Mathematics, 22.10.2019 04:30 2023greenlanden

Consider each of the following compound sentences:
x < 1 and x > -1 x < 1 or x >
does the change of the word from "and" to "or" change the solution set?
use number line graphs to support your answer.

Answers: 2

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Answer from: LuluMathLover101

Yes, see explanation below.

Step-by-step explanation:

First let's solve the first inequality:

x < 1 and x > -1

This means that x needs to fulfill these two inequalities at the same time, therefore, the solution set will be:

-1 < x < 1

The graphic is below and it is the interval (-1,1)

On the other hand the other inequality is

x< 1 or x > -1

So x should fulfill any of these two conditions and the solution set is the interval ( -∞, 1) ∪ (-1, +∞) which is all the real numbers (-∞, +∞)

The graphic line will be all Real numbers

The "and" means we're looking for the intersection of the solution sets. The "or" means we're looking for the union of the solution sets. So, yes, the use of these words change the solution set

Consider each of the following compound sentences: x < 1 and x > -1 x < 1 or x > d

Answer from: Quest

step-by-step explanation:

most likely the answer is a.

alice wins if a total of 5 is rolled. finn wins if a total of 9 is rolled.

Answer from: Quest

x= 1

step-by-step explanation: convert 5\frac{11}{12}5

to improper fraction. use this rule: a \frac{b}{c}=\frac{ac+b}{c}a

ac+b

3.25+0.5x+1\frac{1}{3}+\frac{5}{6}x=\frac{5\times 12+11}{12}3.25+0.5x+1

5×12+11

2 simplify 5\times 125×12 to 6060

3.25+0.5x+1\frac{1}{3}+\frac{5}{6}x=\frac{60+11}{12}3.25+0.5x+1

60+11

3 simplify 60+1160+11 to 7171

3.25+0.5x+1\frac{1}{3}+\frac{5}{6}x=\frac{71}{12}3.25+0.5x+1

4 simplify \frac{5}{6}x

x to \frac{5x}{6}

3.25+0.5x+1\frac{1}{3}+\frac{5x}{6}=\frac{71}{12}3.25+0.5x+1

5 convert 1\frac{1}{3}1

to improper fraction. use this rule: a \frac{b}{c}=\frac{ac+b}{c}a

ac+b

3.25+0.5x+\frac{1\times 3+1}{3}+\frac{5x}{6}=\frac{71}{12}3.25+0.5x+

1×3+1

6 simplify 1\times 31×3 to 33

3.25+0.5x+\frac{3+1}{3}+\frac{5x}{6}=\frac{71}{12}3.25+0.5x+

3+1

7 simplify 3+13+1 to 44

3.25+0.5x+\frac{4}{3}+\frac{5x}{6}=\frac{71}{12}3.25+0.5x+

8 simplify 3.25+0.5x+\frac{4}{3}+\frac{5x}{6}3.25+0.5x+

to \frac{13.75}{3}+\frac{4x}{3}

13.75

\frac{13.75}{3}+\frac{4x}{3}=\frac{71}{12}

13.75

9 simplify \frac{13.75}{3}

13.75

to 4.5833334.583333

4.583333+\frac{4x}{3}=\frac{71}{12}4.583333+

10 subtract 4.5833334.583333 from both sides

\frac{4x}{3}=\frac{71}{12}-4.583333

−4.583333

11 simplify \frac{71}{12}-4.583333

−4.583333 to \frac{4}{3}

\frac{4x}{3}=\frac{4}{3}

12 multiply both sides by 33

4x=\frac{4}{3}\times 34x=

×3

13 simplify \frac{4}{3}\times 3

×3 to \frac{12}{3}

4x=\frac{12}{3}4x=

14 simplify \frac{12}{3}

to 44

4x=44x=4

15 divide both sides by 44

x=1

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Consider each of the following compound sentences:
x < 1 and x > -1 x < 1 or x &g...