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patriot [66]
3 years ago
5

Rewrite the following integer 169 as a base and exponent without using an exponent of 1.

Mathematics
1 answer:
Westkost [7]3 years ago
5 0

Step-by-step explanation:

169 = (13)²

169 is the square of 13.

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Solve for x. <br><br> 5x - 1 = 26<br><br> x = 27/5<br> x = 5<br> x = -5
yan [13]
5x = 26 + 1
5x = 27
x = 27/5

Answer: A) or the first option ✅
4 0
3 years ago
Aye help me please :)
mina [271]

Answer:

there's no solution.

Step-by-step explanation:

if y = 3x -4 then use this instead of y in the second equation

9x -3(3x+4) = 14 ➡9x-9x -12 = 14 9x will eliminate -9x and -12 can't be = 14 so there's no solution

4 0
3 years ago
2(3.4-6)2+80:(17-5.3)2+3.√49
yaroslaw [1]
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8 0
3 years ago
I need help with #6 <br> Can u also show work ?
Anit [1.1K]

Answer:

the answer is a when you do the math while b and c and d all keep going bit a stops and can no loneger be changed

Step-by-step explanation:

1. distribute 2 to what is in the parentheses ( you should get x over 3 +10)

2. combine both x over 3

cancel out the x and get 6

your answer wpuld be 6 = 10

5 0
3 years ago
Recall that a 6-bit string is a bit strings of length 6, and a bit string of weight 3, say, is one with exactly three 1's. How m
strojnjashka [21]

Answer:

1.. Total number of 6 bit strings is 64

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

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

6. Number of string with weight 6 is 1

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

7. Number of string with weight 7 is 0

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.

8 0
3 years ago
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