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sukhopar [10]
3 years ago
7

HELP HELP HELP HELP HELP HELP

Mathematics
1 answer:
dezoksy [38]3 years ago
5 0
10 root 2
10^2 + 10^2 = 200
200 is 10 root 2
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2x - 4y = -8<br> Solve for y <br><br><br> Need help ASAP!
goldenfox [79]

Answer:

y = 1/2x + 2

Step-by-step explanation:

  • 2x - 4y = -8

<u>Solving for y in steps:</u>

  • 2x - 4y + 8 + 4y= 4y
  • 2x + 8 = 4y
  • x + 4 = 2y
  • y = x/2 + 4/2
  • y = 1/2x + 2
7 0
4 years ago
Tony has 8 yards of fabric. how many inches does he have?
hoa [83]
Tony has 288 inches of fabric because there are 36 inches in a yard.
6 0
4 years ago
Read 2 more answers
Where do i start? Show how to apply the order of operation rules as you simplify the following expression. |-5| – 45 ÷ 3
scoundrel [369]
The absolute value of -5 is 5 (absolute value being how many units to 0)
 So we really have  5 -  45 ÷ 3 =
Order of operation says to do the division first (there are no parentheses or exponents).   So now we have 5 - 15=   and the answer to that is -10.
7 0
4 years ago
Solve the system by substitution. Check your solution.
Zepler [3.9K]

Answer:

a.  (15, 15)

Step-by-step explanation:

We start with those two equations:

1) a - 1.2b = -3

2) 0.2b + 0.6a = 12

We'll begin by modifying equation #1 to isolate a:

a = -3 + 1.2b

Then we'll use this value for a in the second equation:

0.2b + 0.6 (-3 + 1.2b) = 12

0.2b - 1.8 + 0.72b = 12

0.92b = 13.8

b = 15

Then we'll place that value of b in the first equation to find a:

a - 1.2 (15) = -3

a - 18 = -3

a = 15

3 0
3 years ago
Suppose that two teams are playing a series of games, each of which is independently won by team A with probability p and by tea
Lady_Fox [76]

Answer:

4 games played = 1/56

5 games played = 5/56

6 games played = 15/56

7 games played = 35/56

Step-by-step explanation:

The probability of each time winning is 0.5

So there are a couple of ways the series could go.

  1. Team A could win all first 4 matches. We can depict this as A-A-A-A
  2. Team A could win 4, while having lost 1.  Lets depict this a A-B-A-A-A
  3. Team A could win 4, while having lost 2. A-B-B-A-A-A
  4. Team A could win 4, while having lost 3. A-B-B-B-A-A-A

These 4 possibilities could be repeated with B winning as well. It should be noted that these are the only ways for the series to end. We find the number of permutations of each possibility above to find their probability.

I advise you to study 'how to permute identical objects' for this.

These permutations are stated below:

1. 4!/4! = 1

2. \frac{5!}{4! * 1!} = 5

3 \frac{6!}{4! * 2!} = 15

4 \frac{7!}{4! * 3!} = 35

These are they ways A or B could win. The total is 1+5+15+35 = 56. The answer given thus reflects the possibilities.

4 0
4 years ago
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