The domain is all the positive values and the range is also all the positive values.
Then the answer is the option D) s>0 and V(s) > 0.
I believe you’re correct.
y = 3x - 6
The linear equation y = -6x + 2; -12x - 2y = -4 has no solution
<h3>Linear equation</h3>
y = -6x + 2
-12x - 2y = -4
- Substitute y = -6x + 2 into (2)
-12x - 2y = -4
-12x - 2(-6x + 2) = -4
-12x + 12x -4 = -4
-12x + 12x = - 4 + 4
0 = 0
y = -6x + 2
y - 2 = -6x
x = (y - 2) / -6
-12x - 2y = -4
-12(y- 2)/ 6 - 2y = -4
(-12y + 24) / 6 - 2y = -4
-2y + 4 - 2y = 4
0 = 0
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