(x4 + y4)/(x + y) = Quotient = x3 - x2y + xy2 - y3 Remainder = - 2y4/(x+y) So, x3 - x2y + xy2 - y3 - 2y4/(x+y)
Assumption 8x2 = eight times x squared Assumption: x3 is x to the third Assumption: y2 is y squared Answer 40 y to the fourth x to the fifth = 40 y4 x5
x=y4 /2
y4
y6 x y2 y4 x y4 y2 x y2 x y4 y2 x y2 x y2 x y2
4x-y4 what = 0
(x4 + y4)/(x + y) = Quotient = x3 - x2y + xy2 - y3 Remainder = - 2y4/(x+y) So, x3 - x2y + xy2 - y3 - 2y4/(x+y)
If x = 3 and y = 4 then the answer is 2
Assumption 8x2 = eight times x squared Assumption: x3 is x to the third Assumption: y2 is y squared Answer 40 y to the fourth x to the fifth = 40 y4 x5
The GCF is y4
x=y4 /2
The simplest equation would be y4 = 0
Use the LCM for their coefficients.Use each vriable with the LARGEST exponent in each term.Ex: Find the LCM of 15 x2y5 and 20 x3y4LCM of 15 and 20 is 60, LCM of x2 and 20 x3 is x3 , LCM of y5 and y4 is y5So the LCM is 60 x3y5
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);
(x2 + y2)(x + y)(x - y) = x4 - y4.
4 + 24 + 16 = 44
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