Solve each compound inequality (show your work!)

Mathematics

1 answer:

zimovet [89]3 years ago

4 0

Answer: -1 < x < 8

x = 3

x ≠ 2

Step-by-step explanation:

Isolate x in the middle. Perform operations to all 3 sides.

-6 < 2x - 4 < 12

+4 +4 +4

-2 < 2x < 16

÷2 ÷2 ÷2

-1 < x < 8

**************************************************************************

Isolate x. Solve each inequality separately. Remember to flip the sign when dividing by a negative.

4x ≤ 12 and -7x ≤ 21

÷4 ÷4 ÷-7 ÷-7

x ≤ 3 and x ≥ 3

Since it is an "and" statement, x is the intersection of both inequalities.

When is x ≤ 3 and ≥ 3? when x = 3

****************************************************************************

Isolate x. Solve each inequality separately.

15x > 30 or 18x < -36

÷15 ÷15 ÷18 ÷18

x > 2 or x < 2

Since it is an "or" statement, x is the union of both inequalities.

When we combine the inequalities, x is every value except 2.

x ≠ 2

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ki77a [65]

1. Sadie should note that $54a^7b^3\ge 0$ and if $a\ge 0$ , then $b\ge 0$ .

2. While simplifying Sadie makes some mistakes (it must be $3^2$ instead of $3^3$ and she missed the sign of absolute value for b). The right way to find the correct answer is:

$\sqrt{54a^7b^3}=\sqrt{3^2\cdot 6\cdot a^2\cdot a^5\cdot b^2\cdot b} =3\cdot a\cdot |b| \sqrt{6\cdot a^5\cdot b}=3ab\sqrt{6a^5b}$ .

3. The simpliest expression is $3ab\sqrt{6a^5b} =3ab\sqrt{6\cdot a^2\cdot a^2\cdot a\cdot b} =3a^3b\sqrt{6ab}$ .