Three whole numbers are each rounded to the nearest 10 The sum of the three rounded numbers is 80
What is the maximum possible sum of the original three numbers?

The maximum sum can be calculated if the number is rounded down and the largest value the ones digit can be is 4. Therefore, 3·4=12; 12 more than 80

Step-by-step explanation:

44 24 24

Step-by-step explanation: those would be the largest.

x =±sqrt(1/3)

step-by-step explanation:

x+4/6x=1/x

multiply each side by 6x to get rid of the fractions

6x*(x+4/6x)=1/x*6x

distribute

6x^2 +4 = 6

subtract 4 from each side

6x^2 +4-4 = 6-4

6x^2 =2

divide by 6

6x^2/6 = 2/6

x^2 = 1/3

take the square root of each side

sqrt(x^2) = ±sqrt(1/3)

x =±sqrt(1/3)

let m = monomial

let t = trinomial

so, m times t = t.

monomial times a trinomial = another trinomial.