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Diano4ka-milaya [45]
3 years ago
11

Can someone please give me the perimeter and area answer SHOW WORK

Mathematics
2 answers:
Usimov [2.4K]3 years ago
7 0

Perimeter

4x-7 plus 4x-7 plus 10x^2 + 4x - 2 plus 10x^2 + 4x -2.

8x - 14 plus 20x^2 + 8x - 4

<u>20x^2 + 16x - 18</u>

(x + 36) (x+10)

x = {36, 7}

gotta go, so I cannot do the rest. I hope that helps a bit.

stich3 [128]3 years ago
6 0
Just add up the lengths of the four sides, and you have the perimeter.

The perimeter is:
(4x-7)*2 + (10x^2+4x-2)*2 =
8x-14 + 20x^2+8x-4 =
20x^2 + 16x - 18
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Point R lies on the directed line segment from L (-8,-10) to M (4,-2) and partitions the segment in the ratio 3 to 5. What are t
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Answer:

R = (- 3.5, - 7 )

Step-by-step explanation:

Using the Section formula

x_{R} = \frac{3(4)+5(-8)}{3+5} = \frac{12-40}{8} = \frac{-28}{8} = - 3.5

y_{R} = \frac{3(-2)+5(-10)}{3+5} = \frac{-6-50}{8} = \frac{-56}{8} = - 7

Thus coordinates of R = (- 3.5, - 7 )

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Answer:

-8/45

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-2/3 ÷ 3 3/4

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Sam walks at a rate of 300 ft per minute. How many inches per second does he walk? (1 foot = 12 inches)
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300/min
300/60 seconds
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When factoring, what should be the greatest common factor of the terms left inside the parenthesis?
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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
3 years ago
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