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guajiro [1.7K]
3 years ago
9

use each of the digits 5 4 3 2 1 exactly once to create two different five digit numbers. Write each number on the line and comp

are the two numbers by using the symbols < > =​
Mathematics
1 answer:
sp2606 [1]3 years ago
8 0

Answer:

12345 < 54321

21435 > 12534

:

Step-by-step explanation:

Given the digits:

1, 2, 3, 4 and 5

We have to use every digit only once and have to make two different five digit numbers.

Using these 5 numbers only once without repetition, we have many numbers possible.

Let us have a look at a few sets and let us compare them.

<u>Set 1:</u> 12345 and 54321

We can see that 12345 is lesser than 54321.

Therefore, we can write (using lesser than sign):

12345 < 54321

<u>Set 2:</u> 21435 and 12534

We can see that 21435 is greater than  12534.

Therefore, we can write (using greater than sign):

21435 > 12534

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Answer:

I believe your answer is going to be roughly 767.8 which if you need to use fractions would be 767 4/5.

Step-by-step explanation:

Hope this helped in some way

7 0
3 years ago
I think I know this much so far for A (but I could be wrong):
Eduardwww [97]
Part A. You have the correct first and second derivative.

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

Part B. You'll need to be more specific. What I would do is show how the quantity (-2x+1)^4 is always nonnegative. This is because x^4 = (x^2)^2 is always nonnegative. So (-2x+1)^4 >= 0. The coefficient -10a is either positive or negative depending on the value of 'a'. If a > 0, then -10a is negative. Making h ' (x) negative. So in this case, h(x) is monotonically decreasing always. On the flip side, if a < 0, then h ' (x) is monotonically increasing as h ' (x) is positive.

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

Part C. What this is saying is basically "if we change 'a' and/or 'b', then the extrema will NOT change". So is that the case? Let's find out

To find the relative extrema, aka local extrema, we plug in h ' (x) = 0
h  ' (x) = -10a(-2x+1)^4
0 = -10a(-2x+1)^4
so either
-10a = 0 or (-2x+1)^4 = 0
The first part is all we care about. Solving for 'a' gets us a = 0. 
But there's a problem. It's clearly stated that 'a' is nonzero. So in any other case, the value of 'a' doesn't lead to altering the path in terms of finding the extrema. We'll focus on solving (-2x+1)^4 = 0 for x. Also, the parameter b is nowhere to be found in h ' (x) so that's out as well. 
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3 years ago
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Butoxors [25]

Answer:

Yes

Step-by-step explanation:

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the frame forms perpendicular lines meaning they are 90 degrees

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777dan777 [17]

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Alla [95]
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