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chancemorris123

  • one year ago

Find the surface area of the composite solid. Round your answer to the nearest hundredth.

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  1. chancemorris123
    • one year ago
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  2. UnkleRhaukus
    • one year ago
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    The total surface area of the composite solid, can be decomposed into three areas: 1 the circular base 2 the lateral surface area of the cylinder (a rectangle) & 3 the lateral area of the cone (a sector)

  3. UnkleRhaukus
    • one year ago
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    The area of a circle (or radius \(r\)) is : \[A_\text{circle} = \pi r^2\]

  4. UnkleRhaukus
    • one year ago
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    The area of a rectangle (of width \(w\), height \(h\)) is: \[A_\text{rectangle} = w\times h\]

  5. chancemorris123
    • one year ago
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    so i tryed it and got c is that right

  6. UnkleRhaukus
    • one year ago
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    i don't know,

  7. chancemorris123
    • one year ago
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    that makes to of us

  8. UnkleRhaukus
    • one year ago
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    The area of the sector (of radius \(r\), and slope \(s\)) is: \[A_\text{sector} = \pi r s\]

  9. UnkleRhaukus
    • one year ago
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    add them all together, what do you get ? (before plugging in the numbers)

  10. chancemorris123
    • one year ago
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    i put c and i only got 30 min :( and got 6 more questions

  11. UnkleRhaukus
    • one year ago
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    i dont know what the answer is, we have to work it out

  12. UnkleRhaukus
    • one year ago
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    \[SA_\text{total} = A_\text{circle}+A_\text{rectangle}+A_\text{sector}\\ \qquad\quad= \quad. . .\]

  13. chancemorris123
    • one year ago
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    https://www.youtube.com/watch?v=Qmb9cnx__hA

  14. chancemorris123
    • one year ago
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    never mind that one i guessed cuz ii need to do the rest befopre time runs out

  15. UnkleRhaukus
    • one year ago
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    NB: the width of the rectangle the circumference of a circle \[w =C = 2\pi r\]

  16. UnkleRhaukus
    • one year ago
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    \[SA_\text{total} = A_\text{circle}+A_\text{rectangle}+A_\text{sector}\\ \qquad\quad= \pi r^2+2\pi r h+\pi rs\\ \qquad\quad= \pi r(r+2h+s)\\ \qquad\quad= \quad. . .\]

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