Marcel used the multiplication table below to determine that 72:24 is equivalent to 6:18. A multiplication table. In the row labeled 3, the numbers 3, 6, 9, 12, 15, 18, 21, 23, and 27 are highlighted. In the row labeled 9, the numbers 9, 18, 27, 36, 45, 54, 63, 72, and 81 are highlighted. What was Marcel’s error? Marcel did not look at the correct rows in highlighted columns. Marcel did not choose numbers from the highlighted columns. Marcel did not put the first and second terms in the same order. Marcel did not use numbers from the same row to create each ratio.

The correct option is;

Marcel did not put the first and second term in the same order

Step-by-step explanation:

From the multiplication table the equivalent ratios of two numbers can be found given that the two numbers have the same factors which is true if the numbers are on the same column in the multiplication table

Therefore, having the numbers ratio presented in the form 72:24 which are on the row labelled 3 and the row labelled 9, we have that the 72 is on row 9 while the 24 is on row 3, for which on the second column, we have 6:18

However the ordering of the ratio gives the value on the 9th column first before the value on the 3rd column, so that the correct result should be 18:6

The error was that Marcel did not put the first and second term in the same order.

∠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.