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robinfr93 Group Title

Can someone help me solve this....

  • 2 years ago
  • 2 years ago

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  1. robinfr93 Group Title
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    |dw:1348871903875:dw| If length of this rectangle be thrice its width and its perimeter be 56, then find the co ordinates of the rest of the sides...

    • 2 years ago
  2. CliffSedge Group Title
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    Can you use that information to set up the two equations you need?

    • 2 years ago
  3. robinfr93 Group Title
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    Ive tried but Im getting two second degree equations!!!

    • 2 years ago
  4. robinfr93 Group Title
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    Actually I'm down with a terrible cold so my brain aint working so well.. So need a lil help!!!

    • 2 years ago
  5. CliffSedge Group Title
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    Woah. Well, "The length is thrice the width" --> L=3W.

    • 2 years ago
  6. CliffSedge Group Title
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    "Perimeter is 56" --> 2L+2W=56.

    • 2 years ago
  7. robinfr93 Group Title
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    yah got that part ryt!!

    • 2 years ago
  8. CliffSedge Group Title
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    Sub the first equation into the second to solve for W.

    • 2 years ago
  9. robinfr93 Group Title
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    length = 18 units and breadth 6

    • 2 years ago
  10. CliffSedge Group Title
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    18 and 6 won't get you a perimeter of 56.

    • 2 years ago
  11. robinfr93 Group Title
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    :P

    • 2 years ago
  12. robinfr93 Group Title
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    ryt!! Sorry!!! 21 and 7!!

    • 2 years ago
  13. CliffSedge Group Title
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    Excellent. Now you can just count over and down from the given point to find all the other coordinates.

    • 2 years ago
  14. robinfr93 Group Title
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    yeah!! But when I try to use the distance formula!!! Im getting two second degree equations!! How m I suppose to solve them when ive got two equation something like ax^2+b^y2+cx+dy+e=0

    • 2 years ago
  15. robinfr93 Group Title
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    I think i made some mistake somewhere. Would you be kind enough to help me out!! Its the darn cold!!

    • 2 years ago
  16. CliffSedge Group Title
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    Forget about the distance formula! Just count.

    • 2 years ago
  17. CliffSedge Group Title
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    The distance formula is so overrated. It is merely the pythagorean theorem, but when you have horizontal and vertical lines, you don't need to solve for a hypotenuse.

    • 2 years ago
  18. robinfr93 Group Title
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    then?? Should I just put in simple increment formula and solve??

    • 2 years ago
  19. robinfr93 Group Title
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    I know its suppose to be darn simple but my heads all frozen up!! @CliffSedge

    • 2 years ago
  20. CliffSedge Group Title
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    |dw:1348872917396:dw|

    • 2 years ago
  21. CliffSedge Group Title
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    |dw:1348872954175:dw| etc.

    • 2 years ago
  22. CliffSedge Group Title
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    No formulas, no solving - elementary arithmetic.

    • 2 years ago
  23. robinfr93 Group Title
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    k got it!! :D Thank man!!! Thanx a ton!! :P hahahahha!! Ryt!!

    • 2 years ago
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