Solve the inequality and check it the graph it?
930501
There needs to be an "x" there somewhere.  If it were\n-4x + 7 > 15, you'd have \n-4x > 8\n  x < -2 (because you switch the sign when you divide by a negative)
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)
1152551
i believe there is no solution.\na has to be positive if it is greater than 0.\nIf a = 1, than a^2+(2/a) = 3,  so a would have to be greater that 1 for a^2+(2/a) to be greater than 3, but it can not at the same time equal less than zero.
746406
8w + 5d <= 80\n2w + 5d <= 50\nw >= 0\nd >= 0\n\nPlot these 4 inequalities on a graph.  The maximum will occur at one of the vertices of the polygon formed.  Plug the coordinates of the vertices into \n\nProfit = 17w + 29d\n\nand see which is largest.
778580
Go to this link:\n\nhttp://www.coolmath.com/graphit/index.html\n\nYou can enter your equation and graph it.
1144014
draw a picture of the graph over the specified interval.  divide the graph into vertical rectangles.  the area under the curve will be the sum of the areas of the rectangles.  saying that youve made "n" rectangles, the only way to get the exact area is to find the sum for the limit as n approaches infinity.  work from that concept.  if you already knew that, or need more help or something, email me and i can give you more steps and a much more detailed description.  that would take a really long time though so im gonna see if this helps first.
859022
x - 2y ≥ 4\n\nGraph x - 2y = 4 (ie y = ½x - 2 Slope = ½, y -intercept = -2)\n\nNow test the origin (0, 0)\n0 - 0 ≥ 4 is FALSE, so the origin is NOT in the region\n\nSo the requred region for this inequality is the half plane bounded by the line NOT containing the origin.\n\nLightly shade this region\n\nx ≤ 4\n\nGraph x = 4 ... a vertical line through 4 on the x-axis\nTest the origin (0, 0)\n0 ≤ 4 is TRUE so the origin is in the region\nSo the required region is the half plane bounded by the line x = 4 that contains the origin\nLightly shade this region differently to the first region\n\nThen the required condition of x - 2y ≥ 4 and x ≤ 4 is the overlap of the two above regions
635075
Graph X=-4 with dotted lines\n\nthen color everything below the line.
476794
This graph helps us to show what happens when a quantity y is inversely proportional to another quantity x.\n\nAgain, it shows us that a computer algorithm implementing y=(1/x) for the order of efficiency of the algorithm..
337327
assign value for your x-variable...say -10, -5, 0, 5, 10 and repesent it in a graph form...
923158
Thinner simple work its cause the 3 is greater than the one other than drwing the grph there is really no work. And if u need the graph then just make an X Y table using values -3 to 3 then solve for y and gragh them
965748
|x - x²| \n= x - x² for x - x² ≥ 0\n= x² - x for x - x² < 0\n\nx - x² = x(1 - x) = 0 when x = 0 and x = 1\n\nSo x - x² ≥ 0 when 0 ≤ x ≤ 1\nand x - x² < 0 when x < 0 and x > 1\n\nSo  dx under the interval [-1,3]\n\n= ∫(x² - x) dx under the interval [-1, 0] +\n∫(x - x²) dx under the interval [0, 1] +\n∫(x² - x) dx under the interval [1, 3]\n\n=  [⅓x³ - ½x² ] [-1, 0] + [½x² - ⅓x³] [0, 1] + [⅓x³ - ½x² ] [1, 3]\n\n= [(⅓0³ - ½0²) - (⅓ (-1)³ - ½ (-1)²) ] + [(½(1)² - ⅓(1)³) - (½0² - ⅓0³)] + [(⅓ 3³ - ½ 3²) - (⅓1³ - ½1²) ]\n\n= [5/6] + [1/6] + [9/2 + 1/6]\n\n= 5 2/3
821475
graph both the graphs as though they were of equality and not inequality and after wards shade the relevant portion
470658
The above answer is basically correct.  Plug in a few more values, such as (21,1543.5), (22,1694), and so on for a few more values of the graph.
41570
A parabola with its base at 0,1.  Go to http://www.coolmath.com/graphit/ Enter x^2+1 and see it for yourself!  :)
168710
http://www.graphcalc.com/
725728
It looks like you solved the inequality correctly.  The line would have a y-intercept at 1/2 (halfway between the origin and 1 on the y-axis).  Since the slope is 3/4, from there you'd count up 3 and over 4 to find another point-----which would be the point (4,3.5).  You are correct that the line is dotted.  Since the final equation says y is LESS THAN the line, you'd shade BELOW that dotted line.