Consider the Following Vectors Below?
824067
The polygon method is where you draw the two vectors on a piece of paper, using the given scale (1cm = 50 units). You need a ruler and a protractor to check your answer.\n\nFor number 1, draw vector B by making a horizontal line from the origin going to the right that is 4 cm (200 units) long. Next, draw a line from the end of vector B that is 2 cm (100 units) long, but goes 70 degrees NE (opposite of 250 degrees because we're subtracting). Now, draw a line from the origin to the end of vector D and you have R. Measure the length of R with a ruler and the angle of R with a protractor to make sure it matches the calculated value.\n\nI hope this helps!
748800
1)c because the forces are equal (a also for the same reason 0 is constant)\n2)b because the box has an unbalanced force in the horizontal direction and to the right\n3)d because the unbalanced force is to the right\n4)d because there is more force to the right\n5)b there is an unblanced force to the right
245176
u*(u-2)+v*(u-2)\n(u+v)*(u-2)
1096302
The electric and magnetic vectors are perpendicular to the direction of propagation.\n\nAlso the electric and magnetic vectors are at right angles to each other.
16382
See Answer 4 on the source page below
853525
Both the above answers are correct, provided the acceleration is horizontal. If it's not then it's a vectors and trig. question.
703485
1) 9 * (t + 1)\n2) 8 * (v + 7)\n3) 25 * (5 - h)
971669
1. 1s2 2s2 2p6 3s1\n2. 1s2 2s2 2p6 \n3. 1s2 2s2 2p4\n4. 1s2 2s2 2p6\n5. 1s2 2s2 2p6 3s2 3p6 3d7 4s2
115853
#include <iostream>\n#include <vector>\nusing namespace std;\n#define rep(i, n) for((i) = 0; (i) < (n); (i)++)\n#define rep2(i, k, n) for((i) = k; (i) < (n); (i)++)\nvoid f21();\nint n;\ndouble mat[100][100];\nint main()\n{\nwhile(true)\n{\ncin >> n;\nif(!n) break;\nmemset(mat, 0, sizeof(mat));\nf21();\n}\nreturn 0;\n}\n\nvoid f21()\n{\ncout << "****************************************" << endl;\ncout << "*          INVERSE OF MATRIX           *" << endl;\ncout << "****************************************" << endl;\n\ncout << "ORDER OF THE MATRIX: " << n << endl;\ncout << "ENTER ELEMENT ROW-WISE" << endl;\nint m = 2*n;\nint i, j, k;\nrep(i, n) {\nrep(j, n) {\ncin >> mat[i][j];\ncout <<"&#92;t";\nprintf("%5.3lf", mat[i][j]);\n}\nrep2(j, n, m) {\nif(i == (j-n)){\nmat[i][j] = 1;\ncontinue;\n}\nmat[i][j] = 0;\n}\ncout << endl;\n}\n\ncout << endl << "INVERSE MATRIX: " << endl;\ndouble pivot, comm;\nrep(i, n) {\npivot = mat[i][i];\nif(mat[i][i] == 0){\ncout << "DIVISION BY ZERO." <<endl;\nreturn;\n}\nrep(j, m) mat[i][j] = mat[i][j]/pivot;\nrep(j, n) {\nif(i == j) continue;\ncomm = mat[j][i];\nrep2(k, i, m) {\nmat[j][k] = mat[j][k] - (comm*mat[i][k]);\n}\n}\n}\n\nrep(i, n) {\nrep2(j, n, m) {\ncout <<"&#92;t";\nprintf("%6.4lf", mat[i][j]);\n}\ncout << endl;\n}\ncout << endl;\n}\n\n/*This is writen in C++ this will output the inverse of a Matrix\ninput:\n4\n1 2 3 4\n2 1 4 2\n2 5 7 12\n1 -1 3 11\n\n5\n1.173 -0.634 0 0 0\n-0.539 1.173 -0.634 0 0\n0 -0.539 1.173 -0.634 0\n0 0 -0.539 1.173 -0.634\n0 0 0 -0.539 1.173\n\n0\n*/
1102516
by any chance is this a question from your maths homework? lol :D\n\nI cant work it out myself because to put it simple, im crap at maths.
105143
Draw a force diagram to help you figure this out.\n\nRemember, perpendicular vectors are independent.\n\nF=ma
140559
Using the abc method… \n[a]FexOy+[b]H2SO4 -->[c]Fe2(SO4)3+[d]SO2 +[e]H2O\nFe-->ax=2c------eq.1\nO-->ya+4b=12c+2d+e----eq.2\nH-->2b=2e----------eq.3\nS-->b=3c+d----------eq.4\n\nLet a=1\n\nFrom eq.1\nax=2c\nx/2=2/2c\nx/2=c-----eq.5\n\nForm eq.3\n2b=2e\n2/2b=2/2e\ne=b-----eq.6 \n\nFrom eq.2\nYa+4b=12c+2d+e\n1y+4b=12(x/2) +2d+b (Form eq.5&6)\nY+3b=6x+2d\n2d=y-6x+3b\nd=(y+3b-6x)/2------------eq.7\n\nFrom eq.4\nb=3c+d (but c=x/2)\nb=3(x/2) +d\nReplacing eq.4 by eq.7 we will get\nb=3x/2+(y+3b-6x) ½\nb= (3x+y+3b-6x) ½\n2[b=(y+3b-3x) ½]\n2b=y+3b-3x\n3x-y=b-------eq.8\n\nReplacing eq.6 by eq.8\ne=b\ne=3x-y\n\nReplacing eq.7 by eq.8\nd=(y+3b-6x)/2\nd=[y+3(3x-y)-6x]/2\nd=[y+9x-3y-6x]/2\nd= (3x-2y)/2\n\nFrom the above calculation we know that:\na=1, b=3x-y, c=x/2, d=(3x-2y)/2 & e=3x-y. \n\n[a]FexOy+[b]H2SO4 -->[c]Fe2(SO4)3+[d]SO2 +[e]H2O\nWhen replacing the values we’ll get:\n[1]FexOy+[3x-y]H2SO4 -->[x/2]Fe2(SO4)3+[(3x-2y)/2]SO2 +[3x-y]H2O
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!
837519
The pre-image of w will be the set of vectors v so that T(v)=w. In other words, you are solving \n\nA*[x;y;z]=[6 4].\n\nThis gives the system of equations\nx+3y+z=6\n2x=4.\n\nClearly, x=2 and 3y+z=4, so the pre-image of w is\n{ (2, y, 4-3y): y is real}
903092
1) b)\n\n2) [H2O]^2 / 2[H2] [O2]\n  \n   [NO]^2 / [N2] [O2]\n\n3) b)\n\n4) b)
851212
hmm im not quite in college yet, but i hope to be help.\ndo you know the values for f, h, v, and a?\nif you do, id substitute it in.\nthen square both sides. \nand then solve.
829822
The vector potential is defined only to within the gradient of a scalar. That is, you can add the gradient of any scalar field to the answer below and get an equally valid expression for the vector potential.\n\nIf B is the constant (vector) value of the magnetic field then the vector potential is:\n\nA(r) = -r x (B/2)\n\nwhere r is the the position vector and x means cross product. In Cartesian coordinates in which the magnetic field is pointing in the direction of the z axis, the three components of A are:\n\nA_x = -(B/2)y\nA_y = (B/2)x\nA_z = 0\n\nwhere B here is a constant number.