A line passes through the point (10,-5) and a slope of -3/2 Write an equation in slope-intercept form for this line.

y = -3/2x + 10

Step-by-step explanation:

The slope = -3/2

Using the slope = -3/2 and the point (10, -5) substitute into y = mx + b and solve for "b"

y = mx + b

-5 = -3/2(10) + b         -3/2 · 10 = -30/2 = -15

-5 = -15 + b

-5 + 15 = b

 10 = b

The equation of the line is y = -3/2x + 10

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

y=3x+4

step-by-step explanation:

given the points a(0,4) and b(-4, -8)

eqn of line ab: (y-4)/(x-0)=(-8-4)/(-4-0)

(y-4)/x=-12/-4=3

y=3x+4