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anonymous

  • one year ago

Determine whether the origin is included in the shaded region and whether the shaded region is above or below the line for the graph of the following inequality: y > three fifthsx − 2 The origin is not included in the shaded region and the shaded area is above the line. The origin is not included in the shaded region and the shaded area is below the line. The origin is included in the shaded region and the shaded area is above the line. The origin is included in the shaded region and the shaded area is below the line.

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  1. anonymous
    • one year ago
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    @queen-of-tokyo

  2. anonymous
    • one year ago
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    @WhateverYouSay

  3. anonymous
    • one year ago
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    @sweetburger

  4. anonymous
    • one year ago
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    @zepdrix

  5. anonymous
    • one year ago
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    @yomamabf

  6. zepdrix
    • one year ago
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    Hey there :) Let's answer the second part of the question first, because it's really simple.

  7. zepdrix
    • one year ago
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    When you have something like this: \(\large\rm y\lt \color{orangered}{2x+3}\) Think of it like this: \(\large\rm y\lt \color{orangered}{\text{the line}}\) The orange part represents the line. So in this example, we want all of the `y values` which are `less than` `the line`. So in this example we would shade `below` the line.

  8. zepdrix
    • one year ago
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    How bout in your problem? y is GREATER THAN some line. So where we gonna shade?

  9. anonymous
    • one year ago
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    above the line correct? @zepdrix

  10. zepdrix
    • one year ago
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    k good :) that already narrows down our choices!

  11. zepdrix
    • one year ago
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    \[\large\rm y\gt\frac{3}{5}x-2\]Recall that the origin is the point (0,0). So let's just plug it in... and find out if the inequality holds true!

  12. zepdrix
    • one year ago
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    \[\large\rm 0\gt\frac{3}{5}\cdot0-2\]After you simplify this a bit, does the inequality hold true?

  13. anonymous
    • one year ago
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    no

  14. anonymous
    • one year ago
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    its the same right?

  15. zepdrix
    • one year ago
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    So after simplifying, we have, 0 > -2

  16. zepdrix
    • one year ago
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    Hmm that appears to hold true, ya? 0 is larger than -2

  17. anonymous
    • one year ago
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    oh yes my bad i was looking at the wrong thing

  18. zepdrix
    • one year ago
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    Ok good! So we figured out that the point (0,0) DOES in fact satisfy our inequality. So it is included in the shaded area.

  19. anonymous
    • one year ago
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    thanks could you help mw with one more

  20. zepdrix
    • one year ago
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    sure, i can try :)

  21. anonymous
    • one year ago
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    Given the system of equations presented here: 3x + 5y = 29 x + 4y = 16 Which of the following actions creates an equivalent system such that, when combined with the other equation, one of the variables is eliminated? Multiply the second equation by −1 to get −x − 4y = −16 Multiply the second equation by −3 to get −3x − 12y = −48 Multiply the first equation by −1 to get −3x − 5y = −29 Multiply the first equation by −3 to get −9x − 15y = −87

  22. zepdrix
    • one year ago
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    You're trying to match up the coefficients on either the x's or the y's, but not both. If we wanted to match up the x's, maybe we would need some kind of 3. If we wanted to match up the y's, it would be a little more difficult. We would multiply the first equation by 4, and the second equation by -5. They would then both have coefficients of 20, one of them being negative. So they would combine nicely. They didn't do that here though. They went the easier route. They're matching up the x's somehow.

  23. zepdrix
    • one year ago
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    We want them to match. Notice the first equation already has a 3, so it doesn't need anything more.

  24. zepdrix
    • one year ago
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    So what do you wanna do to the second equation to match them up?

  25. anonymous
    • one year ago
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    add a 1 to the x?

  26. zepdrix
    • one year ago
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    The first equation has 3x. The second equation has x. If we add 1 to x, hmm I'm not really sure what that will do for us :d We want the same number of x in both equations.

  27. zepdrix
    • one year ago
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    We don't want to add. We want to multiply that second equation by some number that will turn it from x to some kind of 3x.

  28. anonymous
    • one year ago
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    you would multiply it by 3 then

  29. anonymous
    • one year ago
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    @zepdrix

  30. zepdrix
    • one year ago
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    Good good good, we would multiply the second equation 3! That would give us 3x. But in order for them to combine, and disappear, we want one of them to be negative. So let's instead multiply the second equation by negative 3.

  31. zepdrix
    • one year ago
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    boom bam! yay team!

  32. anonymous
    • one year ago
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    so the answer is D yes? @zepdrix

  33. zepdrix
    • one year ago
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    Option D talks about changes to equation 1. We were making changes to equation 2.

  34. anonymous
    • one year ago
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    oh ok i wasn't reading carefully but thank you for your help :) @zepdrix

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