(x-3)2+(y-4)2=24
2x-6+2y=24
2x-14+2y=24
2x=24+14-2y
2x=38-2y
x=19-y
3/4 with the main number divide it
Part A.
Ashwin had a 4-cm cube.
volume of cube = side^3
volume = (4 cm)^3 = 64 cm^3
Ashwin has 64 small cubes.
We need to find the volumes of all prisms in Part A. Only prisms with at most 64 cm^3 volume can be the answer.
A. v = 2 * 4 * 7 = 28
B. v = 4 * 5 * 6 = 120
C. v = 5 * 5 * 4 = 100
D. v = 4 * 7 * 5 = 140
E. v = 3 * 5 * 4 = 60
Part A. answer: A, E
Part B.
Dora had a 5-cm cube.
volume of cube = side^3
volume = (5 cm)^3 = 125 cm^3
Dora has 125 small cubes.
We need to find the volumes of all prisms in Part B. Only prisms with at most 125 cm^3 volume can be the answer.
A. v = 3 * 3 * 8 = 72
B. v = 3 * 4 * 5 = 60
C. v = 6 * 6 * 4 = 144
D. v = 5 * 8 * 4 = 160
E. v = 3 * 5 * 4 = 60
Part B. answer: A, B, E
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.