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Alina [70]
3 years ago
9

Make greatest and smallest possible 7- digit number from 4,7,1,9,0,6,7​

Mathematics
2 answers:
Xelga [282]3 years ago
7 0

Answer:

greatest 9776410 smallest 1046779

Step-by-step explanation:

in smallest we cannot write 0 first because 0 has no units

Marysya12 [62]3 years ago
3 0

Answers:

Largest possible = 9776410

Smallest possible = 1046779

====================================================

Explanation:

To make the greatest 7-digit number, we start with the largest digit at the left-most end. So we start with 9. Then the next digit is the next largest, which is 7, and so on until we get

9776410

If we swap any two distinct digits, then we'll get something smaller than that value above. For instance, swap the 1 and 0 which leads to 9776401 and this is smaller than 9776410. This is simply because 01 is smaller than 10. Every other digit is kept the same. Try out other digit swaps to see that the result is smaller than the value in bold.

------------------------

To make the smallest 7-digit number, we just take the idea mentioned in the first section and we reverse it. We start with the smallest digit. We can't have 0 as the first value or else it won't be a 7-digit number. So we go for 1 instead. Then we can pick 0 next. For the third digit we pick the third smallest (the digit 4), and we build up to get

1046779

If we swap any distinct two digits, then we'll get something larger than this value. Let's say we swap the second 7 and the 9 at the end. We go from 79 at the end to 97. Since 79 < 97, this means 1046779 < 1046797. That's one example of many.

--------------------------

So in short, the largest number possible has all the largest digits to the left (and we decrease going left to right). The smallest number possible is the reverse of this having the smallest digits to the left (then increasing from left to right). The 0 cannot be at the left-most digit position, or else won't have a seven digit number.

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In April, Phillipa rode 13 . 5 + 8 . 7 + 11 . 1 miles on her bicycle. In March, she rode ( 13 . 5 + 8 . 7 + 11 . 1 ) ÷ 2 miles,
bonufazy [111]

Answer:

Phillipa rode the most in May and rode the least in March.

Distance travelled by Phillipa on her cycle in May = 8 × Distance travelled by Phillipa on her cycle in March  

Step-by-step explanation:

Given:

In April, Phillipa rode 13 . 5 + 8 . 7 + 11 . 1 miles on her bicycle. In March, she rode ( 13 . 5 + 8 . 7 + 11 . 1 ) ÷ 2 miles, and in May, she rode 4 × ( 13 . 5 + 8 . 7 + 11 . 1 ) miles.

To find: the month when she rode the most and the month when she rode the least

Solution:

Distance travelled by Phillipa on her cycle in April = 13 . 5 + 8 . 7 + 11 . 1 = 65 + 56 + 11 = 132 miles

Distance travelled by Phillipa on her cycle in March = ( 13 . 5 + 8 . 7 + 11 . 1 ) ÷ 2 = \frac{132}{2}=66 miles

Distance travelled by Phillipa on her cycle in May = 4 × ( 13 . 5 + 8 . 7 + 11 . 1 ) = 4 × 132 = 528 miles

So, she rode the most in May and rode the least in March

Distance travelled by Phillipa on her cycle in May = 8 × Distance travelled by Phillipa on her cycle in March  

8 0
3 years ago
Solve for b.
Dvinal [7]
Expression: <span>a = 9b²c

b</span>² = a/9c

b = √(a/9c)

b = √a/3√c
Or b = 1/3 √(a/c)

In short, Your Answer would be: Option D

Hope this helps!
8 0
3 years ago
Read 2 more answers
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grin007 [14]
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3 years ago
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slava [35]

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