It depends on how complicated you want to make it. The generally accepted answer would be to start at one point, and make a line to the next (a straight line). That's gonna be the answer, say, your teacher might want (sorry if you're an adult :p). The technical answer? Drill a hole through the globe from one point to the other, and your shortest distance would be the straight line. Einstein's answer? A geodesic. Look it up :p
If (p, q) is any point on the line, then the point slope equation is: (y - q)/(x - p) = 2 or (y - q) = 2*(x - p)
Given a straight line with slope m and a point (p,q) on the line, the point-slope formula of the line is (y - q) = m(x - p) It is used to represent a straight line in the Cartesian plane. This allows techniques of algebra to be used in solving problems in geometry.
Complex analysis is a metric space so neighborhoods can be described as open balls. Proof follows a. Assume that the set has an accumulation point call it P. b. An accumulation point is defined as a point in which every neighborhood (open ball) around P contains a point in the set other than P. c. Since P is an accumulation point, I can choose an open ball around P that has a diameter less than the minimum distance between P and all elements of the finite set. Therefore there exists a neighbor hood around P which contains only P. Therefore P is not an accumulation point.
To show that the perpendicular line segment is the shortest among all line segments drawn from a given point not on it, we can use the Pythagorean theorem. Let the given point be P and the line segment be AB, with the perpendicular from P meeting AB at C. By the Pythagorean theorem, the sum of the squares of the two sides of a right triangle is equal to the square of the hypotenuse. In this case, PC is the hypotenuse, and AP and AC are the other two sides. Thus, AC (perpendicular line segment) will always be shorter than any other line segment AB drawn from point P.
3v/ 2 ~ 4.2
3v/ 2 ~ 4.2
1
P = the distance around a figure. A = the surface it takes up.
It depends on how complicated you want to make it. The generally accepted answer would be to start at one point, and make a line to the next (a straight line). That's gonna be the answer, say, your teacher might want (sorry if you're an adult :p). The technical answer? Drill a hole through the globe from one point to the other, and your shortest distance would be the straight line. Einstein's answer? A geodesic. Look it up :p
false
Given a point P = (a,b) and slope m, the equation of a line through P with slope m is (y-b) = m(x-a)
The hyperbolic parallel postulate states that given a line L and a point P, not on the line, there are at least two distinct lines through P that do not intersect L.The negation is that given a line L and a point P, not on the line, there is at most one line through P that does not intersect L.The negation includes the case where there is exactly one such line - which is the Euclidean space.
true
If (p, q) is any point on the line, then the point slope equation is: (y - q)/(x - p) = 2 or (y - q) = 2*(x - p)
Given a straight line with slope m and a point (p,q) on the line, the point-slope formula of the line is (y - q) = m(x - p) It is used to represent a straight line in the Cartesian plane. This allows techniques of algebra to be used in solving problems in geometry.
4v/ 2~5.7