What is the value of x in the figure below? In this diagram, ABD ~ CAD. A. 16/5
B. 16
C. √160
D. √5
E. 8
F. 25/4

F an amazing result.

Step-by-step explanation:

I'm sure I didn't do this the way it was intended, but I think I got it right.

First of all, we have to agree that CAB is marked as a right angle. If you do not agree, leave a note.

Step One

Find the length of AC

BC = 16

AB = 10

AC^2 = BC^2 - AB^2

AC^2 = 16^2 - 10^2

AC^2 = 256 - 100

AC^2 = 156

AC = SQRT(156)

AC = SQRT(4*13 * 3)

AC = 2 SQRT(13)

Step Two

Find the height (AD)

There are two ways to find the area of the large triangle ABC

Area = 1/2 * 16 * AD          and

Area = 1/2 AC * 10            The two legs.

Since these areas must be equal, equate them. You are looking for AD

16 * AD = 2*sqrt(39) * 10   Divide by 16

AD =  20* sqrt(39)/16

AD = 5 * sqrt(39) / 4

Step 3

Find x or DB

DB^2 = 10^2 - [(5 *sqrt(39)/4]^2

DB^2 = 100 - 25*39/16

DB^2 = 100 - 60.9375

DB^2 = 39.0625

DB = SQRT(39.0625)

DB = 6 1/4

6 1/4 is the same thing as 25/6 which is F

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