is this correct way to do this problem?
(7 x1000) + (1 x 100) + (1 x 10) + (1 x1)
or(7 + 1000)+ (1 + 100) + (1 + 10)+ (1 + 1)
2007-02-23 07:21:25 · 8 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
7,000+100+10+1
2007-02-23 07:30:01 · answer #1 · answered by Anonymous · 0⤊ 0⤋
i'm slightly at a loss for words approximately your question, yet i think of it is the respond your searching for: 5 x 10^3 + 2 x 10^2 + a million x 10^a million + 2 x 10^0.5 x 10^3 = 5,000 2 x 10^2 = 2 hundred a million x 10^a million = 10 2 x 10^0 = 2
2016-12-17 17:17:17 · answer #2 · answered by howsare 4 · 0⤊ 0⤋
7 *1 + 1+0.1 + 1*0.01 + 1*0.001
i cant make more out of it. if 7,111...
thus the ones are repeating then 7,1111... = 7 + 1/9
2007-02-23 07:25:39 · answer #3 · answered by gjmb1960 7 · 0⤊ 0⤋
the first one is right a way to check it is to work out the problem you come up with and see if it matches the answer.
2007-02-23 07:30:27 · answer #4 · answered by shamira p 1 · 0⤊ 0⤋
7 groups of throusands
1 group of hundreds
1 group of tens
1 group of ones
2007-02-23 07:27:15 · answer #5 · answered by freeside49 5 · 0⤊ 0⤋
Your first choice is correct.
2007-02-23 07:27:50 · answer #6 · answered by etopro 2 · 0⤊ 0⤋
The first way is correct.
If you did the other way, you would come with 1,121.....which would not equal 7,111.
2007-02-23 07:31:23 · answer #7 · answered by Christopher Y 2 · 0⤊ 0⤋
your first one:
(7x1000)=7,000
7,000+(1x100)=7000+100=7,100
7,100+(1x10)=7,100+10=7,110
7,110+(1x1)=7,110+1=7,111
answer 7,111
2007-02-23 07:30:51 · answer #8 · answered by KatyD 1 · 0⤊ 0⤋