There are different definitions of "whole numbers".
Some define it as an integer (i.e. positive or negative) [some dictionaries]
Some define it as a non-negative integer. [most math definitions]
We will take the math definition, i.e. 0<= whole number < ∞
To find pairs (i.e. two) whole numbers with a sum of 110, we start with
0+110=110
1+109=110
2+108=110
...
54+56=110
55+55=110
Since the next one, 56+54=110 is the same pair (54,56) as 54+56=110, we stop at 55+55=110 for a total of 56 pairs.
You do the implcit differentation, then solve for y' and check where this is defined.
In your case: Differentiate implicitly: 2xy + x²y' - y² - x*2yy' = 0
Solve for y': y'(x²-2xy) +2xy - y² = 0
y' = (2xy-y²) / (x²-2xy)
Check where defined: y' is not defined if the denominator becomes zero, i.e.
x² - 2xy = 0 x(x - 2y) = 0
This has formal solutions x=0 and y=x/2. Now we check whether these values are possible for the initially given definition of y:
0^2*y - 0*y^2 =? 4 0 =? 4
This is impossible, hence the function is not defined for 0, and we can disregard this.
x^2*(x/2) - x(x/2)^2 =? 4 x^3/2 - x^3/4 = 4 x^3/4 = 4 x^3=16 x^3 = 16 x = cubicroot(16)
This is a possible value for y, so we have a point where y is defined, but not y'.
The solution to all of it is hence D - { cubicroot(16) }, where D is the domain of y (which nobody has asked for in this example :-).
(Actually, the check whether 0 is in D is superfluous: If you write as solution D - { 0, cubicroot(16) }, this is also correct - only it so happens that 0 is not in D, so the set difference cannot take it out of there ...).
If someone asks for that D, you have to solve the definition for y and find that domain - I don't know of any [general] way to find the domain without solving for the explicit function).
So the rounded weights would be 3, 5, 2, 6, and 11, as when you round, you look at the number to the right of the one you want to round to. If it's 5 or larger, you add one to the spot you're rounding to, and make the rest of the numbers after into zeroes. If it's below 5, just change the ones after the one you want to round to into zeroes and don't add one.
Then for the estimating of the weight, you would add them together. 3+5+2+6+11. That would equal 27, so the estimate weight is 27 grams.
For the first letter drawn, there are no conditions/restrictions we need to consider. So, the number of ways to draw an i is:
For the second draw, the i is not replaced. That means, we've shrunk the sample space because we're dealing with less elements for the second draw. Firstly, one i disappears since we picked an "i" and there are 10 elements left, not 11.
Thus, to draw the second i, we have:
A postcard is in the shape of a parallelogram. A parallelogram is a quadrilateral with two pair of parallel sides, opposite sides and opposite angles are equal.
Since, the postcard has an area of 12 square inches.
Since, area of parallelogram = 
As area of parallelogram is 12, it means that the product of base and height is 12 square inches.
So, the possible dimensions of postcard are 3 inches and 4 inches and 2 inches and 6 inches.
So, base = 3 inches , height = 4 inches or base = 4 inches , height = 3 inches.
So, base = 2 inches, height = 6 inches or base = 6 inches , height = 2 inches.
