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horsegal244

  • one year ago

Charlie's Computer Company charges $0.65 per pound to ship computers. Part A: Write an equation to determine the total cost, c, to ship p pounds of computers. Use your equation to determine the cost of shipping 2 pounds of computers. Part B: If the company reduces the cost to ship computers by 0.05 per pound, write an equation to determine the total cost, c, to ship p pounds of computers with the reduced cost.

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  1. horsegal244
    • one year ago
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    @eta

  2. horsegal244
    • one year ago
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    @Hero

  3. horsegal244
    • one year ago
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    @Michele_Laino

  4. rainbow_rocks03
    • one year ago
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    plz I need help on a question

  5. horsegal244
    • one year ago
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    Well I need Help too

  6. horsegal244
    • one year ago
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    @kjones0331

  7. Z4K4R1Y4
    • one year ago
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    thin about this... if it costs $0.65 to ship 1 pound of weight how much does it cost to ship 2 pounds?

  8. horsegal244
    • one year ago
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    $1.30

  9. horsegal244
    • one year ago
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    @Z4K4R1Y4

  10. Michele_Laino
    • one year ago
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    it is simple, we have: c=0.65*p

  11. Michele_Laino
    • one year ago
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    since we have to pay $0.65 for one pound

  12. Z4K4R1Y4
    • one year ago
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    so you multiplied the amount of pounds (p) to 0.65 and you got the cost (c) so you have c=0.65p

  13. horsegal244
    • one year ago
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    Ok now what

  14. Z4K4R1Y4
    • one year ago
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    so for part (A) you have 'p' = 2. and for part (B) it's the same equation but instead of 0.65 you have 0.05.

  15. Michele_Laino
    • one year ago
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    part B: the new cost per pound is now: 0.65-0.05=0.60

  16. Z4K4R1Y4
    • one year ago
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    sorry i mad a mistake Michele is right.

  17. Z4K4R1Y4
    • one year ago
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    *made

  18. Michele_Laino
    • one year ago
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    so the new equation, is: c= 0.60*p

  19. horsegal244
    • one year ago
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    ok now what

  20. Z4K4R1Y4
    • one year ago
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    Finished, thats all you need to do for part B.

  21. horsegal244
    • one year ago
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    ok

  22. horsegal244
    • one year ago
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    can u help with 2 more

  23. horsegal244
    • one year ago
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    @Z4K4R1Y4

  24. Z4K4R1Y4
    • one year ago
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    I can try.

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