Simplify each of the problems.

a·a²·a³

(6x²⁵)

(3x³)(3x⁴²)

18c³ ÷ -3c²

9a³b⁵ ÷ -3ab²

(x²)⁷

(-2x⁴)⁵

8x⁰

6x·3x⁵·x⁰

-18

-27

-6

-3

-32

8

6x .3x^{5} . x⁰    = 18 x^{6}

Step-by-step explanation:

-18

-27

-6

-3

-32

8

6x .3x^{5} . x⁰  =   = 18 x^{6}

When we multiply powers then they get added like in the first question.

When they are powered further then they're multiplied.

1) a. a² a³ = a⁶

2) (6 ) (-3 )  =  -18

3)  (3 ) (3 ) (-3 ) =      =-27

4)  18c³ ÷ -3c²  =  -6c

5) 9a³b⁵ ÷ -3ab² =      = -3 a²b³

6) ( )⁷ =  

7) ( -2 [ ] )⁵ =   = -2

8) 8x⁰ = 8

9) 6x .3 . x⁰  =   = 18

 Step-by-step explanation:

1) a . a².a³

= a¹ x a²x a³

= a¹⁺²⁺³

= a⁶

2) (6 ) (-3 )

   = -18

    =  -18

3) (3 ) (3 ) (-3 )

    = -27

    =-27

4) 18c³ ÷ -3c²

    =  - c

     =  -6c

5) 9a³b⁵ ÷ -3ab²

    = -3 a²b³

6) ( )⁷

 =

7) ( -2 [ ] )⁵

 = -2

8) 8x⁰ = 8 x 1 = 8

9) 6x .3 . x⁰

  = 18

9) 6x .3 . x⁰

 = 18

maybe - definition of supplementary angles. im not sure. sorry

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