The area of a rectangle is 52 in². the length of the rectangle is 9 in. longer than the width. what is the width of the rectangle? 3 in. 4 in. 6 in. 8 in.

Givens

Let the width = x

Let the length = x + 9

Area = 52 in^2

Formula

Area = L * W

Solve

52 = (x + 9)(x )            Remove the brackets.

52 = x^2 + 9x             Subtract 52 from both sides

x^2 + 9x - 52 = 0        Factor

You need 2 numbers that add to 9 and multiply to 52. There aren't many

52 and 1

26 and 2

13 and 4

All three will multiply to 52. 52 = 52 + 1 or 52 - 1 (The larger number must be plus). 53 or 51 is not the sum. 52 and 1 is not the answer.

26 and 2 will add or subtract to 28 or 24. Neither 1 is 9.  So that can't be your answer.

13 and 4 will give 9 or 17. There's your answer.

(x + 13)(x - 4) = 0

x = - 13 is not the answer. You cannot have minus distances.

x - 4 = 0

x = 4

The width is 4 inches  <<<< answer.

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