Which is the graph of f(x) = one-fourth(4)x? On a coordinate plane, an exponential function approaches y = 0 in quadrant 1. It curves up to the left and goes through points (2, 1) and (0, 4).

On a coordinate plane, an exponential function approaches y = 0 in quadrant 1. It curves up to the left and goes through points (1, 1) and (0, 4).

On a coordinate plane, an exponential function approaches y = 0 in quadrant 2 and goes through points (3, 2) and (4, 4).

On a coordinate plane, an exponential function approaches y = 0 in quadrant 2 and goes through points (1, 1) and (2, 4).

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intensity of the vector is v= √37 ≈ 6.08 units and make angle ∡α ≈ 9.46° with east direction

step-by-step explanation:

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°

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41*67/2=1373.5

step-by-step explanation:

i just used a calculator

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total of apples - take away four

6-4 = 2

answer is 2 apples