to the nearest hundreth. Im in calc II so it has something to do with cross products
137111
kate1976 has solved the problem using the traditional way, just using the concept of coordinate Geometry.  Usin the cross product of two vectors, the method is different.\n\nDot and Cross Products:\nSuppose [x1,y1] = [ r1 cos(q1), r1 sin(q1)] and \n[x2,y2] = [ r2 cos(q2), r2 sin(q2)] are points in a plane. Then their dot product\n\n[x1,y1].[x2,y2] = x1x2+y1y2  (dot product definition)  ..(1)\n\nand their cross product\n\n[x1,y1].[x2,y2] = x1y2 - y1x2 (cross product definition).(2)\n\nmay be expressed in terms of  trigonometric functions and the angles between the two points, or more precisely their position vectors.\nTo obtain the geometric interpretation,  observe the polar and rectangular ways to multiply the first point by the complex complex conjugate of the second, when both are viewed as complex numbers: That is,\n\n [x1,y1][x2,-y2] = (r1,q1)(r2,-q2)\n\nFrom the equality of two different ways to multiply points in the plane,  observe\n\n[x1x2+y1y2, x1y2 - y1x2]  = (r1r2,q1-q2)\n\nbut\n\n(r1r2,q1- q2) = [ r1r2  cos( q1- q2), r1r2  sin( q1- q2)]\n\nTherefore comparison (equality) of real and imaginary parts yields:\n\nx1x2 + y1y2 = r1r2  cos(q)\n\nand\n\n x1y2 - y1x2 = r1r2  sin( q)\n\n where\n\n q = q1- q2\n\nis the angle between the two points  [x1,y1] = (r1 , q1) and [x2,y2] = ( r2 , q2)\n\n!!!!!!!!!!! Please read q as theta !!!!!!!!!!!\n\nNow proceed using the equations (1) and (2)
391204
Yeah, you described it very poorly.  I don't even know where X should be.  But from what I've read, you're probably dealing with a transversal.  That should be easy.\nSupplements, interior angles, corresponding angles, exterior angles, supplementary/complementary angles  - ring a bell?
170034
The find the lower bound on a/b, you want to find the smallest possible distance a for any fixed short distance b. The closest possible way to fit 2 towns together is just b apart. The closest way to fit a 3rd town is to make an equilateral triangle. The closest way to fit the next 3 towns is to make 3 more equilateral triangles on each side of the first one, resulting a larger equilateral triangle of each side length 2b. The longest road between 2 towns (without another in between) is from one vertex of the triangle to the town in the middle of the opposite side. This makes a right angled triangle with hypotenuse length 2b and a side of length 1b. Use pythagoras' theorem to prove the other side, a, is equal to root 3 times b, and hence a/b is equal to root 3.
481174
take lcm\n(9-10a)/5a^2\nthis is how far it can be done
409328
y = M(n+1)\n   = ½(n+1)²\n   = ½(n²+2n+1)\n\nI think that's the furthest you can go. \nHope this helps=)\napl
533956
http://froogle.google.com/froogle?q=r(x)%3D+x%5E3%2B8/x%5E2%2B4&hl=en&lr=&sa=X&oi=froogle&ct=title
625630
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.
