Find three consecutive integers such that that product of the first and third is one less than 22 times the second integer.

n = the first

n + 1 = the 2nd

n + 2 = the 3rd

n(n + 2) - (n + 1) = 10(n + 2) + 1

n2 + 2n - n - 1 = 10n + 20 + 1

n2 - 9n - 22 = 0

(n - 11)(n + 2) = 0

If two factors = 0, at least one must = 0.

n - 11 = 0

n = 11

11,12,13

n + 2 = 0

n = -2

-2,-1,0

answer: c i think

step-by-step explanation:

dc=6 square root of 3 and ac is 12 square root 3

c and d

step-by-step explanation:

square root of 3/2 = 18/c

cross multiply to get 18*2=√3*c

36/√3 * √3/√3 = c

c= 36*√3/3

c=12√3