Iwill give brainliest for the person who answers these 3 math questions correctly

1. Ok, so we know that this has no solution. To prove that, we're going to first get one variable on a single side of the equation. For that, we'll do the first equations' x.

3x - 6y = -12 : Add 6y to both sides

3x = 6y - 12 : Divide by 3

x = 2y - 4

Ok, so what we get from that is that x IS 2y-4, so we'll plug that in for the other equation.

2y - 4 - 2y = -8 : Combine like terms

0y - 4 = -8 : Add 4

0y = -4

0 = -4

Ok, so there's no solution to that one. There aren't any variables left, therefore it is unsolvable.

The two lines in this system of equations will never intersect.

2.

2x - 3y = 13

x + 2y = -4 : Let's multiply this equation by negative two, so that the x's will cancel out.

2x - 3y = 13

-2x - 4y = 8 : Now we combine these two equations.

-7y = 21 : Divide by -7

y = -3 : Now we'll plug this -3 in for y in whatever equation. I'll choose the second one.

x + 2(-3) = -4

x + -6 = -4

x = 2

So our y is -3, and our x is 2, so our solution is (2, -3).

3 is beyond my knowledge, but I was happy to help you on 1 and 2.

three-fourths of an hour

36÷48=0.75

0.75=

none of the above.

for one of the equations to be used in the process, there has to be a term in the equation since squared is .

since there is only a term in the first equation, we cannot derive a from completing the square.

solving with completing the square