Q: What is the reflection of point P-1 6 across the line y x?

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You use the well-known point-point formula . Here is it: suppose P1 = (x1,y1) and P2 = (x2,y2) An equation for the line containing P1 and P2 is y - y1 = [(y1-y2)/(x1-x2)] (x -x1) Note that the quantity in brackets is the slope of the line. Note also that it does not matter which point is P1 and which is P2.

Let p1 and p2 be the two prime numbers. Because they are prime, their divisors are div(p1) = {1,p1} and div(p2) = {1,p2}. So GCD(p1,p2) = Greatest Common Divisor of p1 and p2 = p1 if p1 equals p2 1 if p1 is different from p2

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.

No. Let p1 be a prime number. Let p2 be a multiple of p1 such that p2 = p1 * k. Then the factors of p2 are: 1, p1, k and p2. ==> p2 is not a prime number. Hence, a multiple of a prime number cannot be a prime number.

Take any two points and form the equation for a straight line. If all the remaining points satisfy the equation, then they lie on astraight line. Else, they don't. Here's an example. Consider n points as P1(x1, y1), P2(x2, y2), ...., Pn(xn, yn). In order to determine if P1, P2, ..., Pn lie on a straight line, form the straight line equation with P1 and P2 as: y-y1= m * (x - x1), where the slope m = (y2-y1)/(x2-x1). Then try to satisfy this equation by the remaining points P3, P4, ..., Pn. That is, verify the following: Is y3-y1= m * (x3 - x1)? Is y4-y1= m * (x4 - x1)? ... Is yn-y1= m * (xn - x1)? If all of the above is true, then the points lie on a straight line.

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You use the well-known point-point formula . Here is it: suppose P1 = (x1,y1) and P2 = (x2,y2) An equation for the line containing P1 and P2 is y - y1 = [(y1-y2)/(x1-x2)] (x -x1) Note that the quantity in brackets is the slope of the line. Note also that it does not matter which point is P1 and which is P2.

Price elasticity demand formula end point formula epd= [q2-q1/q1]/[p2-p1/p1] midpoint formula epd= [q2-q1/(q2+q1)/2] / [p2-p1/(p2+p1)/2]

A pointer can point to address of another pointer. consider the exampleint x=456, *p1, **p2;p1 = &x;p2 = &p1;Copyright Einstein College of EngineeringDepartment of Civil EngineeringTOPprintf("%d", *p1); will display value of x 456.printf("%d", *p2); will also display value of x 456. This is because p2 point p1, and p1 points x.Therefore p2 reads the value of x through pointer p1. Since one pointer is points towards anotherpointer it is called chain pointer. Chain pointer must be declared with ** as in **p2

THE QUESTION IS ACTUALLY WORDED. FIND THE EQUATION OF THE LINE THAT CONTAINS THE POINTS P1(-7,-4) AND P2(2,-8). ALGEBRA

to brioche knit .. cast on then .. you begin row 1 by k1 p1 across the row ending with k1 .. as a foundation row .. you need odd number of stitches and may have to add one stitch to pattern .. it wont hurt it .. row 2: p1 ," k1 below , p1 " repeat " -" across row end with k1 row 3 k1 ," p1 , k1 below" repeat " - " across row end with p1 repeat rows 2 and 3 to k1 below mean to actually grab both the stitch you are knitting and the stitch right below it together at the same time and knit them together .. the one below will actually drop off and become an elongated stitch.

Let p1 and p2 be the two prime numbers. Because they are prime, their divisors are div(p1) = {1,p1} and div(p2) = {1,p2}. So GCD(p1,p2) = Greatest Common Divisor of p1 and p2 = p1 if p1 equals p2 1 if p1 is different from p2

#include<stdio.h> #include<conio.h> #include<string.h> void main() { char str1[]="ravi kant yadav"; char str2[20],*p1,*p2; clrscr(); p1=str1+strlen(str1)-1; //Make p1 point to end of str1. p2=str2; while(p1>=str1) *p2++=*p1--; *p2='\0'; printf("Original string=%s. Reversed String=%s",str1,str2); getch(); }

P1 or parental

P1-e is an expression, not a formula.

In genetics, in a pure-breeding population, the parental generation is the P1 generation. The off-spring of the P1 Generation is called the F1 Generation

If the old population is P1, the new population is P2, and the growth rate is G, G = (P2 - P1) ÷ P1 x 100%

P1: tt F2: tt