Show that one zero of 8x230x27 is the square of the other.

8x2-30x+27 has one root (9/4) the square of the other (3/2)

Step-by-step explanation:

8x2+30x+27 = (2*x+3)*(4*x+9)  => solution = {-3/2, -4/9}

One is not the square of the other because squares are positive.

8x2-30x+27 = (2*x-3)*(4*x-9)  =>

solution = {3/2, 4/9}   since (4/9 = (3/2)^2, therefore the preceding question is correct.

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°

hope this