the rectangular coordinates of a point are x=7, y=-24. find the polar coordinates of the point.?
599995
abuse of yahoo! answers. This is very simple and u SHOULD be able to solve it yourself!!!
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In the polar coordinate system r is always positive. It´s the distance from the origin to the point. \n\nThe transformation to rectangular (or rather carthesian) coordinates is as follows:\n\nx = r cos (theta)\ny = r sin (theta)\n\nSo assuming (r,theta) = (5,52 deg) I find\n\nx = 5 cos (52 deg) = 3.08\ny = 5 sin (52 deg) = 3.94 (both rounded off to two decimals)
123034
If it is y = x^2+4x+6 then proceed as follows\ny = x^2+4x+4-4+6 (add and subtract 4)\ny = (x+2)^2+2\ny-2=(x+2)^2\n\nFor vertex y-2 = 0 i.e. y = 2 and x+2 = 0 i.e. x=-2;\nSo vertex coordinates are (-2,2).
816872
Best to plot this on a piece of graph paper; "+i" is  to the right' , "+j" is upwards, "-i" is to the left, "-j" is downwards, all with respect to the origin.\n\n1. Draw the vectors on the graph paper. "u" is in the second quadrant (i is neg and j is pos). "v" is in the fourth quadrant, with (i is pos, j neg) . There are two answers to this question, since you could go from u to v or from u to v in the opposite direction. I always take the angle from zero to "u" and add it to the angle from 0 to "v".  For "u", the tangent of the angle to the vertical axis is 1/4, then angle = 14%. So the angle from zero degrees to the vert axis = 90, and "u"s angle from zero degrees is 90 + 14 = 104 degrees. For "v", the tangent of the angle to the horizontal axis is 4/3 = 1.333.. and the angle = 53 degrees. Therefore the angle from "u" to "v" is 104 + 53 = 157 degrees. (the other answer is 360 degrees - 157 degrees = 203 degrees if you rotate in the other direction).\n\n2. Again use graph paper. 240 degrees is 30 degrees less that 270 degrees (the vertical axis, down) draw a line 10 graph units long. It will be in the third quadrant where i and j will both be negative. At this point you can use a polar to rectangular converter, or do it by hand. By hand: the i side of the triangle formed with the down axis can be found by saying sin(angle) = i/10, and cos(angle) = j/10. The angle is 30 degrees (from vert downwards). so sin(30) = 0.5, and, cos(30) = 0.866; thus i= 0.5 and j = 0.866 (ignoring pos/neg). Since the vector is in the third quadrant, both i and j are negative.\nAnswer = -0.5i -0.866j\n\nHope this helps!
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X = (0*Y) -5\nY = 0 * X\n\nbut basically just write X = -5 and thats it
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a) Algebraically the x-coordinate of the vertex is x= -b/(2a)\n\nHere that is x = -10/(2 X 1) = -10/2 = -5\n\nPlug that in for x to get the y-coordinate of the vertex.\n\nb) To get the x-intercepts, replace y (here, g(x)) with 0 and solve.\n\n0 = x^2 + 10x + 21 = (x+7)(x+3)\n\nx = -7 or -3 so there are two x-intercepts: -7 and -3.\n\nc) to find the y-intercepts, replace x with zero. \nSo y = 0 + 0 + 21 = 21  \n21 is the y-intercept.
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x+y=-7\nxy=60\n\n(x+y)^2=x^2+2xy+y^2\n49=x^2+y^2+120\n-71=x^2+y^2\nx^2>=0 and y^2>=0\nso I don't think there is a solution\n\nif their sum is -7 and their product is -60... then 5 and -12 are your numbers
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The easiest way to graph this equation is to plug in 0 for x and get one point, then plug in 0 for y and get another, then draw the line that connects the two points.\n4(0) - 3y = 24\n-3y = 24\ny = -8\n(0, -8)\n\n4x -3(0) = 24\n4x = 24\nx = 6\n(6, 0)
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considering both y values are -2 for A and C....\n\nwe know that a straight line exists between these two points, which means that it HAS to be 1 of the 4 sides.\n\ngiven that, the coordinates or K and L HAVE to be (3, infinity (EXCEPT -2)) and (-4, infinity (EXCEPT -2))
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using @ for angle theta:\n\nx = r cos @, y = r sin @, use cos^2 @ + sin^2 @ = 1\n\nr^2 = 1/4*(1-5/4*sin^@)
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(y+1)(y-7)\n=y^2-6y-7
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y=1 is a straight line parallel to x axis\nx=(-1/4) is a straight line parallel to y axis
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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
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Let the coordinates of point P be (x,y). Here, x and y represent general points.\n\n1) Find an expression for the slope of OP. To help you get started, I will tell you the answer directly. It is (y/x).\n\n2) Find an expression for the slope of AP. Try this yourself!\n\n3) Use the statement "slope of OP is 1 less than the slope of AP" to construct an equation in terms of y and x.\n\n4) Carry out some algebraic manipulation to obtain your final answer!\n\n5) Lastly, take a look at your final answer. Are there some coordinates that point P cannot take? For instance, your final answer should be an equation that has (0,0) as 1 of the solutions. But can point P take this set of coordinates? What other coordinates can P not take?
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y= (x^2 - 1)^2\nx^2-1=+/-√y\nx^2=1+/-√y\nx=+/-√(1+/-√y)
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y = (x+3)(x-1)(x-2) = x^3 - 7x + 6
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x=3y+7\nx=3(0)+7\nx=7\nHope this helps! ^^\nalso the y intercept is like this:\nx=3y+7\n0=3y+7\n-7=3y\n-2.23333...=y