kritikamanandhar0 kritikamanandhar0

16-04-2022
Mathematics

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there are 3 numbers in ascending order such that the difference of first two is 10 and that of last two is 12. if the sum of three numbers is 47, find the least number.

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jsimpson11000 jsimpson11000

16-04-2022

Answer:

17/3

Step-by-step explanation:

x + y + z = 47

y - x = 10 x = y - 10

z - y = 10 z = 10 + y

y-10 + y + 10 + y = 47

3 y = 47

y = 47/3 x = 17/3 z = 77/3

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Аноним Аноним

16-04-2022

Answer:

5

Step-by-step explanation:

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