Find the direction angle of the vector b= < -3, 11> . 74.7450 105.2550 164.7450 149.7420

The given vector is b= -3i +11j

b=<-3, 11>

here, base= 3(not -3) (this gives an angle in first quadrant the same angle will foam in second quadrant from -ve x-axis)and perpendicular= 11

Now,

tan(θ')= perpendicular/base = 11/3

θ' = tan⁻¹(11/3)

= 74.7448°

but we require the angle from positive x-axis so,

θ = 180-θ'

= 180-74.7448

= 105.255°

B is the correct answer.

x² + 3x + 7 = 0       [ax² + bx + c]

a = 1

b = 3

c = 7

quadratic formula:

    (+/- is this ±)

plug in the numbers into the variables

your answer is a