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Natasha2012 [34]
3 years ago
12

1/2 x 2/4 x 3/6 x 4/8 x 5/10 x 6/12 simplified

Mathematics
1 answer:
Talja [164]3 years ago
8 0

Answer: 3

Step-by-step explanation: Because, 1/2=2/4=3/6=4/8=5/10=6/12 are all the same. You can test is out by simply factoring out the biggest factor on both sides. For example, 2/4 can be simplified to 1/2, because 2/4=(2/2)/(4/2) you can divide both sides by two. Hope this helps, <3!

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SO anyways :))))
artcher [175]

Answer:

1 / 8², B

Step-by-step explanation:

8 / 8³

8³ = 8*8*8

So, the equation can also be written as:

8 / 8*8*8

divide the numerator and denominator by 8.

1 / 8*8

This is equal to:

1 / 8²

The answer is B.

7 0
2 years ago
What is the area of this quadrilateral?<br> 2 ft<br> 3 ft<br> 8 ft 8 ft<br> 3 ft<br> 2 ft
Alekssandra [29.7K]

Check the picture below.

3 0
2 years ago
Please help me, i have been on this problem for 45 minutes.
docker41 [41]

Answer:

Step-by-step explanation:

The absolute difference is 77-(-8)=85. The only option that is equivalent is

|-8|+77

8 0
3 years ago
I dont get 5 and 6 help me!!!!
Kitty [74]

Answer:

5) A

6) A

Step-by-step explanation:

5) With the sides of the rectangle being dilated, multiply the side lengths by the scale factor

7.5(2.2) and 3(2.2)

16.5 and 6.6

6) Multiply the coordinates by the scale factor as dilations work as (x, y) -> (kx, ky)

6 0
3 years ago
Read 2 more answers
Suppose that you pick a bit string from the set of all bit strings of length ten. Find the probability that a) the bit string ha
emmasim [6.3K]

Answer:

  • 45/1024
  • 1/4
  • 15/128
  • 193/512
  • 9/512

Step-by-step explanation:

There are 2^10 = 1024 bit strings of length 10.

a) There are 10C2 = 45 ways to have exactly two 1-bits in 10 bits

  p(2 1-bits) = 45/1024

__

b) Of the four (4) possibilities for beginning and ending bits (00, 01, 11, 10), exactly one (1) of those is 00.

  p(b0=0 & b9=0) = 1/4

__

c) There are 10C7 = 120 ways to have seven 1-bits in the bit string.

  p(7 1-bits) = 120/1024 = 15/128

__

d) ∑10Ck {for k=0 to 4} = 386 is the total of the number of ways to have 0, 1, 2, 3, or 4 1-bits in the string. If there are more than that, there won't be more 0-bits than 1-bits

  p(more 0 bits) = 386/1024 = 193/512

__

e) The string will have two 1-bits if it starts with 1 and there is a single 1-bit among the other 9 bits. There are 9 ways that can happen, among the 512 ways to have 9 remaining bits.

  p(2 1-bits | first is a 1-bit) = 9/512

6 0
3 years ago
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