7D%5Cimplies%20y%3D%5Ccfrac%7B7x%7D%7B-2%7D%2B%5Ccfrac%7B4%7D%7B-2%7D%20%5C%5C%5C%5C%5C%5C%20y%3D%5Cstackrel%7B%5Cstackrel%7Bm%7D%7B%5Cdownarrow%20%7D%7D%7B-%5Ccfrac%7B7%7D%7B2%7D%7D%20x-2%5Cqquad%20%5Cimpliedby%20%5Cbegin%7Barray%7D%7B%7Cc%7Cll%7D%20%5Ccline%7B1-1%7D%20slope-intercept~form%5C%5C%20%5Ccline%7B1-1%7D%20%5C%5C%20y%3D%5Cunderset%7By-intercept%7D%7B%5Cstackrel%7Bslope%5Cqquad%20%7D%7B%5Cstackrel%7B%5Cdownarrow%20%7D%7Bm%7Dx%2B%5Cunderset%7B%5Cuparrow%20%7D%7Bb%7D%7D%7D%20%5C%5C%5C%5C%20%5Ccline%7B1-1%7D%20%5Cend%7Barray%7D%20%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill)

now, what's the slope of a line parallel to that one above? well, parallel lines have exactly the same slope.
Answer:
<h2>90°</h2>
Step-by-step explanation:
First you must calculate the module or the magnitude of both vectors
The module of u is:

The module of v is:

Now we calculate the scalar product between both vectors

Finally we know that the scalar product of two vectors is equal to:

Where
is the angle between the vectors u and v. Now we solve the equation for 

the answer is 90°
Whenever the scalar product of two vectors is equals to zero it means that the angle between them is 90 °
Answer:
c = 7 (if you have more questions like this in the future, use photomath (an app) or mathpapa algebra calculator (a website))
It should be the second option i’m positive