Nameperstep 3: how tall is a tower of one hundred phow is a tower of one million pennies? taexplain any patterns you see. can you find more than one way to figure this out? height of tower (cm)of pennies1 cmpennies2 cm3 cmpennies4 cmpennies6 cm

points are handed out whenever you post answers.

brainliest answers are handed out by the person asking the question, so whether your answer is picked depends entirely on whether the asker bothers to pick

if you're asking about your rank, it depends on having a certain number of points and a certain number of brainliest answers. for example, at the time of this writing, i have 44338 points out of 15000 needed for the next (nearly triple the requirement! ) but i only have 24 out of 50 required brainliest answers. this number decays, meaning after a certain amount of time (i think about a month), the total number of brainliest answers you've earned doesn't count toward this total. (i've been stuck at the same rank for a long

the lesson to learn here is that you shouldn't worry about points or rank. chances are, no matter how good an answer you may provide, your efforts will go mostly unnoticed unless you focus on the newest/most active questions that get posted.

∠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.