for what values of b are the vectors < -12,b,3 > and <b,b^2, b> orthogonal? There might be multiple answers
340285
-12b + b^3 + 3b = 0 (multiplying the two vectors together)\n\nb(b^2 - 9) = 0\n\nb(b + 3)(b - 3) = 0\n\nso b=0 or b=3 or b=-3
967876
1) 2a^3 - 13a^2b + 11ab^2 + 14b^3\n\n2) (2a-4b) (4a^2+8ab+16b^2)\n\n3)  (b-8)(b-1)=0\n      b=8   or    b=1\n\n4) 4m^3 - 26m^2 - 14m=0\n2m(2m^2 - 13m - 7)=0\n2m=0  or  2m^2 - 13m -7=0\nm=0  or    (2m+1)(m-7)=0\n                  m=-0.5   or  m=7\n\n so the final answer is m=0 and m=-0.5 and m=7
245176
u*(u-2)+v*(u-2)\n(u+v)*(u-2)
844003
HI: I tried to answer your email but it said that your email address has not been confirmed and so the email didn't go through.\n\nalpha and beta are scalers for the vector space V.  This problem is actually so easy that people don't believe what they've done.  You really just need to show that Span S behaves like V.  \n\nSpan S must be a vector space itself.  Two elements from the span is \nu=a1*v1 + a2*v2...ak*vk\nand \nw=b1*v1 + b2*v2...bk*vk\n\nshow that alpha*u+beta*w is in Span S where alpha, beta are scalers for your vector space.\n\nalpha*u+beta*w = (alpha*a1+beta*b1)*v1 + ....\nis a linear combination of the vectors in Span S, thus it's in Span S.  \n\nSo, Span S is not empty and is closed, thus its a subspace.
955805
That simplifies to 3[2x-1]>12\n\n[2x-1]>4\n\n2x-1>4    2x-1<-4\n\nx>5/2  or x<-3/2\n\nWhen you have absolute values, you have to look at both ends of the spectrum
976254
The strategy is as follows:\n\n1.  Find two vectors in the plane.  One can be found using the line.  Another can be found using the point A and any point you like on the line.  \n\n(If (x,y,z) and (u,v,w) are two points in the plane, then the vector (x-u)i + (y-v)j + (z-w)k lies in the plane; or more accurately, this vector is parallel to the plane.)\n\n2. Find the cross product of the two vectors in part 1.  Call this new vector N (for normal).\n\n3. The equation of the plane is m(x-a) + n(y-b) + p(z-c) = 0, where N = mi + nj + pk and where A = (a,b,c).\n\n(Why does step 3 work?  A point B = (x,y,z) lies in the plane if and only if the vector AB is orthogonal to N.  In other words, their dot product is zero.  If you write this out in coordinates, you get exactly the condition that m(x-a) + n(y-b) + p(z-c) = 0.)
865354
You’ve got 3 vectors:\nU=[-128, 0] – anti parallel to X-axis\nV=[+64, 0]  - parallel to X-axis\nW=[128*(cos(30), 128*sin(30)] – having both X & Y components.\n If you have to find a Sum S of 2 vectors A & B, A being =[Ax, Ay], B being [Bx, By],  then do as Sx=Ax+Bx  &  Sy=Ay+By, that is  S=[Ax+Bx, Ay+By].\nNow F=U+V+W=[-128+64+128*cos(30), 0+0+128*sin(30)], (cos(30)=sqrt(3)/2, sin(30)=1/2)).\nSo F=[-64+128*sqrt(3)/2, 128/2]=[128*(sqrt(3)-1)/2, 128/2]=\n=64*[sqrt(3)-1, 1]. \nNow how to draw it in your picture?\nAbs(F)=sqrt((F*F))=64*sqrt((sqrt(3)-1)^2+1^2)=\n=64*sqrt(3-2*sqrt(3)+1+1)=\n=64*sqrt(5-2*sqrt(3))=64*1.5359=98.3N – this is absolute value of force.\nNow let’s find its components Fx & Fy, or better say angle of F to X-axis.\nAs F=64*[sqrt(3)-1, 1], then tan(alpha)=sin/cos=1/(sqrt(3)-1)=1.366\nalpha=Atan(1.366)=0.939rad=0.939*180/3.1416=53.8deg \nSo it shall look like a stick of 98.3N long with angle of 53.8deg to X-axis.\n(for Swimming Champs only!)
