The sum of three numbers is 62. the second number is equal to the first number diminished by 4. the third number is four times the first. what are the numbers? .

  11, 7, 44

Step-by-step explanation:

Let x represent the first number. Then the second is (x-4) and the third is (4x). Their sum is ...

  x +(x -4) +(4x) = 62

  6x = 66 . . . . . . . . . . . . add 4, collect terms

  x = 11 . . . . . . . . . divide by 6

  x -4 = 7 . . . . . . . find the second number

  4x = 44 . . . . . . . find the third number

The three numbers are 11, 7, and 44.

step-by-step explanation:

in the absence of parentheses, i will have to make some assumptions:   i interpret your (3x+5/y^6)(4y^3/5)​ to mean:

3x + 5     4y^(3/5)       (3x + 5)(4)

* =

    y^6             1               y^(6-3/5)

if this is not what you expected, repost your question using parentheses to indicate whether you meant

4y^3

    or   4y^(3/5)

    5

d

step-by-step explanation: