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grin007 [14]
2 years ago
11

Find the perimeter. Simplify your answer.

Mathematics
2 answers:
Kisachek [45]2 years ago
8 0

Answer:

4s + 2

Step-by-step explanation:

The way we find the perimeter is we add up all the sides

s+1 + s + s+1 + s

Now we add up light terms and we get 4s+2 as our answer

larisa [96]2 years ago
7 0

Answer:

4s+2

Step-by-step explanation:

For perimeter we just have to add up all the sides.

So the equation would look like this:

s + s + (s+1) + (s+1)

There are 4 s variables and 1+1=2

So this can be simplified into :

4s+2

Hope this helps!

(Pls mark as brainiest)

Thanks!

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dybincka [34]
-1.5 is the answer because it is 75
3 0
3 years ago
What is the absolute value of Point Clabeled on the number line?
Yuki888 [10]

Answer:

  2 1/4

Step-by-step explanation:

An absolute value cannot be negative, so the negative answer choices are eliminated. The point labeled C is 5 tick marks to the right of 1. We know from the other numbers on the line that 4 tick marks constitutes one unit, so C is 5/4 = 1 1/4 units to the right of 1. Its value is 2 1/4.

5 0
3 years ago
Is -1/8 greater than -1/10
Alika [10]

Answer:

false

Step-by-step explanation:

-1/8 = -0.125

-1/10 = -0.1<u>00</u>

as it is negative  

-0.125 < -0.1

so it is false

8 0
3 years ago
There are 14 juniors and 16 seniors in a chess club. a) From the 30 members, how many ways are there to arrange 5 members of the
Dmitrij [34]

Answer:

a) 17,100,720

b) 4,717,440

c) 10,920

d) 2821

Step-by-step explanation:

14 juniors and 16 seniors = 30 people

a) From the 30 members, how many ways are there to arrange 5 members of the club in a line?

As it is a ordered arrangement

30.29.28.27.26 = 17,100,720

b) How many ways are there to arrange 5 members of the club in a line if there must be a senior at the beginning of the line and at the end of the line?

16.28.27.26.15 = 4,717,440

c) If the club sends 2 juniors and 2 seniors to the tournament, how many possible groupings are there?

Not ordered arrangement. And means that we need to multiply the results.

C₁₄,₂ * C₁₆,₂

C₁₄,₂ = <u>14.13.12!</u> = <u>14.13 </u>= 91

           12! 2!            2    

C₁₆,₂ = <u>16.15.14!</u> = <u>16.15 </u>= 120

           14! 2!            2    

C₁₄,₂ * C₁₆,₂ = 91.120 = 10,920

d) If the club sends either 4 juniors or 4 seniors, how many possible groupings are there?

Or means that we need to sum the results.

C₁₄,₄ + C₁₆,₄

C₁₄,₄ = <u>14.13.12.11.10!</u> = <u>14.13.12.11 </u>= 1001

                  10! 4!               4.3.2.1    

C₁₆,₄ = <u>16.15.14.13.12!</u> = <u>16.15.14.13 </u>= 1820

                  12! 4!               4.3.2.1    

C₁₄,₄ + C₁₆,₄ = 1001 + 1820 = 2821

7 0
4 years ago
Someone can you please help Me out ?
viktelen [127]
There were 16 oat meal

40 divided 5/3 = 24

40-24 = 16
Hope this helps!
3 0
4 years ago
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