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jennyrlz

  • one year ago

Hey i need some algebra help :)

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  1. jennyrlz
    • one year ago
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    so a friend of mine is asking for help but i don't really remember much of my algebra days :/

  2. whpalmer4
    • one year ago
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    so, what's the question?

  3. jennyrlz
    • one year ago
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    ok so first he was asked to draw a flag one sec let me draw it

  4. jennyrlz
    • one year ago
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    |dw:1433290028954:dw|

  5. jennyrlz
    • one year ago
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    |dw:1433290054568:dw|

  6. jennyrlz
    • one year ago
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    ok so lets get to the problem

  7. jennyrlz
    • one year ago
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    he os asked to show these points in y=mx+b and in standard form

  8. jennyrlz
    • one year ago
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    but if we already have all the x y's what do i do ; ;

  9. jennyrlz
    • one year ago
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    |dw:1433290221695:dw|

  10. whpalmer4
    • one year ago
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    maybe he's supposed to write equations for those lines that make up the diagram?

  11. jennyrlz
    • one year ago
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    hmm how would i go about doing this? he isnt really that familiar with the material :/

  12. whpalmer4
    • one year ago
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    Well, the vertical and horizontal lines are pretty easy. For example, the bottom line is a horizontal line that passes through (-5,-3) and (5,-3). We know the slope is 0, so it is just \[y=k\] for some value of \(k\), right? Doesn't matter what value of \(x\), \(y\) is always -3 for that line.

  13. whpalmer4
    • one year ago
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    Similarly, any vertical lines are just \[x=k\]for some value of \(k\), because all the \(x\) values are the same, and only the \(y\) value changes.

  14. jennyrlz
    • one year ago
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    1. draw the flag out in graphing paper, graph must include coordinates and labled increments plot each point at the end of each line segment on the flag for each line creat a table with four points write an equation in slope intercept and standard and list the domain and range

  15. jennyrlz
    • one year ago
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    those are the instructions

  16. whpalmer4
    • one year ago
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    Now, the "slanted" lines are probably what the problem author is interested in...there we know two points the line passes through, and from that, we can determine the equation of the line.

  17. whpalmer4
    • one year ago
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    okay, yeah, looks like what I guessed. Let's take that line from \((-5,-3)\) to \((0,0)\) as an example. The domain is going to be -5 to 0, inclusive; those are the values that \(x\) is allowed to take. \[-5\le x\le 0\]is another way you could write that. The range is the range of values \(y\) can have over the domain. At the minimum, \(y=-3\) and at the maximum, \(y = 0\), so we could write the range as \[-3\le y \le 0\] Any question about that?

  18. jennyrlz
    • one year ago
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    umm

  19. jennyrlz
    • one year ago
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    let me show you what i just did

  20. jennyrlz
    • one year ago
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    ahh |dw:1433290930006:dw|

  21. jennyrlz
    • one year ago
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    correct?

  22. jennyrlz
    • one year ago
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    and i still have trouble with domain and range

  23. whpalmer4
    • one year ago
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    yes, put 4 points in that table, different values of x, all the same value of y, and you have your table. Now domain is the set of allowed values of x (the independent variable), and range is the set of resulting values of y (the dependent variable). For this line segment, how would you describe the allowed values of x?

  24. jennyrlz
    • one year ago
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    hmm you lost me xD. but the allowed values reffer to the x then wouldnt they be -5 to 5 because thisline goes from -5 to 5?

  25. whpalmer4
    • one year ago
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    exactly. "domain" is just a fancy term to confuse you :-)

  26. jennyrlz
    • one year ago
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    xD

  27. jennyrlz
    • one year ago
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    and for range?

  28. whpalmer4
    • one year ago
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    isn't there only one possible y value for this line?

  29. jennyrlz
    • one year ago
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    what do you mean?

  30. jennyrlz
    • one year ago
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    i think i just made sense of it, the range would be 5 xD

  31. jennyrlz
    • one year ago
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    correct?

  32. whpalmer4
    • one year ago
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    no, the range is 3, isn't it? isn't y = 3 for all values of x for that line?

  33. jennyrlz
    • one year ago
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    oh yea i was refering to his table xD

  34. jennyrlz
    • one year ago
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    i made my own so i can help him xD

  35. jennyrlz
    • one year ago
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    Thank you SOOOOOOOOOO much.

  36. whpalmer4
    • one year ago
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    glad I could help you help your friend :-)

  37. whpalmer4
    • one year ago
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    I find that explaining something to someone else is a really good way to reinforce your own knowledge.

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