Can you please explain the diagonal rule?
917414
You apparently know what the diagonal rule looks like, but the link below gives a quite nice picture of it.\n\nBasically it is a summary of the order in which electron shells will be filled up. Follow the arrows round to discover where the electrons will go, starting at the 1s level.\n\nSo for 15:\n2 electrons will go into 1s, filling that up\n1s2\nthe next electrons will go into 2s2, 2p6, 3s2 and then the remaining 3 electrons into the 3p orbital, giving us a structure of:\n1s2 2s2 2p6 3s2 3p3\nRemeber, on your diagram, the first two characters (2p, 4d etc) are the names of the orbital, and the final digit is the number of electrons in that orbital. On your diagram the final digit shows the maximum number of electrons on the orbital; when you write out the structure the first few orbitals will be full (1s2 2s2 2p6 3s2) and the last one will be part full of however many electrons you have left (3p3.)
723712
One way to solve this is to draw it on graph paper and count up the squares.  Howevr, I think I have a solution that doesn't resort to graphing:\n\nThe units 15 and 42 have a common divisor of 3.  So, if the diagonal starts at (0,0), it will also intersect (5,14) and (10,28).  Therefore, we can reduce this problem to testing a 5 by 14 rectangle and multplying by 3.\n\nFor the 5 by 14 rectangle, since this diagonal won't hit any vertices of the unit squares, just figure you will pass through 14 squares as you make your way across, and you will move up 4 squares.  So, 4 + 14 = 18 squares.  \n\nMultiple by 3 to get your answer for the 15 by 42 rectangle: 18 * 3 = 54.
4370
A step by step method to understand cramer's rule:\n\nhttp://www.purplemath.com/modules/cramers.htm\n(basically its the first result found by google; you can do more googling)
579155
The first one isn't, because it's undefined when x=4 (assuming the (x-4) is the denominator, not just the x)\n\nThe second one is; it's a horizontal line with no breaks or holes.
583405
The area of a kite is half the product of the diagonals. Note: This formula works for the area of a rhombus as well, since a rhombus is a special kind of kite. Note that the diagonals of a kite are perpendicular.\n\nJoe is right that you can do it as two triangles, but you'll get the same result.\n\nLook at the image in the link below.  Notice how the outside rectangle has double the area of the inside kite.  Notice also how the height and width are the same.  So it is exactly half the height times the width.
732315
This one's a little tricky.\n\nAllow me to use "t" instead of "delta"\n\n\nint ((cos(2t))^3)dt\n\nSince (cos(2t))^2 = 1 - (sin(2t))^2\n\nSubstitute.  this becomes:\n\nint ((1 - (sin(2t))^2) * cos(2t))dt\n\nSubstitute u = sin(2t).  Then du = cos(2t)dt.\nSo dt = du / cos(2t)\nThe key here is when you plug this in, the cos(2t) values cancel.\nSo we get:\n\nint (1 - u^2)du\n= u - (1/3)u^3 + C\n= sin(2t) - ((sin(2t))^3)/3 + C
911904
using the rule\n\ndf/dx = df/dy * dy/dx and y = 7x +(49x^2-16)^1/2\n\nwe get df/dy = 1/y = 1/(7x+49x^2-16)^1/2 ...1\n\ndy/dx = 7+ 1/2(49x^2-16)^(-1/2) d/dx(49x^2-16)\n= 7+ 1/(2*49x^2-16)^(1/2) * 98 x ...2\n\nusing d/dx(y^n) = n y^(n-1) dy/dx where y = 49x^2-16\n\nfrom 1and 2 by multiplying\ndf/dx =  (7+ 98x/(2*49x^2-16)^(1/2))/(7x+49x^2-16)^1/2 ...
169724
It's simple Any Symmetric matrix must satisfy aij=aji where aij is the element is ith row and j column\nSo for the diagonal akk must be positive for symmetric positive definit matrix
964284
The question apparently is "How many diagonals does a polygon with n sides have?"  \n\nThe first formula is n(n-3)/2.  \nOne way to see this is to notice that you can draw (n-3) diagonals from every vertex of the polygon. This is because there are (n-1) other vertices, but two of them are adjacent vertices and so don't count towards making diagonals. This seems to give n(n-2) diagonals, but this way of counting counts every diagonal twice since each diagonal connects to two vertices, so you have to divide by 2.\n\nBy a recursive formula, we mean a way of expressing the answer for n vertices in terms of the answer for n-1 vertices. Suppose, for example, that you already know the answer for a polygon with n-1 vertices. Now if you add another vertex between two of the vertices of the original polygon, then all the diagonals of the original polygon will still be diagonals of the new polygon, and so will the side joining the two vertices that you added a new vertex between, and so will the line segments joining the new vertex to all the other vertices of the original polygon. (Got all that?) So if we let diag(n) be the number of diagonals for a polygon with n sides, we get the formula:\n\n        diag(n) = diag(n-1) + n - 3 + 1 or\n\n                  diag(n-1) + n - 2\n\nThe first formula is better, since it actually gives you the answer.  \nBut sometimes it's easier to get a recursive formula first and use \nthat to get an explicit formula (your first formula is an explicit one \nsince you only need the number of vertices in the polygon to get the number of diagonals in that polygon).  This is called "solving the recursion."  Sometimes a recursive formula is the best you can do because there simply is no explicit formula.\n\nI hope this helps.
