Mathematics, 28.07.2019 11:30 skyhighozzie

What is the total number of digits required in numbering the pages of a book, which has 1,450 pages?

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Answer from: Falconpride4079

The total number of digits required in numbering the pages of a book, which has 1,450 pages is 4693.

Step-by-step explanation:

The Total number of pages in the book is 1,450 pages.

Divide the amount 1450 into one-digit numbers, two-digit numbers, three-digit numbers, and four digits numbers.

Therefore,

One-digit numbers will be 1,2,3,4......9.

Count=9.

Two-digit numbers will be 10,11,12,13.......99.

Count=90.

Three-digit numbers will be 100,101,102,..............999.

Count=900.

Four-digit numbers will be 1000,1001,1002.......1450.

Count=451.

The Total number of words in printing the one-digit numbers is 9.

The Total number of words in printing the two-digit numbers is 90 multiply by 2 which is 180.

The Total number of words in printing the three-digit numbers is 900 multiply by 3 which is 2700.

The Total number of words in printing the four-digit numbers is 451 multiply by 4 which is 1804.

The total number of digits required in numbering the pages of a book, which has 1,450 pages will be the addition of words used in printing the one, two, three, and four digits numbers.

Therefore,

$\rm{Printing\;words}=9+180+2700+1804\\\rm{Printing\;words}=4693.$

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Answer from: Damagingawsomeness2

9 digits for 1-9
180 digits for 10-99
2700 digits for 100-999
451*4 = 1804 digits for 1000-1450

a total of 4693 digits

Answer from: Quest

answer:

step-by-step explanation:

Answer from: Quest

step-by-step explanation:

the original amount doesn't ''re just looking for the time it takes for 1/2 of the substance to we have

(1/2) =) take the ln of both sides

ln (1/2) = ) and we can write

ln(1/2) = ) lne and lne = 1 so we have

ln(1/2) = -.1092t divide both sides by -.1092

t = ln(1/2)/-.1092 = about 6.347 days = about 6.3 days (rounded)

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