Answer:
No solutions
Step-by-step explanation:
Solve for x the system of two inequalities:

First, solve each inequality separately:
1.

2.

Now, we get

and

There are no common values for x, for which both inequalities hold, so the system of two inequalities has no solutions.
<span>{(1, 16), (2, 64), (3, 144), (4, 256), (5, 400)} Hope this helps </span>
Answer:
8 new chicks can be fitted in the coop.
Step-by-step explanation:
A chicken coop holds 10 hens or 20 chicks.
That means space in the coop for 10 hens = space for 20 chicks
Or space for 1 hen = space for 2 chicks
Now
th of the hens were removed from the coop.
So, number of hens removed from the coop = 
= 4 hens
And space for 4 hens in the coop = space for the 8 chicks
Therefore, 8 new chicks can be fitted in the coop.
9|⁻¹/₃z| + 6 < 30
9|⁻¹/₃z| + 6 < ±30
9|⁻¹/₃z| + 6 < 30 or 9|⁻¹/₃z| + 6 > -30
9(¹/₃z) + 6 < 30 or 9(¹/₃z) + 6 > -30
3z + 6 < 30 or 3z + 6 > -30
- 6 - 6 - 6 - 6
3z < 24 or 3z > -24
3 3 3 3
z < 8 or z > -8
Solution Set: {z|z < 8 or z > -8} and (-∞, 8) or (-8, ∞)