C. 100 is the aswer i thinck
Answer:
Scales are used to represent real distances and proportions into maps and planes, that way we can have a bigger perspective about something just having in paper.
One direct use of scales are seen in architecture plans, there, the architect uses scales to represent the construction he is designing. Also, the architect can have a more complete point of view about the principial elements on his work.
Note: In option A, it should be "+" instead of "(".
Given:
The expression is

To find:
The equivalent expression.
Solution:
We have,

Using distributive property, we get


On combining like terms, we get


The expression
is equivalent to the given expression.
Therefore, the correct option is A.
Answer:
The Answer is C .............
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).