Given:

And

Required:
To find the two possible values of c.
Explanation:
Consider

So

And also given

Now from (1) and (2), we get


Now put a in (1) we get

We can interpret that either of a or b are equal to 3 or 5.
When a=3 and b=5, we have

When a=5 and b=3, we have

Final Answer:
The option D is correct.
31 and 41
Answer:
Step-by-step explanation:
There aren't any choices, but I'd guess that she'd use pint or cup.
Well, to write the definition of "Perpendicular lines form right angel" using "if and only if". the sentence would become :
<em>
</em><em>Two lines are perpendicular if and only if they form a right angle
</em><em>
</em><em />Hope this helps