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Arisa [49]
3 years ago
10

In this problem, substitute (x+1) in place of y in the second equation.​

Mathematics
1 answer:
netineya [11]3 years ago
3 0

Answer:

x = 8

Step-by-step explanation:

x + 2(x+1)=8

x+2x+2=8

x+2x=6

3x=6

x=6/3

x=2

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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
What does 256 to the power of 0 equal too?
morpeh [17]

Answer:

1

Step-by-step explanation:

Anything to the power of 0 is 1

Hope I helped :)

Please consider Brainliest :)

8 0
3 years ago
What is -3 times x squared?
kvasek [131]
What is -3 times x squared?-9x 
4 0
3 years ago
I need to know what is the chance the driver still got in an accident
Yakvenalex [24]

Answer:

answer is 35

Step-by-step explanation:

5 0
3 years ago
What is the 40th term of the arithmetic sequence 13, 26, 39, 52
Diano4ka-milaya [45]

Answer:

your answer would be 520

Step-by-step explanation:

13 x 40 = 520

7 0
2 years ago
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