Given vector A with rightwards harpoon with barb upwards on top has a magnitude of 5 meters directed 35 degrees above the positive x-axis, and vector B with rightwards harpoon with barb upwards on top has a magnitude of 8 meters directed 70 degrees above the positive x-axis, calculate A with rightwards harpoon with barb upwards on top times B with rightwards harpoon with barb upwards on top . A. 32.8 B. 13 C. 22.9 D. 40

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This did i unseated, sorry

W= fd
where f is an applied force and d is the distance over which it is applied. when a parachuting skydiver is at constant velocity, he is experiencing "terminal velocity", the point at which his acceleration down is canceled out by air resistance. that means that the resisting force of the air is equal to his weight, but in the opposite direction. with this, we can solve for the work done by the force.
w = fd
w = (92.0)(9.81 m/s^2) (325)
w= 2.93 x 10^5 j
the value for work in this case is positive because it is pushing up on the skydiver, who has a downward force of gravity.
therefore the answer is 2.93 x 10^5 j