The x-axis brainliest plz :)
2x - 6y = 12
so we pick any number for x and then solve for y
lets say x = 0
2(0) - 6y = 12
-6y = 12
y = -12/6
y = -2....so when x = 0, y = -2....(0,-2) <== one point
lets say x = 1
2(1) - 6x = 12
2 - 6x = 12
-6x = 12 - 2
-6x = 10
x = -10/6
x = - 5/3...so when x = 1, y = -5/3....(1,-5/3) <== another point
Lets say x = 2
2(2) - 6y = 12
4 - 6y = 12
-6y = 12 - 4
-6y = 8
y = -8/6
y = - 4/3....so when x = 2, y = -4/3.....(2,-4/3) <== another point
lets say x = 3
2(3) - 6y = 12
6 - 6y = 12
-6y = 12 - 6
-6y = 6
y = -6/6
y = -1....so when x = 3, y = -1.....(3,-1) <== another point
now there is 4 points.
Answer:
take the answer on picture which I attached with your question
Answer:
Step-by-step explanation:
√13.7/(3.05^2+2.6)
Answer: 
<u>Step-by-step explanation:</u>
![\text{Use the distance formula: }d_AB=\sqrt{(x_A-x_B)^2+(y_A-y_B)^2}\\where\ (X_A, y_A)=(-3, -2)\\and\ (x_B,y_B)=(4, -7)\\\\\\d_AB=\sqrt{(-3-4)^2+[-2-(-7)]^2}\\\\.\quad =\sqrt{(-7)^2+(5)^2}\\\\.\quad =\sqrt{49+25}\\\\.\quad =\boxed{\sqrt{74}}](https://tex.z-dn.net/?f=%5Ctext%7BUse%20the%20distance%20formula%3A%20%7Dd_AB%3D%5Csqrt%7B%28x_A-x_B%29%5E2%2B%28y_A-y_B%29%5E2%7D%5C%5Cwhere%5C%20%28X_A%2C%20y_A%29%3D%28-3%2C%20-2%29%5C%5Cand%5C%20%28x_B%2Cy_B%29%3D%284%2C%20-7%29%5C%5C%5C%5C%5C%5Cd_AB%3D%5Csqrt%7B%28-3-4%29%5E2%2B%5B-2-%28-7%29%5D%5E2%7D%5C%5C%5C%5C.%5Cquad%20%3D%5Csqrt%7B%28-7%29%5E2%2B%285%29%5E2%7D%5C%5C%5C%5C.%5Cquad%20%3D%5Csqrt%7B49%2B25%7D%5C%5C%5C%5C.%5Cquad%20%3D%5Cboxed%7B%5Csqrt%7B74%7D%7D)
