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serg [7]
3 years ago
13

Plzzzzzzzzzzzzzzzzzzzzzzzzzzzzz help

Mathematics
1 answer:
nadezda [96]3 years ago
5 0
<h2>\large\mathfrak{{\pmb{\underline{\red{Given}}{\red{:}}}}}</h2>
  • ∠RSU and ∠USV are complementary angles
  • ∠RSU = (4x – 15)°
  • ∠USV = 20°

<h2>\large\mathfrak{{\pmb{\underline{\color{red}{To~find}}{\color{red}{:}}}}}</h2>
  • The value of x

<h2>\large\mathfrak{{\pmb{\underline{\orange{Solution}}{\orange{:}}}}}</h2>
  • ∠RSU and ∠USV are complementary angles

\sf\implies{ \angle RSU + \angle USV = 90 \degree }

\sf\implies{ (4x-15) + 20 = 90 }

\sf\implies{ 4x-15+20=90}

\sf\implies{4x+5=90}

\sf\implies{4x=90-5}

\sf\implies{4x=85}

\sf\implies{x={\dfrac{85}{4}}}

\sf\implies{ {\orange{\underline{\boxed{\sf{\pmb{x=21.25}}}}}}}

<h2>\large\mathfrak{\therefore{\pmb{\underline{\color{gold}{Required~answer}}{\color{gold}{:}}}}}</h2>
  • \sf{ {\underline{\underline{\color{gold}{\sf{\pmb{x=21.25}}}}}}}
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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
Can somebody help me please
serious [3.7K]

Answer:

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Step-by-step explanation:

y-2y=7 (sub x=7 to the equation)

-y=7

y=-7 (not sure of its correct)

if y=-7,

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Answer:

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