Bis the midpoint of ac and d is the mid point of ce solve for x given bd=3x + 5 and ae= 4x+20

triangle BCD is similar to triangle ACE (side, angle C, side)

AC = 2 BC

therefore AE = 2 BD

therefore

4x+20 = 2 (3x+5) solve that

Step-by-step explanation:

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

is there any photos or whatever?