exponent exponent
algo diferente htt-ps://oke-.io/xn-lZlfp
factor I (fibrinogen), factor II (prothrombin), factor III (tissue thromboplastin), factor IV (calcium), factor V (proaccelerin), factor VI (no longer considered active in hemostasis), factor VII (factor-vii), factor VIII (antihemophilicfactor), factor IX (plasma thromboplastincomponent; Christmas factor), factor X (stuart-factor-stuart-prower-factor), factor XI (plasma thromboplastinantecedent), factor XII (factor-xii), factor XIII (fibrin stabilizing factor).
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factor pair = 36,1 factor pair = 18,2 factor pair = 12,3 factor pair = 9,4 factor pair = 6,6
xn- yn=(x - y)(xn-1 + xn-2y +xn-3y2 +. . .+x2yn-3+xyn-2 + yn-1)
The answer is 28.
What is the only perfect number of the form Xn + Yn
What is the question ? This is the equation of a line in 3 dimensions. Every point on the line satisfies the equation. There's no question here that needs an answer.
Andrew Wiles solved/proved Fermats Last Theorem. The theorem states Xn + Yn = Zn , where n represents 3, 4, 5,......... there is no solution.
Theorem: If two similar triangles have a scalar factor a : b, then the ratio of their perimeters is a : bBy the theorem, the ratio of the perimeters of the similar triangles is 2 : 3.For rectangles, perimeter is 2*(L1 + W1). If the second rectangle's sides are scaled by a factor S, then its perimeter is 2*(S*L1 + S*W1) = S*2*(L1 + W1), or the perimeter of the first, multiplied by the same factor S.In general, if an N-sided polygon has sides {x1, x2, x3....,xN}, then its perimeter is x1 + x2 + x3 + ... + xN. If the second similar polygon (with each side (labeled y, with corresponding subscripts) scaled by S, so that y1 = S*x1, etc. The perimeter is y1 + y2 + ... + yN = S*x1 + S*x2 + ... + S*xN = S*(x1 + x2 + ... + xN ),which is the factor S, times the perimeter of the first polygon.
exponent exponent
(xn+2-1)/(x2-1)ExplanationLet Y=1+x2+x4+...+xn. Now notice that:Y=1+x2+x4+...+xn=x2(1+x2+x4+...+xn-2)+1Y+xn+2=x2(1+x2+x4+...+xn-2+xn)+1Y+xn+2=x2*Y+1Y+xn+2-x2*Y=1Y-x2*Y=1-xn+2Y(1-x2)=1-xn+2Y=(1-xn+2)/(1-x2)=(xn+2-1)/(x2-1)
If the relationship can be written as y = ax + b where a and b are constants then it is a linear transformation. More formally, If f(xn) = yn and yi - yj = a*(xi - xj) for any pair of numbers i and j, then the transformation is linear.
xn+1 = 1/2 ( xn + N/xn )
the answer would be exponentthe n in x indicating the number of factor of x is exponent
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