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anonymous

  • one year ago

Question down below please I need help!!

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  1. anonymous
    • one year ago
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  2. freckles
    • one year ago
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    what trouble are you having?

  3. freckles
    • one year ago
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    identifying the coordinates for C and D?

  4. anonymous
    • one year ago
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    Yes I don't know how to solve the problem overall

  5. freckles
    • one year ago
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    |dw:1437368521661:dw| y=b is a horizontal line any point falling on it will have it's y coordinate as b y=0 is a horizontal line any point falling on it will have it's y coordinate as 0

  6. freckles
    • one year ago
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    |dw:1437368652986:dw| what I'm saying is I know 2a comes after a sometime you could have picked something else I'm sure calling it x=c would have been fine we would just assume c>a

  7. freckles
    • one year ago
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    anyways x=a is a vertical line any point falling on it will have it's x coordinate as a x=2a is a vertical line any point falling on it will have it's x coordinate as 2a

  8. freckles
    • one year ago
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    so what coordinates are you going to assign to C?

  9. freckles
    • one year ago
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    notice the lines goes through point C above on the picture are x=2a and y=b and I did say any point on x=2a will have x-coordinate 2a and I also said any point on y=b will have y-coordinate b so we could give the coordinates of C as?

  10. anonymous
    • one year ago
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    2a?

  11. anonymous
    • one year ago
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    c has cordinates of 2a, b?

  12. freckles
    • one year ago
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    yes C has coordinate (2a,b) or that is what we are assigning C anyways it is appropriate since 2a does come after a but as I said before we could have pick another number like c and make the assumption c is greater than a but anyways |dw:1437369322715:dw| now we have to assign coordinates to D

  13. freckles
    • one year ago
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    notice D lays on the horizontal line y=0 any point falling on y=0 has y-coordinate 0

  14. freckles
    • one year ago
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    |dw:1437369388989:dw| what number are you going to put in place of that ?

  15. anonymous
    • one year ago
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    umm 2a squared

  16. freckles
    • one year ago
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    I guess you could say 2a^2 the spacing between B's x-coordinate and C's x-coordinate seem pretty close to the spacing between C's x-coordinate and D's x-coordinate I think a better number might be 3a but I think 2a^2 would be great if we don't care about scales

  17. freckles
    • one year ago
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    by the way that is according to your picture I have drawn my D a little closer to my C than the pic actually shows

  18. freckles
    • one year ago
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    |dw:1437369720333:dw|

  19. anonymous
    • one year ago
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    ok

  20. freckles
    • one year ago
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    now we want to verify that is a trapezoid

  21. freckles
    • one year ago
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    So we need to show the slope of line segment BC is the same as the slope of the line segment AD

  22. freckles
    • one year ago
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    This is because a trapezoid is a 4-sided flat shape with straight sides that has a pair of opposite sides that are parallel. To show two lines are parallel you just show that their slopes are equal.

  23. freckles
    • one year ago
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    \[\text{ slope of a line } y=mx+b \text{ with coordinates } (x_1,y_1) \text{ and } (x_2,y_2) \\ \text{ is } \frac{y_2-y_1}{x_2-x_1}\]

  24. freckles
    • one year ago
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    so line BC has points (a,b) and (2a,b) so it's slope is?

  25. anonymous
    • one year ago
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    I don't know actually.

  26. freckles
    • one year ago
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    you can also just observe from the picture the slope of line segment BC is zero since it is a horizontal line segment but you can also do it algebraically using the formula I gave you have: \[(x_1,y_1)=(a,b) \text{ and } (x_2,y_2)=(2a,b) \\ \text{ so } x_1=a \\ y_1=b \\ x_2 =2a \\ y_2=b \\ \text{ so entering into the the formula } \\ \frac{y_2-y_1}{x_2-x_1}=\frac{b-b}{2a-a}\] we know that b-b=0 and 2a-a=a and 0/a is 0 (since we also know a is not 0)

  27. anonymous
    • one year ago
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    oh ok

  28. freckles
    • one year ago
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    anyways we also need the slope of the side that is opposite to BC

  29. freckles
    • one year ago
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    if we can show that the side opposite to BC has the same slope I would say we are done since obviously the shape is 4-sided flat shape

  30. freckles
    • one year ago
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    What did I say the slope of any horizontal line is?

  31. anonymous
    • one year ago
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    0?

  32. freckles
    • one year ago
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    so slope of BC is 0 and slope of AD is 0 |dw:1437370464744:dw| the opposite sides called BC and AD are parallel since they both have the same slope I think they asked you to find the coordinates to use the slope formula though (but honestly that is kind of weird since the line segments BC and AD obviously appear horizontal ) \[\text{ slope of } BC=\frac{b-b}{2a-a}=\frac{0}{a}=0 \\ \\ \text{ slope of } AC=\frac{0-0}{3a-0}=...\] I will let you finish that if you want to.

  33. freckles
    • one year ago
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    anyways that is done we have shown this 4-sided flat shape does consist of opposite sides that are parallel

  34. anonymous
    • one year ago
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    0/3a

  35. freckles
    • one year ago
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    and 0/(3a) is 0 since 0/anything =0 (exlcuding the anything from being 0) (we know a isn't 0 since according to the picture a is greater than 0)

  36. freckles
    • one year ago
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    in short 0/(3a)=0

  37. anonymous
    • one year ago
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    thank you

  38. freckles
    • one year ago
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    np do you have anymore question about this question?

  39. freckles
    • one year ago
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    I'm about to vanish into thin air

  40. anonymous
    • one year ago
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    no thank you though

  41. freckles
    • one year ago
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    abracadabra

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