Question

3 2/3+2.3 repeating

Answer 1

Answer: 6

Explanation:

`3(2)/(3)+2.3^-`

Let's begin by converting the repeating decimal to a fraction and then to a mixed number.

Take the number:

`2.3^-`

Let x (our result) be equal to that number.

`x=2.3^-`

Multiply by 1 followed by as many zeros as repeating decimals; in this case 1. Therefore, 10.

`10x=2.3^-*10`

This basically moves the decimal point one number to the right and since we have infinite 3's, we simply move 1 three and add another.

`10x=23.3^-`

Subtract this and the original equation (
`x=2.3^-`)

`10x=23.3^-\\-x=2.3^-`

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10 - 1 = 9 and we have the same infinite repeating number 3, therefore they cancel out, leaving 23-2 = 21

`9x=21`

Divide by 9

`x=(21)/(9)`

Simplify by 3.

21/3=7

9/3=3

`x=(7)/(3)`

Now, convert to a mixed number. To do this, divide.

7/3=2

6

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1

The 2 is the whole number, 1 is the numerator and 3 the denominator.

`2(1)/(3)`

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Now we can solve this;

`3(2)/(3)+2(1)/(3)`

Add the whole numbers and the fractions separately.

`(3+2)+((2)/(3)+(1)/(3))`

`5(3)/(3)`

3 and 3 can be simplied.

3/3=1

3/3=1

`5(1)/(1)`

The result is 1, add this to the 5.

`5+1=6`

Answer 2

6.

Explanation:

2.3 repeating = 2 1/3

So 3 2/3 + 2 1/3

= 5 + 1

= 6.