1123602
All diagonals of a perfect cuboid are integers\n\nAssume the cuboid has height h, width w and depth d and h>w>d.\n\nThe lengths of the face diagonals are:\n\n    * fd12 = h2 + w2\n    * fd22 = h2 + d2\n    * fd32 = w2 + d2\n\nThe length of the space diagonal is:\n\n    * sd2 = h2 + w2 + d2\n\nIf h,w and d are all even, then every fd and the sd are divisible by 2, and so we can divide h,w and d by 2.\n\nThe sum of 2 odd integers squared is never a square, and so the only possiblity is for exactly one of h,w,d to be odd.\n\nIn turn, this implies that two of the fdi are odd, the other even. Consider fd1 to be even.\n\nThe aim of this proof is to demonstrate that any such fd1 must contain an infinite power of 2 as a factor.\n\nIf we expand the fd equations, using a,b,c as parameters (h=2a, w=2b, d=2c+1), then we arrive at:\n\n4[a2 + b2 + c2 + c] + 1 = sd2\n\na and b must be of the same parity, as 4k+1 is only square for even k.\n\nHowever a2 + b2 = fd12/4\n\nAnd so fd12/4 must be even, i.e. h=4a, w=4b.\n\nIf we repeat the process with the new parameterization, then we arrive at:\n\n4[8a2 + 8b2 + c2 + c] + 1 = sd2\n\nBut now fd12/16 must be even, i.e. h=8a, w=8b.\n\nThis is obviously absurd.
46479
Oh boy. I'm not sure that i can explain why the rules are what they are, but i know what the rules are, and i can tell you:\n\n1. 45-45-90\n -The two legs are equal\nLeg 1=a\nLeg 2=b\na=b\n -The hypotenuse is 2 times the square root of a (or b, since they are equal)\n\n2.30-60-90\n -The key to this is the small leg, "a". The hypotenuse is equal to 2a (2 times "a").\n -The large leg is equal to 3 times the square root of the small leg, "a".\n\nHope this helps!\n\n-incarnadyne49\n\nThe first guy is wrong, i'm sure!!! I would know, i am a calc student, and access to a geometry textbook!!
703879
This sounds like plane geometry where you have an n sided polygon and want to know how many diagonals you can draw.  From each vertex, you can draw a diagonal to every other vertex except to yourself and the two next to you (those two line make sides of the polygon not diagonals).  So for n sides (which means n vertices), you can draw n-3 diagonals at each vertex.  This gives (n-3)*n.  But this just counts the starting end of each line. Whenever you draw one of these lines, it goes to another vertex.  If we drew (n-3)*n lines, we would traces over each diagonal twice, once from each end so there are only half as many diagonals as that.\n\nThe answer then is: (n-3)*n/2
1166733
dy/dx = {(3x^2-2)(4x-3)^4 - (x^3-2x)(4)(4x-3)^3(4)}/(4x-3)^8\n= {(3x^2-2)(4x-3)^4 - 16(x^3-2x)(4x-3)^3}/(4x-3)^8\n\nUsing the quotient rule.  d/dx f(x)/g(x) = {f'(x)g(x) - g'(x)f(x)}/g(x)^2
352749
By the chain rule:\n[e^(3x^(2)-7x)]*(6x-7)
826187
I am guessing you want the vector projection onto one of the 6 sides rather than one of the 12 edges.\n\nThe diagonal vector is (1, 1, 1).  Projecting it onto any side simply removes the coordinate in the direction of projection.  Projecting it onto the side in the xy plane gives (1, 1, 0). \n\nIf you did mean an edge (the line where 2 sides meet), then the projection is always two 0s and a 1.  The one is in the direction of the edge.
1169011
Power Rule:\nd/dx (x^n) = n*x^(n-1)\nProduct Rule:\nd/dx (f(x)g(x)) = f'(x)g(x) + f(x)g'(x)\nChain Rule:\nd/dx (f(g(x)) = f'(g(x))*g'(x)\n\n1) y = (- 9x^2)(e^- 3x)\nApply the poduct rule and chain rule.\n(-9x^2)'(e^-3x) + (-9x^2)(e^-3x)'\n= (-18x)(e^-3x) + (-9x^2)(e^-3x)(-3)\n= -18x(e^-3x) + 27x^2(e^-3x)\n2) y = ln(3x^2 - 5x)\nApply the chain rule\nd/dx (ln(g(x))) = g'(x)/g(x)\n(3x^2 -5x)'/(3x^2 -5x)\n= (6x -5)/(3x^2 -5x)\n3) y = x^3 - 4x^2 \nJust apply the power rule.\n3x^(3-1) - 4(2)x^(2-1)\n= 3x^2 - 8x\n4) y = 7x^(5/6) \nApply the power rule again.\n7(5/6)x^(5/6 -1)\n= (35/6)x^(-1/6)\n= 35/(6x^(1/6))\n5) y = 3x^2(4x - 5)^9\nUse the product rule and chain rule\n(3x^2)'(4x-5)^9 + (3x^2)((4x-5)^9)'\n= (6x)(4x-5)^9 + (3x^2)((4x-5)^8)(4)\n= 6x(4x-5)^9 + 12x^2(4x-5)^8\nThis can be simplified\nFactor 6x(4x-5)^8 out.\n6x(4x-5)^8 [(4x-5) + 2x(1)]\n= 6x(4x-5)^8 [6x-5]\n= 6x(6x-5)(4x-5)^8
309782
You are attempting to find the diagonal of a square with 5 on a side. The answer is obviously 5*√(2). C, the hypotenuse of a right triangle, is the longest side of the said triangle. Therefore, C must be larger than 5 in your particular problem. Hence, the first two answers are incorrect.\n\nExample:\nIn general, a square with sides of length x has a diagonal of length x*√(2).\n\na^2 + b^2 = c^2\nx^2 + x^2 = c^2\n2*x^2 = c^2\n√(2*x^2) = √(c^2)\n√(x^2)*√(2) = c\nx*√(2) = c\n\nIf x = 5, then c = 5*√(2).