Evaluate the following, using the distributive property. (i) – 39 × 99 (ii) (– 85) × 43 + 43 × ( – 15)

(iii) 53 × ( – 9) – ( – 109) × 53

(iv) 68 × (–17) + ( –68) × 3

(v) 26 x (-48) + (-48) x 36

(vi) (-41) x 102

(vii)(-57) x (-19) + 5

(i) - 33 × 99

= - 3267

(iii) 53 × ( - 9) - ( - 109) × 53

= - 477 + 109 × 53

= - 477 + 5777

= + 5300

(ii) (-85) × 43 + 43 × ( - 15)

= - 3655 + 43 - 645

= - 3655 - 645 + 43

= - 4300 + 43

= - 4257

∠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.