What is the reflection of the point (3,4) in the line 3x+4y=50?
962974
the reflection of point (3,4) will be on a line perpendicular to 3x+4y=50 and passing by the given point, and at equal dist from the line to that of the point(3,4) and the line.\n\nanyway I'll try to show u how to solve that\n\nfirst we have to write the equation of the line passing thru (3,4) and perpendicular of y= -3/4x + 50/3\nm1 is the slope of the equation y=-3/4x + 50/3 so m1=-3/4\nm2 is the slope of the equation perpendicular to the above one\nm1 * m2 = -1\nso m2=4/3\n\nthen we will have y=4/3x + b\nreplacing x and y in the above equation we get\n4 = 4/3 *3 +b so b = 0\nthen the equation passing thru point A(3,4) and perpendicular to the equation 3x+4y=50 will be:\ny = 4/3 x\n\nNow to find the distance between point A(3,4) and the equation\n y=-3/4x +50/4 we should get the intersection point of the two perpendicular lines\nto do that we set the two y coordinates equal\n4/3x = -3/4x + 50/4\n(4/3 + 3/4)x = 50/4\n25/12 x = 50/4\nx = 6\nand replacing x in either equation we get y=4/3 * 6 = 8\nthen the intersection point will be I(6,8)\nLet point B(xb,yb) be the reflection of A\nso I midpoint of AB\nthe xi = (xb + xa)/2 => 2xi - xa = xb = 2*6 - 3 = 9\nand yi = (yb + ya)/2=> yb = 2yi -ya = 2*8 - 4 = 12\n\nso the reflection of point A(3,4) is point B(9,12)
738757
3x - 4y + 17 = 0\n\n3x - 4y + 17 - 3x = 0 - 3x\n\n- 4y + 17 = -3x\n\n- 4y + 17 - 17 = - 3x - 17\n\n-4y = - 3x - 17\n\n-4y/-4 = - 3x/- 4 - 17/- 4\n\ny - 3x/4 + 17/4\n\n- - - - - - - - - - - - - -  \n\n4x + 3y + 6 = 0\n\n4x + 3y + 6 - 4x = 0 - 4x\n\n3y + 6 = - 4x\n\n3y + 6 - 6 = - 4x - 6\n\n3y = - 4x - 6\n\n3y/3 = - 4x/3 - 6/2\n\ny = -  4x/3 - 2\n\n- - - - - - - - - - - - - -\n\n3x + 4y - 41 = 0\n\n3x + 4y - 41 - 3x = 0 - 3x\n\n4y - 41 = - 3x\n\n4y - 41 + 41 = - 3x + 41\n\n4y = -3x + 41\n\n4y/4 = - 3x/4 + 41/4\n\ny = -3x/4 + 41/4\n\n- - - - - - - -s-
182913
The slope of the line is 3/4.
167128
There's an infinite number of answers.\n\nIf you do it all out, they both equal the same line...\nI multiplied the second equation by a -4 and got 0=0.\n\nIf you just multiply it by a 4 you'd see that it's the same exact line.
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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))
384702
The first 3 are all straight lines, so you just need to get 2 points in each.\n1. y = |x| + 3 for x = 1 then y = 4, for x = 2 y = 5.\n\n2. y = 3/2x - 2 for x = 2 then y = 1, for x = 4 y = 4\n\n3. y = x - 5 for x = 5 y = 0 for x = 6 y = 1\n\nFor #4. This is a curve shaped like and upside down U that passes through 0,0, for all values of x, y will always be negaive.\n\nThe best way is to substitute values of x and get values of y and sketch.\n\nHope I was some help
183640
20x +  4y = 32    the first line *4\n-20x - 20y = -80   the second line *5\n________________   add lines \n       -16y = -48 => y=3 =>x=1
945033
Derivative is y=cos x -5. Substituting in 0 gives you the slope -4. Now plug in the coordinates (0,3) to get the line, which is y=-4x+3
941047
The answers are:\na)35deg\nb)50deg\n\nThings to remember: The angle of incidence is the angle between the beam of light and the line drawn perpendicular to the mirror, NOT between the beam of light and the surface of the mirror. Also, angle of incidence = angle of reflection, so, for part a), the beam reflects towards M2 at an angle of 35deg. Form a triangle with the two intersecting mirrors and this reflected beam (internal angles in a triangle add to 180 deg). You will find that the angle of incidence onto M2 is also 35deg.\n\nFor part b): the angle of reflection from M1 is 60deg, and again forming a triangle with the intersection of M1 and M2, you will find that the angle of incidence for M2 is 50deg.
40850
-3 <4x\n x < -3/4\n solution set : {- infinity ,.............,-3/4}
781563
y>(12-3x)/4\n\nSteps:\n\nmultiply the 12 on both sides\n3x+4y>12\n\nminus the 3x isolating the y variable\n4y>12-3x\n\ndivide by 4 on both sides to get y by itself\ny>(12-3x)/4 or...\n\nsimplified 3-(3x/4)
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)
417188
y=Asin(2x)+Bcos(2x)+(x/4)\n\nNow you get to determine why.
269336
To solve this problem, you have to substitue the coordinates into point-slope form using the slope (-5) as m.  The y1 refers to the y coordinate, while the x1 refers to the x coordinate.\n \ny-y1=m(x-x1)\ny+12=-5(x+12)\ny+12=-5x-60\ny=-5x-72\n\nslope=-5\ny-int.=-72\n\nThe person above me has everything right except when she had -12=-5(-12)+b.\n-12=-5(-12)+b\n-12=60+b\n-72=b\nb=y-int=-72
800260
4y+3=2y\n      3=-2y\n-1 1/2= y\n\n\nslope- 3
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a) x int when y=0\n-3x+8y=24\n-3x+8x0=24\n-3x=24\n24/-3=x\nx= -8\n\ny int when x=0\n-3x+8y=24\n-3x0+8y=24\n8y=24\n24/8=y\ny=3\n\nso then on your graph put a point at -8 on the x axis and a point at 3 on the y axis.\n\nget the idea? i hope this helps. missy_sven@hotmail.com if you need more help. maths a b*tch!! here one more:\n\nc)1x-4y=32\nx int when y=0\n1x-4x0=32\n1x=32\n32/1=x\nx=32\n\n\ny int when x=0\n1x-4y=32\n1x0-4y=32\n-4y=32\n32/-4=y\ny= -8\n\nthen graph it the same way
973940
A previous answer was incorrect. It is\n\n-4x(x+3y)(x-2y)