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tw101

  • one year ago

Please help! I have an assignment that I have been trying to do but i'm a bit stuck. I have done the first questions but need help on the last ones. Help would be much appreciated. (i will put pictures below)

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  1. tw101
    • one year ago
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  2. tw101
    • one year ago
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    This is what I have. And I have already done questions 1,2,and 3, but only got halfway on 6&7. I drew the triangle and rectangle but I'm not sure how to get the length of the sides.Im also not sure how to do 4&5. I know that for parallel the slopes are the same and for perpendicular they're opposite reciprocals, but do I have to draw two new lines to accommodate the given points?

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  3. phi
    • one year ago
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    to do Q4, we need to know the equation of the line for Q2 (for example) a parallel line will have the same slope I don't know what slope you got for Q2, but pretend it was 2. then the equation for Q4 would be y = 2x + b (same slope, but we need to find b) they say this line goes through the point (1,-3) (i.e. when x is 1, y is -3) so use that info to get -3 = 2*1 + b or -3 = 2+ b to find b, add -2 to both sides and simplify but remember we need the slope for Q2 (which is probably not 2)

  4. phi
    • one year ago
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    for Q5 a perpendicular slope to slope "m" is -1/m for example if m is 2/3 then the perpendicular slope is -3/2 (flip and negate) or: -1/2 and 2 or 2 and -1/2 or 3 and -1/3 say you choose the line from Q2. get its slope and "flip it" then multiply by -1 after you get the slope, it is the same problem as Q4... you have to find b

  5. phi
    • one year ago
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    for Q6, I would pick a "simple triangle" (a right triangle with legs that go straight across and straight up/down) we don't have to, but it is easier. the length of the straight across line is easy: count the number of squares (or subtract the x values) for a slanted line you have to use the distance formula do you have that in your notes?

  6. phi
    • one year ago
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    I don't know if you have to plot the lines for Q4 or Q5 but it can't hurt. once you have the equation, find two points on the new line and plot them... then connect the dots

  7. tw101
    • one year ago
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    THank you so much for helping me! sorry i didnt respond right away! I didnt realize that someone was actually looking at my question. I had given up haha.

  8. phi
    • one year ago
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    try to make progress if you post your work someone (me? if I'm here) will check it.

  9. tw101
    • one year ago
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    okay. for 4 and 5 what do you mean by plot lines?

  10. phi
    • one year ago
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    you wrote *** but do I have to draw two new lines to accommodate the given points?*** which I assumed meant: do I have to plot the new lines which will be parallel or perpendicular to the original line? and I thought it would not hurt.

  11. tw101
    • one year ago
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    oh okay, I see.

  12. tw101
    • one year ago
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    I was going to do the line I drew over the left side of the triangle since its goes through the points it gave me. would that work?

  13. phi
    • one year ago
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    what two points does it go through. I'm not sure which line you mean

  14. tw101
    • one year ago
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    one second. Ill show the updated picture

  15. tw101
    • one year ago
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    I cut the triangle in half to make it how you said it would be easier.

  16. tw101
    • one year ago
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  17. tw101
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    the line that runs alongside the left side of the triangle

  18. phi
    • one year ago
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    ok, but first we have to fix that line. it goes through (2,5) and (4,8) its slope is change in y divided by change in x (8-5)/(4-2) or 3/2 we can say y = (3/2)x + b now use (2,5): 5 = (3/2)*2 + b 5= 3+b b= 2 and the equation is y= (3/2)x + 2

  19. tw101
    • one year ago
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    okay, so where would I go from there

  20. phi
    • one year ago
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    notice, when y= 19 (top of the outer box) 19= (3/2)x + 2 add -2 to both sides 17 = 3/2 x multiply both sides by 2/3 34/3 = x or x = 11.333 (11 and 1/3) not x=12 as you have. in other words your line on the graph is a little off

  21. tw101
    • one year ago
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    okay, so would i move it up just a little?

