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pashok25 [27]
3 years ago
6

which statement is the contrapositive of p ? p: if two angles are complementary, then the sum of their measures is 90

Mathematics
1 answer:
Over [174]3 years ago
5 0

Answer:  If the sum of the measures of two angles is not 90°, then they  are not complementary angles.

<u>Step-by-step explanation:</u>

Contrapositive of p → q    is    ~q → ~p  where p is the hypothesis and q is the conclusion.

Hypothesis (p) = Two angles are complementary

                  ~p = Two angles are not complementary

Conclusion (q) = The sum of their measures is 90°

                  ~q = The sum of their measures is not 90°

~ p → ~q = If the sum of the measures of two angles is not 90°,

                then they are not complementary angles.

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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

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assuming this is a typo and you meant to say g(-7).

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