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rosijanka [135]
3 years ago
12

How do we show where an inequality is true on the number line! 50 points!

Mathematics
1 answer:
Vinvika [58]3 years ago
3 0
Usually, we use the number line to solve inequalities with the symbols, < , ≤ , >  , and ≥ .(the second and last one was rather hard to find on my keyboard) In order to solve an inequality using the number line, though, just turn the inequality sign to an equal sign. Then, solve the equation. Next step, graph the point on the number line (remember to graph as an open circle if the original inequality was <, or >). The number line should now be divided into 2 regions, one to the left of the graphed point, and one to the right of said point.

After that, pick a point in both regions and "test" it, check to see if it satisfies the inequality when plugged in for the variable. If it does, draw a darker line from the point into that region, with an arrow at the end. That is the solution to the equation: if one point in the region satisfies the inequality, the entire region will satisfy the inequality.

I had to check back in an old textbook to remember all of that. Sorry about the earlier answer. That was rather foolish to do so without actually understanding the question.



You might be interested in
Solve the inequality |6x + 2| &lt; 10 and graph its solution.
mars1129 [50]
This is a modulus inequality.
First part: when (6x + 2) is positive
6x + 2 < 10
6x < 10 - 2
6x < 8
x < 8/6
x < 4/3

Second part: when (6x + 2) is negative.

-(6x + 2) < 10   Divide both sides of inequality by -1 and change the sign.

  (6x + 2) > -10
   6x + 2 > -10
   6x > -10 - 2
   6x > -12        Divide both sides by 6.
     x > -12/6
     x > -2.

Combined solution:   x < 4/3  and x > -2

  -2 < x < 4/3.

Graph is a line on the number line between -2 and 4/3.  
 
-2 and 4/3 are excluded from solution.
8 0
3 years ago
PLEASE HELP
Alecsey [184]

Answer:

The answer to your question is letter B

Step-by-step explanation:

Process

1.- Find two points of each line

Line A    (-2, 0)   (-1, 2)

Line B    (-1. - 5)   (-6, 0)

2.- Find the slope and equation of each line

Line A

             m = \frac{2 - 0}{-1 + 2}

             m = \frac{2}{1}

                    m = 2

             y - 0 = 2(x + 2)

            y = 2x + 4

Line B

             m = \frac{0 + 5}{-6 + 1}

             m = \frac{5}{-5}

                    m = -1

             y - 0 = -1(x + 6)

             y = -x - 6

3.- Find the inequalities

Line A, we are interesteed in the lower area of the line, so the inequality is

                      y ≤ 2x + 4

Line B, we are also interested in the lower area of the line so the inequality is

                      y ≥ - x - 6

5 0
3 years ago
I need help with both of these questions please
tankabanditka [31]

Answer:

\frac{22}{30} = \frac{11}{15}

and

\frac{176}{7} = 25 \frac{1}{7}

Step-by-step explanation:

First make all the fractions into improper fractions

1\frac{5}{6} = \frac{11}{6} \\\\2\frac{1}{2} = \frac{5}{2} \\\\3\frac{1}{7} = \frac{22}{7}

After doing that put them back into the problems

Problem 1)

1\frac{5}{6}  ÷ 2\frac{1}{2} , plug in the improper fractions

\frac{11}{6} ÷ \frac{5}{2}

to divide, you need to flip the second fraction and multiply

\frac{11}{6}· \frac{2}{5} = \frac{22}{30}

then reduce

\frac{22}{30} = \frac{11}{15}

Problem 2)

3\frac{1}{7}÷ \frac{1}{8}\\

Plug in improper fraction

\frac{22}{7} ÷ \frac{1}{8}\\

Then flip second fraction and multiply

\frac{22}{7} · \frac{8}{1}

Multiply

\frac{176}{7}

Reduce, or make it a mixed number

\frac{176}{7} = 25 \frac{1}{7}

3 0
3 years ago
For the equation, decide if it is always true or never true.<br><br> 2(x + 3) = 5x + 6 − 3x
SCORPION-xisa [38]

Answer:

always true

Step-by-step explanation:

Given

2(x + 3) = 5x + 6 - 3x ← distribute left side

2x + 6 = 2x + 6

Since both sides are equal then any real value of x is a solution.

Thus the equation is always true

3 0
2 years ago
0000003.09+0000002.1​
Ivanshal [37]

Answer:

5.19

Step-by-step explanation:

6 0
3 years ago
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