the answer would be A.
7x × X gives you 7x^2
7x × 2 gives you 14 X
so the answer is the square root of 7x^2 + 14 X
Answer:
L.C.M of 432 and 486: 3888
H.C.F of 432 and 486: 54
Step-by-step explanation:
(x - 5)^2 =
(x - 5)(x - 5) =
x^2 - 5x - 5x + 25 =
x^2 - 10x + 25 <===
Answer: There are 32 students who like brussels sprouts but dislike beans.
Step-by-step explanation:
Since we have given that
Number of students eat in the cafeteria = 120
Fraction dislike lime beans = 
Fraction who dislike lime beans and brussles sprouts both = 
Fraction who dislikes lime beans but not brussles is given by
So, Number of students who like brussels sprouts but dislike lima beans is given by
Hence, there are 32 students who like brussels sprouts but dislike beans.
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.
