which is the equation of a hyperbola with directrices at x = ±3 and foci at (4, 0) and (−4, 0)?
x squared over 16 minus y squared over 4 equals 1
x squared over 12 minus y squared over 4 equals 1
y squared over 16 minus x squared over 4 equals 1
y squared over 12 minus x squared over 4 equals 1

answer: the answer is ann

4 candies

step-by-step explanation:

7 x 28 = 196

200 - 196 = 4 candies remaining.

answer: (c) m∠qpo + (2x + 16)° = 180°

step-by-step explanation:

the opposite angles of a quadrilateral are supplementary.

so, ∠o + ∠q = 180°   and   ∠p + ∠r = 180°

since ∠r = 2x + 16°, we can use substitution as follows:

∠p + ∠r = 180°

∠p + (2x + 16)° = 180°

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