374338
Working above is good until this step:\n\n2 + 0.8x >0.9(x + 6)\n2 + 0.8x >0.9x + 5.4)\n0.1x > -3.4 ← Symbol error in this line\n\nShould read 0.1x < -3.4\nie x < -34
74049
For the distance problem, you simply use the distance formula:\nd=sqrt[(x1-x2)^2 + (y1-y2)^2]  \nwhere (x1,y1) is your first set of quardinates:(2,-1)\n(ie x1=2 and y1=-1)\nand where (x2,y2) is your secound set of quardinates:(-3,8)\n(ie x2=-3 and y2=8.)\n\nSo we would have:\n\nd= sqrt[(x1-x2)^2 + (y1-y2)^2]\nd= sqrt[(2-(-3))^2 + (-1-8)^2]\nd= sqrt[(5)^2 + (-9)^2]\nd= sqrt[25 + 81]\nd= sqrt(106)\nd= 10.296\n\nFor the system of equations, you would solve for x in the first equation, and then substitute that value into the second equation.\n\nlike this:\nx-y=3\n +y +y\nx=3+y\n\nthen you take the second equation\n2x + 3y=11\nand substitute your answer for x (x=3+y) into the x in the second equation. \n(x gets replaced by 3+y)\n2x+2y=11\n2(3+y) + 2y=11\nthen you solve for y\n2(3+y) + 2y=11\n6+2y + 2y=11\n6+4y=11\n4y=5\ny=4/5 or 0.8 which ever you like best, but remember fractions are our friends!\n\nanswering these on the calculator would be difficult and require some work by hand anyway, so its easier just to solve these by hand.\n\nI hope that helps.
192095
ok, u have to use distance formula\n\nsqrt[(-4-(-7))^2+(c-7)^2]=5   \nsquare both sides\n[(-4-(-7))^2+(c-7)^2]=25      \nadd the stuff in the parentesies\n[(-3)^2+(c-7)^2]=25           \nexponents\n9+(c-7)^2=25                  \nsubtract from both sides\n(c-7)^2=16                    \nsquare root both sides\nc-7= +/-4                     \nadd to both sides\nc=11 or c=3
390495
I think the best answer would\nbe to give y in terms of x.\nThis is :\ny = (6302869x^2 - 384932345355.22x - 59096) / (231689354x + 23431653.57948)
495470
I know you want something more accurate than rÅvi suggested.  It's a long time since I dealt with this stuff, but do you do cross product and dot product?  If so, work out the cross product of two of them (i don't remember exactly how to do this and would have to play with it for maybe half an hour to work it out for myself), as this gives a vector perpendicular to the two vectors.\nThen work out the dot product (I do remember this one!  It's \nx1*x2 + y1*y2 + z1*z2)\nand if it's zero then the third vector is also perpendicular to the cross product, so the three of them are coplanar.\n\nIn brief, a condition for v1, v2, v3 to be in the same plane is\nv1.(v2xv3) = 0  where I've used . to represent dot because I don't know how to type a dot here.
31396
http://mathforum.org/library/drmath/sets/high_calculus.html
551234
a common way of handling this type of problem is the least squares method. Instead of finding the minimum of the distance distance to each point (difficult because of the absolute values) find the minimum of the squares of the distances (always a possitive value) So the formula you want to diferentiate is:\n\nD=(sqrt((x-5)^2+(y-5)^2)-4)^2 + ...\n\nthe more complex problem is what is meant by proximity to the circles. from your equation, it seems that it is only to the nearest point on each circle, but it could also mean to the centers of each circle (significantly easier) or to the area of each circle.\n\none of the first things you should do is plot the circles and eyeball guess. This gives you an idea of what the correct answer should be. With 20+ circles it is going to be difficult to eyeball. It also doesn't make much sense because there will be circles hiden behind other circles.\n\nDepending on how accurate you need to be, there are several ways to guess:\n\na really quick way would be to do a center of mass calculation (this would give you the closest point to the areas of the circle and could be done very easily on a spreadsheet or code.
527973
Use the distance formula:\n\nD = √ [ (y2-y1) ² + (x2-x1) ² ]\n\n1)  D = √ [ (-2+4) ² + (-2+4) ² ] = √(4+4) = √8 = 2√2\n\n2)  D = √ [ (3+1) ² + (-2+7) ² ] = √(16+25) = √41\n\n3)  D = √ [ (4-5) ² + (6+3) ² ] = √(1+81) = √82\n\nNRT
465157
A. Just multiply CF and then subract the resulting matrix from C. Is there something specific about the required operations you don't understand?
420918
http://www-math.mit.edu/~sipser/book.html\n\nI hope this may be of some help.