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Please check the explanation.

Step-by-step explanation:

Part a)

Given

It is clear that m∠ABD and m∠CBD lie on a straight line.

Thus, the sum of m∠ABD and m∠CBD will be 180°.

Hence,

m∠ABD and m∠CBD are supplementary angles.

Part b)

As m∠ABD and m∠CBD are supplementary angles, so the sum of m∠ABD and m∠CBD will be 180°.

Thus,

m∠ABD + m∠CBD = 180°

substitute m∠ABD = (7q-46)° and m∠CBD = (3q + 6)°

(7q-46)° + (3q + 6)° =  180°

Part c)

Given the equation

(7q-46)° + (3q + 6)° =  180°

Combining the like terms

7q + 3q - 46 + 6 = 180

10q - 40 = 180

10q = 180 + 40

10q = 220

divide both sides by 10

10q/10 = 220/10

q = 22

Therefore, the value of q = 22

Part d)

As

m∠ABD = (7q-46)°

substituting q = 22

m∠ABD = 7(22) - 46

             = 154 - 46

             = 108°

also

m∠CBD = (3q + 6)°

substituting q = 22

m∠CBD = 3(22) + 6

              = 66+6

              = 72°

Therefore,

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