The figure shos a vector a in the xyplane and a vectore b in the direction of k.  Their lenghts are |a| =5....
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a cross b is a*b*sin theta (the angle between them)\nIn your case a and b are perpendicular theta=90 sin theta=1\na x b = 5x2x1 = 10 c where c is a vector perpendicular to their plane but the problem says |a x b| and this modulus means the algebraic answer only without vectors hence the answer is 10\nI hope this helps
899386
The x's will all be the same, and the y's will be the difference between 4 & the number:\n\nA=(2,5)\nB=(0,3)\nC=(5,2)\n\nAnother way to do this is to actually draw the triangle and the line y=4. This will give you a visual of what should happen.
1159510
the four angles add up to 360 degrees
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the graph of |x - 2| + |y - 5| = 5 consists of four lines which are graphs of the following functions:\nx-2+y-5=5 which is  line A: x+y=12\n-x+2-y+5=5 which is line B: x+y=2\n-x+2+y-5=5 which is line C: x-y=-8\nx-2-y+5=5 which is line D: x-y=2\nnext thing to do is graph these four lines. the easiest way is to find the 0s. \nline A: x+y=12\n        x=0 y=12 point A1(0,12)\n        y=12 x=0 point A2(12,0)\nline B:  x+y=2\n         x=0 y=2 point B1(0,2)\n         x=2 y=0  point B2(2,0)\nline C: x-y=-8\n         x=0 y=8 point C1(0,8)\n         x=-8 y=0 point C2(-8,0)\nline D: x-y=2\n        x=0 y=-2  point D1(0,-2)\n        x=2 y=0   point D2(2,0)\nthese four lines enclose a figure that is a rectangle. lets find the length of each side. after you plot the four lines you will notice that line x+y=12 intersects with line x-y=2 lets call the intersection point I1 to find I1 coordinates we rearrange y=12-x and y=x-2 and from y=y we get x-2=12-x or 2x=14 or x=7 if x=7 y=12-7=7-2=5 the point I1 is I1(7,5).\nline x+y=12 also intersects with line x-y=-8. Lets call this intersection point I2. to find coordinates we have y=12-x and y=x+8 y=y or 12-x=x+8 or 2x=4 or x=2 for x=2 y=12-2=2+8=10 so the point is I2(2,10) \nside I1I2 of the rectangle length is side a. to find length of a we do\na^2=(7-2)^2+(10-5)^2=25+25=50\na=sqrt(50)\nwe need the second side, lets call it b.\nto find length of side b we need at least one more point. lets find intersection of x-y=2 and x+y=2 and call it point I3. y=x-2 and y=2-x or x-2=2-x or 2x=4 or x=2. for x=2 y=2-2=2-2=0 point I3(2,0)\nside b:\nb^2=(7-2)^2+(5-0)^2=50\nb=sqrt(50)\nit means a=b or our rectangle is a square!\nso the area of the square is Area=side^2 since our square&#92;s side is a=b=sqrt(50), the area is 50.\nI hope this helps, I really don't have time to explain it in more details.
456084
angle b and angle c.\nthey are both on the same line of the x- axis and collide at a central point with respect to A.
925243
e is the correct answer\n\ncoplaner means all in the same plane. the magnitude of the three can result in a, b, or c.\n\nd is impossible
921006
Area = πab where a and b are the semi-major and semi-minor axis length
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Draw the two points (a,b) and (c,d) on a graph so that the segment connecting them is NOT horizontal or vertical. Then make a right triangle with one vertical leg and one horizontal leg where the connecting segment is the hypotenuse.  \n\nThe length of the horizontal leg is c-a\nThe length of the vertical leg is d-b\n\nSo the length of the hypotenuse, which is the distance from (a,b) to (c,d), is\n\nsquare root of (c-a)^2+(d-b)^2
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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
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When dealing with simple shapes, an area is spanned up by two vectors. The area shape will have a normal vector indicating how it is oriented in space, resulting from the cross product of the two linear vectors with the length indicating the area units.
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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.)
35459
You need to apply vectors.
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polyeig(-5,-2,4,2) \n\nwill solve the equasion.  To learn about interpelations type:\n\nhelp interp1
636683
i think the 2 characteristics of a vector are-\n\nit has a definite direction and it follows the vector addition. i hope thats correct!!
773761
I believe the answer is biconvex and elliptical.
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The triangle will have 2 sides of length x, and the remaining side will be length P-2x making the parameter x + x + P - 2x = P.\nBegin by drawing a line from the vertex where the two sides of length x meet, to the center of the side which is length P-2x. The line should be perpendicular to the side P-2x thereby splitting the triangle in half, into 2 right triangles. One of the right triangles has hypotenuse = x and one leg = (1/2)(p-2x). So the remaining leg has length S.R. (x^2 - (p/2 -x)^2). We calculate the area of this right triangle, and double it since there are two triangles:\nArea = (2)(1/2){(1/2)(p-2x)S.R.(x^2 - (p/2-x)^2)\n         = (1/2)(p-2x)S.R.{px - (p^2)/4}\n         = (1/4)(p-2x)S.R.(4px-p^2)\nNote that your given answer is wrong - it has an x^2 where it should have a p^2....
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I drew the picture as you described it and it looks like a parallelogram with its diagonals drawn in.  MX is half of one diagonal and MR is one side of the parallelogram.  If this is indeed the case, there is no reason why MX is congruent to MR.  So, either you need a better description of what the figure looks like, or you left out a given, such as DN is congruent to XN or something like that.