Colin has a pad with x pieces of paper on it. For his first class, he wrote on 5 fewer than half of the pieces of paper in the pad. He used 2 more sheets in his second class than in his first. How many sheets are left for his third class? ill give brainliest to the first answer

Colin has 8 sheets left for his third class.

Step-by-step explanation:

Given that:

Total Number of pieces of papers =

Number of pieces of papers used for 1st class = 5 fewer than half of the pieces in the pad

Writing the equation:

Also, Given that number of pieces of papers used for the 2nd class are 2 more than that of papers used in the 1st class.

Now, number of pieces of papers left for the third class = Total number of pieces of papers in the pad - Number of pieces of papers used in the first class - Number of pieces of papers used in the first class

So, the answer is:

Colin has 8 sheets left for his third class.

341

step-by-step explanation:

all you have to do is 9x31 (which equals 279) and then 20x31 (which equals 620). then you have to subtract. 620-279=341.

∠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.