Q: What is the answer 1 7?

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2 1/7 7/7 + 7/7 + 1/7 = 15/7 1 + 1 + 1/7 = 2 1/7

1/7 of 7 = 1

15/72 1/7:= [(7 * 2)+1]/7= [14 + 1]/7= 15/7 in improper fraction

7 - 1 = 66

It will depend on where you put your parentheses. Root 7 -( 1/root 7) is different from (root 7-1)/root 7. * * * * * True, but a more helpful answer: [sqrt(7) - 1]/[sqrt(7) + 1] - [sqrt(7) + 1]/[sqrt(7) - 1] Multiplying the numerator and denominator of the first fraction by [sqrt(7) - 1] and the second fraction by [sqrt(7) + 1] = [sqrt(7) - 1]2/[7 - 1] - [sqrt(7) + 1]2/[7 - 1] =[7 - 2*sqrt(7) + 1]/6 - [7 + 2*sqrt(7) + 1]/6 = 16/6 = 8/3

Related questions

(1 + 1/7) = (7/7 + 1/7) = 8/7

2 1/7 7/7 + 7/7 + 1/7 = 15/7 1 + 1 + 1/7 = 2 1/7

7 ÷ 1/7 = 7 × 7/1 = 49.

1/7 of 7 = 1

It is 1/7. When you multiply (1/7) by 7 you get 1 as a result.

Yes since 7+1 > 7 1+7 > 7 7+7 > 1 Those are all true. So Yes.

7*(7 + 7) + 7/7 + 7/7 = 7*14 + 1 + 1 = 98 + 1 + 1 = 100

15/72 1/7:= [(7 * 2)+1]/7= [14 + 1]/7= 15/7 in improper fraction

7 - 1 = 66

It will depend on where you put your parentheses. Root 7 -( 1/root 7) is different from (root 7-1)/root 7. * * * * * True, but a more helpful answer: [sqrt(7) - 1]/[sqrt(7) + 1] - [sqrt(7) + 1]/[sqrt(7) - 1] Multiplying the numerator and denominator of the first fraction by [sqrt(7) - 1] and the second fraction by [sqrt(7) + 1] = [sqrt(7) - 1]2/[7 - 1] - [sqrt(7) + 1]2/[7 - 1] =[7 - 2*sqrt(7) + 1]/6 - [7 + 2*sqrt(7) + 1]/6 = 16/6 = 8/3

7

4 * 7/1 = 28 ---------( means there are 28 1/7's in 4 )