A. Find the length of the midsegment of an equilateral triangle with side lengths of 12.5 cm. b. Given that UT is the perpendicular bisector of line segment AB, where T is on AB, find the length of AT given AT = 3x + 6 and TB = 42 - x.

c. Given angle EFG has angle bisector FH, where EF = GF, find the value of y if EH = 5y + 10 and HG = 28 - y.

a.the length of midsegment of an equilateral triangle with side lengths of 12.5 cm.=12.5/2=6.25cm

b.

sinceUT is the perpendicular bisector of line segment AB, where T is on AB, find the length of AT given AT = 3x + 6 and TB = 42 - x.

than AT=BT

3x+6=42-x

3x+x=42-6

x=36/4=9

AT=3×9+6=27+6=33

c.

since angle EFG has angle bisector FH, where EF = GF, find the value of y if EH = 5y + 10 and HG = 28 - y.

EH=GH

5y+10=28-y

6y=28-10

y=18/6=3

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