These are the only combinations of exactly 3 tiles that add to 33.
5,7,21
5,9,19
5,11,17
5,13,15
7,9,17
7,11,15
9,11,13
All tiles with numbers above 21 do not help you. 23, 25, 27, 29, 31, 33, 35, 37, 39, 41, 43, 45. There are 12 of them.
Every three-number combination must have one of the numbers 5, 7, or 9, so if the six numbers from 13 to 21 are picked, in addition to the 12 higher numbers mentioned above, you already picked 18 tiles, and you still have no solution. To obtain the solutions 5,7,21; 5,9,19; 7,9,17; he needs two more numbers in addition to the 18 he already has, so he needs 20 tiles in total to be guaranteed three of them add to exactly 33.
Answer: 20 tiles
Answer: Choice D

=====================================================
Explanation:
The left portion is the interval (-∞, -2)
This is a shorthand way of saying 
The curved parenthesis says "do not include this endpoint as part of the solution set". Note the open hole at x = -2 in the diagram.
In contrast, the value x = 4 is included (due to the filled in circle), so we use a square bracket for this endpoint. Therefore, the right-hand portion is represented by [4, ∞) which translates to 
Negative and positive infinity will always use a parenthesis, and never a square bracket. This is because we can only approach infinity but never reach it, so we cannot include it as an endpoint.
All of this builds up to the full interval notation to be 
The only square bracket is near the 4; everything else is a curved parenthesis. This is why choice D is the final answer.
I am pretty sure the value of ‘n’ is 2.5!
It would be 2/4. 2/4 simplifies to 1/2.
The equation of the tangent line at x=1 can be written in point-slope form as
... L(x) = f'(1)(x -1) +f(1)
The derivative is ...
... f'(x) = 4x^3 +4x
so the slope of the tangent line is f'(1) = 4+4 = 8.
The value of the function at x=1 is
... f(1) = 1^4 +2·1^2 = 3
So, your linearization is ...
... L(x) = 8(x -1) +3
or
... L(x) = 8x -5