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Ber [7]
3 years ago
9

If A > B and C > D, then A + C > B + D.

Mathematics
1 answer:
steposvetlana [31]3 years ago
4 0
The answer is Always true. because A is greater than B so put it this way if A=2 and B =1 then 2>1 same goes for C>D so if A+C would have to be greater than B+D for instance 2+2>1+1 or 4>2 so
<span>If A>B and C>D, then A+C>B+D Always  true.
</span>
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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
10% for infomercials in a 24 got day, how many hours of infomercials are there​
Free_Kalibri [48]

Answer:

40 yes it is i got 100 on this

Step-by-step explanation:

8 0
3 years ago
Which expression represents rational numbers? Select all that apply
Vladimir79 [104]

Answer:

1. \sqrt{100} \sqrt{100}

2. 13.5 + \sqrt{81}

3. \sqrt{9} + \sqrt{729}

6. \frac{3}{5} + 2.5\\\frac{3+12.5}{5}\\\frac{15.5}{5}\\= 3.1

These four options are rational;


Step-by-step explanation:

1. \sqrt{100} \sqrt{100} equals to 10 * 10 = 100 which is rational

2. 13.5 + \sqrt{81} equals 13.5 + 9 = 22.5

3. \sqrt{9} + \sqrt{729} -- 3+27 = 30

6. \frac{3}{5} + 2.5\\\frac{3+12.5}{5}\\\frac{15.5}{5}\\= 3.1

Option 4 and 5 are irrational because they include  \sqrt{353}  and \sqrt{216} which are not a perfect square and their answers will be non recurring and non terminating decimal fraction.

5 0
3 years ago
Convert 3,200,000 to scientific notation
Crank

Answer:

Step-by-step explanation:

3.2x10^6

4 0
2 years ago
Read 2 more answers
What is the value of m in the equation 4m + 2(m + 1) = 9m + 5
ratelena [41]
M=-1 

4m+2(m+1)=9m +5
4m+2m+2=9m+5
6m+2=9m+5
6m+2-6m= 9m+5-6m
2-5=3m+5-5
-3/3=3m/3
m=-1
3 0
2 years ago
Read 2 more answers
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