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Math Help - Two problems with solutions...help!!

  1. #1
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    Two problems with solutions...help!!

    Sergio plans to paint the 4 walls of his room with 1 coat of paint. The walls are rectangular, and, according to his measurements, each wall is 10ft by 15ft. He will not need to paint the single 3-foot-by-5-foot rectangular window in his room and the 3.5 foot-by-7-foot rectangular door. Sergio knows that each gallon of paint covers between 300 and 350 square feet. If only 1-gallon cans of paint are available, what is the minimum number of cans of paint Sergio needs to buy to paint his walls?

    The book says the solution is 2 but I can't figure it out??

    Also, I need to solve the following:

    8^(2x + 1) = 4^(1 - x)

    The solution is -1/8...not sure why though...any help would be appreciated on these.
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  2. #2
    Super Member wingless's Avatar
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    There are 4 10x15 ft walls. A walls measures are 10x15 ft and its area is 150 ft². 4 walls will make 150x4 = 600 ft². As he won't paint the 3x5 ft (15 ft²) window and the 3.5x7 ft (24.5 ft²) door, we can subtract them from the total area of walls.
    600 - 15 = 585
    585 - 24.5 = 560.5 ft²

    If 1 gallon can of paint can cover 300-350 ft², and the question asks for the best situation, we will take it as 350 ft². So he will need 2 cans to paint 560.5 ft².

    ------------

    8^{2x + 1} = 4^{1 - x}

    (2^3)^{2x+1} = (2^2)^{1-x}

    2^{6x+3} = 2^{2-2x}

    6x + 3 = 2 - 2x

    8x = -1

    x = - \frac{1}{8}
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  3. #3
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    Thanks so much wingless--can't believe I didn't think of that for both problems!
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  4. #4
    Super Member wingless's Avatar
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    You're welcome. Don't worry, as many questions as you solve, as much as you think on them, you'll get better at 'seeing things' in questions
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