  22. phi
    • one year ago
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    not up. I would move the point at (12,19) to (11 1/3 , 19) you will have to estimate 1/3 of box over from 11 then redraw the line to go through the 3 points (2,5), (4,8) and (11 1/3, 19)

  23. tw101
    • one year ago
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    Okay. The (4,8) is the lower left part of the rectangle..

  24. phi
    • one year ago
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    ok. after you fix the line we can do Q4

  25. tw101
    • one year ago
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    Okay:)

  26. phi
    • one year ago
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    btw, y= (3/2) x + 2 will be an answer for Q2 . a line with positive slope and you have the work up above on how we found the equation

  27. phi
    • one year ago
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    Q4 ... the question is a bit muddled but I assume it is the new line that goes through (1,-3) and they want it parallel to y= (3/2) x + 2 parallel means the same slope. so write the new equation. can you do that ?

  28. tw101
    • one year ago
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    so would it just be y=(1/-3)x+2?

  29. phi
    • one year ago
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    start with y = m x + b (the "skeleton" that we fill in) we want this line to be parallel to y= (3/2)x + 2 what slope should we use for m ? (parallel means *same* slope)

  30. tw101
    • one year ago
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    okay, so would the (3/2) be the same then?

  31. phi
    • one year ago
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    Q1 is making the graph. you mean Q2 and Q3? for the "old line" we can use the line you used for Q2 what was it ?

  32. tw101
    • one year ago
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    yes, i already did 1,2, and 3

  33. tw101
    • one year ago
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    You have to choose 2 locations that make a line and are either positive or negative. I

  34. phi
    • one year ago
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    what did you choose?

  35. tw101
    • one year ago
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    i chose (2,5) (12,19) for positive (21,5) (12,19) for negative.

  36. tw101
    • one year ago
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    so for Q4, would it still be y=(3/2) +b

  37. tw101
    • one year ago
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    but we still have to find b

  38. phi
    • one year ago
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    if you change to (2,5), (4,8) we will get 3/2 for the slope (a much nicer number than if we use (12,19)

  39. tw101
    • one year ago
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    okay

  40. phi
    • one year ago
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    and change (21,5) (12,19) for negative. to (21,5), (19, 8) (same reason..a nicer slope)

  41. tw101
    • one year ago
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    make both of them negative?

  42. phi
    • one year ago
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    no. for Q2 use the points (2,5) and (4,8) (slope will be 3/2) for Q3 use the points (21,5), (19,8) (slope will be -3/2)

  43. tw101
    • one year ago
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    Ohhh okay haha

  44. phi
    • one year ago
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    I hope that gets straightened out. but for Q4, we want a line parallel to the line from Q1 which has a slope of 3/2 we start with y= mx + b we replace m with 3/2 y= (3/2) x + b we want this new line to go through (1,-3) replace x and y with those numbers -3 = (3/2)*1 + b can you find b ?

  45. tw101
    • one year ago
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    would i solve for b

  46. phi
    • one year ago
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    yes. can you do that?

  47. tw101
    • one year ago
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    I mean do i isolate it

  48. tw101
    • one year ago
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    yes

  49. tw101
    • one year ago
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    okay so i got (-3/1)-(3/2)

  50. phi
    • one year ago
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    yes, but people would combine that either put -3 over the common denominator of 2 (i.e. write it as -6/2 ) or change 3/2 to 1.5 and figure out -3 - 1.5

  51. tw101
    • one year ago
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    okay so it would be -4.5?

  52. phi
    • one year ago
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    yes, or -9/2 (same thing)

  53. phi
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    so the answer for Q4 is y = (3/2) x - 9/2

  54. tw101
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    y=(3/2)*1+(-9/2)?

  55. tw101
    • one year ago
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    ok you already said that.

