Outside a home, there is a 10-key keypad with the letters A, B, C, D, E, F, G, H, I, and J that can be used to open the garage if the correct ten-letter code is entered. Each key may only be used once. How many codes are possible? The number of possible codes is:

3,628,800 (aka "10!")

Step-by-step explanation:

The first letter in the code can be 10 possible letters.

Since every letter can only be used once, the second letter can only be 9 possible letters.

E.G.: If the first letter is A, the second letter is B-J.

The third letter would be 8 possible letters, than the fourth would be 7, etc.

In math form, figuring out the possible combinations would be written out as so:

10 × 9 × 8 × 7 × 6 × 5 × 4 × 3 × 2 × 1

An easier way to write this out would be using "10!", which stands for "10 Factorial"

The result of the above equals 3,628,800.

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