Boris chooses three different numbers the sum of the three numbers is 36 one of the numbers is a cube number the other two numbers are factors of 20 find three numbers chosen by boris. write the numbers in ascending order

3,4,5

Step-by-step explanation:

Here, we need to asume that the values are integers, as if not, there will be infinite solutions.

Lets go by parts:

The 1st number is a cubic, lets call it A

The second is a factor of 20, lets call it B. The same for the 3rd, call it C.

So, A + B+ C = 36

There are no many factors of 20, those are the numbers that when multiplied gives as the value of 20. Those are 1, 2, 4, 5, 10 and 20. So, B and C are some of these numbers.

Then, we know A if less than 36, other way the whole sum will be greater than 36. How many cubic number with integer cubic roots are less than 36? Well, lets guess:

1^3 = 1 -> valid, as it is less than 36

2^3 = 8 -> valid, as it is less than 36

3^2 = 27 -> valid, less than 36

4^3 = 64 -> not valid, as it is greater than 36

So, A ir 1, 2 or 3.

If A is 1, then B+C needs to be 35. But, from the factors of 20 that we listed, there are no combinations of 2 numbers that sum 35. So, A CAN'T be 1.

If A = 2, then we have:

8 + B + C = 36

B + C = 28

And again, there are not combinations of two factors of 20 that sum 28 (try yourself).

If A=3:

27 + B + C = 36

B +C = 9

And we have a winner!!! If B=4 and C=5 (or viceversa C=4 and B=5)

27 + 4 + 5 = 36

So, A=3, B=4 and C=5 or A=3, C=4 and B=5 are solutions.

hope it

step-by-step explanation:

x^(1/6) · x^(1/6) · x^(1/6)   ·x^(1/6) = x^(4/6)   since we add the exponents

4/6 can be simplified to 2/3

the simplified form is x^(2/3)

y > 1/3x

y < = -x + 4

hope this : )