16x2 - 128x + 256 = 16(x2 - 8x + 16) = 16(x - 4)2, or [4(x - 4)]2, or (4x - 16)2
5x-6+3x+128x-6+128x+6
To be rational, the numerator has to be a factor of 52 (the last term), and the denominator has to be a factor of 3 (the coefficient of the first term). Try all the combinations, including negative numbers (52/1, -52/1, 52/3, -52/3, 26/1, 26/3, etc.), to see whether one of them is a zero.
11 = (42 - 4) - (4 / 4) 12 = (4 + 4) + (√4 + √4) 13 = (42 - 4) + (4 / 4) 14 = (4 + 4 + 4 + √4) 15 = (4 * 4) - (4 / 4) 16 = (4 + 4 + 4 + 4) 17 = (42 + √4) - (4 / 4) 18 = (42 + 4) - (4 - √4) 19 = (42 + 4) - (4 / 4) 20 = (4 * 4) + (√4 + √4)
Here is one set of solutions. The answers here are not unique. 1 = (4*4)/(4*4) 2 = 4/4 + 4/4 3 = (4+4+4)/4 4 = (4-4)*4 + 4 5 = (4*4 + 4) / 4 6 = 4 + (4+4)/4 7 = 4 + 4 - 4/4 8 = 4 + 4 + 4 - 4 9 = 4 + 4 + 4/4 10 = (44 - 4)/4
16x2 - 128x + 256 = 16(x2 - 8x + 16) = 16(x - 4)2, or [4(x - 4)]2, or (4x - 16)2
5x-6+3x+128x-6+128x+6
all you have to do is 128x.20 and u get 25.6
No, 1TB is 1024GB, 128x larger than 8GB.
112x9y9 128x2y7 -80x7y2 The GCF is 16x2y2
(x+2)6 (x2 + 4x + 4)(x+2)4 (x2 + 4x + 4)(x2 + 4x + 4)(x+2)2 (x2 + 4x + 4)(x2 + 4x + 4)(x2 + 4x + 4) (x4 + 4x3 + 4x2 + 4x3 + 16x2 + 16x + 4x2 + 16x + 16)(x2 + 4x + 4) (x4 + 8x3 + 24x2 + 32x + 16)(x2 + 4x + 4) (x6 + 4x5 + 4x4 + 8x5 + 32x4 + 32x3 + 24x4 + 96x3 + 96x2 + 32x3 + 128x2 +128x + 16x2 + 64x + 64) (x6 + 12x5 + 60x4 + 160x3 + 240x2 + 192x +64) the coefficeints are 1, 12, 60, 160, 240, and 192
The second integer is 64 (128/2), so the first is 62 and the third is 66.OrLet the 1st integer be x and the 3rd integer be y.Since even integers differ by 2, then we have:y = x + 4x + y = 128x + x + 4 = 1282x = 124x = 62So the integers are 62, 64, and 66, and their sum is 192
Let the first consecutive integer be x. So that:the second integer is x + 1,the third integer is x + 2, andthe fourth integer is x + 3.We have:(x + 1) + (x + 3) = 1322x + 4 = 1322x = 128x = 64 the first integerThus, the four consecutive integers are 64, 65, 66, and 67.
Remember for Right Angled Triangles. First using Pythagoras the area(A) of triangles is equal to half the base multiplied to the perpendicular height. side 'a' = 2x + 4 side b = 8x + 8 side h = 10x First using Pythagoras find the value of 'x' . Hence (10x)^2 = ( 2x + 4)^2 + ()8x + 8)^2 Hence 100x^2 = 4x^2 + 8x + 16 + 64x^2 + 128x + 64 Collect like terms and form a quadratic equation. 100x^2 - 4x^2 - 64x^2 -8x - 128x - 16 - 64 = 0 32x^2 - 136x - 80 = 0 Factor out '8' 4x^2 - 17x - 10 = 0 Use Quadratic Eq'n x = {--17 +/- sqrt[(-17)^2 - 4(4)(-10)]} / 2(4) x = {17 +/- sqrt[289 + 160]} / 8 x = {17 +/- sqrt[449]} / 8 x = {17 +/- 21.189....} / 8 x = 38.189... / 8 = 4.773.... & x = - 4.189 / 8 ; Unresolved because , philosophically you cannot have a negative length. Using 4.773... substitute for 'x' in to the two shorter sides. Hence A = (0.5)(2(4.773...) + 4)(8(4.773...) + 8) A = (0.5(9.547... + 4)(38.184... + 8) A = (0.5)(13.547...)(46.184...) A = 312.827.... units^2
If you go to minecraftforum.net, they have a thread dedicated to texture packs. The packs are sorted by resolution, so you can see if there's a 32x or 64x pack that has the right look and feel for you. I don't know of any texture packs that exactly mirror the original texture in a higher resolution. :(
To be rational, the numerator has to be a factor of 52 (the last term), and the denominator has to be a factor of 3 (the coefficient of the first term). Try all the combinations, including negative numbers (52/1, -52/1, 52/3, -52/3, 26/1, 26/3, etc.), to see whether one of them is a zero.
That would be 64 x 64 = 4096 If you know calculus, this is a maxima-minima problem. X+ Y = 128 Y = 128 - X PRODUCT = P = XY = X ( 128 -X) = 128X - X^2 Take first derivative with respect to X and set = 0 - that is slope zero where value is either minimum or maximum dP/dX = 128 - 2X = 0 X = 64 Y = 64