  56. tw101
    • one year ago
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    sorry haha

  57. tw101
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    right, sorry

  58. tw101
    • one year ago
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    that was disgusting

  59. tw101
    • one year ago
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    wow. okay. im sorry you guys had to go through that.

  60. tw101
    • one year ago
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    i blocked them

  61. phi
    • one year ago
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    ready for Q5? let's use the same line with slope (3/2) and we want a perpendicular line y= mx + b any idea what we use for m ?

  62. tw101
    • one year ago
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    (-3/2)?

  63. tw101
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    since theyre reciprocals?

  64. phi
    • one year ago
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    we want "flip and negate" you negated but did not flip

  65. tw101
    • one year ago
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    (-2/3)

  66. phi
    • one year ago
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    yes so we have y = (-2/3) x + b and they want this to go through point (-2,7)

  67. phi
    • one year ago
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    to find b, replace x and y with those numbers

  68. tw101
    • one year ago
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    okay so it would be 7=(-2/3)*-2+b

  69. phi
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    yes

  70. tw101
    • one year ago
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    then solve for b again right

  71. phi
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    yes

  72. tw101
    • one year ago
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    7/-2- (-2/3)=b

  73. tw101
    • one year ago
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    sorry i didnt put parenthesis on the first pair

  74. phi
    • one year ago
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    that looks a bit peculiar. start with 7=(-2/3)*-2+b can you do -2 * -2/3 ?

  75. tw101
    • one year ago
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    (-4/6)?

  76. tw101
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    (4,6)

  77. tw101
    • one year ago
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    sorry forgot a negative times a positive for a second. haha

  78. phi
    • one year ago
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    it is \[ -2 \cdot \frac{-2}{3} = \frac{-2}{1} \cdot \frac{-2}{3}\] when you multiply fractions you multiply top * top and bottom * bottom

  79. tw101
    • one year ago
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    4/3?

  80. phi
    • one year ago
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    yes. so you have 7 = 4/3 + b now add -4/3 to both sides

  81. tw101
    • one year ago
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    (7/-2)+(4/3)=b?

  82. tw101
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    wait whoops

  83. tw101
    • one year ago
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    hold on, sorry

  84. tw101
    • one year ago
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    7+4/3? i was looking at my old equation

  85. phi
    • one year ago
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    close. but added +4/3 we want -4/3 (because on the right side we want 4/3 - 4/3 which cancels)

  86. tw101
    • one year ago
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    im sorry, thank you for being so patient with me. math really isnt my stronest subject... at all.

  87. tw101
    • one year ago
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    okay, gotcha

  88. tw101
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    so just 7+(-4/3)

  89. phi
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    b= 7 - 4/3 change 7/1 to 21/3 (multiply top and bottom by 3)

  90. phi
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    and yes you can write it 7 + (-4/3)

  91. tw101
    • one year ago
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    so 25/3?

  92. tw101
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    21/3+4/3=25/3

  93. phi
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    if you did 21/3 + 4/3 we have 21/3 - 4/3

  94. tw101
    • one year ago
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    oh right ints negative

  95. tw101
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    17/3

  96. phi
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    and the equation is ?

  97. tw101
    • one year ago
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    y=(-2/3)x+17/3?

  98. phi
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    yes

  99. tw101
    • one year ago
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    okay great

  100. phi
    • one year ago
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    for Q6, did you change the top point from (12,19) to (11 1/3 , 19) ?

  101. tw101
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    of the triangle?

  102. phi
    • one year ago
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    yes (otherwise you will get marked off (if they notice!) because (12, 19) is not really on the line from (2,5) to (4,8)

  103. tw101
    • one year ago
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    yep

  104. tw101
    • one year ago
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    just draw a side straight down from that point now

  105. phi
    • one year ago
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    ok, now figure out the slope of the straight up/down line can you do that ?

