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-4 + x = 6
x - 6 = -12
x/5 + 3 = 2
5x - 7 = -12
Solve for r: 3r = 12
Solve for x: x ÷ 2 = 3
7 = x + 12
x - 2 = 5
r/5 - 6 = -1
-5x + 2 = 67
5y - 10 = -25
x/5 = 1
4 + x/7 = 2
2 - 4x = 14
Solve for the variable. 8 = y + 3
Solve for the variable. 8 = y - 3
4 - 2x = 10

step-by-step explanation:

we can use this expression to calculate the energy gain as a particle moves with a field or the energy loss as it moves against it. (note that it is also used to define the electron-volt, a unit of energy equal to that gained by an electron as it falls through a pd of 1 v, i.e. w = 1 ev = 1.6 10-19 c 1 v = 1.6 10-19j).in fact this expression is really nothing more than the statement that potential is the potential energy per unit charge. in this case, v= w/qis simply stating that the change in potential energy(or work done) per unit charge is equal to the change in potential (or potential difference), v.student questions: uniform electric fields2electron beamcathode+5 kvanode0 v

the answer to your question is minor arc.

i think the answer is a.( -3, -1 )