How many integers between 2020 and 2400 have four distinct digits arranged in increasing order?

 E=15

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.

Sorry if it is incorrect

when studying polynomials, you often hear the terms zeros, roots, factors and x x the multiplicity of a zero is important because it tells us how the graph of the let's look at the graph of a function that has the same zeros

step-by-step explanation:

the smallest possible measurement of the width is 14 feet

step-by-step explanation:

let's assume length of parallelogram is 'l'

width of parallelogram is 'w'

the perimeter of a parallelogram must be no less than 40 feet

so, we get

we are given

length of parallelogram is 6 feet

so,

l=6

we can plug it

now, we can solve for w

subtract both sides by 12

divide both sides by 2

and we get

so,

the smallest possible measurement of the width is 14 feet