This seems to be referring to a particular construction of the perpendicular bisector of a segment which is not shown. Typically we set our compass needle on one endpoint of the segment and compass pencil on the other and draw the circle, and then swap endpoints and draw the other circle, then the line through the intersections of the circles is the perpendicular bisector.
There aren't any parallel lines involved in the above described construction, so I'll skip the first one.
2. Why do the circles have to be congruent ...
The perpendicular bisector is the set of points equidistant from the two endpoints of the segment. Constructing two circles of the same radius, centered on each endpoint, guarantees that the places they meet will be the same distance from both endpoints. If the radii were different the meets wouldn't be equidistant from the endpoints so wouldn't be on the perpendicular bisector.
3. ... circles of different sizes ...
[We just answered that. Let's do it again.]
Let's say we have a circle centered on each endpoint with different radii. Any point where the two circles meet will then be a different distance from one endpoint of the segment than from the other. Since the perpendicular bisector is the points that are the same distance from each endpoint, the intersection of circles with different radii isn't on it.
4. ... construct the perpendicular bisector ... a different way?
Maybe what I first described is different; there are no parallel lines.
A. or "Company A"
The reason is because its the biggest and longest box on the table
38.5 mm is greater than 10 cm.
HOPE THIS HELPED
:)
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Answer:
- (f×g)(x) = 2x^2 +2x
- (f×g)(x) = 6y^2 -11y -35
Step-by-step explanation:
The distributive property applies.
1. (fg)(x) = f(x)·g(x) = (2x)·(x +1)
(f×g)(x) = 2x^2 +2x
__
2. (fg)(x) = f(x)·g(x) = (2y -7)·(3y +5) = 2y(3y +5) -7(3y +5)
(f×g)(x) = 6y^2 -11y -35
_____
<em>Additional comment</em>
Please note in problem 2 that the function argument is x and the variable in the given expressions is y. This means the function value is exactly and only <em>6y^2 -11y -35 for any value of x</em>. It does not change when x changes. (We suspect a typo.)
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