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vfiekz [6]
3 years ago
10

HELPP!!

Mathematics
1 answer:
SSSSS [86.1K]3 years ago
7 0

Answer:

There are 15 integers between 2020 and 2400 which have four distinct digits arranged in increasing order.

Step-by-step explanation:

This can be obtained by after a simple counting of number from 2020 and 2400 as follows:

The first set of integers are:

2345, 2346, 2347, 2348, and 2349.

Therefore, there are 6 integers in first set.

The second set of integers are:

2356, 2357, 2358, and 2359.

Therefore, there are 4 integers in second set.

The third set of integers are:

2367, 2368, and 2369.

Therefore, there are 3 integers in third set.

The fourth set of integers are:

2378, and 2379.

Therefore, there are 2 integers in fourth set.

The fifth and the last set of integer is:

2389

Therefore, there is only 1 integers in fifth set.

Adding all the integers from each of the set above, we have:

Total number of integers = 6 + 4 + 3 + 2 + 1 = 15

Therefore, there are 15 integers between 2020 and 2400 which have four distinct digits arranged in increasing order.

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3 years ago
25 pts to who ever can help with this question:
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Answer:

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B. 8s+8

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Step-by-step explanation:

Is difficult to read the exact expressions. If a mistake is present then fix the number and use the same method.

A. The perimeter of the square is all 4 equal sides added together. The length of a side is \frac{5s+2}{4}. Since there are four sides then:

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B. The perimeter of the rectangle is all four sides added together. A rectangle has two pairs of equal side so P = 2l+2w.

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8(7)+8-5(7)-2

56+8-35-2

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3584

Step-by-step explanation:

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