Given: 1 = 2 3 = 4 d midpoint of bc = de prove: a = e 1. ∠1=∠2, ∠3=∠4, d is midpoint of segment be, bc = de cpcte 2. bd = de given 3. bc = bd substitution 4. triangle abd congruent to triangle ebc definition of midpoint 5. ∠a = ∠e asa

Answers:

1. <1=<2, <3=<4, D is midpoint of segment BE, BC=DE → Given

2. BD=DE → Definition of midpoint

3. BC=BD → Substitution property

4. Triangle ABD congruent to triangle EBC → ASA

5. <A=<E → CPCTE

Solution:

1. <1=<2, <3=<4, D is midpoint of segment BE, BC=DE

All this information is given by the problem.

2. BD=DE

If D is the midpoint of segment BE, D divides this segment into two congruent parts BD and DE, then BD must be equal to DE: BD=DE by definition of midpoint.

3. BC=BD

The problem says that BC=DE (1)

And by point 2 we know that BD=DE→DE=BD (2)

Then by substitution property if we can replace DE in equation (1) by BD (because of equation (2) ):

(1) BC=DE and (2) DE=BD → (1) BC=BD

4. Triangle ABD congruent to triangle EBC

The triangles ABD and EBC have a congruent side (BD in triangle ABD and BC in triangle EBC) and the two adjacent angles congruent too (<1 in triangle ABD with <2 in triangle EBC; and <3 in triangle ABD with <4 in triangle EBC), then by Angle Side Angle (ASA) the two triangles must be congruent.

5. <A=<E

If the triangles are congruent (by point 4), all its parts must be congruent too (Corresponding Parts of Congruent Triangles are Equal: CPCTE), then the third angle in triangle ABD (<A) must be equal to the third angle in triangle EBC (<E):

<A=<E

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

see explanation

step-by-step explanation:

a

(6r + 7) + (13 + 7r) ( remove brackets and collect like terms )

= 6r + 7 + 13 + 7r

= 13r + 20

b

remove brackets and collect like terms

13 - r - 1 + r

= 12 - r

c

remove brackets and collect like terms

= - 8 - r + 2r - 4

= - 12 + r

d

remove brackets and collect like terms

= 7r - - - 6r

[-   - = - - = - ]

= r -