Hope this helps! Don’t mind the part I crossed off.
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The answer is option
D) r < 5 or r > – 1.
I'm going to graph each inequality below on a number line.
A) r > 5 or r > – 1.

The result is found just by joining those two intervals together. Actually that compound inequality only implies
r > – 1which does not represent all real numbers.
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B) r > 5 or r < – 1.

Numbers between – 1 and 5 (including them) are not included in the union, so you don't have all real numbers represented there either.
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C) r < 5 or r < – 1.

Numbers that are greater or equal to 5 are not in the union. So it does not represent all real numbers.
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D) r < 5 or r > – 1.
Now
all real numbers are included in the union. So this is the right choice.
Answer:
option D) r < 5 or r > – 1.
I hope this helps. =)
Answer:
790
Step-by-step explanation:
You put 5 instead of x and you calculat
6*5^3 +8*5= 750+40=790
Hello:<span>
the equation is : y = ax+b
the slope is a : a×(-3/2) = -1......(
perpendicular to a line with a slope of -3/2)
a = 2/3 y=(2/3)x+b
the line that passes through (-2, -2) :- 2 =
(2/3)(-2)+b
b = -2/3
<span> the equation is : y = (2/3)x-2/3</span></span>
<span><span>y =(2/3)(x-1)</span></span>