find the equation of the line passing through (1,3), parrallel to x=5
121201
When x= 5 ,this meant slope is equal to zero..\n\nwhen m= 0 \n\nand the value of y can be ignored....\n\nso the answer would be x= 1
408464
Notice the first equation only involves x.\nx - 5 = 6\nAdd 5 to both sides\nx = 11\nNow replace x with 11 in the second equation\n3(11) - 2y = 5\n33 - 2y = 5\n-2y = 5 - 33\n-2y = -28\ny = -28/-2\ny = 14\nAnswer: (11,14)
371295
3x = -5 simplifies to x=-5/3\nThis is simply a vertical parallel to the y-axis i.e. at every point y, x=-5/3. \n\nThis means that a parallel line is also one that is constant for x. If it has to go through (9, 6) then it must be the line x=9.
502078
-b/(2a)\n5(2*3)\n5/6\n\nsub 5/6 back in the equation to get y
361189
when you multiply the 2 through on the right side of the equation you get:\ny = 2x-2+3\ny = 2x+1\nthe slope of this line is the number in front of the x.
556889
Well you know that perpendicular slopes are opposite reciprocals so in this case the slope of the line you are finding is: (1/3)\n\nThen you have to plug it into the equation with the given points.\n\ny = m(x) + c\n5 = (1/3)(0) + c\n5 = c\n\nSo then your final equation is: y = (1/3)x + 5
564366
a) - Just substitute the numbers, so the problem becomes:-\n\n5+ 3(4-2) - (Do the sum in brackets first then multiply the answer by 3 then add the 5)\n\nb) - Y=3(x-5)(x-2) - Lose the brackets first by multiplyiny what's inside by eachother - (x-5)(x-2) - First multiply the xs, then the x in the first bracket by the 2 in the 2nd, then the minus 5 in the first bracket by the x in the 2nd, then the minus 5 in the 1st by the minus 2 in the 2nd. Add together the 4 terms and multiply by the 3 outside the brackets, and you should get to one of the options given.\n\n\n3) - Should be fairly obvious from the options given - Y = x+5 so Y is 5 greater than X \n\nHope this helps
263281
You need to get the y alone on one side of the equation. Subtract 5x from both sides and you get:\n-3y=-5x-3\nDivide each variable by -3 to get:\ny=(5/3)x+1\nNow you can graph using this equation.\nThe y-intercept is 1 and the slope is 5/3.\nHope this helps.
133486
The midpoint of that segment is (3, -2) - all you do is take the average of the x coordinates and the average of the y coordinates.\n\nThe slope of that segment is found by\n\nm = (y2 - y1)/(x2 - x1)\n= (1 - (-5))/(4 - 2)\n= 6/2\n= 3\n\nSince you're looking for a line perpendicular to this one, take the slope, invert it, and change the sign.  So the slope of the new line will be -1/3.\n\nNow you have a point and a slope for the new point: m = -1/3, and point (3, -2).  Use the point-slope formula to find the equation of the line:\n\ny - y1 = m(x - x1)\ny - (-2) = (-1/3)(x - 3)\ny + 2 = (-1/3)x + 1\ny = (-1/3)x - 1
367865
problem1\n\nsince you have said P (4,-5) is a point on the function, then the corresponding point will be (4,-5) .. :)\n\nproblem2\n\nfor a line having y intercept 3 will have following form\n\ny =mx+3\n\nwhere m is the slope of the line. Now it is said to be perpendicular to y =2/3x-4. So slope for this is 2/3\n\nsince slopes of two lines perpendicular to each other if multiplied yields -1\n\nm X 2/3 = -1\n\nor m = -3/2\n\nso the required line will be\n\ny = -3/2 x+3\n\nproblem3\n\nf(x) = 1/x^2 -2x-8\n\nto make the f(x0<0, let us try with some arbitary values of x\n\nif x = 1, f(x) = -9\nif x=2, f(x) = -11.75\nif x = 0, f(x) = infinite (due to 1/x^2)\nif x = -1, f(x) = -5\nif x= -2, f(x)= -3.75\nif x=-3, f(x) = -1.88889\nif x=-4, f(x) = 0.0625\n\nHope this helps
438025
find the equation of the line that passes through the point of intersection of the lines x-3y=4 and 3x+y=2 and is perpendicular to the line 3x+4y=0. Give your final answer in the form ax+by+c=0 where a,b and c are integers to be determined.\n \nfirst we find the point of intersection\n\nx-3y=4,3x+y=2\n\n>>>>x=1,y= -1\n\ntherefore, point of intersection of two lines\nis(1,-1)\n\n3x+4y=0\ny=(-3/4)x\n\nthe gradient of line perpendicular to this line\n= -1/gradient\n\nperpendicular line passes through (1,-1) and \nhas a gradient of 4/3\n\nequation of perpendicular is\n\ny=4/3x+c  say\n\nsubstitute values at point(1,-1)\n\n-1= 4/3+c\n\n>>>>>c= -7/3\n\nequation of line is y=4/3x-7/3\n\n>>>>>3y-4x+7=0 \n\ntherefore, the equation of the line that passes through the point of intersection of the lines x-3y=4 and 3x+y=2 and is perpendicular \nto the line 3x+4y=0 is\n\n3y-4x+7=0\n\nwhen doing these problems,it helps to draw \na rough graph of the situation\n\nanother answer could be\n\n4x-3y-7=0 or 300y-400x+700=0\n\ni hope that this helps
411872
Equation of a line is y=mx+b  m=slope  b=y intercept.\n\nPlug the first set into this equation:\n\n3 = 2(4) + b\n\nb= -5 \n\nNext plug in 7 for y and solve for x\n\n7= 2x - 5\n\n2x = 12\n\nx = 6
441789
y=-1/3x + 6 is your line\ny=-1/3x+1 is a parallel line
376210
C is correct but not complete... only the 3 is factored out... the real answer would be:\n\n3(x-1)(x+4)
204486
y=(x+8)^3-5\ny+5=(x+8)^3\n(y+5)^(1/3)=x+8\nx=(y+5)^(1/3)-8
85988
move everything to one side...\n\nx^3+5x^2-x-5=0\n(x^2-1)(x+5)=0\n\nsolve each parenthesis\n\n1). x^2-1=0\nx^2=1\nx=+1 and -1\n\n2). x+5=0\nx=-5\n\nso the answers are 1, -1, -5
387478
The parabola y = (x+3)(x-5) crosses the x axis at -3 and 5.  It opens upward, so its vertex must be below the x axis.  So y < 0 for x in (-3, 5).