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True or false p lies in plane b the line containing p and q must lie in plane b?
False. In order for the line PQ to lie in plane B, then both P and Q must lie in plane B.
If a and b are two points in the plane the perpendicular bisector of Ab is the set of all points equidistant from a and b?
The perpendicular bisector of the line segment connecting points ( A ) and ( B ) in the plane is a line that divides the segment into two equal parts at a right angle. Every point on this line is equidistant from points ( A ) and ( B ). This means that if you take any point ( P ) on the perpendicular bisector, the distance from ( P ) to ( A ) will be the same as the distance from ( P ) to ( B ). Thus, the perpendicular bisector is the locus of points satisfying this equidistance condition.
Where do all the points on a plane equidistant from the endpoints of ab lie?
All points on a plane that are equidistant from the endpoints of a line segment ( ab ) lie on the perpendicular bisector of the segment. This line bisects ( ab ) at a right angle and includes all points that are the same distance from both endpoints ( a ) and ( b ). Therefore, any point on this line is equidistant from ( a ) and ( b ).
Why does b times ab equal ab squared and not equal ab plus b squared?
b*ab = ab2 Suppose b*ab = ab + b2. Assume a and b are non-zero integers. Then ab2 = ab + b2 b = 1 + b/a would have to be true for all b. Counter-example: b = 2; a = 3 b(ab) = 2(3)(2) = 12 = ab2 = (4)(3) ab + b2 = (2)(3) + (2) = 10 but 10 does not = 12. Contradiction. So it cannot be the case that b = 1 + b/a is true for all b and, therefore, b*ab does not = ab + b2
Given two points A and B in the three dimensional space what is the set of points equidistant from A and B?
A plane is the set of all points in 3-D space equidistant from two points, A and B. If it will help to see it, the set of all points in a plane that are equidistant from points A and B in the plane will be a line. Extend that thinking off the plane and you'll have another plane perpendicular to the original plane, the one with A and B in it. And the question specified that A and B were in 3-D space. Another way to look at is to look at a line segment between A and B. Find the midpoint of that line segment, and then draw a plane perpendicular to the line segment, specifying that that plane also includes the midpoint of the line segment AB. Same thing. The set of all points that make up that plane will be equidistant from A and B. At the risk of running it into the ground, given a line segment AB, if the line segment is bisected by a plane perpendicular to the line segment, it (the plane) will contain the set of all points equidistant from A and B.
True or false p lies in plane b the line containing p and q must lie in plane b?
False. In order for the line PQ to lie in plane B, then both P and Q must lie in plane B.
If a and b are two points in the plane the perpendicular bisector of Ab is the set of all points equidistant from a and b?
The perpendicular bisector of the line segment connecting points ( A ) and ( B ) in the plane is a line that divides the segment into two equal parts at a right angle. Every point on this line is equidistant from points ( A ) and ( B ). This means that if you take any point ( P ) on the perpendicular bisector, the distance from ( P ) to ( A ) will be the same as the distance from ( P ) to ( B ). Thus, the perpendicular bisector is the locus of points satisfying this equidistance condition.
What is the geometrical proof of a plus b whole square?
In view of the graphic capabilities of this site, you will need to use a fair amount of imagination! Here goes: <------a+b------> <----a----><-b-> +-----------+-----+ |. . . . . . . .|. . . .| |. . . P. . . .|. .Q. | |. . . . . . . .|. . . .| +-----------+-----+ |. . . R. . . .|. .S. | |. . . . . . . .|. . . .| +----------+-----+ Where P = a*a = a2 Q = a*b R = b*a = a*b S = b*b = b2 (a + b)2 = P + Q + R + S = a2 + ab + ab + b2 = a2 + 2ab + b2
What is the geometrical proof of a-b whole square?
Given the graphic capability of this site, you are going to have to use some imagination! <---------a---------> <---a-b---><--b--> +-----------+-------+ |...............|..........| |.......P......|....Q...| |...............|..........| +-----------+-------+ |.......R......|....S....| |...............|..........| +-----------+-------+ In the above graphic, P, S and the whole figure are meant to be squares. The total area is P+Q+R+S = a2 P = (a-b)2 Q = b*(a-b) = (a-b)*b = a*b - b2 R = (a-b)*b = a*b = a*b - b2 and S = b2 Now, P = {P+Q+R+S} - Q - R - S = a2 - ab + b2 - ab + b2 - b2 = a2 - 2ab + b2
There are two inclined planes ab and bc making a triangle of angles 45-90-45 90 degree is at b which has a pulley the string over it connects two masses m on ab and 2m on bc the friction coefficie?
(Coefficient of friction of plane ab) + 2(Coefficient of friction of plane bc) = 1 Coefficient of friction of plane ab = Coefficient of friction of plane bc = 1/3 = 0.33333......
Where do all the points on a plane equidistant from the endpoints of ab lie?
All points on a plane that are equidistant from the endpoints of a line segment ( ab ) lie on the perpendicular bisector of the segment. This line bisects ( ab ) at a right angle and includes all points that are the same distance from both endpoints ( a ) and ( b ). Therefore, any point on this line is equidistant from ( a ) and ( b ).
What blood type can donate donate to all blood?
Ab+ universal receiver o- universal donor blood types: can donate to: can receive from: ab+: ab+: ab+ ab- a+ a- b+ b- o+ o- ab-: ab+ ab-: ab- b- a- o- a+: a+ ab+: a+ a- o+ o- a-: a+ a- ab+ ab-: a- o- b+: b+ ab+: b+ b- o+ o- b-: b- b+ ab- ab+: b- o- 0+: o+ a+ b+ ab+: o- o+ o-: o+ o- a+ a- b+ b- ab+ ab-: o-
What is ab times b?
(a x b)^b =ab x b^2 =ab^3
What is the equation of ab?
a+b(a+B)=ab
What is the principal behind solubility product?
For the molecule AB the solubilty product is: p = [A] x [B], A and B being the species solved in the solution.
Why does b times ab equal ab squared and not equal ab plus b squared?
b*ab = ab2 Suppose b*ab = ab + b2. Assume a and b are non-zero integers. Then ab2 = ab + b2 b = 1 + b/a would have to be true for all b. Counter-example: b = 2; a = 3 b(ab) = 2(3)(2) = 12 = ab2 = (4)(3) ab + b2 = (2)(3) + (2) = 10 but 10 does not = 12. Contradiction. So it cannot be the case that b = 1 + b/a is true for all b and, therefore, b*ab does not = ab + b2
Given two points A and B in the three dimensional space what is the set of points equidistant from A and B?
A plane is the set of all points in 3-D space equidistant from two points, A and B. If it will help to see it, the set of all points in a plane that are equidistant from points A and B in the plane will be a line. Extend that thinking off the plane and you'll have another plane perpendicular to the original plane, the one with A and B in it. And the question specified that A and B were in 3-D space. Another way to look at is to look at a line segment between A and B. Find the midpoint of that line segment, and then draw a plane perpendicular to the line segment, specifying that that plane also includes the midpoint of the line segment AB. Same thing. The set of all points that make up that plane will be equidistant from A and B. At the risk of running it into the ground, given a line segment AB, if the line segment is bisected by a plane perpendicular to the line segment, it (the plane) will contain the set of all points equidistant from A and B.
