How do you graph linear inqualities ninthgrade math ?
402272
1.   First solve for the Y (e.g.  Y > 2X + 1)\n\n2.   Then graph the linear equation (convert the inequality to an equation by changing the inequality sign to = and graph the equation\n   so   Y > 2X + 1 --------------  Y = 2X + 1\n\n3.   Get any point that is not on the line you drew\n\n   Let us say (0,2). Substitute the coordinates of this  \n   point to the corresponding X and Y of the inequality\n   Y > 2X + 1 \n   2 > 2(0) + 1\n   2 > 1, since this a TRUE statement and the point (0,2)   \n   is above the line, everything above the line is part of \n   the graph and must be shaded.\n\n4.   Consider the point (0,0). Verify this point as in Step 3.\n   Y > 2X + 1 \n   0 > 2(0) + 1\n   0 > 1 is FALSE, and the point (0,0) is below the line.\n   This verifies that step 3 is correct.\n(Otherwise just shade the upper portion of the line if the inequality is  in the form \n" Y > ........"  and shade the lower portion if the inequality is in the form \n" Y < ........"\n    \nNote that the graph of the line Y = 2x + 1 is not part of  the graph of the inequality Y > 2X + 1, because this is a Strict Inequality, so we when we draw the line we usually use broken lines or dashed lines to indicate so.\n\nInequalities like "less than or equal to" and "greater than or equal to" are not Strict Inequalities and their graphs include the lines.  We use a full line to draw the graph of the line and this indicates that the line is part of the graph and the solution set of the inequality.
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!
775797
http://www.coolmath.com/graphit/
816764
This is equivalent to y < -3.  Since y = -3 would be a HORIZONTAL (not vertical) line, you'd graph a dotted line going straight across at -3 on the y-axis.  Since you want the y-values that are LESS than that, you'd shade below that line.
148719
You just draw a graph yourself.\n\n! ain't gonna graph for you. Rather ! love calculus.
514202
All you have to do is combine like terms and then solve for x or y.  Then for the solving and graphing, refer to your textbook.  I don't feel like telling you HOW to graph it.
833077
In each, get y by itself.  I'll do #1 for you\n\nadd the absolute value part to both sides to get y = |9x-36| - 10\n\nAs long as x >= 4, the absolute value does not affect the outcome so the equation is simply y = 9x - 36 - 10 or y = 9x - 46.\n\nSo draw that graph starting with x = 4, then to the right.\n\nIf x<4 the absloute value of 9x-36 becomes its opposite which is -9x + 36.  So you have y = -9x + 36 - 10 or y = -9x + 26\n\nSo draw that graph starting at 4 and going left.  You should end up with a V-shaped graph with the point of the V at (4,-10)
621346
y=mx+b (this is known as the standard form of linear equations)\n\nm=slope\nx=x values\ny=y values\nb= y-intercept\n\nso the answer for you question is:\n\nslope = -1\ny-intercept = -3\n\nEXPLANATION:\n\ny = mx + b\ny = -x-3\n\nm= -1   and    b = -3\n \nslope = m    and   y-intercept = b\n\nthus, the slope is -1 and the y-intercept is -3\n\n\nNote: the slope is rise/run (rise over run) -- to make a graph use the y-intercept and the slope -- once you have the slope and the y-intercept, making this graph should be easy because it's a linear graph -- but if you still don't get it ask your teacher\n\n\nthus, the slope is not a-1 -----there is no "a"
891120
One easy way is to plug in a number for x.  From there, find numbers for y that work with that x.\n\nStart with x = 0 and find a y that works for both inequalities.\nx - 6y < 6\n-x + y < 4\n\nx - 6y < 6\nPlugging in 0 for x, we get\n-6y < 6\ny > -1\n\n-x + y < 4\nPlugging in 0 for x, we get\ny < 4\n\nWhen x is 0, y has to be greater than -1 and less than 4.\nThat gives us four solution points already.\n(0,0), (0,1), (0,2), (0,3)
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Generally, you break an inequality such as this into two parts and graph each separately.  So you'd be graphing x=-1 and x=3.\n\nThe type of line depends on the sign: if it's a <= or >=, then use solid lines to indicate anything on the line satisfies the original inequality.  If it's just < or >, then use a dashed line.  (This is similar to open and closed circles on a number line.)  Then, you pick points in the different regions that you created with the lines and determine if they satisfy the inequality.  If so, the region gets shaded.\n\nThere is no y dependency, so you basically have two vertical lines for your graph for this problem.  Checking the regions, only the area between the lines satisfies the inequality, so that's the only region that gets shaded.
