Simplify to create an equivalent expression. \qquad{-4(z+3)-4(5-4z)}−4(z+3)−4(5− 4z)minus, 4, left parenthesis, z, plus, 3, right parenthesis, minus, 4, left parenthesis, 5, minus, 4, z, right parenthesis Choose 1 Choose 1 (Choice A) A -24z-32−24z−32minus, 24, z, minus, 32 (Choice B) B 12z+812z+812, z, plus, 8 (Choice C) C 12z-3212z−3212, z, minus, 32 (Choice D) D 12z+3212z+32

D) 12z -32

Step-by-step explanation:

Simplify to create an equivalent expression.

-4(z+3)-4(5-4z)}−4(z+3)−4(5−4z)

Choose 1

A) A -24z-32

B) B 12z+8

C) C 12z-32

D) D 12z+32

-4(z+3)-4(5-4z)

= -4z -12 -20 + 16z

= 12z -32 (D)

0.02

step-by-step explanation:

time of 9 192 631 770 periods of the radiation corresponding to the transition between the 2 hyperfine levels of the ground state of the caesium 133 atom at 0 k (but other seconds are sometimes used in astronomy)

i dont get the question

step-by-step explanation: