Answer this pl... 1.) 2³-3×(8-6)=
a.9 b.2 c.8 d.10
2.) 12÷2×(8÷2)=
a.24 b.20 c.18 d.28
3.) 20+(12×19)=
a.267 b.258 c. 250 d.248
4.) 21+4×12=
a.69 b.59 c.65 d.49
5.) 15×13+25×6=
a.125 b.265 c.345 d.575

1.c

2.b

3.d

4.a

5.b

Step-by-step explanation:

sana po maka tulong

There are mistakes in the last answer to your question! Here is the correct solution.

1. b

2. a

3. d

4. a

5. c

∠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.

lockers 1, 4, 9, 16, 25, 40, and 49 are open

i ran through it by the other answer is wrong