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

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ash2326
 3 years ago
Best ResponseYou've already chosen the best response.2Can you find the area of right triangle with base 5 mi and height 5 mi?

ash2326
 3 years ago
Best ResponseYou've already chosen the best response.2Do you know the formula for area of right triangle and\or circle?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Circle Area = π × r2 Circumference = 2 × π × r r = radius

ash2326
 3 years ago
Best ResponseYou've already chosen the best response.2Good, 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
 3 years ago
Best ResponseYou've already chosen the best response.0wat do you mean by that

ash2326
 3 years ago
Best ResponseYou've already chosen the best response.2put b=5 and h=5, then solve it to find the area of triangle

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0so ur saying 5x5x3.14

ash2326
 3 years ago
Best ResponseYou've already chosen the best response.2Nope, \[\text{Area of figure}=\text{Area of triangle+ Area of circle}\] \[\text{Area of triangle}=\frac 1 2 \times 5 \times 5\ mi^2\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0oh ok i get 12 1/2 tho

ash2326
 3 years ago
Best ResponseYou've already chosen the best response.2Yes, 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
 3 years ago
Best ResponseYou've already chosen the best response.0A semicircle 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 semicircle is half of a circle. Therefore the area of a semicircle will be: area of a semicircle = 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
 3 years ago
Best ResponseYou've already chosen the best response.2What's the radius here?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0First, let's look at the 2 dimensional region (base of solid) Each crosssection perpendicular to the xaxis is a "strip" of length e^x  0 = e^x Now we can look at 3 dimensional solid Each crosssection perpendicular to the xaxis is a semicircle with diameter = e^x So each crosssection 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
 3 years ago
Best ResponseYou've already chosen the best response.2From where are you posting this??? It's not relevant here

amistre64
 3 years ago
Best ResponseYou've already chosen the best response.0@preetha we need to get shadow to clean up the database .....

amistre64
 3 years ago
Best ResponseYou've already chosen the best response.0@shadowfiend might need your assistance in cleaning up some database material

amistre64
 3 years ago
Best ResponseYou've already chosen the best response.0please check the users profile page for some rather inappropriate content.

amistre64
 3 years ago
Best ResponseYou've already chosen the best response.0they also seem to have a habit of editing the content out of their questions :/
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