The base of a three-dimensional figure is bound by the line y = 4 - x on the interval [-2, 2]. Vertical cross-sections that are perpendicular to the x-axis are right triangles with height equal to 4. Algebraically, find the area of each triangle. 3-2-1 1 2 3 4 5 6 7 ○ A(x) = (4 − x) ○ A(x) = (4 − x)² O A(x) = 2 (4 − x) O A(x) = 2 (4 + x)
The base of a three-dimensional figure is bound by the line y = 4 - x on the interval [-2, 2]. Vertical cross-sections that are perpendicular to the x-axis are right triangles with height equal to 4. Algebraically, find the area of each triangle. 3-2-1 1 2 3 4 5 6 7 ○ A(x) = (4 − x) ○ A(x) = (4 − x)² O A(x) = 2 (4 − x) O A(x) = 2 (4 + x)
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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![The base of a three-dimensional figure is bound by the line y = 4 - x on the interval [-2, 2]. Vertical cross-sections that are perpendicular
to the x-axis are right triangles with height equal to 4.
Algebraically, find the area of each triangle.
-3-2-11- 1 2 3 4 5 6 7
X
o A(x) = (4x)
○ A(x) = (4x)²
O A(x) = 2 (4 − x)
O A(x) = 2 (4 + x)](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F0c2bad51-6a1d-4d43-aa13-d2d93d3d6c59%2F1f6fd0b8-cd5f-49e8-a3ac-255689a369a9%2F1a8b0lb_processed.png&w=3840&q=75)
Transcribed Image Text:The base of a three-dimensional figure is bound by the line y = 4 - x on the interval [-2, 2]. Vertical cross-sections that are perpendicular
to the x-axis are right triangles with height equal to 4.
Algebraically, find the area of each triangle.
-3-2-11- 1 2 3 4 5 6 7
X
o A(x) = (4x)
○ A(x) = (4x)²
O A(x) = 2 (4 − x)
O A(x) = 2 (4 + x)
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