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AfilCa [17]
2 years ago
7

Which place value first determines which of the numbers 6.399 or 6.400 is the larger number?

Mathematics
1 answer:
kupik [55]2 years ago
8 0

Comparing the numbers, it is found that the tenths value first determines which of the numbers 6.399 or 6.400 is the larger number.

The decimal numbers are 6.399 and 6.400.

  • The ones value for each is 6.
  • For the first value, the tenths digit is of 3, while for the second is of 4.

The tenths digit is the <u>first in which there is a difference</u>, hence, it determines which of the numbers 6.399 or 6.400 is the larger number.

A similar problem is given at brainly.com/question/17248958

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In this question, p q r is a prime number. Most of the prime number is an odd number. If p q r all odd number, it wouldn't be possible to get 73 since
odd x odd + odd= odd + odd = even
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There is only one odd prime number which is 2. If you put 2 in the r the result would be:
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The least possible value of p+q+r would be achieved by founding the highest q since its coefficient is 2 times r. Maximum q would be 73/2= 36.5 so you can try backward from that. Since q= 31, q=29, q=23 and q=19 wouldn't result in a prime number r, the least result would be q=17
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