Answer:
if im correct i think the answer is b
Step-by-step explanation:
-3/8 = -0.375
-5/8 = -0.625
-1/8 = -0.125
1/4 = 0.25
<span>0.5 = 0.5
</span>
Therefore 1/4 & 0.5 & -1/8 > -3/8
If you're using the app, try seeing this answer through your browser: brainly.com/question/3000586——————————
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.
—————
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.
—————
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.
—————
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. =)
???????????I don't know exactly what you mean
Given:
The vertices of the rectangle ABCD are A(0,1), B(2,4), C(6,0), D(4,-3).
To find:
The area of the rectangle.
Solution:
Distance formula:

Using the distance formula, we get




Similarly,





Now, the length of the rectangle is
and the width of the rectangle is
. So, the area of the rectangle is:




Therefore, the area of the rectangle is 20 square units.