What is the solutions set of y=x^2+2x+7and y=x+7

when x=0 ,y=7

when x=-1, y=6

Step-by-step explanation:

Taking as equation 1 and 2 respectively

Inserting equation 2 in 1,

getting rid of 7 since its in both the sides with the same + signs,

when x=0 ,

y=7

when x=-1 , y=6

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

answer: 12   they all have at least 28 days

step-by-step explanation: haha