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3 years ago

13

when writing a repeating decimal as a fraction, why does the fraction always have only 9s or 9s and 0s as digits in the denomina

tor

1 answer:

Vaselesa [24]3 years ago

4 0

thats not necessarily true. 0.8333... is just 5/6 as a decimal. I suppose it depends on the number.

You might be interested in

On a single roll of a pair of dice, what are the odds against rolling a sum of 12?

o-na [289]

Answer:

$\frac{1}{35}$

Step-by-step explanation:

On a single roll of a pair of dice. When a pair of dice are rolled the possible outcomes are as follows:

(1,1) (1,2) (1,3) (1,4) (1,5) (1,6)

(2,1) (2,2) (2,3) (2,4) (2,5) (2,6)

(3,1) (3,2) (3,3) (3,4) (3,5) (3,6)

(4,1) (4,2) (4,3) (4,4) (4,5) (4,6)

(5,1) (5,2) (5,3) (5,4) (5,5) (5,6)

(6,1) (6,2) (6,3) (6,4) (6,5) (6,6)

The number of outcomes that gives us 12 are (6,6). There is only one outcome that gives us sum 12.

Total outcomes = 36

Odd against favor = $\frac{non \ favorable\ outcomes}{favorable \ outcomes}$

Number of outcomes of getting sum 12 is 1

Number of outcomes of not getting sum 12 is 36-1= 35

odds against rolling a sum of 12= $\frac{1}{35}$

3 0

3 years ago

Luden [163]

Is this just free points or is there a question

7 0

3 years ago

The bakery makes 6 pies. Each slice is V8 of the whole pie. How many slices are there

ANEK [815]

Answer:

6

Step-by-step explanation:

4 0

3 years ago

Resolve to me these exercises please?

Dima020 [189]

A)4^(n+3)=8^14
2^(2×(n+3))=2^(3×14)
2^(2n+6)=2^42
2^2n=2^36
n=18

b) (assuming a : is divide)
3^(2n+1)=9^17/3^3
3^(2n+1)=3^(2×16)/3^3
3^(2n+1)=3^29
3^2n=3^28
2n=28
n=14

d) (6^n)^4×36=216^10
6^4n×6^2=6^(3×10)
6^(4n+2)=6^30
6^4n=6^28
4n=28
n=7

e)7^(n^2)÷7=49^24
7^(n^2-1)=7^(2×24)
7^(n^2)=7^49
n^2=49
n=7

g)15^(n+4)÷5^(n+4)=81^6
3^(n+4)×5^(n+4)÷5^(n+4)=3^(4×6)
3^(n+4)=3^24
n=20

h)81^n÷9^n+9^(n+2)÷9=90÷9^6
9^2n÷9^n+9^(n+2)÷9=9*10/9^6
9^n+9^(n+1)=10/9^5
I don't know where to go from here

I)what?

7 0

3 years ago

Working alone Ryan can dig a 10ft by 10ft hole in five hours castel can dig the same same hole in six how long would it take the

Sergeeva-Olga [200]

Answer:

The time taken by both when they work together is 2 hours 42 minutes .

Step-by-step explanation:

Given as :

The time taken by Ryan to dig 10 ft by 10 ft hole = x = 5 hours

The time taken bu Castel to dig 10 ft by 10 ft hole = y = 6 hours

Let The time taken by both when they work together = z hours

<u>According to question</u>

The time taken by both when they work together = time taken by Ryan to dig 10 ft by 10 ft hole + time taken bu Castel to dig 10 ft by 10 ft hole

i.e $\dfrac{1}{z}$ = $\dfrac{1}{x}$ + $\dfrac{1}{y}$

Or, $\dfrac{1}{z}$ = $\dfrac{1}{5}$ + $\dfrac{1}{6}$

<u>Taking LCM</u>

Or, $\dfrac{1}{z}$ = $\dfrac{6 + 5}{30}$

Or, $\dfrac{1}{z}$ = $\dfrac{11}{30}$

Or, z = $\dfrac{30}{11}$ hours

∴ z = 2 hours 42 minutes

So, The time taken by both when they work together = z = 2 hours 42 min

Hence, The time taken by both when they work together is 2 hours 42 minutes . Answer

3 0

3 years ago