During the Vietnam War there were the following classifications: Class 1, 2, 3, and 4. 1-A, 1-A-O, 1-C, 1-D, 1-H, 1-O, and 1-W. 2-A, 2-C, 2-D, and 2-S. 3-A. 4-A, 4-B, 4-C, 4-D, 4-F, 4-G, and 4-W. 1-A meant the man was available for military service. 1-W meant he was a Conscientious Objector. 4-F meant the man was unfit for military service.
First substitute the coordinates of (x1, y1) into the equation, then simplify the equation so it has y in terms of x. y - y1 = m(x - x1) y - y1 = mx - mx1 y = mx - mx1 + y1 y = mx + (y1 - mx1) y = mx + (C)
Each element is the mean of the corresponding elements. Thus, the mean of (x1, y1) and ( x2, y2) is [( x1 + x2)/2, (y1 + y2)/2]
If you mean: 3x-y = 1 then it is a straight line equation
There are many calculations that could be done: =SUM(Y1:Y10) =AVERAGE(Y1:Y10) =MAX(Y1:Y10) =MIN(Y1:Y10) =COUNT(Y1:Y10)
5x-y1 = 4
for a linear function m= rise/ run if given 2 points on the graph m= (y2-y1)/(x2-x1) remember to simplify (lowest terms)! the 2 behind the x and the y is are not exponents! They mean the range in the second bracket: ex. (5,6)<--- 6= x2 5= x1 (7,8) <--- 8= y2 7=y1
The expression "5x y1 3x 2y2" seems to lack operators between the terms, making it unclear how to interpret it. If you meant to add or multiply these terms, please clarify. For instance, if you meant to represent a polynomial, you might need to include operators like "+" or "×". Please provide more details for a precise answer.
It shows the relationship of y in terms of x. [y = (yIntercept) + ((slope)*(x))] [slope = (y2 - y1)/(x2 - x1)]
#include<stdio.h> #include<graphics.h> #include<math.h> #include<conio.h> #include<dos.h> #include<alloc.h> #include<stdlib.h> #define RAD 3.141592/180 class fig { private: int x1,x2,y1,y2,xinc,yinc; public: void car() { xinc=10;yinc=10; x1=y1=10; x2=x1+90;y2=y1+35; int poly[]={x1+5,y1+10,x1+15,y1+10,x1+20,y1,x1+50,y1,x1+60,y1+10,x1+90,y1+17,x1+90,y1+20,x1+5,y1+20,x1+5,y1+10}; setfillstyle(SOLID_FILL,LIGHTGRAY); setlinestyle(SOLID_LINE,1,2); setcolor(4); drawpoly(9,poly); line(x1+15,y1+10,x1+60,y1+10); line(x1+20,y1+10,x1+20,y1); line(x1+35,y1+10,x1+35,y1); line(x1+50,y1+10,x1+50,y1); floodfill(x1+18,y1+8,4); floodfill(x1+28,y1+8,4); floodfill(x1+36,y1+8,4); floodfill(x1+52,y1+8,4); setfillstyle(SOLID_FILL,4); floodfill(x1+18,y1+12,4); setfillstyle(SOLID_FILL,BLUE); bar(x1+5,y1+20,x1+90,y1+25); setcolor(DARKGRAY); circle(x1+20,y1+25,8); circle(x1+20,y1+25,6); setfillstyle(1,8); floodfill(x1+21,y1+25,8); circle(x1+70,y1+25,8); circle(x1+70,y1+25,6); floodfill(x1+71,y1+25,8); int size=imagesize(x1,y1,x2,y2); void far *buf=farmalloc(size); getimage(x1,y1,x2,y2,buf); while(!kbhit()) { putimage(x1,y1,buf,XOR_PUT); x1+=xinc;x2+=xinc; if(x2<(getmaxx()-10)) putimage(x1,y1,buf,COPY_PUT); else { cleardevice(); x1=10;x2=x1+90; y1+=yinc;y2+=yinc; if(y2<(getmaxy()-10)) { putimage(x1,y1,buf,COPY_PUT); } else {y1=10;y2=y1+35;} } delay(200); } farfree(buf); getch(); } } } } void main() { int gd=DETECT,gm; initgraph(&gd,&gm,"d:\\cplus"); fig f; f.car(); cleardevice(); closegraph(); }
GIven 2 distance points (x1,y1) and (x2,y2) we can draw a line between those point and the midpoint formula finds the midpoint of that line. Call the midpoint (m1,m2) then it is [ (x1+x1 )/2 , (y1+y2/2)]
Line (x1, y1, x2, y1); Line (x2, y1, x2, y2); Line (x2, y2, x1, y2); Line (x1, y2, x1, y1);
{3x +y =1 {x+y= -3