The code "6H301 Y-1" on a boat motor typically indicates the model and serial number of the engine. The "6H301" part often refers to the specific model or type of the motor, while the "Y-1" may indicate a production batch or a specific manufacturing location. This information is essential for identifying parts, service, and compatibility with various boat systems. Always consult the manufacturer's documentation for precise details about the motor.
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
Tangent(theta) is sine over cosine, or y over x. x is 120. Theta is 32 and 37. y1 is height of cliff, and y2 is height of cliff plus flagpole.Tan(32) = y1 / 120, so y1 = 120 tan(32) = 75.Tan(37) = y2 / 120, so y2 = 120 tan(37) = 90.Height of flagpole is y2 - y1 = 90 - 75 = 15.All results rounded to nearest integer.
#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(); }
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
if we take the (x1,y1),(x2,y2) as coordinates the formula was (x-x1)/(x2-x1)=(y-y1)/(y2-y1)
You can use Pythagoras to solve this to get the co-ordinates. from Pythagoras we have a^2 + b^2 = c^2 For this a and b will be the x and y distance and c the distance between the 2 points so we get (7-2) ^2 + (8-y1)^2 = 13^2 So we need so solve this to get a correct solution which is (8-y1)^2 = 144 so 8 - y1 = + or - 12 so y1 = -4 or y1 = 20
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)
slope is the steepness of a line, it is defined by the change in the y values divided by the change in the x values of any two points on a line (x1, y1) and (x2, y2) slope = (y1 - y2)/(x1 - x2)