Answer:
1947
Step-by-step explanation:
This is the answer
The squared term and the function are symmetrical about the value of x that makes (x+5)^2 = 0. That value is ...
D. x = -5
Not to fear, BubbleSmartie11 is here!
Ok, lets get going. I like how you want explanation, not answer. Great job trying to figure it out on your own, now just going for the answer. Ok, first, we're going to do this.
.45 x 21.95. Whatever your answer is, don't forget to readjust the decimal point. Text it to me for me to correct and check it. Have fun!
From One Smartie to Another- BubbleSmartie11
Total Surface Area: 310 sq.in.
50+50+35+35+70+70 (areas of all separate rectangles) = 310.
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.
