PX and PY are tangent to center C from exterior point P.\nProve Cp is the perpendicular bisector of XY.
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You have to draw the lines and mark the points.\nIf Cp has a 90 degree angle with the line you just draw and crosses it right in the middle, then you have graphic proof that it is the perpendicular (which means it has a 90 degree angle) bisector (which means that it splits another line in two halfs).
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it's the the angular bisector of the angle between the two sides.\n\nKemmy, I always thought the perpendicular bisectors intersect at a point (called the epicentre) which happens to be equidistant from the vertices of the scalene triangle, as opposed to being equidistant from two sides of it. \n\nDo you have a link that proves it?
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csc (theta) cot (theta) -3, P(pi/2, 0)\n\n∫[cscθ cotθ - 3]dθ\n= -cscθ - 3θ + c\n\nWhen θ = π/2 expression = 0\nie -cscπ/2 - 3π/2 + c = 0\nSo c = 3π/2 + 1\n\nThus particular integral sought is -cscθ - 3θ + 3π/2 + 1\n\nMean Value Theorem states that in [a, b] there exists c such that \nf'(c) = [f(b) - f(a)]/[b - a] if f(x) is continuous on the closed interval [a, b] and differentiable on the open interval (a, b)\n\nf(x) = x^⅓; a = -1, b = 5\n[f(5) - f(-1)]/[5 - -1]\n= [5^⅓ + 1]/6\n\nf'(x) = 1/[3x^⅔]\n\nSo f'(c) = 1/[3c^⅔]\n\nNow solve 1/[3c^⅔] = [5^⅓ + 1]/6\nSo 3c^⅔ = 6/[5^⅓ + 1]\nie c^⅔ = 2/[5^⅓ + 1]\nc ≈ 0.6340 which is in the interval [-1, 5]\n\nHowever while f(x) is continuous on the closed interval [-1, 5] it is NOT differentiable at 0 which lies on the open interval (-1, 5) so the theorem falls over on this count
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1) B\n\nPolar Coordinates are in the from (r, theta) where,\n\nr = sqrt(x^2+y^2)\ntheta = arctan(y/x)\n\nSince x = -1, y = -sqrt(3), plug in these values:\n\nr = sqrt(1 + 3) = 2\ntheta = arctan(-1/-sqrt(3))\n\nYou can deduce the angle by figuring out which quadrant the point is in.  Since both x and y are negative, the point lies in the 3rd quadrant.  You know that the range of angles in this quadrant are 180->270 degrees.  Therefore, out of your choices, only B fits this constraint.\n\n2) B\n\nPlug in x = 0 and solve for f(x):\n\nf(x) = 3(0)^2 + 2(0) + 5 -----------> f(x) = 5\n\n----------\n\nHope this helps
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Hi,\nThe answer is C.\n\nYou could make a ray-diagram and see for yourself...
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There is a theorem (I think you remember it better). I just put it with my own words. Let points A, B, C & D belong to the same circle. Then the angle ABC=(Arc ADC)/2\nThus the Sum of 7 Arcs under 7 Angles will be = to 2pi radians, that means that the Sum of those Angles = pi radians, the main demand being they belong to the circle.
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it is C as in use the formula for CI for one sample proprtion from your book(i better get the best answer choice cause this is the fifth one i am answering)\nand the above guy is a moron
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Orthocentre = point where altitudes intersect (An altitude is the line from the vertex to the opposite side (extended if necessary))\n\nIncentre = centre of the incircle = point where perpendicular bisectors of each of the sides intersect. \nRadius of incircle = √[(s-a)(s-b)(s-c)/s] where s = semiperimeter\n\nCentroid = point where medians meet (Median is the line joining a vertex to the midpoint of the opposite sides) The centroid is the centre of gravity of the triangle and the centroid divides each median in the ratio 2:1 On a numberplane the centroid of a triangle is the point which is at the average of the sum of the three x- and y-coordinates of its vertices)\n\nOne other is:\nCircumcentre = centre of the circumcircle = point where angular bisectors meet.\nDiameter of circumcircle = a/sinA = b /sinB = c/sinC
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Let the trianglas be ABC & BCD\nAD forms 2 triangles, ABD & ACD\nAB=AC since ABC is isisceles\nBC+BD since BCD is isosceles.\nangle ABC=angle ACB\nangle DBC=angls DCB so angls ABD=angle ACD so ABD is congruent to ACD by SAS\n\ncall the point where AD crosses BC E\n\nso both vertex angles are bisected.\nlook at ABC.\nAE bisects BAC & AE is a commin side for BAE & CAE.\nsince AB=AC, the triangles are congruent by SAS\ntherefore BE=CE\n& angle AEB= angle AEC which are supplements so they must be right angles.\nQED
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(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)])
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You finally have xy=0 so x=0 or y=0 (unless it is a problem of modular algebra...)\n\n----\n\nNotice that xy doesn't "cancel": you have \n45xy=148xy\n\nBring all terms on the left:\n(45-148)xy = 0 ----> -103xy = 0\n\ndivide by -103:\nxy=0\n\nby the law of cancellation, the last equation is true for\n(x=0 and y whatever) OR (x whatever and y=0)\n\nSo there are infinite solutions for the couple (x,y)\n\n\nGraphically, the solutions are given by all the points of the x and y axes
