Answer:
A non-equilateral rhombus.
Step-by-step explanation:
We can solve this graphically.
We start with square:
ABCD
with:
A = (11, - 7)
B = (9, - 4)
C = (11, - 1)
D = (13, - 4)
Only with the vertices, we can see that ABCD is equilateral, as the length of each side is:
AB = √( (11 - 9)^2 + (-7 -(-4))^2) = √( (2)^2 + (3)^2) = √(4 + 9) = √13
BC = √( (11 - 9)^2 + (-1 -(-4))^2) = √13
CD = √( (11 - 13)^2 + (-1 -(-4))^2) = √13
DA = √( (11 - 13)^2 + (-7 -(-4))^2) = √13
And we change C by C' = (11, 1)
In the image you can see the 5 points and the figure that they make:
The figure ABCD is a rhombus, and ABC'D is also a rhombus, the only difference between the figures is that ABCD is equilateral while ABC'D is not equilateral.
B. 17.16 represents a credit of $17.16; "credits" mean that you're receiving something, so it's added. A is incorrect because it shows the money being taken away and that isn't what a credit does. C and D aren't even the same amount of money in the question, so they aren't your answer either.
Answer:
Option B.
Step-by-step explanation:
It is given that
A = {The Rationals}
B = {The Irrationals}
We need to find the set A∪B.
If we have two sets X and Y then union of these sets (X∪Y) contains all the elements of set X, of set Y or both.
It is given that A is the set of rations and B is the set of irrational, so the union A∪B is the combined set of all rational or irrational numbers.
A∪B = {The Rationals} + {The Irrationals}
A∪B = {The Reals}
Therefore, the correct option is B.
