Calculate / Ss (4xy + e- )dS, where S is the triangle in Figure 1 with vertices (0, 0, 3), (1,0, 2) and (0, 4, 1). (0, 0, 3) (1,0, 2) (0, 4, 1)

Compute ∮c (ydx+zdy+xdz) where C is the triangle with vertices (3,0,0), (0,0,2), and (0,6,0) traversed in this order.

What are the vertices of ΔA'B'C produced by T−3, 6 (ΔABC) = ΔA'B'C?   A. A′(0, 6), B′(0, 4), C′(−3, 3)   B. A′(6, 6), B′(6, 4), C′(3, 3)   C. A′(0, −6), B′(0, −8), C′(−3, 9)   D. A′(6, −6), B′(6, −8), C′(3, 9)

Draw the triangle with vertices A = (2, 2), B = (-1, 3), and C = (0, 0). By regarding it as half of a parallelogram, explain why its area equals area (ABC) = 1 ¦det [_ 2 2 -1 3 Move the third vertex to C = (1, -4) and justify the formula 1 x1 У1 1 2 2 1 1 area (ABC) = det x2 y2 1 = det -1 3 2 2 X3 Уз 1 1 -4 1 }] Hint: Subtracting the last row from each of the others leaves 2 2 1 1 60 16 det -1 3 1 = det -2 7 0 = det -2 7 1 -4 1 1 -41 Sketch A' = (1, 6), B' = (-2, 7), C' = (0, 0) and their relation to A, B, C.

2. Triangle ABC with vertices (0, 7), B(7, 3), and C(1, 4): (x, y) → (x – 3, y – 4) A' (_ , B’(__ C' (_ 111

3. Triangle with vertices P(- 13, - 3) , Q(- 7, 3) , and R(-6,-2; 270 degrees counterclockwise rotation about C(- 5, - 4)

16. Triangle CDE with vertices C(2, 4), D(6, 7), and E(7, 3): 90° counterclockwise about the origin C'(_ D'(__, E' (_,

A triangle has vertices at R(1, 1), S(-2,-4), and T(-3,-3). The triangle is transformed according to the rule Ro 270. What are the coordinates of S'? O (4, 2) O (2, 4) O (2, 4) O (4,-2)