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anonymous

  • one year ago

Have half done, need help with other half. A company manufactures aluminum mailboxes in the shape of a box with a half-cylinder top. The company will make 1173 mailboxes this week. If each mailbox has dimensions as shown in the figure below, how many square meters of aluminum will be needed to make these mailboxes? In your calculations, use the value 3.14 for π , and round up your answer to the next square meter.

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  1. anonymous
    • one year ago
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  2. anonymous
    • one year ago
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    2(0.35*0.55)+2(0.35*0.23)+(0.55*0.23)=0.6725 2(3.14)(0.115)^2+(0.55)2(3.14)(0.115) Help with the rest please

  3. anonymous
    • one year ago
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    0.1679115+0.6725 = 0.8404115 / 1173 = 1398 ???

  4. campbell_st
    • one year ago
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    my question is whether the semi-circular to is closed or open..? you need the surface area of the mailbox front and back rectangle A = 2 x 0.23 x 0.35 = left and right rectangle A = 2 x 0.55 x 0.35 = rectangular base A = 0.23 x 0 55 = so find those areas now the curved top. closed cylinder = radius is 0.115 find the \[ A = \frac{1}{2} \times( 2 \times 3.14 \times 0.115^2 + 2 \times 3.14 \times 0.115\times 0.55)\] if the cylindrical top is open... delete \[2 \times 3.14 \times 0.115^2\] from the calculation. when you get the surface area of 1 mail box, multiply by 173 and then round the answer up... to the nearest square metre

  5. anonymous
    • one year ago
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    so 0.2401315 / 1173 ?

  6. anonymous
    • one year ago
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    is the cylinder open?? idk

  7. campbell_st
    • one year ago
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    when you have the surface area of 1 mailbox multiply by 1173... then round the answer

  8. anonymous
    • one year ago
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    0.2401315 * 1173 so is the answer 282?

  9. campbell_st
    • one year ago
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    I haven't done the calculation, I just explained how you should go about finding the surface area of 1 box then finding the total area of metal needed.

  10. anonymous
    • one year ago
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    oh...ok

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