Istep is coming up and im not !
okay so im in 7th grade but i really should be in 6th. i don't really understand the < with lines and all that. also the arrows with the filled or closed dots pointing htis way or that
the "solve for" problems also i'm not good with
could somebody me

> and < with the lines under it mean that they can also be equal to the constant on the other side (greater/less than or equal to)

So x <= 3 can be anything less than 3 and 3.

*the = in this inequality is the equivalent of a line

Now vice versa.

x => 3 can be anything greater than 3 and 3.

Now for the dots.

The open dots in signs are (< and >).

Basically they start at that point they are on but don't include it.

So an open dot at 3 pointing to the left means that the value is greater than 3 but does not include 3.

The closed dots in signs are (<= and =>).

They start at the point and include it.

So a closed dot a 3 pointing to the left means that the value is greater than 3 and includes 3.

Arrows.

Arrows pointing to the left mean that the values are getting bigger.

Arrows pointing to the right mean that the values are getting smaller.

ie : An arrow pointing to the left with a starting point at 3 = 3 and anything bigger, such as 4.

ie : An arrow pointing to the right with a starting point at 3 = 3 and anything smaller, such as 2.

Solve for problems.

The main concept is isolate the variable.

This basically means make the variable be lonely ^^

Use +, -, *, and /

Ex : 6x + 2 > 3

First, we would subtract 2 from both sides.

6x > 1

Then, we would divide each side by 6, leaving the x alone.

x > 1/6

And voila! x is anything greater than 1/6 but it does not include 1/6.

Hope this helps! ^^

∠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.