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Sedaia [141]
4 years ago
8

Consider any positive four-digit integer that has all of its digits distinct and none equal to zero. What is the largest possibl

e difference between such an integer and any integer that results from rearranging its digits
Mathematics
1 answer:
andrew11 [14]4 years ago
6 0

Answer:

<h2>Check the explanation.</h2>

Step-by-step explanation:

The highest four digit number is 9999 and lowest 4 digit number is 1000. The difference between them is (9999 - 1000) = 8999.

8999 cannot be difference according to the question, as here all the digits should be distinct rather than 0.

The difference between the 9 and 1 is 8, which is the highest difference among the unit digits rather than 0.

The 2nd lowest number after 9 is 8 and the 2nd highest unit digit after 1 is 2.

(8 - 2) = 6.

The difference will be highest if difference is calculated between the highest and lowest number.

With the 4 digits, 9, 8, 2, 1 the lowest number is 1289 and the highest number is 9821.

The difference is (9821 - 1289) = 8532.

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Dmitrij [34]

Answer:

V=9\pi\sqrt{3}

Step-by-step explanation:

In order to solve this problem we must start by graphing the given function and finding the differential area we will use to set our integral up. (See attached picture).

The formula we will use for this problem is the following:

V=\int\limits^b_a {\pi r^{2}} \, dy

where:

r=6-(6-3 sec(y))

r=3 sec(y)

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so the volume becomes:

V=\int\limits^\frac{\pi}{3}_0 {\pi (3 sec(y))^{2}} \, dy

This can be simplified to:

V=\int\limits^\frac{\pi}{3}_0 {9\pi sec^{2}(y)} \, dy

and the integral can be rewritten like this:

V=9\pi\int\limits^\frac{\pi}{3}_0 {sec^{2}(y)} \, dy

which is a standard integral so we solve it to:

V=9\pi[tan y]\limits^\frac{\pi}{3}_0

so we get:

V=9\pi[tan \frac{\pi}{3} - tan 0]

which yields:

V=9\pi\sqrt{3}]

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and with the straight edge, you could easily divide the angle into two equal halves

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