Write an equation perpendicular to the line y = 3x + 5 through the point (6, -5). State your answer as y=mx+b.

y=(-1/3) x - 3

Step-by-step explanation:

A line that is perpendicular to it has a slope that is an opposite reciprocal, which in this case is -1/3.  Now you use point slope form which is y-y1=m(x-x1) that would be y+5=-1/3(x-6).  Distribute the -1/3 and get y+5= -1/3 x+ 2.  Subtract 5 from both sides and get y=-1/3 x - 3

∠1≅∠2 by the alternate exterior angles theorem.

step-by-step explanation:

given, a ∥ b and ∠1 ≅ ∠3 .we have to prove that e ∥ f

we know that ∠1≅∠3 and that a || b because they are given. we see that by the alternate exterior angles theorem. therefore, ∠2≅∠3 by the transitive property. so, we can conclude that e || f by the converse alternate exterior angles theorem.

we have to fill the missing statement.

transitivity property states that if a = b and b = c, then a = c.

now, given ∠1≅∠3 and by transitivity property ∠2≅∠3 .

hence, to apply transitivity property one angle must be common which is not in result after applying this property which is ∠1.

the only options in which ∠1 is present are ∠1 and ∠2, ∠1 and ∠4

∠1 and ∠4 is not possible ∵ after applying transitivity we didn't get ∠4.

hence, the missing statement is ∠1≅∠2.

so, ∠1≅∠2 by the alternate exterior angles theorem.

l.jnl.j/n.m,lm,

step-by-step explanation: