Which compound inequality has no solution? x + 5 < 5 and > 1 3(x + 4) < 12 or 3x + 1 > 10 10 < 3(x – 2) + 1 or < 2 11 ≤ 5(x + 3) – 9 and 5(x + 3) – 9 < 21

Checking the compound inequalities the first one has no solution:

x + 5 < 5 and > 1

Step-by-step explanation:

1) x+5<5 and >1

Subtracting 5 both sides of the inequality:

x+5-5<5-5 and >1

x<0 and >1

Solution = (-Infinite, 0) ∩ (1, Infinite)

Solution = ∅ = { } (Empty set)

This compound inequality has no solution

2) 3(x + 4) < 12 or 3x + 1 > 10

3(x) +3(4) < 12 or 3x + 1-1 > 10-1

3x +12 < 12 or 3x > 9

3x +12-12 < 12-12 or 3x/3 > 9/3

3x < 0 or x > 3

3x/3 < 0/3 or x > 3

x < 0 or x > 3

Solution = (-Infinite, 0) U (3, Infinite)

3) 10 < 3(x – 2) + 1 or < 2

10 < 3(x) – 3(2) + 1 or < 2

10 < 3x – 6 + 1 or < 2

10 < 3x – 5 or < 2

10+5 < 3x – 5+5 or < 2

15 < 3x or < 2

15/3 < 3x/3 or < 2

5 < x or < 2

x>5 or <2

Solution =  (5, Infinite) U (-Infinite, 2)

Solution = (-Infinite, 2) U (5, Infinite)

4) 11 ≤ 5(x + 3) – 9 and 5(x + 3) – 9 < 21

11 ≤ 5(x) + 5(3) – 9 and 5(x) + 5(3) – 9 < 21

11 ≤ 5x + 15 – 9 and 5x + 15 – 9 < 21

11 ≤ 5x + 6 and 5x + 6 < 21

11 -6 ≤ 5x + 6 -6 and 5x + 6 - 6 < 21 -6

5 ≤ 5x  and 5x < 15

5/5 ≤ 5x/5  and 5x/5 < 15/5

1 ≤ x  and x < 3

x ≥ 1  and x < 3

Solution = [1, Infinite) ∩ (-Infinite, 3)

Solution = [1, 3)

Checking the compound inequalities the first one has no solution:

x + 5 < 5 and > 1

Step-by-step explanation:

question

why are the following expressions not monomials?

1) 3cd^x

a. the variable is in the denominato

2) x+2w

c. the expression is not a product.

3) 3/h

a. the variable is in the denominator

4) ab^-1

c. the expression is not a product.

hope this !