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x2 + y4 + x4 +y2 = x6 + y6
unless you know what x and y are.
* * * * *
x2 + y4 + x4 + y2 ??
I don't believe that this expression can be factorised or otherwise simplified.
It certainly does not equal x6 + y6,
for all x and all y:
for example, if x = y = 1, then
x2 + y4 + x4 + y2 = 4, whilst
x6 + y6 = 2;
thus, they are two manifestly unequal quantities.
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How do you express x4 plus y4 if given x plus y equals a xy equals b?
(x + y)2 = x2 + 2xy + y2 So x2 + y2 = (x + y)2 - 2xy = a2 - 2b Then (x2 + y2)2 = x4 + 2x2y2 + y4 So x4 + y4 = (x2 + y2)2 - 2x2y2 = (a2 - 2b)2 - 2b2 = a4 - 4a2b + 4b2 - 2b2 = a4 - 4a2b + 2b2
What is y to the 8th power as an product in four different was with only positive exponents?
y6 x y2 y4 x y4 y2 x y2 x y4 y2 x y2 x y2 x y2
Factor completely 32y4 -2x4 y4?
2y4(16 - x4) = 2y4 (4 + x2)(4 - x2) = 2y4(4 + x2)(2 + x)(2 - x)
X4 plus y4 divided by x plus y?
(x4 + y4)/(x + y) = Quotient = x3 - x2y + xy2 - y3 Remainder = - 2y4/(x+y) So, x3 - x2y + xy2 - y3 - 2y4/(x+y)
How do you solve y4-84 plus 5y2?
You cannot solve it since only one side of an equation is given. If the equation was y4 + 5y2 - 84 = 0 then y4 + 12y2 - 7y2 - 84 = 0 or y2*(y2 + 12) - 7*(y2 + 12) = 0 or (y2 - 7)*(y2 + 12) = 0 then y2 = 7 or y2 = - 12 y = +or- sqrt(7) and, if you are in the complex domain, also y = +or- i*sqrt(12) where i is the imaginary square root of -1.
How can you do x4-y4?
(x2 + y2)(x + y)(x - y) = x4 - y4.
How do you express x4 plus y4 if given x plus y equals a xy equals b?
(x + y)2 = x2 + 2xy + y2 So x2 + y2 = (x + y)2 - 2xy = a2 - 2b Then (x2 + y2)2 = x4 + 2x2y2 + y4 So x4 + y4 = (x2 + y2)2 - 2x2y2 = (a2 - 2b)2 - 2b2 = a4 - 4a2b + 4b2 - 2b2 = a4 - 4a2b + 2b2
What is the square root of 100 x4 y4?
sqrt(100x4y4) = sqrt(100)*sqrt(x4)*sqrt(y4) = 10*x2*y2
What is the answer for x plus y to 4th power?
(x+y)4 = (x2+2xy+y2)2 = x4+4x3y+6x2y2+4xy3+y4
What are the factors of x2 - y4?
-2
Can you completely factor x32 - y32?
0
Write a program in c to implement scan-line polygon filling algorithm?
#include<GL/glut.h> float x1,x2,x3,x4,y1,y2,y3,y4; void draw_pixel(int x,int y) { glColor3f(0.5,0.0,1.0); glPointSize(0.5); glBegin(GL_POINTS); glVertex2i(x,y); glEnd(); } void edgedetect(float x1,float y1,float x2,float y2,int *le,int *re) { float temp,x,mx; int i; if(y1>y2) { temp=x1,x1=x2,x2=temp; temp=y1,y1=y2,y2=temp; } if(y1==y2) mx=x2-x1; else mx=(x2-x1)/(y2-y1); x=x1; for(i=int(y1);i<=(int)y2;i++) { if(x<(float)le[i]) le[i]=(int)x; if(x>(float)re[i]) re[i]=(int)x; x+=mx; } } void scanfill(float x1,float y1,float x2,float y2,float x3,float y3,float x4,float y4) { int le[500],re[500],i,j; for(i=0;i<500;i++) le[i]=500,re[i]=0; edgedetect(x1,y1,x2,y2,le,re); edgedetect(x2,y2,x3,y3,le,re); edgedetect(x3,y3,x4,y4,le,re); edgedetect(x4,y4,x1,y1,le,re); for(j=0;j<500;j++) { if(le[j]<=re[j]) for(i=le[j];i<re[j];i++) draw_pixel(i,j); } } void display() { x1=250.0;y1=200.0;x2=150.0;y2=300.0;x3=250.0; y3=400.0;x4=350.0;y4=300.0; glBegin(GL_TRIANGLES); glVertex2f(x1,y1); glVertex2f(x2,y2); glVertex2f(x3,y3); glVertex2f(x4,y4); glEnd(); scanfill(x1,y1,x2,y2,x3,y3,x4,y4); glFlush(); } void init() { glClearColor(1.0,1.0,1.0,1.0); glMatrixMode(GL_PROJECTION); glLoadIdentity(); gluOrtho2D(0.0,499.0,0.0,499.0); } void main(int argc,char **argv) { glutInit(&argc,argv); glutInitDisplayMode(GLUT_SINGLE|GLUT_RGB); glutInitWindowSize(500,500); glutCreateWindow("scanline"); glutDisplayFunc(display); init(); glutMainLoop(); }
Does it offend you yeah poster?
javascript:R=0; x1=.1; y1=.05; x2=.25; y2=.24; x3=1.6; y3=.24; x4=300; y4=200; x5=300; y5=200; DI=document.images; DIL=DI.length; function A(){for(i=0; i-DIL; i++){DIS=DI[ i ].style; DIS.position='absolute'; DIS.left=Math.sin(R*x1+i*x2+x3)*x4+x5; DIS.top=Math.cos(R*y1+i*y2+y3)*y4+y5}R++}setInterval('A()',5); void(0);
What is y to the 8th power as an product in four different was with only positive exponents?
y6 x y2 y4 x y4 y2 x y2 x y4 y2 x y2 x y2 x y2
How do you measure a rectangular land with 4 different dimensions?
If there are 4 different dimensions to a rectangular land, post in on FB...and tell your local news...I have only seen a 2 dimensional land (even normal objects have only 3 dimensions).Now to get serious [if you are that serious ignore all the [] like this, but do look out for ()]. If you mean coordinates, it isn't hard either. After all, you already said it is rectangle. Find the length of 2 adjacent sides [touching each other on one point] and multiply them together to get the area [if you want to measure just take a measuring tape and go to the points to do it].If it isn't rectangular but just 4 sided, arrange the points in this formula (if you dont understand it, wikipedia has the possibly clearer explanation, search for "shoelace formula")x1 x2 x3 x4 x1y1 y2 y3 y4 y1where the coordinates of the points are (x1,y1) (x2,y2) (x3,y3) (x4,y4) linked to the previous point by a single line. From now it might get confusing so good luck(x1 * y2) + (x2 * y3) + (x3 * y4) + (x4 * y1) - (y1 * x2) - (y2 * x3) - (y3 * x4) - (y4 * x1)Remove the negative (if there is) and divide the result by 2. You will get the answer
Factor completely 32y4 -2x4 y4?
2y4(16 - x4) = 2y4 (4 + x2)(4 - x2) = 2y4(4 + x2)(2 + x)(2 - x)
X4 plus y4 divided by x plus y?
(x4 + y4)/(x + y) = Quotient = x3 - x2y + xy2 - y3 Remainder = - 2y4/(x+y) So, x3 - x2y + xy2 - y3 - 2y4/(x+y)
