Tenisha solved the equation below by graphing a system of equations.
log35x = log (2x+8)
Which point approximates the solution for Tenisha's system of equations?
(0.9, 0.8)
(1.0, 1.4)
(2.3, 1.1)
(2.7, 13.3)

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

b. would be the answer : )

step-by-step explanation:

step 1: add -16x to both sides.

16x−4y+−16x=64+−16x  

−4y=−16x+64

step 2: divide both sides by -4.

−4y /-4 = -16x + 64/-4

answer:   y=4x−16

hope i'm right