Question 1:
To start off this question, we can tell that this is a square because it has 4 right angles and 4 congruent sides.
A square has four parallel sides and 4 congruent sides, so a square is a rhombus and parallelogram.
A square has 4 right angles, so it's also a rectangle.
A square has 4 sides, so it's also a quadrilateral.
The first choice is your answer.
Question 2:
Not all quadrilaterals are rectangles, so A is incorrect.
Not all quadrilaterals are squares, so B is incorrect.
All rectangles are types of quadrilaterals, so C is correct.
Not all quadrilaterals are parallelograms, so D is incorrect.
Thus, C is your answer.
Question 3:
The first choice will not work because a rhombus will satisfy those conditions, and a rhombus is not always a square.
The second choice will work because only a square will satisfy that condition because only squares have 4 congruent sides along with equal diagonals.
Thus, the second choice is your answer.
Have an awesome day! :)
Answer:

Explanation:
The <em>end behavior</em> of a <em>rational function</em> is the limit of the function as x approaches negative infinity and infinity.
Note that the the values of even functions are the same for ± x. That implies that their limits for ± ∞ are equal.
The limits of the quadratic function of general form
as x approaches negative infinity or infinity, when
is positive, are infinity.
That is because as the absolute value of x gets bigger y becomes bigger too.
In mathematical symbols, that is:

Hence, the graphs of any quadratic function with positive coefficient of the quadratic term will have the same end behavior as the graph of y = 3x².
Two examples are:

Answer:
1st and 3rd and 5th
Step-by-step explanation:
These are like terms, meaning that they are completely the same disregarding the coefficient.
To combine like terms, add the coefficients.
16+(-12)=4
Final answer: 4x
Answer:

Step-by-step explanation:
Expression for the rectangular area and perimeter are, respectively:


After some algebraic manipulation, area expression can be reduce to an one-variable form:


The first derivative of the previous equation is:

Let the expression be equalized to zero:


The second derivative is:

According to the Second Derivative Test, the critical value found in previous steps is a maximum. Then:

The maximum area is:

