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pishuonlain [190]
2 years ago
6

A very large book has pages numbered from 1 to 11521. How many times is the digit 3 printed? Helppp

Mathematics
1 answer:
-Dominant- [34]2 years ago
7 0

Answer:

The total number of pages from page X to page Y is Y minus X plus 1.

So from 1 to 9, there are 9 minus 1 is 8 plus 1 is 9.

Step-by-step explanation:

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The Naturally Made Bath and Body store pays $550 a month for rent and utilities. The average cost for its products to be manufac
Lilit [14]

Answer:

220 products need to be sold to break it even.

Step-by-step explanation:

550 + 3.00x = 5.50x

550 = 2.50x

220 = x

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Alexander's dividing oranges into eighths he has 5 oranges.how many eights will be have
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Ther will be 40 eights. Hope this helps!
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at a restaurant, Trevor decides to tip the server 16% of the bill. If he leaves $4.00 tip, which proportion could be used to fin
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4\25 is the proportion of the percentage


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Members of the pep club were selling raffle tickets for $1.50 and $5. The number of $1.50 tickets sold was two less than four ti
marysya [2.9K]

Answer:

see below

Step-by-step explanation:

1.5x + 5y = 1152    

x = 4y – 2

We can substitute the second equation into the first equation

Which one-variable linear equation can be formed using the substitution method?

1.5(4y-2) +5y = 1152  

Distribute

6y -3 +5y = 1152

Combine like terms

11y-3 = 1152

Add 3 to each side

11y-3+3 = 1152+3

11y = 1155

Divide each side by 11

11y/11 = 1155/11

y = 105

How many $5 raffle tickets were sold?

105 5 dollar tickets were sold

Now we need to find the number of 1.50 tickets

Which equation can be used to determine how many $1.50 raffle tickets were sold?

x = 4y – 2

x = 4(105) -2

   =420-2

   = 418

How many $1.50 raffle tickets were sold?

418    $1.50 tickets were sold

3 0
2 years ago
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A specimen of aluminum having a rectangular cross section 10 mm 12.7 mm (0.4 in. 0.5 in.) is pulled in tension with 35,500 N (80
miskamm [114]

Answer:

\epsilon = 3.958\times 10^{-3}

Step-by-step explanation:

The Young's Module of Aluminium is E = 69\times 10^{9}\,Pa. The axial stress on the specimen is:

\sigma = \frac{F}{A_{t}}

\sigma = \frac{35500\,N}{(0.01\,m)\cdot (0.013\,m)}

\sigma = 2.731\times 10^{8}\,Pa

The strain is derived of following expression:

\epsilon = \frac{\sigma}{E}

\epsilon = \frac{2.731\times 10^{8}\,Pa}{69\times 10^{9}\,Pa}

\epsilon = 3.958\times 10^{-3}

4 0
2 years ago
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