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hoa [83]
2 years ago
8

The area of the shape shown below?

Mathematics
1 answer:
krek1111 [17]2 years ago
3 0

Answer:

The answer is 12 units^2.

#Hope it helps uh.........

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5,367➗4= help please
butalik [34]

Answer: 1,341 3/4

Step-by-step explanation:

7 0
2 years ago
Read 2 more answers
142 students are going on a field trip. There will be six drivers, and two different types of vehicles. A bus can hold 51 people
otez555 [7]

Answer: 2 buses and 4 vans would be needed

Step-by-step explanation:

Let x represent the number of buses that would be needed.

Let y represent the number of vans that would be needed.

There will be six drivers, and two different types of vehicles. This means that

x + y = 6

142 students are going on a field trip. A bus can hold 51 people while a van can hold 10. This means that

51x + 10y = 142 - - - - - - - - - - - 1

Substituting x = 6 - y into equation 1, it becomes

51(6 - y) + 10y = 142

306 - 51y + 10y = 142

- 51y + 10y = 142 - 306

- 41y = - 164

y = - 164/ - 41

y = 4

x = 6 - y = 6 - 4

x = 2

8 0
3 years ago
Mental Math
Dimas [21]

Answer:

288 I think

Step-by-step explanation:

24 hours each month. Each year has 12 months so 24x12=288

4 0
3 years ago
Read 2 more answers
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
Someone help me with this please !!!!!:(
Zielflug [23.3K]

Answer:

8.1.

Step-by-step explanation:

a^2+b^2=c^2

8^2+2^2=c^2

64+4

68

and you square root it and get 8.2 so i giras is 8.1

hope it helps.

7 0
2 years ago
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