Two question pls solve very fast step by step

(1594/29)  - (-233)/14

Step-by-step explanation:

Simplify the following:

(6×7 - 1 + 16 - 1) - 1/29 - 1

6×7 = 42:

(42 - 1 + 16 - 1) - 1/29 - 1

42 - 1 + 16 - 1 = (42 + 16) - (1 + 1):

((42 + 16) - (1 + 1)) - 1/29 - 1

| 4 | 2

+ | 1 | 6

| 5 | 8:

(58 - (1 + 1)) - 1/29 - 1

1 + 1 = 2:

(58 - 2) - 1/29 - 1

| 5 | 8

- | | 2

| 5 | 6:

(56) - 1/29 - 1

(56) - 1/29 - 1 = (56 - 1/29 - 1):

(56 - 1/29 - 1)

Put 56 - 1/29 - 1 over the common denominator 29. 56 - 1/29 - 1 = (29×56)/29 - 1/29 - 29/29:

((29×56)/29 - 1/29 - 29/29)

| | 5 | 6

× | | 2 | 9

| 5 | 0 | 4

1 | 1 | 2 | 0

1 | 6 | 2 | 4:

(1624/29 - 1/29 - 29/29)

1624/29 - 1/29 - 29/29 = (1624 - 1 - 29)/29:

((1624 - 1 - 29)/29)

1624 - 1 - 29 = 1624 - (1 + 29):

((1624 - (1 + 29))/29)

| 1 |  

| 2 | 9

+ | | 1

| 3 | 0:

((1624 - 30)/29)

| | 5 | 12 |  

| 1 | 6 | 2 | 4

- | | | 3 | 0

| 1 | 5 | 9 | 4:

 (1594/29)

Simplify the following:

-7 - 10/(-28) - 10

The gcd of -10 and -28 is -2, so (-10)/(-28) = (-2×5)/(-2×14) = (-2)/(-2)×5/14 = 5/14:

-7 + 5/14 - 10

Put -7 + 5/14 - 10 over the common denominator 14. -7 + 5/14 - 10 = (14 (-7))/14 + 5/14 + (14 (-10))/14:

(14 (-7))/14 + 5/14 + (14 (-10))/14

14 (-7) = -98:

(-98)/14 + 5/14 + (14 (-10))/14

14 (-10) = -140:

(-98)/14 + 5/14 + (-140)/14

-98/14 + 5/14 - 140/14 = (-98 + 5 - 140)/14:

(-98 + 5 - 140)/14

-98 + 5 - 140 = 5 - (98 + 140):

(5 - (98 + 140))/14

| 1 | |  

| 1 | 4 | 0

+ | | 9 | 8

| 2 | 3 | 8:

(5 - 238)/14

5 - 238 = -(238 - 5):

(-(238 - 5))/14

| 2 | 3 | 8

- | | | 5

| 2 | 3 | 3:

(-233)/14

i believe the answer is letter c) 8

step-by-step explanation:

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