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timurjin [86]
2 years ago
7

Find sin of i need help can u help me

Mathematics
1 answer:
astraxan [27]2 years ago
6 0

Answer:

The answer is

Step-by-step explanation:

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In how many ways can we seat 3 pairs of siblings in a row of 7 chairs, so that nobody sits next to their sibling
monitta

Answer:

1,968

Step-by-step explanation:

Let x₁ and x₂, y₁ and y₂, and z₁ and z₂ represent the 3 pairs of siblings, and let;

Set X represent the set where the siblings x₁ and x₂ sit together

Set Y represent the set where the siblings y₁ and y₂ sit together

Set Z represent the set where the siblings z₁ and z₂ sit together

We have;

Where the three siblings don't sit together given as X^c∩Y^c∩Z^c

By set theory, we have;

\left | X^c \cap Y^c \cap Z^c  \right | = \left | X^c \cup Y^c \cup Z^c  \right | =  \left | U  \right | - \left | X \cup Y \cup Z  \right |

\left | U  \right | - \left | X \cup Y \cup Z  \right | = \left | U  \right | - \left (\left | X \right | +  \left | Y\right | +  \left | Z\right | -  \left | X \cap Y\right | -  \left | X \cap Z\right | -  \left | Y\cap Z\right | +  \left | X \cap Y \cap Z\right | \right)

Therefore;

\left | X^c \cap Y^c \cap Z^c  \right | = \left | U  \right | - \left (\left | X \right | +  \left | Y\right | +  \left | Z\right | -  \left | X \cap Y\right | -  \left | X \cap Z\right | -  \left | Y\cap Z\right | +  \left | X \cap Y \cap Z\right | \right)

Where;

\left | U\right | = The number of ways the 3 pairs of siblings can sit on the 7 chairs = 7!

\left | X\right | = The number of ways x₁ and x₂ can sit together on the 7 chairs = 2 × 6!

\left | Y\right | = The number of ways y₁ and y₂ can sit together on the 7 chairs = 2 × 6!

\left | Z\right | = The number of ways z₁ and z₂ can sit together on the 7 chairs = 2 × 6!

\left | X \cap Y\right | = The number of ways x₁ and x₂ and y₁ and y₂ can sit together on the 7 chairs = 2 × 2 × 5!

\left | X \cap Z\right | = The number of ways x₁ and x₂ and z₁ and z₂ can sit together on the 7 chairs = 2 × 2 × 5!

\left | Y \cap Z\right | = The number of ways y₁ and y₂ and z₁ and z₂ can sit together on the 7 chairs = 2 × 2 × 5!

\left | X \cap Y \cap Z\right | = The number of ways x₁ and x₂,  y₁ and y₂ and z₁ and z₂ can sit together on the 7 chairs = 2 × 2 × 2 × 4!

Therefore, we get;

\left | X^c \cap Y^c \cap Z^c  \right | = 7! - (2×6! + 2×6! + 2×6! - 2 × 2 × 5! - 2 × 2 × 5! - 2 × 2 × 5! + 2 × 2 × 2 × 4!)

\left | X^c \cap Y^c \cap Z^c  \right | = 5,040 - 3072 = 1,968

The number of ways where the three siblings don't sit together given as \left | X^c \cap Y^c \cap Z^c  \right |  = 1,968

5 0
3 years ago
Use square ABCD for this problem<br><br> IF AC=26,find BC.
katen-ka-za [31]

Answer:

x=\sqrt{338}

Step-by-step explanation:

a^2+b^2=c^2

x^2+x^2=26^2

2x^2=26^2

2x^2=676

/2           /2

x^2=338

\sqrt{x^2} =\sqrt{338}

x=\sqrt{338}

8 0
3 years ago
Read 2 more answers
PLEASE HELP!! WILL MARK BRAINLIEST!! PLZ!
Svet_ta [14]

Some points are (7,3) and (-7,-5).

7 0
3 years ago
Read 2 more answers
Please help me. I don't know what the answer is thank you!
nlexa [21]
Hello,

mes angle AEC=(69+105)/2=87
mes angle AED=(74+112)/2=93

Answer A
4 0
3 years ago
HEYYYYYYYYYYY I NEED HELPPPPPP WILL GIVE BRAINLIST
tino4ka555 [31]

Answer:

1/3

Step-by-step explanation:

There are six triangles

2/6 of them are green

We can simplify this to 1/3

4 0
3 years ago
Read 2 more answers
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