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When using the slope formula does it have to be (y2-y1)(x2-x1) or can it be (y1-y2)(x1-x2)?
If you mean: (y2-y1)/(x2-x1) and (y1-y2)/(x1-x2) then either works out the same.
How do you find the mean of ordered pair?
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]
What formula with functions can you enter to calculate the range of values in Y1 to Y10?
There are many calculations that could be done: =SUM(Y1:Y10) =AVERAGE(Y1:Y10) =MAX(Y1:Y10) =MIN(Y1:Y10) =COUNT(Y1:Y10)
What is 3x-y1?
If you mean: 3x-y = 1 then it is a straight line equation
What is 5x-y1?
5x-y1 = 4
Angle of elevation of top of vertical cliff as seen from boat 120m away is 32degrees. Angle of elevation of top of flagpole at edge of cliff as seen from boat is 37degrees. Find the height of flagpole?
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.
C program to design a car?
#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(); }
How do you draw a square using line command?
Line (x1, y1, x2, y1); Line (x2, y1, x2, y2); Line (x2, y2, x1, y2); Line (x1, y2, x1, y1);
What is the answer to 3x y1 x y1?
{3x +y =1 {x+y= -3
How do you find linear equations with just coordinates?
if we take the (x1,y1),(x2,y2) as coordinates the formula was (x-x1)/(x2-x1)=(y-y1)/(y2-y1)
The distance between the points 2 y1 and 7 8 is 13 What is a possible value of 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
How do you transfer point slope form into slope intercept form?
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)
