Answer:
I think this is right? The instructions seem a bit vague to me...
Condition A: Many
Condition B: One
Condition C: One
Condition D: Many
Condition E: Many
Condition F: One
Step-by-step explanation:
So Condition A: A rectangle with 4 right angles is going to be in many quadrilaterals because squares are also a rectangle with 4 right angles. So condition A has 2 or maybe more that I forgot about.
Condition B: A square with one side measuring 5 inches is one quadrilateral. Squares are one specific type of quadrilateral so they're probably going to go into one quadrilateral
Condition C: A rhombus can sometimes also be a square, but since in this case, it's angle is only 43°, it should be one quadrilateral.
Condition D: There are a lot of parallelograms. Just because one angle is 32°, it doesn't mean there is just one. There are rhombuses and trapezoids that have 32° angles that would be parallelograms. This one is many.
Condition E: This one was hard for me to visualize, but I'm pretty sure this one is also many. There are at least 2. One side could be 6 inches and another could be 8 inches and then, it could be vice versa, so there are probably more than one. This one is many.
Condition F: Is this even possible? I think it's one.
The possible digits excluding 0 are the following:
1,2,3,4,5,6,7,8,9. From the possible digits, the only pair with a
difference of 8 is 1 and 9 but since it is said that the tens digit is larger
so the number must be 91.
Answer:
nah im just gonna take ur points fam
Step-by-step explanation:
Answer:
Question 3: (D) All real numbers. Question 4: (B) y > 3.
