The sum of two consecutive odd integers is 20 less than four times the smaller of the two integers. Which of the following equations could be used to find the smaller integer, n?
(1)(n+1)+(n+3) = 20 - 4n
(2) n+ (n + 2) = 4n-20
(3) (n+3)+n=20-4n
(4) n+(n+ 2) = 4(n-20)

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)

you have orange juice on wednesday. (idk why this is the answer)

step-by-step explanation:

i have no clue what you just said. write a question in english?