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3 years ago

Will give brainliest!

Mathematics

2 answers:

Alex3 years ago

7 0

Answer:

1. 1 1/3

2. 6/7

3. 1 1/4

Step-by-step explanation:

For the first one, the question is to reduce 16/12. So we need to see how many 12's there are in 16 --> 1. Because 16 - ( 12 · 1 ) = 4. We can't fit another 16 in four! Therefore the answer is 1 4/12 because the remainder that we solved above is 4. It can be simplified to 1 1/3.

5/7 + 1/7 is 6/7 because the denominator of the fractions are the same all you have to do is add the numerators together and leave the denominator 7:

5 1 6

----- + ------ = -------

7 7 7

To reduce 25/20 all you have to do is see how many 25's are in 20. Which is 1. 25 - ( 20 · 1 ) = 5. But there is still 5 left over so it would be: 1 5/20. Which can be simplified as 1 1/4

HOPE THIS HELPS :)

aleksandr82 [10.1K]3 years ago

5 0

Answer:

1.) 1 1/3

2.) 6/7

Step-by-step explanation:

1.) 16/12 = 4/3 = 1 1/3

2.) Add the numerators if the denominators are the same already

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3 years ago

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4 years ago

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RideAnS [48]

Answer:

$\text{C. }60$

Step-by-step explanation:

Question from image:

"The digits 1, 2, 3, 4, and 5 can be formed to arrange 120 different numbers. How many of these numbers will have the digits 1 and 2 in increasing order? For example, 14352 and 51234 are two such numbers."

Let's start by taking a look with the number 1. There are four possible places 1 could be, because there needs to be space for the 2 after it.

Checkmarks mark where the 1 can be, the x marks where it cannot be.

$\underline{\checkmark}\:\underline{\checkmark}\:\underline{\checkmark}\:\underline{\checkmark}\:\underline{X}$

Let's start with the first position:

$\underline{\checkmark}\:\underline{\#}\:\underline{\#}\:\underline{\#}\:\underline{\#}$

There are four places the 2 can be. For each of these four places, we can arrange the remaining 3 digits in $3!=6$ ways. Therefore, there are $4\cdot 6=\boxed{24}$ possible numbers when 1 is the first digit of the number.

Continue this process with the remaining possible positions for 1.

Second position:

$\underline{X}\underline{\checkmark}\:\underline{\#}\:\underline{\#}\:\underline{\#}$

There are three places the 2 can be, since the 2 must be behind the 1. For each of these three places, the remaining 3 digits can be arranged in $3!=6$ ways. Therefore, there are $3\cdot 6=\boxed{18}$ possible numbers when 1 is the second digit of the number.

The pattern continues. Next there will be 2 places to place the 2. For each of these, there are $3!=6$ ways to rearrange the remaining 3 digits for a total of $2\cdot 6=\boxed{12}$ possible numbers when 1 is the third digit of the number.

Lastly, when 1 is the fourth digit of the number, there is only 1 place the 2 can be. For this one place, there are still $3!=6$ ways to rearrange the remaining three numbers. Therefore, there are $1\cdot 6=\boxed{6}$ possible numbers when 1 is the fourth digit of the number.

Thus, there are $24+18+12+6=\boxed{60}$ numbers that will have the digits 1 and 2 in increasing order, from a set of 120 five-digit numbers created by the digits 1 through 5, where no digit may be repeated.

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