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VMariaS [17]
3 years ago
6

How many times does the digit 7 appear among the terms of the sequence of consecutive integer numbers 7, 8, 9, ...., 777?

Mathematics
2 answers:
harina [27]3 years ago
7 0

Answer:

It appears 234  times

Step-by-step explanation:

For first 6 hundreds i.e, 17-107, 117-207, 217-307, 317-407, 417-507, 517-607, there are 20 sevens for each. Which gives a total of 20*6 = 120 sevens.

The number '7' along with this makes a total of 121 sevens.

From 617-707, there are 28 sevens. TOTAL = 121 + 28 = 149

From 708-716, there are 9 sevens.

Now, from 717-777, we have a total of 76 sevens

Adding these all makes total= 149+9+76 = 234

Dahasolnce [82]3 years ago
4 0

Answer:

234 times

Step-by-step explanation:

<u>Number of times the number 7 appears in a hundred</u>

7 as units digit (07-17-27 ..... 97): 10 times

7 as tens digit (70-71-72..... 79): 10 times

20 times the digit 7 appears in first one hundred (0-100)

Let's calculate how many times 7 would be as units or tens in 7 hundreds

20X7 = 140 times digit 7 appears until number 699

<u>Now, from 700 to 777</u>

7 as hundreds digit (700-701-702 .... 777): 78 times

7 as tens digit (770-771-772 .... 777): 8 times

7 as units digit (707-717-727....777): 8 times

78 + 8 + 8 = 94 times the digit 7 appears in the range 700 - 777. Plus 140 times

140 + 94 = 234 times

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Step-by-step explanation:

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3 0
2 years ago
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3 years ago
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zlopas [31]

Well, following the order of PEMDAS, I got choice B. 52

For instance, when you plug in 5 for x, you get F(5)=2(5)^2+2.

Moreover, following PEMDAS, you're supposed to solve what's inside the parenthesis, but since there is no operation going on inside the parenthesis, then you simple move on to the exponent.

In this case, you square the number 5, which gives you F(5)=2(25)+2

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