  106. tw101
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    i think so

  107. tw101
    • one year ago
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    jsut a minute:)

  108. tw101
    • one year ago
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    okay so the two points are (11 1/3,19) and (11 1/3,4) so then i did the y1-y2/x1-x2 thing and got (0,15) not sure what to do now. would i substitute that into the equation? y=mx+b

  109. phi
    • one year ago
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    they don't want the equation of the line. they want its slope change in y divided by change in x notice you will get (19-4)/ (11.333- 11.333) = 15/0 and divide by 0 is not allowed. we say the slope of a vertical line is undefined.

  110. phi
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    they also want the length of that side. it is 19-4 or 15

  111. tw101
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    okay so slope:undefined length:15

  112. phi
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    yes. now do the bottom side

  113. tw101
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    ok

  114. phi
    • one year ago
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    though in your picture it looks like you only go down to y=5 (not y=4)

  115. phi
    • one year ago
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    I would try to use the point (2,5) in this triangle because we know the slope of the slant side will be 3/2 (we did that work up above)

  116. tw101
    • one year ago
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    youre right. 14 then

  117. tw101
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    sorry my internet is being really weird

  118. phi
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    yes. and change the number where you show your work.

  119. tw101
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    okay so for the bottom side i got (11 1/3,5) (2,5) then when subtracted i got -1/7 1/3

  120. phi
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    ok, now you can do the bottom line (2,5) to (11 1/3, 5) change in y is (5-5) change in x is 11 1/3 - 2 = 9 1/3 and 0/ 9.333 is 0

  121. tw101
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    ugh no that is not what i got

  122. phi
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    if you have a calculator, it will help

  123. tw101
    • one year ago
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    sorry my computer is freezing up

  124. phi
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    the length will be 9 1/3 (9 and 1/3) slope will be 0

  125. phi
    • one year ago
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    as a check, you can count how long it is. but they want you to "show the work" so you should also show 11 1/3 - 2 = 9 1/3

  126. tw101
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    would the slope be 0?

  127. tw101
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    oh okay you said that haha sorry, again , slow computer

  128. phi
    • one year ago
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    yes , slope is 0 (horizontal lines have slope 0)

  129. phi
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    now the slant side (2,5) to (11 1/3, 19) first find change in y and change in x

  130. phi
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    hint we did this already (see length of the vertical and horizontal sides)

  131. tw101
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    yes, sorry i had to reset my connection

  132. phi
    • one year ago
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    you should get 14 and 9 1/3

  133. tw101
    • one year ago
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    yes, i did. im so sorry idk what going on with my wifi

  134. tw101
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    change of x 19-5 change of y 11 1/3-2

  135. tw101
    • one year ago
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    14 and 9 1/3 yes.

  136. phi
    • one year ago
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    ok on your numbers, but x goes sideways... change in x is 9 1/3 the slope will be 14 / (9 1/3) to do that, write 9 1/3 as an improper fraction 28/3 when you divide by a fraction, you can "flip it" and multiply so 14 * 3/28 which is 3/2

  137. tw101
    • one year ago
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    ok

  138. phi
    • one year ago
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    slope is 3/2 length we use the distance formula \[ \sqrt{ (y_2-y_1)^2 + (x_2-x_1)^2 } \] which is \[ \sqrt{ 14^2 + 9.3333^2} = \sqrt{196+87.11}= \sqrt{283.11}= 16.8\]

  139. phi
    • one year ago
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    because the triangle has a right angle formed by the vertical and horizontal lines, it is a right triangle

  140. tw101
    • one year ago
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    oh okay i see.

  141. phi
    • one year ago
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    for Q7 I would use the "mall" as the quadrilateral the slopes will be 0 for the top and bottom (horizontal lines) and undefined for the sides the lengths are easy (I hope... count how long each side is) because the shape has 4 90 degree angles and the opposite sides are equal it is a rectangle.

  142. phi
    • one year ago
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    I have to go.

  143. tw101
    • one year ago
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    9 ad 15

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