Find the center, vertices, and foci of the ellipse with equation x squared divided by one hundred plus y squared divided by sixty four = 1.

Step-by-step explanation:

The general equation of an ellipse is:

(x − h)²/a² + (y − k)²/b² = 1

where (h, k) is the center.

If a > b, the ellipse is horizontal, so the vertices are at (h ± a, k), and the foci are at (h ± c, k), where c² = a² − b².

If b > a, the ellipse is vertical, so the vertices are at (h, k ± b), and the foci are at (h, k ± c), where c² = b² − a².

x²/100 + y²/64 = 1

The center is (0, 0).

This is a horizontal ellipse, so the vertices are at (-10, 0) and (10, 0).

The distance between the foci and the center is c = √(100 − 64) = 6, so the foci are at (-6, 0) and (6, 0).

Graph: desmos.com/calculator/kd6z9qvhi0

the correct answer is 5

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