There will be only 6 cubes with only one side painted red.
Cube : A cube is a 3D shape having 6 equal square sides. it has 6 faces , 8 vertices and 12 edges.
According to the question
A 3 cm cube is divided into one centimeter cube.
Every face will have 9 cube each and from that 9 cube there will be only one 1 cube that will have one side painted red.
Since there are 6 faces,
cubes painted one side red = 6x1=6.
Therefore , There will be only 6 cubes with only one side painted red.
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It is in between
And
There for it is in between 5 and 6
Additionally, it is closer to root 25, meaning that it is closer to 5
Rounded to the nearest hundredth is
5.29
There we have an information of two functions 
Using this two functions , we need to find the composition of functions (h\circ g)(t).
The composition of two functions h and g is the new function , by performing g first and then performing h.
Composition of h and g (t) = 
First plugin the value of 
We know that , we need to find h(3t+3),
That is, to replace t by 3t+3,
Now distribute 2 into 3t+3,
Now plug in 
Thus the solution is (D). 
-- He must have at least one of each color in the case, so the first 3 of the 5 marbles in the case are blue-green-black.
Now the rest of the collection consists of
4 blue
4 green
2 black
and there's space for 2 more marbles in the case.
So the question really asks: "In how many ways can 2 marbles
be selected from 4 blue ones, 4 green ones, and 2 black ones ?"
-- Well, there are 10 marbles all together.
So the first one chosen can be any one of the 10,
and for each of those,
the second one can be any one of the remaining 9 .
Total number of ways to pick 2 out of the 10 = (10 x 9) = 90 ways.
-- BUT ... there are not nearly that many different combinations
to wind up with in the case.
The first of the two picks can be any one of the 3 colors,
and for each of those,
the second pick can also be any one of the 3 colors.
So there are actually only 9 distinguishable ways (ways that
you can tell apart) to pick the last two marbles.
