for line segmet PQ one endpoint is P(-2,-6) and midpoint is M(-4,1). Find the coordinated of endpoint Q.?expla
363113
The midpoint of two points P(a,b) and Q(x,y) is\nM((a+x)/2,(b+y)/2).  That is, you add the x coordinates together and divide by 2 and then add the y-coordinates together and divide by 2.\nUsing P(-2,-6), Q(x,y) and M(-4,1), you have the two equations\n(-2+x)/2 = -4 and (-6+y)/2 = 1\nSolving for x and y you get\nx = -6 and y = 8\nThus, the other point is (-6,8).
669580
-E-N-G-L-A-N-D- 2 POINTS
826437
(x/a)²+(y/b)² = 1 (← Note my variation! I am using a², b² in place of your a, b)\n\nx = acos t , y = bsint is its parametric equation, a, b >0\n\nThen if T1(acost, bsint), (0 ≤ t ≤ π/2),  is the point on the ellipse in the 1st quadrant, the point on the ellipse vertically below it is\nT2 (acost, -bsint).\nThen the point horizontally to the left of that on the ellipse is\nT3 (-acost, - bsint)\nThe final point is vertically above on the ellipse namely\nT4 (acost, -bsint) which is also horizontally to the left of the point T1\n\nSo perimeter P = |T1T2| + |T2T3| + |T3T4| + |T4T1|\n= 2bsint + 2acost + 2bsint + 2acost\n= 4(bsint + acost)\n\ndP/dt = 4(bcost - asint)\n= 0 for stationary points\nSo asint = bcost\nie sint/cost = b/a\nSo t = arctan(b/a) (0 ≤ t ≤ π/2) (ie a,b>0)\n\nd²P/dt² = -4(acost + bsint)\n< 0 for (0 ≤ t ≤ π/2)\nand therefore P is maximum when t = arctan(b/a) (0 ≤ t ≤ π/2)\n\nIf t = arctan(b/a) then tan t = b/a \nThus cost = a/[√(a² + b²)] and sint = b/[√(a² + b²)]\n\nThus T1 ≡(a²/[√(a² + b²)], b²/[√(a² + b²)]), \nT2 ≡(a²/[√(a² + b²)], -b²/[√(a² + b²)]),\nT3 ≡(-a²/[√(a² + b²)], -b²/[√(a² + b²)]),\nT4 ≡(-a²/[√(a² + b²)], b²/[√(a² + b²)])
907364
Almost.  You would actually include the endpoints 0 and -3 (so you need to use a bracket instead of parentheses).\nD = [0, infinity)\nR = [-3, infinity)
587217
First, always make the side with the squared term a perfect square so the equation is in standard form.  This is done by moving the constant term (25) to the left side so you have x - 25 = y^2 - 14y.  Now, complete the square on the left and your equation now becomes\nx - 25 + 49 = y^2 - 14y + 49 => x + 24 = (y - 7)^2.  It is of the form (x - h) = 4p(y - k)^2.  Whenever a parabola is written in this form, the vertex is always (h,k), or in this case, (-24,7).  For your equation, 4p = 1, so p = 1/4.\nSince the parabola opens to the right, the focus is 1/4 unit to the right of the vertex, so its coordinates would be (-24 + 1/4,7) = (-23 3/4,7).  The axis of symmetry is found by setting the squared term in your equation to 0 and solving for the variable.  In this case, 4p(y-7)^2 = 0 => y = 7 is the axis of symmetry.  The latus rectum is always 4p units long and in this case, 4(/14) = 1.  The latus rectum is a chord that goes through the focus and is perpendicular to the axis of symmetry.  Finding the direction of opening is simple.  Write the equation in standard form as I did above and make sure linear term is positive, then observe the following.  If the equation looks like (x - h) = 4p(y - k)^2 and 4p is positive it opens to the right, if 4p is negative, it opens to the left.  If the equation looks like (y - k) = 4p(x - h)^2 and 4p is positive it opens up, if 4p is negative, it opens down. In all cases, the focus is always p units from the vertex and the latus rectum is always 4p units long and the vertex is always (h,k).  Let me know if this is not clear.
187542
Answer:\ny=-1.25x + 3\n\nProof:\nWe are given two points to work with:\nPoint 1: (x1=0, y1=3)\nPoint 2: (x2=4, y2=-2)\n\nSo, use the "two point equation" to solve this problem.\ny-y1 = (y2-y1)/(x2-x1) * (x-x1)\n\ny-3 = (-2-3)/(4-0) * (x-0)\ny-3 = -5/4 * x\ny = -5/4 * x + 3\n\ny=-1.25x + 3
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the sides are 2 units (3-1=2) \nand the ends are 10 units. (6-(-4)=10)\n\narea = side times end = 2*10 = 20\nperimeter = 2 * side + 2 * end = 2*2 + 2*10=4+20=24
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I think that you want x element of [-6,4)\n\nYou wanted closed on the left and open on the right, is that true?\n\nJust remember that for intervals:\nopen   ( )\nclosed [ ]\n\nAnd you can match them up: [-6,4)
55111
The formula for a midpoint is (X1+X2)/2 for X and (Y1+Y2)/2 for Y.  For your specific problem, the answer would be: (6,9)
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What is n?  what is m?  What is q?  How can we come up with an answer without this data?\n\nI'm going to assume that q is the midpoint the segment or segments, and that m and n are also equidistant from q.\nThen you set up a right triangle, you're looking for the hypoteneuse length and the legs measure 10 and 16.\n\nIf that's so, then lm = √(10²+16²)\nlm = 2√(5²+8²)\nlm = 2√89
794182
What nicknamr said, and also, the vertex would be located at the x-value found for the axis of symmetry, and the y-value for that x-value.  So, -b/(2a) = -12/(2 x -1) = 6.  Then, y= -(6)^2 + 12(6) -32 = -36 + 72 -32 = 4.  The vertex comes out to be (6,4).  (I would have answered all the other stuff too, but he already did it, and I found it rather redundant to repeat it all, so I just added on to it =)  )
930136
Use the slope formula to get the slope.  All parallel lines will have the same slope.\n\nm = (y2-y1) / (x2-x1)\n\nSo, for your points you have the following:\n\nm = (2-1) / (4-15) = 1/(-11) = -1/11\n\nTherefore all lines parallel to this line will also have a slope of -1/11.\n\nHope that helped!\n\n~ ♥ ~
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I should tell all answers above are wrong or doesn't have\nproper claim for it.\n\nEven though 1/3 pi r*r * h \nwhere h is the height of midpoint looks correct.
279710
I did a search here's where I found the answer\nhttp://www.jcu.edu/math/isep/Quincunx/Quincunx.html
876350
Sorry, but your question doesn't make sense.  Maybe you mean the midpoint of one of the sides of a right triangle.
510711
Look at the triangle formed by the line segment AB.  The diagonal side(I'm assuming it's AB) is the hypotenuse.  Label the side across from θ as D.  Label the straight side next to θ as E.  \ncscθ equals the hypotenuse divided by D\ncscθ=(AB)/D\nI don't know what m is supposed to be in the equation, so I can't help you there.\n\nEdit: To find m, take the inverse tangent of D/E.
745771
first find the midpoint of AB which is C(9,-2).\nNow obviously the one-fourth distance will be the midpoint of AC which is ...well I let you do this. This is simple knowing the formula for the midpoint .\nMathguy did well too but you are supposed to apply, I think, midpoint formula.