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| 07 Feb 2017 01:51 PM |
i have 2 solve systems of inequalities but im dumb so idk if im doing this stuff right
k |
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| 07 Feb 2017 02:28 PM |
year but theres 2 of those things and we have to shade them in and stuff thats what i dont get
k |
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ebenton95
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| Joined: 06 Nov 2011 |
| Total Posts: 9671 |
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| 07 Feb 2017 02:50 PM |
| the shading is on a graph right? |
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| 07 Feb 2017 02:58 PM |
If >, shade above the line If < shade below the line |
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| 07 Feb 2017 03:00 PM |
theres 2 equations/lines though so do i shade in the area where it is true for both?
k |
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| 07 Feb 2017 03:01 PM |
| Here's what I do. I draw little lines above or below the link e depending on > or <. Then I shade it the spot where both lines are and that's the solution |
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| 07 Feb 2017 03:02 PM |
here ill show u what i did hold on a minute ik im dumb lol
k |
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| 07 Feb 2017 03:02 PM |
here ill show u what i did hold on a minute ik im dumb lol
k
k |
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ebenton95
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| Joined: 06 Nov 2011 |
| Total Posts: 9671 |
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| 07 Feb 2017 03:13 PM |
| i believe all you do is treat the inequality like an equality to find what the lines are going to be, and then shade in/use a dotted or solid line depending on what the inequality sign is (shade above for greater, shade below for less, solid line for greater/less than or equal, dotted for just greater/less than) |
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| 07 Feb 2017 03:20 PM |
Shading is really easy. Here's a trick my teacher taught.
Ex: y<4x+9
Once you graph the line, you plug in (0,0) for the x and y values, giving you 0<0+9. Because this is a true statement (0<9), you would shade the side of the line which has the (0,0) point on. If you have a problem like y<6x, you couldn't use that trick as the y-point is already 0; use (1,1), and do the same method.
Ex:
y<6x
1<1
This is false (greater than or equal to (then vice versa) would be true. So shade the part of the graph which point (1,1) is NOT on.
It's easier to hear it than to read it, so sorry. |
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| 07 Feb 2017 03:47 PM |
| ### ARE U UNDER THE AGE OF 13 LOL ALGEBRA 1 |
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