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Yasminelove1

  • 3 years ago

What is the approximate perimeter of the composite figure shown? Use three point one four for pi.

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  1. Yasminelove1
    • 3 years ago
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  2. ash2326
    • 3 years ago
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    Can you find the area of right triangle with base 5 mi and height 5 mi?

  3. Yasminelove1
    • 3 years ago
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    no

  4. ash2326
    • 3 years ago
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    Do you know the formula for area of right triangle and\or circle?

  5. Yasminelove1
    • 3 years ago
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    A=1/2xbxh or A=bxh/2

  6. Yasminelove1
    • 3 years ago
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    Circle Area = π × r2 Circumference = 2 × π × r r = radius

  7. ash2326
    • 3 years ago
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    Good, you're right here for the triangle b=5 mi h=5 mi \[\frac 1 2 b\times h\] Just plugin the values, you'll get the area of triangle

  8. Yasminelove1
    • 3 years ago
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    wat do you mean by that

  9. ash2326
    • 3 years ago
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    put b=5 and h=5, then solve it to find the area of triangle

  10. Yasminelove1
    • 3 years ago
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    ok

  11. Yasminelove1
    • 3 years ago
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    so ur saying 5x5x3.14

  12. ash2326
    • 3 years ago
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    Nope, \[\text{Area of figure}=\text{Area of triangle+ Area of circle}\] \[\text{Area of triangle}=\frac 1 2 \times 5 \times 5\ mi^2\]

  13. Yasminelove1
    • 3 years ago
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    oh ok i get 12 1/2 tho

  14. ash2326
    • 3 years ago
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    Yes, you're right. Now we need to find the area of semi circle, It's diameter is 7 mi Can you find the area?

  15. Yasminelove1
    • 3 years ago
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    A semi-circle is half of a circle and the area of a circle of radius r is pi r2. We know that the area of a circle is: area of a circle = pi r2 A semi-circle is half of a circle. Therefore the area of a semi-circle will be: area of a semi-circle = 1/2 pi r2 We know that the radius is 9ft. Pi is approximately 3.14 or 22/7. Now you just have to plug in the numbers and you will find the area!

  16. ash2326
    • 3 years ago
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    What's the radius here?

  17. Yasminelove1
    • 3 years ago
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    First, let's look at the 2 dimensional region (base of solid) Each cross-section perpendicular to the x-axis is a "strip" of length e^x - 0 = e^x Now we can look at 3 dimensional solid Each cross-section perpendicular to the x-axis is a semi-circle with diameter = e^x So each cross-section has area = 1/2 (π r²) = 1/2 π (e^x/2)² = π/8 * e^(2x) Each very thin cross section of width dx has volume = π/8 * e^(2x) dx Integrating from x = -1 to x = 1, we get: Volume = ∫₋₁¹ π/8 * e^(2x) dx . . . . . . = π/16 * e^(2x) |₋₁¹ . . . . . . = π/16 (e^(2) - e^(-2)) . . . . . . = 1.4242647516

  18. ash2326
    • 3 years ago
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    From where are you posting this??? It's not relevant here

  19. amistre64
    • 3 years ago
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    @preetha we need to get shadow to clean up the database .....

  20. amistre64
    • 3 years ago
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    @shadowfiend might need your assistance in cleaning up some database material

  21. shadowfiend
    • 3 years ago
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    ?

  22. amistre64
    • 3 years ago
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    please check the users profile page for some rather inappropriate content.

  23. amistre64
    • 3 years ago
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    they also seem to have a habit of editing the content out of their questions :/

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