A triangle has vertices at B(-3,0), C(2, -1), D(-1,2). Which transformation would produce an image with vertices B"(1, -2), C"(0, 3), D"(3, 0)? (x, y) - (x + 1, y + 1). (x, y) - (y, x):
(x, y) - (x + 1, y + 1), (x, y) + (-x, y),
O (X, Y) - (x, -y). (x, y) - (x + 2 y + 2)
O(x, y) + (-x, y). (x, y) - (x + 2 y + 2)

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

step-by-step explanation:

you divide the denominator by the numerator on both sides and if the quotients are equal then the fractions are equivalent