anonymous
  • anonymous
What is the approximate perimeter of the composite figure shown? Use three point one four for pi.
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
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ash2326
  • ash2326
Can you find the area of right triangle with base 5 mi and height 5 mi?
anonymous
  • anonymous
no

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ash2326
  • ash2326
Do you know the formula for area of right triangle and\or circle?
anonymous
  • anonymous
A=1/2xbxh or A=bxh/2
anonymous
  • anonymous
Circle Area = π × r2 Circumference = 2 × π × r r = radius
ash2326
  • ash2326
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
anonymous
  • anonymous
wat do you mean by that
ash2326
  • ash2326
put b=5 and h=5, then solve it to find the area of triangle
anonymous
  • anonymous
ok
anonymous
  • anonymous
so ur saying 5x5x3.14
ash2326
  • ash2326
Nope, \[\text{Area of figure}=\text{Area of triangle+ Area of circle}\] \[\text{Area of triangle}=\frac 1 2 \times 5 \times 5\ mi^2\]
anonymous
  • anonymous
oh ok i get 12 1/2 tho
ash2326
  • ash2326
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?
anonymous
  • anonymous
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!
ash2326
  • ash2326
What's the radius here?
anonymous
  • anonymous
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
ash2326
  • ash2326
From where are you posting this??? It's not relevant here
amistre64
  • amistre64
@preetha we need to get shadow to clean up the database .....
amistre64
  • amistre64
@shadowfiend might need your assistance in cleaning up some database material
shadowfiend
  • shadowfiend
?
amistre64
  • amistre64
please check the users profile page for some rather inappropriate content.
amistre64
  • amistre64
they also seem to have a habit of editing the content out of their questions :/

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