What is the sum of two vectors?
465594
The word that fills in the blank is "resultant."
93533
Well, the sum of two terms thing is stupid, but oh well. (You can immediatly make it (x-5)(x-9))\nx^2-9x-5x+45=x(x-9)-5(x-9) --> (x-5)(x-9)
749461
u = 3i - 5j\nv = i - 2j\n\nso in order to find vector 2u - 3v, u have to find out 2u and 3v first, then subtract them.\n\nso, 2u = 2(3i - 5j) = 6i - 10j\nand 3v = 3(i-2j) = 3i - 6j\n\nnow, 2u - 3v = 6i - 10j - (3i - 6j) \n\n2u - 3v = 6i - 10j - 3i + 6j\n2u - 3v = 3i - 4j\n\nso your answer is: 3i - 4j\nhope that helped!
119559
its (x+2)(3x+1) now thats rite lol
594084
if these are vectors then,\nvector c has a direction of (i+j+k)/√3\nand since vector b is along c vector b is 3(i+j+k)/√3 =     √3(i+j+k).\n Now finding a dot b  and a cross b will be easy.\na.b=4√3   and    a*b=√3(i-k)  by normal vector multiplication.
854360
S=(c-b)/(b-a)\nSb-Sa-c+b=0\nc=Sb-Sa+b\nc=S(b-a)+b\nHowever you like it.
390709
Two angles whose sum is 90 degrees.
856009
false,\nuse the identity and -identity,\nboth are invertible, but the sum is the zero matrix, that is NOT invertible.....
324333
Look at this link.\n\nUse the first two "sum to product" identities.\n\nOn the right you will have:\n\n= (2 / 2) (sin a / cos a) (cos b / sin b)\n\n= 1 (tan a) (cot b)\n\na = (x + y) / 2,  b = (x - y) / 2\n\nso\n\n= (tan (x+y)/2) (cot (x-y)/2)\n\nOh, darn!  You're right, the signs are reversed from this.  I wonder if the problem was given to you incorrectly.  I don't see how you could go from our result to the problem the way it was given.  Ugh!\n\nI'm sorry, but I can't go any further with this.  Good luck, and let us know if the problem was stated wrong, please.
869931
You need to show that the only way to make a combination of them sum to zero is for all the multipliers to be zero.  For instance, say you have three vectors x, y, and z .  Then you need to show that if \n\nAx + By + Cz = 0 \n\nis true, then A, B, and C must ALL be equal to zero.  That is, the only way for the above equation to be true is if all the coefficients are zero.  If you can find a set of A, B, and C that satisfy the above equation, then the vectors are NOT linearly independent.
609611
Dear amit.wizkid, thou little knoweth what an abysmal error thou maketh.\n\ntut_einstein, you are better off noting that the sum of the first 'n' terms of an AP, whose first term is 'a' and the difference between terms is 'd' is :\n\nS = (n/2)[2a + (n-1)d]\n\nYou state that \nS1/S2 = (n+1)/(n+3)\n\nwhich is \n\n[2a1 + (n-1)d1] :[2a2 + (n-1)d2] = (n+1) : (n+3) ..... (I)\n\nRemember that what you desire to know is\n(a1 + 5d1) : (a2 + 5d2) \n\nPut n = 11 in (I)\n\n[2a1 + 10 d1] :[2a2 + 10 d2] = 12:14\n\nwhich gives\n\n[a1 + 5d1] : [a2 + 5d2] = 6:7\n\nwhich is what you desire. \n\nShould suffice.
970698
(a + b)(c + d + e)\n= a(c + d + e) + b(c + d + e)\n= ac + ad + ae + bc + bd + be\n\nOR\n\n(c + d + e)(a + b)\n= c(a + b) + d(a + b) + e(a + b)\n= etc.
724685
r-s = t\nadd s to both sides\nr = t+s
868541
the truth about this is find the right formula for this equasion, then set it with ur need of what space you put it in to example : =sum(a1:a28)
672754
Take out the five so you have:\nx^2 + 8 + 16\nthen you can change that to\n(x + 4)^2 b/c you have to find a number that is the product of the a and c and the sum of b\nwhere you have: ax^2 + bx + c
652953
I got:\n1. 6x^2 - 36x +48\n2. 6x + 22
314857
you should try to learn reading any book about vector analysis.\nSomething about vectors:\nhttp://mathworld.wolfram.com/topics/VectorAlgebra.html