Description
Explanation & Answer
Hi there! Thank you for the opportunity to help you with your question!
Dividing decimals can be very hard at first! But with a little bit of practice things work out! :)
Whenever you have a division involving decimals, one technique to deal with it is to try to get rid of as many decimal points as we can. So when we have:
7| 7.77
One way we can get rid of all decimal points is to multiply both numbers by 100 (because when we do that we get 100*7.77 = 777 which has no decimals!) One way to notice that this is actually an ok thing to do is to think of the following problem: if I had two dollars to divide between two people, each person would get a dollar. Now, if I multiply both numbers by 100, it would mean I have $2*100 = $100 to divide between 2*100 = 200 people. In the end, each person would still get a dollar!
So, if we multiply both numbers by 100 we get:
700 | 777
Now this can be done in the standard way. We start by asking "how many times does 700 fit into 777", which is 1
Then we subtract:
777-700 = 77
Since 700 does not fit into 77, we need to add a zero, and include a decimal point, so our answer so far is 1.
700 | 770
Now, 700 fits exactly once into 770. Then we subtract:
770 - 700 = 70.
So our answer so far is 1.1
700 | 700
Now 700 fits exactly into 700, with no remainder!
700 - 700 = 0
So we get our answer to be 1.11
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From the image, I can't see the problems very well. It is also very hard to explain this concept in 20 minutes, but I do hope this helps a little bit. I will give a few more examples:
1.1 | 6.82
Multiplying both numbers by 100, we get rid of all decimals:
110 | 682
110 fits 6 times into 682:
682 - 110*6 = 682 -660 = 22.
Since 110 is bigger than 22, we need to add a decimal point 6. and add a zerp
110 |220
Now 110 fits twice in 220:
220-110*2 = 220 - 220 = 0
So our final anwser is 6.2
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2.3 | 7.36
You can use same method, get answer 3.2.
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2.9 |12.76
is 4.4
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I can't really see the other problems, but I hope this helps a bit
Please let me know if you need any clarification. Always glad to help!