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Draw the line y = 1; this is a horizontal line through y = 1.\nThink about all the points below this line (since you have y < 1).  \n\nDraw the line y = 2x - 3; this has a y-intercept of -3 and a slope of 2, so it goes through (0, -3) and goes upward to (1, -1), (2, 1), etc.)\nThink about all the points above this line (since you have y > 2x - 3).\n\nBecause the question says *and* you should only consider the points that are in both regions (below y = 1 and above y = 2x - 3).  Shade in the intersection of these two regions.\n\nI've attached a graph of what it should look like...
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linear: graph is a straight line that is not vertical.  table shows values steadily increasing or decreasing.\n\nquadratic: graph looks like a "U" - either up (v) or down (^) and either wide opening or small opening.  table shows values decreasing/increasing rapidly, then slowly, then the opposite direction, such as:\nx:  1  2 3 4 5 6 7 8  9\ny: 16 8 4 2 1 2 4 8 16\n\nexponential: similar to the quadratic, but it increases more rapidly, has y=1 when x=0, and does not "go the opposite direction" like the quadratic does.  it starts out very little, like a fraction, then shoots up and doesn't come back.  the table is like that: starts out little, increases a little, increases more and more rapidly, then its too big.\n\nabsolute value: looks like a "V", not like a quadratic that looks like a "U".  absolute values have an indifferentiable point, or in other words the bottom tip of the "V".  it's like a linear equation that is positive on one side where it should be negative.  the table is like a linear, but always positive where it should be negative.\n\nsquare root: looks like half a "U" turned 90 degrees clockwise on its side and only positive.  starts along the x-axis with y=0 at some x, then increases rapidly for a short time, then increases very slowly afterwards.\nx: 1  2  3   4    5    6      7     8\ny: 0 .5 .9 1.2 1.4 1.5  1.55 1.57\n\nnote that the sample tables are not representing real functions, they just graph a general idea of what it should look like.
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You need to get y by itself.\n\nadd -3x to both sides so you get 2y> 3x+4\nnow divide 2 --> y>3/2x+2\n\ngraph that as you would if it were y=3/2x+2\n(starting at 2 on the y axis. then move up 3 points, and 2 to the right.)\n\nMake sure your line is DOTTED. It is only solid if the inequalilty is less/greater than or equal to.\n\nto check which side is to be shaded, plot points on both sides. try (0,0) if the line doesn't cross through that.\n\n:)\n\nIf the answer is correct, shade the region where (0,0) or the coordinates that work are.
891649
(y-3x) cannot be graphed because there is no equality...\nyou could graph y=3x... that would be a line that passes through (0,0) with a slope of 3
152511
Which function, Subsitution, Linear COmbination, Graph?
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Draw yourself a cartesian graph, vertical line is the y axis, horizontal is the x axis.\n\npick points on the x axis like 1, 2, 3, etc.  plug thos values into the equation and solve for y. then pair the numbers, place points in the graph by taking the x value going straight up (or down if y is a negative number) and place the dot accross from the y value.  do this for each set of points then connect the dots with a line or a curve. for instance in your equationstart with x=1\n\ny= 7(1-5)(1-2)^4(1+.5)(1+2.5)^3\ny= 7(-4)(-1)^4(1.5)(3.5)^3\ny=7(-4)(1)(1.5)(42.875)\ny=-1800.75\n\nso in XY format the first point would go at (1, -1800.75)
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is it asking you to graph that. otherwise it is not solvable. if it's asking you to graph it then you have to plug into your graphing calculator y=x-9