u need to be more specific but I would guess 10- 18
Let the point P have coordinates (p, q, r) and let the equation of the plane be ax + by +cz + d = 0Then the distance from the point to the plane is abs(ap + bq + cr) / sqrt(a^2 + b^2 + c^2).
Suppose the square is ABCD. Draw the diagonal AC.Mark one point on the diagonal, P (not the midpoint of AC), at a distance x from A. Mark another point, Q, also on the diagonal, at the same distance from C.Then,PBQD is a rhombus,ABPD and BCDQ are arrowheads.Suppose the square is ABCD. Draw the diagonal AC.Mark one point on the diagonal, P (not the midpoint of AC), at a distance x from A. Mark another point, Q, also on the diagonal, at the same distance from C.Then,PBQD is a rhombus,ABPD and BCDQ are arrowheads.Suppose the square is ABCD. Draw the diagonal AC.Mark one point on the diagonal, P (not the midpoint of AC), at a distance x from A. Mark another point, Q, also on the diagonal, at the same distance from C.Then,PBQD is a rhombus,ABPD and BCDQ are arrowheads.Suppose the square is ABCD. Draw the diagonal AC.Mark one point on the diagonal, P (not the midpoint of AC), at a distance x from A. Mark another point, Q, also on the diagonal, at the same distance from C.Then,PBQD is a rhombus,ABPD and BCDQ are arrowheads.
If we assume a 2D topography with XX and YY axes, we can write any two points on that surface as p(x,y) and P(X,Y) where x, X, y, and Y are the coordinates of each point. So the distance between p and P is S = sqrt(Sx^2 + Sy^2) where Sx = X - x and Sy = Y - y. EX: Assume p(2,4) and P(4,9); then we have S = sqrt((4 - 2)^2 + (9 - 4)^2) = 5.39 units. ANS.
61
The gradient, at any point P:(x, y, z), of a scalar point function Φ(x, y, z) is a vector that is normal to that level surface of Φ(x, y, z) that passes through point P. The magnitude of the gradient is equal to the rate of change of Φ (with respect to distance) in the direction of the normal to the level surface at point P. Grad Φ, evaluated at a point P:(x0, y0, z0), is normal to the level surface Φ(x, y, z) = c passing through point P. The constant c is given by c = Φ(x0, y0, z0).
Displacement is how far the object is from the starting point, while distance traveled is the how far the object traveled all together. -Eric P
\sqrt(9.8)~ 3.13
3v/ 2 ~ 4.2
It the UK the mint does: the British 50 pence and 20 p coins are both heptagonal.
The distance between the above places is approximately equal to 3421 nautical miles. To convert miles to nautical miles, multiply the miles by 0.86. This is point to point straight distance. The actual distance will change according to the route.
3v/ 2 ~ 4.2
6.7 mi
Let the point P have coordinates (p, q, r) and let the equation of the plane be ax + by +cz + d = 0Then the distance from the point to the plane is abs(ap + bq + cr) / sqrt(a^2 + b^2 + c^2).
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.
zero Half the distance between them would be 4 units; so 3 units from P would not be close enough to Q to be equidistant.
It is the Pythagorean distance formmula.If P = (x1, y1) and Q = (x2, y2) thenDistance between P and Q = sqrt[(x1 - x2)2 + (y1 - y2)2]It is the Pythagorean distance formmula.If P = (x1, y1) and Q = (x2, y2) thenDistance between P and Q = sqrt[(x1 - x2)2 + (y1 - y2)2]It is the Pythagorean distance formmula.If P = (x1, y1) and Q = (x2, y2) thenDistance between P and Q = sqrt[(x1 - x2)2 + (y1 - y2)2]It is the Pythagorean distance formmula.If P = (x1, y1) and Q = (x2, y2) thenDistance between P and Q = sqrt[(x1 - x2)2 + (y1 - y2)2]
The parallax method can be used to calculate the distance between planets and other celestial bodies. The formula for this is d(pc) = 1/p where p is parallax measured in arcseconds.