The vertex of this parabola is at (4,-3). which of the following could be its equation?

Answer with Step-by-step explanation:

The equation with general formula

x-h=4a(y-k)²

has vertex at (h,k)

The  equation with vertex at (4,-3) can be written as:

x-4=4a(y+3)²

x=4a(y+3)²+4

Let 4a=-3

x=-3(y+3)²+4

Hence, the equation of parabola with vertex at (4,-3) can be:

x=-3(y+3)²+4

Hence, option A is correct

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what is the image?

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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