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mylen [45]
3 years ago
13

What is .02 repeating decimal as a fraction?

Mathematics
2 answers:
Alexxx [7]3 years ago
6 0

Answer:

2/99

Step-by-step explanation:

x=.02020202...

100x=2.020202..

-x       -02020202..

99x=2

99x/99  

<em>2/99</em>

spin [16.1K]3 years ago
4 0
2/9=0.2222222
2/90=0.02222222


THE FRACTION is 2/90 or 1/45
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3 years ago
Let an be the sum of the first n positive odd integers.
sesenic [268]

Answer:

A) The first 4 terms of the sequences are: a_{1} =16, a_{2} =24, a_{3} =32 and a_{4} =40.

B) An explicit formula for this sequence can be written as: a_{n} =8*(n+1)

C) A recursive formula for this sequence can be written as:

\left \{ {{a_{1} =16} \atop {a_{n} =a_{n-1}+8}} \right.

Step-by-step explanation:

A) You can find the firs terms of this sequence simply selecting an odd integer and summing the consecutive 3 ones:

a_{n} = Odd_{n}+Odd_{n+1}+Odd_{n+2}+Odd_{n+3} (a.1)

a_{1}=1+3+5+7=16

a_{2}=3+5+7+9=24

a_{3}=5+7+9+11=32

a_{4}=7+9+11+13=40

B) Observe the sequence of odd numbers 1, 3, 5, 7, 9, 11, 13(...).

You can express this sequence as:

Odd_{n}=(2*n-1) (b.1)

If you merge the expression b.1 in a.1, you obtain the explicit formula of the sequence:

a_{n} = Odd_{n}+Odd_{n+1}+Odd_{n+2}+Odd_{n+3} (a.1)

a_{n} = (2*n-1)+((2*(n+1)-1))+((2*(n+2)-1))+((2*(n+3)-1)) (b.2)

a_{n} = 8*n+8 (b.3)

a_{n} =8*(n+1) (b.s)

C) The recursive formula has to be written considering an initial term and an N term linked with the previous term. You can see an addition of 8 between a term and the next one. So you can express each term as an addition of 8 with the previous one. Therefore, if the first term is 16:

\left \{ {{a_{1} =16} \atop {a_{n} =a_{n-1}+8}} \right. (c.s)

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3 years ago
What is the image point of (2, 4) after a translation left 4 units and down 1 unit?
alexandr402 [8]

Answer:

-2,3

Step-by-step explanation:

because you are going to the left you would end up getting -2 for the x and since you are going one down you would have 3 for the y.

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