Find the missing factor 5y^2-4y-9=(y+1)

y = {1,2}

Step-by-step explanation:

5y^2 - 4y - 9 = y + 1

5y^2 - 5y - 10 = 0

5(y^2 - y - 2) = 0

5 (y - 2) (y - 1) = 0

y = {1,2}

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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