No, because the second line is not defined.
No because for them to be parallel with each other they must have the same slopes and their slopes are 3 and -1 respectively
The question is curiously vague. Do the two lines exist in the same plane? If they do, then they must intersect somewhere -- unless they are parallel. For non-parallel lines, the distance between the two lines at the point of intersection is zero. For parallel lines, the shortest distance between them is the length of the line segment that is perpendicular to both. For intersecting lines, there is an infinite number of distances between the infinite number of pairs of points on the lines. But for any pair of points -- one point on line A and another on line B -- the shortest distance between them will still be a straight line. Given two lines in 3D (space) there are four possibilities # the lines are collinear (they overlap) # the lines intersect at one point # the lines are parallel # the lines are skew (not parallel and not intersecting) The question of "shortest distance" is only interesting in the skew case. Let's say p0 and p1 are points on the lines L0 and L1, respectively. Also d0 and d1 are the direction vectors of L0 and L1, respectively. The shortest distance is (p0 - p1) * , in which * is dot product, and is the normalized cross product. The point on L0 that is nearest to L1 is p0 + d0(((p1 - p0) * k) / (d0 * k)), in which k is d1 x d0 x d1.
First of all draw a line segment that is about 2 cm long between two points P0 and P1. At the one of the outer points, draw another line that is at an angle of 90 degrees from the first line segment. This will cause the new line segment to stand straight on the first segment. Draw another line segment between the not used endpoint of the new line segment, let's call it P2, and the not used endpoint of the first line segment. This will create a triangle. Now on the P2 endpoint, draw another line segment that is again at 90 degrees angle. Repeat the previous steps and you will have created a root spiral.
10log10 (p1 /p0 ) where the pi are power values.
Compounding of simple interest can be used by the following formula:P = P0 * (i+1)nwhere i is the ineterest rate; n the number of years; P0 initial principleP = 10,000 * (0.03+1)30= 10,000*(2.427262)= $24,272.62
No because for them to be parallel with each other they must have the same slopes and their slopes are 3 and -1 respectively
asawqtgrbikn[p0-[p0-[p0-[p0-[p0-[p0-[p0-[p0-
The question is curiously vague. Do the two lines exist in the same plane? If they do, then they must intersect somewhere -- unless they are parallel. For non-parallel lines, the distance between the two lines at the point of intersection is zero. For parallel lines, the shortest distance between them is the length of the line segment that is perpendicular to both. For intersecting lines, there is an infinite number of distances between the infinite number of pairs of points on the lines. But for any pair of points -- one point on line A and another on line B -- the shortest distance between them will still be a straight line. Given two lines in 3D (space) there are four possibilities # the lines are collinear (they overlap) # the lines intersect at one point # the lines are parallel # the lines are skew (not parallel and not intersecting) The question of "shortest distance" is only interesting in the skew case. Let's say p0 and p1 are points on the lines L0 and L1, respectively. Also d0 and d1 are the direction vectors of L0 and L1, respectively. The shortest distance is (p0 - p1) * , in which * is dot product, and is the normalized cross product. The point on L0 that is nearest to L1 is p0 + d0(((p1 - p0) * k) / (d0 * k)), in which k is d1 x d0 x d1.
pu = p0 + u(p1 - p0)
Addresses 80 - 87H are assigned to the P0 port
If the patient is still pregnant, a patient who is G1 and P0 is a primigravida.
#include<reg51.h> sbit RS=P2^0; sbit E=P2^1; void cmddiply() { RS=0; E=1; E=0; } void datadiply() { RS=1; E=1; E=0; } void delay() { int i; for(i=0;i<30000;i++); } void main() { while(1) { P0=0x38; cmddiply(); delay(); P0=0x0E; cmddiply(); delay(); P0=0x06; cmddiply(); delay(); P0=0x01; cmddiply(); delay(); P0=0x80; cmddiply(); delay(); P0='1'; datadiply(); delay(); P0='7'; datadiply(); delay(); } }
((Q1-Q0)/average of Q0 and Q1) over ((P1-P0)/average of P0 and P1)
First of all draw a line segment that is about 2 cm long between two points P0 and P1. At the one of the outer points, draw another line that is at an angle of 90 degrees from the first line segment. This will cause the new line segment to stand straight on the first segment. Draw another line segment between the not used endpoint of the new line segment, let's call it P2, and the not used endpoint of the first line segment. This will create a triangle. Now on the P2 endpoint, draw another line segment that is again at 90 degrees angle. Repeat the previous steps and you will have created a root spiral.
barrier potential P0=(kT/q)*ln(Na*Nd/Ni^2) when T ↑, P0↑.
The aneroid measures elevation and air pressure has an airtight chamber that is sensetive to changes in the air pressures. Hope this helps.
The loudness of a shout or sound pressure level is found by this form: ptotal = p0 + posc ..where... p0 = local ambient pressure posc = sound pressure deviation