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.
Answer:
(2a +b)·(13a^2 -5ab +b^2)
Step-by-step explanation:
The factorization of the difference of cubes is a standard form:
(p -q)^3 = (p -q)(p^2 +pq +q^2)
Here, you have ...
so the factorization is ...
(3a -(a -b))·((3a)^2 +(3a)(a -b) +(a -b)^2) . . . . substitute for p and q
= (2a +b)·(9a^2 +3a^2 -3ab +a^2 -2ab +b^2) . . . . simplify a bit
= (2a +b)·(13a^2 -5ab +b^2) . . . . . . collect terms
No, they aren't. 3/3+ 1 whole
Answer: 61.92751306414704
Step-by-step explanation:
F(g(-1))
g(-1) = 3(-1) = -3
f(-3) = 2(-3) -1 = -6 -1 = -7