This is urgent
(02.03 mc) solve for x: −3|x − 3| = −6 (5 points)
select one: a. x = 1, x = −1 b. x = 1, x = 5 c. x = 0, x = 5 d. no solutions

A. x = 1, x = -1

Step-by-step explanation:

Given the inequality

−3|x − 3| = −6

The equation will give us two solutions due to the nodulus sign. The modulus sign means that the function x- 3 can take both positive and negative value.

If |x-3| is positive, we have:

-3(x+3) = -6

-3x-9 = -6

Moving -9 to the other side:

-3x = -6-(-9)

-3x = -6+9

-3x = 3

x = 3/-3

x = -1

Similarly if |x-3| is negative, we have:

-3{-(x-3)} = -6

-3(-x+3) = -6

3x-9 = -6

3x = -6+9

3x = 3

x = 3/3

x = 1

The solutions are x = -1 and x = 1

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