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I know how to do it, but it's going to cost me a lot of work, and sorry I don't have the time!\n\nI can give you some tips to start off though:\n\nDraw an x,y diagram where the center of the circle is the origin. Draw AB as a horizontal baseline which is 21 below the origin.\n\nThen you will have to create functions for the circle and then for the lines that have the tangents with the circle that together form the triangle. Particularly the formulas for the triangle lines will be hard.\n\nYou will have to calculate at which other points than P the triangle lines touch the circle. For that you will need the derivative of the function for the circle.\n\nEnglish is not my native language, so I hoped I used the correct English terminology. Good luck!\n\nP.S.:  aytu_y, what you say about equilateral triangles is correct, but it doesn't work in this case because here point P is not in the middle! (PA = 23 while PB = 27)
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There are a number of ways to do it. Remember that a vector can be moved around to have its origin start anywhere.  \n\nDo not confuse the vector with the Cartesian coordinate of a point.  The notation is very similar, but they are different things altogether!\n\nSo to do this graphically, draw an xy plane.  Place U at the point (0,0) it ends at (0,4).  now V starts at (0,4) and extends 3 units horizontally, and -5 units vertically. so the ending point is (0+3, 4+-5)= (3,-1)\n\nNow you wish to find the vector equivalent to the two vectors placed end to end.  This vector needs to start at (0,0) and end at (3,-1), so it is <3,-1>.  We could have added the vectors in the opposite order, and come to the same answer.\n\nPlease note that we did NOT need to start at (0,0) but anywhere in our plane.  We only chose it because it was a convenient place to start.\n\nThis was called the "Parallelogram rule", at least they called it that 30 years ago!\n\nIf you do this for several problems by hand on graph paper, you will see why we can simply add the corresponding values of the two vectors.
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No picture to be found.  \n\nAre M, N, O, P all on the same line?  \n\nIf so, NOQ is isoceles, which makes MNQ and QOP congruent angles.  Since two sides and the included angle uniquely define the third side, so MQ and QP are congruent. and MQP is isoceles.
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look std 10 th NCERT mathematics text book... its in the part 2 of  mathematics in geometry section.
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let line 1 make an angle x with the x-axis\nso the line perpendicular toit will make an\nangle of 90*+x with the xaxis\nm1=tanx\nm2=tan(90+x)\n=-cotx\nm1m2=tanx*(-cotx)=-1\nhence proved
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The orthocentre of a triangle is the point where all the altitudes meet\n\nSo if you have ΔABC one altitude is the line through A perpendicular to BC, another is the line through B perpendicular to AC and the last is the line through C perpendicular to AB\n\nUse these facts with the coordinates of A, B, and C and the slopes of AB, AC and BC (Find where two altitudes meet and if you are correct that point will also lie on the 3rd altitude)\n\neg Find the orthocentre of the ΔABC where A ≡(1, -3), B ≡(-1, -5), C ≡(3, 1)\n\nSlope AB = Δy/Δx\n= 2/(2) \n= 1\n\nSlope of perpendicular = -1 (-1 * 1 = -1)\nAltitude to AB goes through C\nSo equation of altitude of form y = -x + b\n(3, 1) lies on altitude\nTherefore 1 = - 3 + b\nb = 4\nThus altitude through C is y = -x + 4\nie x + y = 4 ............... (1)\n\nSlope AC = Δy/Δx\n= 4/2\n= 2\n\nSlope of perpendicular = -1/2 (2 * -1/2 = -1)\nAltitude to AC goes through B\nSo equation of altitude of form y = -1/2 x + b\n(-1, -5) lies on altitude\nTherefore -5 = -1/2 * -1 + b\nb = -11/2\nThus altitude through C is y = -1/2 x - 11/2\nie x + 2y = -11 ............... (2)\n\nSlope BC = Δy/Δx\n= 6/(4) \n= 3/2\n\nSlope of perpendicular = -2/3 (3/2 * -2/3 = -1)\nAltitude to BC goes through A\nSo equation of altitude of form y = -2/3 x + b\n(1, -3) lies on altitude\nTherefore -3 = -2/3 * 1 + b\nb = -7/3\nThus altitude through C is y = -2/3 x - 7/3\nie 2x + 3y = -7 ............... (3)\n\nie we have:\n x + y = 4 ................. (1)\n x + 2y = -11 .................. (2)\n2x + 3y = -7 ............... (3)\n\nEquation (2) - equation (1)\ny = -15\nFrom equation (1) x = 19\n\nCheck in equation (2) \nx + 2y = 19 + -30 = -11 YESSSSSSSSS!!!!!!!!!!!!!\n\nCheck in equation (3) to ensure that this point lies on this altiude also\n2x + 3y = 38 + -45 = -7 YESSSSSSSSS!!!!!!!!!!!!!\n\nSo the orthocentre of ΔABC is (19, -15)