Which compound equalities have x = 2 as a solution? Check all that apply. 4 < 5x – 1 < 10 4 < 5x – 3 < 10

4 < 5x – 7 < 10

4 < 2x + 1 < 10

4 < 2x + 3 < 10

4 < 2x + 6 < 10

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

my answer got deleted but it's a

step-by-step explanation: