Arock climber's elevation above sea level changed −8212 feet in 214 minutes as she descended. what was the climber's average change in elevation per minute? enter your answer as a mixed number, in simplified form, in the box. in ft
i think is g(1)=-1.
hope that i can
intensity of the vector is v= √37 ≈ 6.08 units and make angle ∡α ≈ 9.46° with east direction
required vector is consists of the two componentsvx= 2+4=6 units and vy= 1 unit and vx ⊥ vywe will use pythagorian theorem to find intensity of the vector vv∧2 = vx∧2 + vy∧2 => v = √vx∧2 + vy∧2 = √6∧2 + 1∧2 = √36+1 = √37 ≈ 6.08 unitsthe angle ∡α between vector and east direction we wil find with tanαtanα = 1/6 => α = arc tanα = arc 1/6 => α ≈ 9.46°
elevation per minute=-8212/214=
-38 and 40/107
answer is -38 and 40/107 ft per minute
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