Brian, Chris, and Damien took a math test that had 20 questions. The number of questions Brian got right is 14 more than the number of questions Chris got right. Damien correctly answered 2 less than the number of questions Chris answered correctly. If Brian and Damien have the same score, which statement is true?

C. Brian and Damien both answered 2 more questions correctly than Chris did.

Step-by-step explanation:

A. Brian and Damien both answered 2 fewer questions correctly than Chris did.

B. Brian and Damien both answered 4 more questions correctly than Chris did.

C. Brian and Damien both answered 2 more questions correctly than Chris did.

D. Brian and Damien both answered 4 fewer questions correctly than Chris did.

Let

B = number of questions Brian got right

C = number of questions Chris got right

D = number of questions Damien got right

The number of questions Brian got right is 14 more than 1/4 the number of questions Chris got right.

B = 14 + (1/4) * C

Damien correctly answered 2 less than 5/4 the number of questions Chris answered correctly.

D = (5/4) C - 2

If Brian and Damien have the same score,

14 + (1/4) * C = (5/4) C - 2

14 + 1/4C = 5/4C - 2

14 + 2 = 5/4C - 1/4C

16 = 5C-C/4

16 = 4/4C

C = 16

B = 14 + (1/4) * C

= 14 + 1/4*16

= 14 + 4

= 18

D = (5/4) C - 2

= 5/4*16 - 2

= 5*4 - 2

= 20 - 2

= 18

12 servings. 4 times 3= 12

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