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kirza4 [7]
2 years ago
6

7TH GRADE MATH PLEASE HELP PLEASE WILL GIVE BRAINLIEST AND FOLLOW !!!!!!!!

Mathematics
1 answer:
Nookie1986 [14]2 years ago
5 0

Answer: I would say it’s (0,-2) for the answer.

Step-by-step explanation:

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Simplify &cm-4) - :m+5)-(m - 6)-(n+7)
Mekhanik [1.2K]

Answer:

m-1-n

Step-by-step explanation:

7 0
3 years ago
6 tractors take 7 days to collect
Lyrx [107]

Answer:

= 1.75

Step-by-step explanation:

7/6 divided by 6

= 7/36

7/36 = x/9

Cross multiply fractions:

36x = 63

Divide each side by 36:

x = 7/4 or 1.75

3 0
3 years ago
Four darts are thrown at this dartboard. If all four darts hit the board, how many different point totals are possible. (Dartboa
jonny [76]

Answer:

Different point total are possible = 497

Step-by-step explanation:

Given - Four darts are thrown at this dartboard. Dartboard regions are  

             1, 4, 7, 10 points

To find - If all four darts hit the board, how many different point totals are

              possible.

Proof -

Are given there are 4 regions -

1st region - 1 point

2nd region - 4 points

3rd region - 7 points

4th region - 10 points

Case I :

If all the darts hit the different region -

Total points possible are - 1 + 4 + 7 + 10 = 22 points

Case II :

If all the darts hit the same region -

It means either they hit 1 region or 2nd region or 3rd region or 4th region

If they all 4 hit first region , points are - 1+1+1+1 = 4

If they all 4 hit second region , points are - 4+4+4+4= 16

If they all 4 hit third region , points are - 7+7+7+7 = 28

If they all 4 hit fourth region , points are - 10+10+10+10 = 40

So,

the Total points possible are - 4 + 16 + 28 + 40 = 88

Case III :

If 3 darts hit the same region -

Sub-case 1 :

If 1 dart hit 1st region, other 3 dart hit 2nd region

Points are - 1 + 4 + 4+ 4 = 13

Sub-case 2 :

If 1 dart hit 1st region , other 3 dart hit 3rd region

Points are - 1 + 7 + 7 + 7 = 22

Sub-case 3 :

If 1 dart hit 1st region , other 3 dart hit 4th region

Points are - 1 + 10 + 10 + 10 = 31

Sub-case 4 :

If 1 dart hit 2nd region , other 3 dart hit 1st region

Points are - 4 + 1 + 1 + 1 = 7

Sub-case 5 :

If 1 dart hit 2nd region , other 3 dart hit 3rd region

Points are - 4 + 7 + 7 + 7 = 25

Sub-case 6 :

If 1 dart hit 2nd region , other 3 dart hit 4th region

Points are - 4 + 10 + 10 + 10 = 34

Sub-case 7 :

If 1 dart hit 3rd region , other 3 dart hit 1st region

Points are - 7 + 1 + 1 + 1 = 10

Sub-case 8 :

If 1 dart hit 3rd region , other 3 dart hit 2nd region

Points are - 7 + 4 + 4 + 4 = 19

Sub-case 9 :

If 1 dart hit 3rd region , other 3 dart hit 4th region

Points are - 7 + 10 + 10 + 10 = 37

Sub-case 10 :

If 1 dart hit 4th region , other 3 dart hit 1st region

Points are - 10 + 1 + 1 + 1 = 13

Sub-case 11 :

If 1 dart hit 4th region , other 3 dart hit 2nd region

Points are - 10 + 4 + 4 + 4 = 22

Sub-case 12 :

If 1 dart hit 4th region , other 3 dart hit 3rd region

Points are - 10 + 7 + 7 + 7 = 31

So,

Total points possible are - 13+22+21+7+25+34+10+19+37+13+22+31 = 254

Case IV :

If 2 darts hit the same region -

Sub-case 1:

If 2 darts hits 1st region , other 2 darts hit 2nd region

Points are - 1 + 1 + 4 + 4 = 10

Sub-case 2:

If 2 darts hits 1st region , other 2 darts hit 3rd region

Points are - 1 + 1 + 7 + 7 = 16

Sub-case 3:

If 2 darts hits 1st region , other 2 darts hit 4th region

Points are - 1 + 1 + 10 + 10 = 23

Sub-case 4:

If 2 darts hits 2nd region , other 2 darts hit 3rd region

Points are - 4 + 4 + 7 + 7 = 22

Sub-case 5:

If 2 darts hits 2nd region , other 2 darts hit 4th region

Points are - 4 + 4 + 10 + 10 = 28

Sub-case 6:

If 2 darts hits 3rd region , other 2 darts hit 4th region

Points are - 7 + 7 + 10 + 10 = 34

So,

Total points possible are - 10 + 16 + 23 + 22 + 28 + 34 = 133

Case V :

If 1 darts hit the same region -

This case is include in the Case III.

∴ we get

Different point total are possible = Points in Case I +  II +  III +  IV

                                                       = 22 + 88 + 254 + 133

                                                       = 497

⇒Different point total are possible = 497

4 0
2 years ago
The question is in the photo. it is asking for 2 answers
Alex777 [14]

9514 1404 393

Answer:

  • 2nd force: 99.91 lb
  • resultant: 213.97 lb

Step-by-step explanation:

In the parallelogram shown, angle B is the supplement of angle DAB:

  ∠B = 180° -77°37' = 102°23'

Angle ACB is the difference of angles 77°37' and 27°8', so is 50°29'.

Now, we know the angles and one side of triangle ABC. We can use the law of sines to solve for the other two sides.

  BC/sin(A) = AB/sin(C)

  AD = BC = AB·sin(A)/sin(C) = (169 lb)sin(27°8')/sin(50°29') ≈ 99.91 lb

 AC = AB·sin(B)/sin(C) = (169 lb)sin(102°23')/sin(50°29') ≈ 213.97 lb

7 0
3 years ago
What do you call an agarrangment of reciving goods or serviced
olasank [31]
The answer to this is credit.<span />
7 0
3 years ago
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