508006
x^2 + 4bx + 2b=0\n(x+2b)^2 - 4b^2 + 2b = 0\n(x+2b)^2 = 4b^2 - 2b\n(x+2b) = SQRT(4b^2 - 2b)\nx = [SQRT(4b^2 - 2b)]-2b\n\n3x^2 -bx +1=0\n3(x^2 - (b/3)x + (1/3)) = 0\nx^2 - (b/3)x + (1/3) = 0\n(x-(b/6))^2 -(b/6)^2 + (1/3) = 0\n(x-(b/6))^2 = (b/6)^2 - (1/3)\nx-(b/6) = SQRT[b^2/36 - 12/36]\nx-(b/6) = [SQRT{^2 - 12)] / 6\nx = {[SQRT{^2 - 12)]+b} / 6\n\nb^2x^2 - 2bx + c^2= 0\n(bx)^2 - 2(bx) + c^2 = 0\n(bx-1)^2 - 1 + c^2 = 0\n(bx-1)^2 = 1 - c^2\nbx-1 = SQRT(1-c^2)\nbx = SQRT(1-c^2)+1\nx = [SQRT(1-c^2)+1]/b
966048
(3ab^2)(3ab)^2\n(3ab^2)(9a^2b^2)\n27a^3b^4
965314
Not sure why u can't do these yourself but 1) = -360 \n2) -12\n3)-56.8\n4)-45\n5)1328\n6)-144\n7)-.5\n8)25\n9)11\n Those should be right
481076
Since you already know what vectors are, take two vectors from this plane as a basis for all vectors in the plane. The easiest basis is just to take two sides from the triangle formed by the 3 given points. \nFrom your study of linear algebra, you may remember that one way of checking whether a set of n vectors was linearly independent was to take the n-volume of their parallelotope (their determinant). If the n-volume was 0, then obviously the set of vectors makes less than an n-dimensional parallelotope, so they can't span an n-dimensional space. We also define a vector v to be orthogonal to a vector w if their inner product is 0.\nThus, we are looking for the vector w = (w1, w2, w3) such that the determinant (u, v, w) is non-zero and u*w=0 and v*w=0. If we expand the determinant above along the unknown vector w, choosing the n-volume to be the same as the area of the parallelegram spanned by u and v and expand the dot products, we find we have system of 3 linear equations in 3 variables. The solution vector of this system is called cross product of u and v, the formula for which you can find in the source below. It is a vector perpendicular to u and v that has as magnitude the area of the parallelegram spanned by u and v.
1008143
Use the Dot product as for 2D vectors\nA . B = |A||B|cos(angle)\ncos(angle) = dot product divided by the product of the magnitude of the vectors\nUse inverse cosine to find the angle.\nExample:\nA = 4i + 4j + 2k (i,j, and k are unit vectors in x,y,z respectively)\nB = 2i + 3j + 5k\n\nA.B = (2)(4) + (4)(3) + (2)(5) = 8 + 12 + 10 =20\n|A| = sqrt of (4x4 + 4x4 + 2x2) = sqrt of (36) = 6\n|B| = sqrt of (2x2 + 3x3 + 5x5) = sqrt of (38) =~ 6.16\n\ncos(angle) =  20/(6x6.16) = 20/36.96 = 0.54\nThe angle = arccos(.54) = 57 degrees\n\nThere is plenty of stuff here:\nhttp://mathworld.wolfram.com/topics/VectorAlgebra.html\n\nor start with Wikipedia \nhttp://en.wikipedia.org/wiki/Dot_product
475786
Represent these binomials in Vector Notation to do the calculations. The Vector Space will be in 7 dimensions, So choose a 7 dim vestor to represent each polynomial.\n\nLet V1: 9x^5 -2x^3 +6x^2 + 3 \nLet V2: 5x^4 - 7x^2 + 6 \nLet V3: 3x^6 - 5x^5 -x^2\n\nMatix Vector Form for Vector 1=V1 etc..\nWe are in 7 space, so express our vector with (&) and seven\ndigits. Let V=V1+V2+V3\nV1= (0, 9, 0, -2, 6, 0, 3)\nV2= (0, 0, 5, 0, -7, 0, 6)\nV3= (3,-5, 0, 0, -1, 0, 0)\n\nV = (3, 4, 5, -2,-2, 0, 9)\nIn algebraic representation\nAns is:\n3x^6+ 4x^5 +5X^4 -2x^3 -2x^2 +9\n\n---\nVectors in 2 space, ex.\nV1=(x1, x2)\nV2=(y1, y2)\nV1+V2=(x1+y1, x2+y2)\nThat is what I did for the 7 space vector above. I expressed the 6 th degree binmial with a 7 dim vector,\nto make calcualting the answer faster. On paper, you could line up all of your equations so that each binomial eqs line up. All x^6 lined up under all x^6, all x^5 lined up under x^5,...etc. But using vectors makes this quicker and faster to calculate the binomials.\n\nMy sources: Two MS's in Mathematics that I received many, many years ago.
550482
-2a(b+2)-6b=\nor\n-2(ab+2a+3b)
787986
I'll do some of this.\n\nPart 1.  Vector AB = -8i + 8j  (magnitude 8 sqrt 2)\nVector AC = 4i + 3j  (magnitude 5)\nVector BC = 12i - 5j  (magnitude 13)\nPerimeter = sum of magnitudes = 18 + 8 sqrt 2\n\nThe median of a triangle goes from a vertex to the midpoint of the opposite side.  You can do that.  Also, I don't know what transversal you're talking about.\n\nPart 2.  I assume you mean a = 5i + 12j and b = -i + 40j\nI don't know what the "opposite vector" means.\n\ns = a + b = 4i + 52j\nr = a - b = 6i - 28j\n\nPart 3.  I guess you mean a = -3i - 4j and b = -3i + 2j\nThese are linearly independent because one is not a multiple of the other.\n\nPart 4.  u dot v = 6 v1 - 9(72) = 0\nv1 = 9(12) = 108\nv = (108, 72) in your notation.
753754
I presume you mean\n\n[(6x^2) / (4x^2y-12xy)] * [(x^2+x-12) / (3x^2+12x)]\n\nThere seems to be something wrong. Are you sure about those "y" terms?
150854
i bet your thinking, my god, it has 3 values (x,y,z), how the heck does one attack this\n\nbelieve it or not, its simple\n\njust like you would in R^2\n\nsqrt[ (x-x1)^2 + (y-y1)^2 + (z-z1)^2]\n\nso \n\nsqrt [ (3-1)^2 + (-3- -4)^2 + (2-7)^2]\n\nso\n\nsqrt[ (2)^2 + 1^2 + (-5)^2]\n\nso\n\nsqrt [ 4 +1+25]\n\nso\n\nsqrt [30]\n\nhope that helps