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loris [4]
3 years ago
14

What is the work for 100 - 6x = 160 -10

Mathematics
2 answers:
nikdorinn [45]3 years ago
4 0
If you would like to solve the equation 100 - 6 * x = 160 - 10, you can calculate this using the following steps:

100 - 6 * x = 160 - 10
100 - 160 + 10 = 6 * x
- 50 = 6 * x     /6
x = - 50 / 6
x = - 25 / 3
x = - 8 1/3

The correct result would be - 8 1/3.
Darya [45]3 years ago
3 0
First you make the X by it's self so you minus 100 by each side
100-6x=160-10
- -
100 100
So the problem would be -6x=-10 then you divide each side by 6.
-6/6=1
-10/-6=60
I think you can figure it out from there. :-)
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<em>The question is not clearly readable, but I'm assuming the expression which makes more sense, so you can have a clue</em>

Answer:

i^{97}-i=0

Step-by-step explanation:

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The imaginary unit i is defined as

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The first powers of i are

i^0=1

i^1=i

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i^4=i^2.i^2=1

And so on the cycle repeats every four numbers. To find the value of i^{96} we can find the remainder of 96/4=0. So i^{96}=i^0=1

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i(i^{96}-1)

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The required value is 0

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3 years ago
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strojnjashka [21]

Answer:

1.. Total number of 6 bit strings is 64

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3. Number of 6-bit strings with weight of 1 is 6

4. Number of 6-bit strings with weight of 3 is 20

5. Number of 6-bit strings with weight of 5 is 6

6. Number of 6-bit strings with weight of 6 is 1

7. Number of 6-bit strings with weight of 7 is 0

Step-by-step explanation:

A bit string is a string that contains 0 and 1 only

1. Total number of 6 bit strings is 2^6 = 64

2. Number of 6 bit strings with weight 0 is 1

Explanation

Weight 0 means a string with no occurrence of 1

Here, we are only interested in occurrence and not order of occurrence

We apply combination formula for this

nCr = n!/(n-r)!r!

n = 6 and r = 0 i.e. no occurrence of 1

6C0 = 6!/(6-0)!0!

6C0 = 6!/6!0!

6C0 = 1

Hence, the number of string with weight 0 (i.e. no occurrence of 1 ) is 1

3. Number of string with weight 1 is 6

Explanation

Weight 0 means a string with exactly 1 occurrence of '1'

Here, we are only interested in occurrence and not order of occurrence

We apply combination formula for this

nCr = n!/(n-r)!r!

n = 6 and r = 1

6C1 = 6!/(6-1)!1!

6C1 = 6!/5!1!

6C1 = 6

Hence, the number of string with weight 6

4. Number of string with weight 3 is 20

Explanation

n = 6 and r = 3

6C3 = 6!/(6-3)!3!

6C3 = 6!/3!3!

6C3 = 20

Hence, the number of string with weight 3 is 20

5. Number of string with weight 5 is 6

Explanation

n = 6 and r = 5

6C5 = 6!/(6-5)!5!

6C5 = 6!/1!5!

6C5 = 6

Hence, the number of string with weight 5 is 6

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Explanation

n = 6 and r = 6

6C6 = 6!/(6-6)!6!

6C6 = 6!/0!6!

6C6 = 1

Hence, the number of string with weight 6 is 1

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Weight of 7 means that a string that has 7 occurrence of 1

The total length of a 6 bit is 6

Since 6 is less than 7, there's no way a bit of weight 7 can occur.

So, the right answer for this is 0.

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Answer:

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this leaves you with

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