How many real solutions does j2+75=0
1 answer:
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<h3>Answer: Choice C</h3>
{x | x < -12 or x > -6}
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Explanation:
Let's solve the first inequality for x.
(-2/3)x > 8
-2x > 8*3
-2x > 24
x < 24/(-2)
x < -12
The inequality sign flips when we divide both sides by a negative value.
Let's do the same for the second inequality.
(-2/3)x < 4
-2x < 4*3
-2x < 12
x > 12/(-2)
x > -6
The conclusion of each section is that x < -12 or x > -6 which points us to <u>choice C</u> as the final answer.
Side note: The intervals x < -12 and x > -6 do not overlap in any way. There's a gap between the two pieces. We consider these intervals to be disjoint. The number line graph is below.
In this question, we have to write the set of points from -6 to 3 but excluding -2 and 3 as a union of intervals .
For union, we use U.
For the points that we have to exclude , we put parenthesis on there side that is ().
Therefore the required interval form is
As we see that () are used with -2 and 3 . And that's the required interval form .