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anonymous

  • one year ago

check attachment below thanks!

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  1. anonymous
    • one year ago
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  2. UsukiDoll
    • one year ago
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    First. we need this equation in y = mx+b form Next, pick a test point.. say (0,0) letting x = 0 and y = 0 and plug it into the inequality . If the inequality holds, then it's true and (0,0) is shaded. If the inequality doesn't hold, then it's false and (0,0) won't be shaded Third, let x = 1,2,3, and solve for y to gain more points to plot Finally, compare the graph to the picture

  3. UsukiDoll
    • one year ago
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    ok. let's have x = 0 and y = 0 plug those values into the choices you are given which of these inequalities are true?

  4. anonymous
    • one year ago
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    @UsukiDoll I'm sorry i just went full retard. I'm reading everything youre telling me, but nothing is making sense I'm so sorry

  5. UsukiDoll
    • one year ago
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    I'll do an example picking a test point of (0,0) and the inequality x-3y>5 we have 0-3(0) > 5 0-0>5 0>5 and that doesn't make sense at all. so throw that inequality out

  6. UsukiDoll
    • one year ago
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    there's one more inequality that needs to go. which one is it?

  7. anonymous
    • one year ago
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    x - 3y < 5

  8. UsukiDoll
    • one year ago
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    no that inequality holds true 0 is less than 5

  9. marigirl
    • one year ago
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    yes

  10. UsukiDoll
    • one year ago
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    dude . x-3y < 5 is a true inequality when the test point is (0,0) so that stays. It's just that there are two of them out there and we need to know which inequality is the same as the picture

  11. UsukiDoll
    • one year ago
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    so we either have x-3y <5 or x-2y <5 we know that the test point (0,0) is true so that test point is shaded.

  12. anonymous
    • one year ago
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    wait wouldn't the answer be what i just mentioned? x - 3y < 5

  13. UsukiDoll
    • one year ago
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    prove that it's correct.... I'm not even done explaining things and this is out of the blue.

  14. anonymous
    • one year ago
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    oh. no i didn't do anything. i just have strong feelings for that choice. somethings just telling me thats it.

  15. UsukiDoll
    • one year ago
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    so when you have this on a test, what are you going to do? You can't just guess.

  16. anonymous
    • one year ago
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    @UsukiDoll thats true. I'm sorry for interrupting you. can you help me figure it out??

  17. UsukiDoll
    • one year ago
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    anyway changing it to y = mx+b form x-3y =5 \[-3y=5-x\] \[y = \frac{5}{-3}+\frac{-x}{-3} \] \[y = \frac{x}{3} +(\frac{-5}{3})\]

  18. UsukiDoll
    • one year ago
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    and for x-2y < 5 just change signs to = and then we need y =mx+b form again x-2y=5 -2y=5-x \[y=\frac{-5}{2}+\frac{x}{2}\]

  19. UsukiDoll
    • one year ago
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  20. UsukiDoll
    • one year ago
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    so which colored line is above the -2 in the y-axis?

  21. UsukiDoll
    • one year ago
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    it was x-3y<5 lucky guess

  22. anonymous
    • one year ago
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    are you serious?! wow thanks. i had a feeling lol

  23. UsukiDoll
    • one year ago
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    yeah but I've outlined the process for manual solving (except for graphing because there is no way that I was doing that by hand)

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