Cual de las siguientes rectas son perpendiculares a la recta de la ecuación 4x+6y-6=0

Ok, i will answer in English.

The first step is to know that when we have a linear equation like:

y = a*x + b

any line with a slope equal to -1/a will be perpendicular to our line.

So first let's write our equation in the standard form:

4x + 6y - 6 = 0

6y = -4x + 6

y = (-2/3)*x + 1

then the slope is (-2/3)

then the slope of the perpendicular line will be:

-(1/(-2/3)) = 3/2

So the only thing any line with the equation:

y = (3/2)*x + c

where c can be any real number, is perpendicular to 4x+6y-6=0

step-by-step explanation:

i have done this all on a piece of paper but idk how to put it on here cuz it needs to be drawn

answer: (c) m∠qpo + (2x + 16)° = 180°

step-by-step explanation:

the opposite angles of a quadrilateral are supplementary.

so, ∠o + ∠q = 180°   and   ∠p + ∠r = 180°

since ∠r = 2x + 16°, we can use substitution as follows:

∠p + ∠r = 180°

∠p + (2x + 16)° = 180°