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erik [133]
3 years ago
14

Select the order pair(s) that are solutions to this system of inequalities:

Mathematics
1 answer:
Phoenix [80]3 years ago
5 0
You need to substitute the ordered pairs into the system of equations and pick the ones that makes the system of equations true: 

0 < -3(0) + 2  (True)
0 ≥ 0 - 1         (True) 
----------------
-4 < -3(0) + 2 (True)
-4 ≥ 0 - 1        (False) 
----------------
4 < -3(1) + 2  (False)
4 ≥ 1 - 1         (True) 
----------------
3 < -3(-3) + 1 (True)
3 ≥ -3 - 1        (True)
----------------

The Answer is (0,0) and (-3,3)
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Answer:

points (0,2) and (5,5)

Step-by-step explanation:

2 points define a line.

the line is in the equation format, y=mx+b

b is the y intercept; in this case, the y intercept is 2 (0,2).

Make your first point (0,2).

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If you're confused, you could also substitute any number you want for x, like, for example, x=5.

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7 0
3 years ago
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8 0
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Read 2 more answers
Which number produces an irrational number when multiplied by 1/3?
Georgia [21]

A number multiplied by <span>13</span> only produces an irrational number if it is itself an irrational number.

Explanation:

Suppose there were some rational number which when multiplied by <span>13</span> generated an irrational number.

If the number is rational then, by definition, it can be expressed as <span>ab</span> for some pair of integers a and b.

<span><span>ab</span>×<span>13</span>=<span><span>a×1</span><span>b×3</span></span>=<span>a<span>3b</span></span></span>

Since a is an integer and 
since if b is an integer then <span>3b</span> is also an integer,
then
by definition <span><span>a3</span>b</span> is a rational number, contrary to our supposition that some rational number <span>ab</span> when multiplied by <span>13</span> could generate an irrational number.


4 0
4 years ago
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