Point Q'Q ′
Q, prime is the image of Q(-7,-6)Q(−7,−6)Q, left parenthesis, minus, 7, comma, minus, 6, right parenthesis under the translation (x, y)\to(x+12,y+8)(x, y)→(x+12,y+8)left parenthesis, x, comma, y, right parenthesis, \to, left parenthesis, x, plus, 12, comma, y, plus, 8, right parenthesis.
What are the coordinates of Q'Q
′
Q, prime?
((left parenthesis
,,comma
))

let's solve your equation step-by-step.

4x+2=2x+12

step 1: subtract 2x from both sides.

4x+2−2x=2x+12−2x

2x+2=12

step 2: subtract 2 from both sides.

2x+2−2=12−2

2x=10

step 3: divide both sides by 2.

2x

2

=

10

2

x=5

step-by-step explanation:

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

idk

step-by-step explanation: