Consider the expression. 6 · 60 · 6–3
Which statements are true about the expression? Check all that apply.
The 6 without an exponent is equivalent to the 6 having a 0 exponent.
The sum of the exponents is 2.
Multiply the exponents to simplify the expression.
The expression has a value of StartFraction 1 Over 36 EndFraction.
An equivalent expression is 65 · 6-7.

4. The expression has a value of 1/36 . True

5. An equivalent expression is (6)^5. (6)^- 7  True

Step-by-step explanation: Not true. The 6 without an exponent is equivalent to the 6 having a 0 exponent.

Any number having an exponent zero has a value equal to 1.

6 · (6)^0 · (6)^–3

= 6.1.1/216

= 1/36

= (6)^-2

2. The sum of exponents is 2.  

False ( the sum of exponents is -2)

3.Multiply the exponents to solve the expression .

False when the bases are the same the exponents are added or subtracted.

4. The expression has a value of 1/36 .

True

5. An equivalent expression is (6)^5. (6)^- 7

True

the correct answers are D and E. I just got this right. hope it helps

∠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.

so, converting firstly the mixed fractions to improper fractions and then "adding".