Simplify the following 3√28-5√63+4√112

7√7

using the ' rule of radicals '

• √a × √b ⇔ √ab

simplifying the radicals

√28 = √(4 × 7 ) = √4 × √7 = 2√7

√63 = √(9 × 7) =√9 × √7 = 3√7

√112 = √(16 × 7 ) = √16 × √7 = 4√7

substituting into the expression

3(2√7) - 5(3√7) + 4(4√7) = 6√7 - 15√7 + 16√7 = 7√7

First, simply each term.  Then, add the coefficients of the like terms (same radicals).

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chad should not have included the values (1, 2) and (3, 6) because that would mean he only had half a healing card.

∠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.