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anonymous

  • one year ago

can you check my work?

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  1. anonymous
    • one year ago
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    To which graph does the point (2, 4) belong? y ≥ x + 3 y ≥ −x + 8 y ≥ 4x − 5 <--- my answer choice y ≥ −2x + 9

  2. anonymous
    • one year ago
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    @hartnn @surjithayer

  3. nincompoop
    • one year ago
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    how did you solve for it?

  4. anonymous
    • one year ago
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    my brother showed me the way and i was seeing if its correct then if someone could show me the steps because he didn't explain it clearly

  5. nincompoop
    • one year ago
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    so how did you pick that option ? I just want to know if you guessed or you did real math.

  6. anonymous
    • one year ago
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    no my brother did the math and told me the answer but im not sure because when he was explaining it it didn't make sense...

  7. anonymous
    • one year ago
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    plug every point one by one and see which satisfy

  8. anonymous
    • one year ago
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    im also looking for someone to show me how to do it

  9. anonymous
    • one year ago
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    okay give me a sec

  10. nincompoop
    • one year ago
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    okay. There are different ways to do it, but it all starts with identifying the intercepts. Do you need a tutorial in identifying intercepts or do you already know it?

  11. anonymous
    • one year ago
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    should i try doing it like surji said for now but give me one sec plz

  12. nincompoop
    • one year ago
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    you can do it, but it seems that you need to understand beyond plugging in values.

  13. anonymous
    • one year ago
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    (2, 4) y ≥ x + 3 = 4 ≥ 2 +3 = 4 ≥ 5 which is false so its not A y ≥ −x + 8 = 4 ≥ -2 + 8 = 4 ≥ 6 which is false so not B y ≥ 4x − 5 = 4 ≥ 4(2) - 5 = 4 ≥ 8-5 = 4 ≥ 3 which is correct so it is C y ≥ −2x + 9 = 4 ≥ -2(2) + 9 = 4 ≥ -4 + 9 = 4 ≥ 5 which is false so not D

  14. anonymous
    • one year ago
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    okay so now that i have done it that way can you show me your way? nincompoop?

  15. nincompoop
    • one year ago
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    my way is learning linear equation.

  16. anonymous
    • one year ago
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    can you teach me if you don't mind>

  17. nincompoop
    • one year ago
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    since your inequality has greater than or EQUAL to, we can start with equality \(y = mx + b \)

  18. anonymous
    • one year ago
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    then what>|?

  19. nincompoop
    • one year ago
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    The m is your slope. e b is your y-intercept - point where the value of the coordinate is (0, y). It means that the value of x is just zero (x = 0) and the value of y is anywhere -infinity and +infinity. So, pretty much the value of b is your actual y intercept |dw:1433602860664:dw|

  20. anonymous
    • one year ago
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    so so do you plug in like (0,4) then (2,0)?

  21. nincompoop
    • one year ago
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    |dw:1433602878082:dw|

  22. nincompoop
    • one year ago
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    brb. work-related

  23. ganeshie8
    • one year ago
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    |dw:1433602934519:dw|

  24. anonymous
    • one year ago
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    she wants to teach me the way of doing it without plugging them in, and i want to know how to do it that way too

  25. ganeshie8
    • one year ago
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    sounds awesome! :)

  26. anonymous
    • one year ago
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    yupp but i have more questions i guess you guys can help with later

  27. nincompoop
    • one year ago
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    No, we cannot just do it like that. Next, is to identify the x-intercept. It is the point where it touches the x-axis, which also tells you that the value of y is zero (y=0). What this means analytically is to set your y to zero and solve for the rest of the equation. if you have \(y = x+3 \rightarrow 0 = x + 3 \) then solve for x, which in the example I showed you becomes: 0-3 = x + 3 - 3 -3 = x and your b is 3 so your intercepts are (0,3) and (-3, 0) |dw:1433604051743:dw|

  28. nincompoop
    • one year ago
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    |dw:1433604121994:dw|

  29. nincompoop
    • one year ago
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    |dw:1433604186698:dw|

  30. nincompoop
    • one year ago
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    suppose this is an inequality so that instead of y = x+3 we have \(y \ge x + 3 \) our linear equation is still the same, but the values of the points now are from anywhere the line lies and also above (greater). |dw:1433604445639:dw|

  31. nincompoop
    • one year ago
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    if instead the inequality is \(y \le x+3 \) then we shade the area of the region from where the line lies and the one below (less) it. |dw:1433604556647:dw|

  32. anonymous
    • one year ago
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    so thats how y <= x + 3 looks in a graph?

  33. nincompoop
    • one year ago
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    correct! So, what we have covered so far are equality and inequality with greater than or equal to and less than or equal to. now, we need to do greater than or less than.

  34. nincompoop
    • one year ago
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    the good thing about this is that the still use the same LINE! meaning, that learning about linear equation in the slope-intercept form y = mx+b is quite helpful tool. now instead of including the line itself, we just need anywhere above (greater) or below (lesser). |dw:1433604945236:dw|

  35. nincompoop
    • one year ago
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    |dw:1433605013888:dw|

  36. nincompoop
    • one year ago
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    Now that you have an idea how the graph is like with linear equation and inequalities. We can determine if (2,4) is a point in the equation \(y\ge x+3 \) analytically. You can easily do this by "plugging in the values of x and y and see if it returns a TRUE value. Clearly, 4 is not greater than or equal to 5. And you do with the rest of the options. You can also do this graphically, but only if you have the patience to properly graph the equations and inequalities given. Never rely only on analytical solutions, because there will be problems where graphical understanding gives you a better intuition and idea how to attack a problem, and this is why I took the time to teach you the concept.

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