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Karolina [17]
3 years ago
12

Factor each expression below. a. 16x − 4 = b. −10x 5x2 = c. 30y − 24x =

Mathematics
1 answer:
Artyom0805 [142]3 years ago
4 0

A) Answer:

4 (4x - 1)

B) Answer:

-5x ( 2 + x )

C) Answer:

6 (5y - 4x)

<u>hope</u><u> </u><u>it</u><u> </u><u>helped</u><u>.</u>

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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
There are 29 pencils in a basket. 7 of these pencils are green. The rest of them are yellow.
Sever21 [200]
A) 22:7
B) 29:22



29 - 7 = 22, so there are 22 yellow pencils.

There are 22 yellow pencils and 7 green pencils, so the ratio of yellow pencils to green pencils is 22:7.

There are 29 pencils, and 22 of them are yellow, so the ratio of all pencils to yellow pencils is 29:22.



Please consider marking this answer as Brainliest to help me advance.
5 0
3 years ago
M/3-4&lt;-1 <br> answer this question<br> please show your work
Len [333]

Answer:

m<9

Step-by-step explanation:

m/3 - 4 < -1

        +4  +4

______________

m/3<3

 × 3 ×3

__________

m<9

Hope this Helps!

3 0
3 years ago
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A rate simplified so that it has a denominator of 1?
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\sf\frac{28}{7} = \frac{7\times 4}{7\times 1}=\frac{4}{1}\\\\\frac{28}{7} ~is ~a ~rate~ that~ when~ simplified ~has ~1~ in~ the ~denominator.
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3 years ago
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In a college library there are 4 times as many nonfiction books as fiction books. How many times the number of nonfiction books
balu736 [363]
Four? Isn't it stated in the question? Unless I am assuming wrong. 
8 0
3 years ago
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