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Blababa [14]
3 years ago
14

Pls help question on the photo 50 points

Mathematics
2 answers:
Katarina [22]3 years ago
4 0

Answer:

10

Step-by-step explanation:

Each box is worth 2 units, because the x and y axis are counting by 2s

ki77a [65]3 years ago
3 0
The answer would probably be 10
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How many unique triangles can be drawn with given side lengths of 8 inches,10.3 inches,and 13 inches?​
goldfiish [28.3K]

Answer:

3 unique triangles

Step-by-step explanation:

If you use the Triangle Inequality Theorem, it states that the sum of 2 sides of the triangle would equal more than the third side. So three triangles can be made with those side lengths.

3 0
3 years ago
Which of the following is NOT a monomial?
Sergio [31]
<span>3m + n is not a monomial since it is a binomial</span>
6 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

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3 years ago
Four markers cost $7.04
Mariulka [41]

Answer:

1.76 for each marker

Step-by-step explanation:

7 0
3 years ago
Read 2 more answers
(Will give Brainliest, these are angle relationships) Please help, and show work, what relationship is it and set up an equation
kozerog [31]
Supplementary angles
x+20+5x+10=180
6x+30=180
6x=150
x=25
7 0
3 years ago
Read 2 more answers
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