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DETAILS LARLINALG8 6.R.030. Rotate the triangle with vertices (2, 3), (3, 2), and (2, 0) counterclockwise...
DETAILS LARLINALG8 3.R.068. Use a determinant to find the area of the triangle with the given vertices. (-2, 0), (2, 0), (0,5)
DETAILS LARLINALG8 3.R.027. Find JAl and A-11. 1 0 -2 А 0 3 = 2 - 5 7 6 (a) JA (b) |A-1,
6. (1 point) Use Stokes' Theorem to find the line integral /2y dx + dy + (4-3x) dz, where C is the boundary of the triangle with vertices (0,0,0), (1,3,-2), and -2,4,5), oriented counterclockwise as viewed from the point (1, 0, 0) 6. (1 point) Use Stokes' Theorem to find the line integral /2y dx + dy + (4-3x) dz, where C is the boundary of the triangle with vertices (0,0,0), (1,3,-2), and -2,4,5), oriented counterclockwise as viewed from the...
DETAILS LARLINALG8 2.R.003. Perform the matrix operation. 1 2 9 -2 8 6 -5 8 ] 5 0 0 9 0
Viewing Saved Work Revert to Last Response 8. DETAILS LARLINALG8 3.R.027. Find (Al and A-11. 1 0 -2 A= 03 2 -5 7 6 (a) Al (b) A-11 9. DETAILS LARLINALG8 3.R.068.
Problem 3. (15 points) T3 finite element is defined over AABC (in physical coordinates). The vertices of this triangle have the following coordinates: A(-3, -5), B(2,-1), and C(-6, 4) f(x, y)dS ДАВС where f(x, y) 3 2х?-5у? + 3ху+x — у. Bonus problem (5 extra points) a) Solve Problem 3 using 3 point integration rule. b) Which rule (1 point or 3 point) gives more accurate result? c) What is the integration error, if 3 point rule is used? Problem...
(1 point) Let T'be the region inside the triangle with vertices (0, 0), (3, 0) and (3, 2), and let f(x, y) be the function which is 0 outside of T and f(x, y)-38 SK y Ior (x, y) inside T. 3888 Then E(XY)- 4.8
4. Let - xy’i +3yj , and let C be the counterclockwise oriented triangle whose vertices are 0(0,0), P(2,0) and Q(2,8). Using Green's Theorem, ايثار a) Sf(-2x)dyex b) ſj(-2xy)dydx c) ff(xv* + 3y Mdvdx 0555(xv? +3 y Ddydx e) none of these 00 0 0
DETAILS LARLINALG8 4.R.062. Find the coordinate matrix of x in R' relative to the basis B'. B' = {(1, -1, 2, 1), (1, 1, -4,3), (1, 2, 0, 3), (1, 2, -2, 0)}, x = (6,5, -8,2) [x]g: = Hill 11
DETAILS LARLINALG8 7.1.009. Determine whether x is an eigenvector of A. A= 6 2 2 3 (a) x = (1, 0) x is an eigenvector. O x is not an eigenvector. (b) x = (1, 2) x is an eigenvector. x is not an eigenvector. (C) x = (2, 1) x is an eigenvector. x is not an eigenvector. (d) x = (1, -2) x is an eigenvector. x is not an eigenvector.