Answer:
1/12
Step-by-step explanation:
There is only number between 1 and 12 that is both a multiple of 5 and 2, that being 10. 10 has an equal chance out of all the other numbers that can be spun, and there are 12 numbers, so 10 has a 1/12 chance of being spun.
The range is the ouutput from inputing the input
basically
25=k²+2k+1 and 64=k²+2k+1
the values that satisfy both equations (not at the same tim) are the valuess that are the domain
solve each
25=k²+2k+1
minus 25 both sides (or recognize the perfect square trinomial, but anyway)
0=k²+2k-24
factor
0=(k+6)(k-4)
set to zero
k+6=0
k=-6
k-4=0
k=4
k=-6 or 4
64=k²+2k+1
minus 64 both sides
0=k²+2k-63
facor
0=(k-7)(k+9)
set to zer
k-7=0
k=7
k+9=0
k=-9
k=-9 or 7
so the domain has the numbers
-9,-6,4,7
it seems we only want the positive square roots so
answer is {4,7} is the domain
MrBillDoesMath!
Answer to #4: 81/256 * s^8 * t^ 12
Comments:
(7x^3) ^ (1/2) = 7 ^ (1/2) * x^(3/2) where ^(1/2) means the square root of a quantity. The answer written (7x^3) is NOT correct.
---------------------
(1) (27s^7t^11)^ (4/3)
= 27^(4/3) * (s^7)^(4/3) * (t^11)^ (4/3)
As 27 = 3^3, 27 ^(4/3) = 3^4 = 81
(2) (-64st^2)^ (4/3) = (-64)^(4/3) * (s^4/3) * t(^8/3)
As 64 = (-4)^3, (-64)^(4/3) = (-4)^4 = +256
So (1)/(2) =
81 * s^(28/3)* t^(44/3)
------------------------------- =
256 s^(4/3) * t^((8/3)
81/256 * s ^ (28/3 - 4/3) * t^(44/3 - 8/3) =
81/256 * s^(24/3) * t (36/3) =
81/256 * s^8 * t^ 12
MrB
R is 4 if I'm wrong sorry
