Find three numbers such that their sum is 12, the sum of the first, twice the second, and three times the third is 31, and the sum of the third and nine times the second is 1

3,-1,10

Step-by-step explanation:

let the numbers be x,y,z.

x+y+z=12

x+2y+3z=31

z+9y=1

z=1-9y

subtract 1st from 2nd

y+2z=19

y+2(1-9y)=19

y+2-18y=19

-17 y=17

y=-1

z=1-9(-1)=1+9=10

x-1+10=12

x=12+1-10=3

42

step-by-step explanation:

256/6=42

x =-.20752

step-by-step explanation:

4^(x+2)=12

4^(x+2)=12

take the log base 4 on each side

log 4 (4^(x+2) = log 4(12)

we know log b(a^y) = y log b (a)

(x+2) log 4(4) = log 4 (12)

x+2 = log 4 (12)

we know   logb c = loga c/loga b

we want to convert to base 10   so a = 10

by default, we do not write the 10   c = 4   and b = 10

log4 10 = log 12/log 4

log4 10 = log 12/ log 4

x+2 = log 12/ log 4

subtract 2 from each side

x +2 -2 = log 12/ log 4 -2

x = 1.79248125 -2

x =-.20752