Move constant to other side
add 1
4c^2-8c=1
divide by 4 to make leading coeficient 1
c^2-2c=1/4
take 1/2 of linear coeficient and square it
-2/2=-1, (-1)^2=1
add that to both sides
c^2-2c+1=1/4+1
factor perfect squaer and add
(c-1)^2=5/4
square root both sides
c-1=+/-(√5)/2
add 1
c=1+/-(√5)/2
c=2.12 or -0.12
A) 1
B) 10/3
C) 6
D) 18/5
So only C is bigger than 4
$15.6375, but if rounded up, $15.64, and $15.63 if rounded down
If two dices are rolled once, you would have a total number of 6*6=36 possibilities.
To get a sum greater than 10:
A) one of your dices could be showing a 5, the other a 6. That’s two possibilities. Dice A being the 5 or dice B being the 5.
B) both of your dices could be showing 6s.
That’s one possibility.
So your overall possibility to get a sum greater than 10 is (1+2)/36 3/36=1/12
One twelfth.
The answer to this question is c=3