If EBCD is a parallelogram, EB = 16, ED = 25, BF = 11, EC = 34, mZBED = 55°, mZCDB = 67°, and mZBCE = 24º, find each missing measure.

B

с

BC =

mZEDC =

BD =

mZEBD =

F

FC =

E

D

mZBEC =

mZDBC =

CD =

1. BC = 25

2. m∠EDC = 125°

3) BD = 22

4) m∠EBD = 67°

5) FC = 17

6) m∠BEC = 24°

7) CD = 16

8) m∠DBC =  58°

Step-by-step explanation:

The dimensions of parallelogram EBCD are;

EB = 16, ED = 25, BF = 11, EC = 34, m∠BED = 55°, m∠CDB = 67°

1. BC = ED = 25 Opposite sides of a parallelogram are equal

2. m∠BCD = m∠BED = 55 opposite angles of a parallelogram are equal

m∠BCD + mEDC = 180° adjacent sides of a parallelogram are supplementary

m∠EDC = 180° - m∠BCD

∴ m∠EDC = 180° - 55° = 125°

m∠EDC = 125°

3) BD = BF × 2 Diagonals of a parallelogram bisect each other

∴ BD = 11 × 2 = 22

BD = 22

4) m∠EBD = m∠CDB = 67° from alternate angles between parallel lines

5) FC = EC/2 from diagonals of a parallelogram bisect each other

∴ FC = EC/2 = 34/2 = 17

FC = 17

6) m∠BEC = m∠BCE = 24° Alternate angles between parallel lines

m∠BEC = 24°

7) CD = EB = 16 Opposite sides of a parallelogram are equal

8) m∠EBC = m∠EDC = 125° from opposite angles of a parallelogram are equal

m∠DBC + m∠EBD = m∠EBC from angle addition postulate

∴ m∠DBC = m∠EBC - m∠EBD

m∠DBC = 125° - 67° = 58°

m∠DBC =  58°

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