Use the work shown below to write the equation for a line that has a slope of Negative two-fifths and passes through (15, –2). y = mx + b 1. Substitute the known values: negative 2 = negative two-fifths (15) + b. 2. Solve for b: negative 2 = negative 6 + b. b = 4. What is the equation of the line in slope-intercept form? Negative 2 = negative two-fifths x + 4 y = negative two-fifths (15) + 4 y = negative two-fifths x minus 4 y = negative two-fifths x + 4

Step-by-step explanation:

The formula:

The steps:

The result

What is the equation of the line in slope-intercept form?

the answer is d

Step-by-step explanation:

just did it on edg

∠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.