Advanced Placement (AP), 27.04.2021 17:30 catherineguyen3216

SHOW ALL YOUR WORK. REMEMBER THAT PROGRAM SEGMENTS ARE TO BE WRITTEN IN JAVA. Assume that the classes listed in the Java Quick Reference have been imported where appropriate.
Unless otherwise noted in the question, assume that parameters in method calls are not null and that methods are called only when their preconditions are satisfied.
In writing solutions for each question, you may use any of the accessible methods that are listed in classes defined in that question. Writing significant amounts of code that can be replaced by a call to one of these methods will not receive full credit.

A set of classes using inheritance is used to represent animals observed at a wildlife sanctuary. A portion of the class hierarchy is shown in the following diagram.

The figure presents a diagram of a class hierarchy with three levels. The highest level has one class labeled Animal. The middle level has two classes labeled Carnivore and Herbivore. The lowest level has one class labeled Elephant. Three arrows are drawn: one from Elephant to Herbivore, one from Herbivore to Animal, and one from Carnivore to Animal.
All Animal objects have the following attributes.

A String variable indicating whether the animal is a carnivore or a herbivore
A String variable representing the animal species (e. g., lion, giraffe, zebra)
A String variable representing the name of the individual animal (e. g., Lisa, Gary, Percy)
The Animal class also contains a toString method that indicates the state of an animal.

The following table shows the intended behavior of the Animal class.

Statement Result
Animal lisa = new Animal("carnivore", "lion", "Lisa"); A new Animal object is created.
lisa. toString(); The string "Lisa the lion is a carnivore" is returned.
(a) Write the complete Animal class. Your implementation must meet all specifications and conform to the behavior shown in the table.

Question 2
The Herbivore class is a subclass of Animal. The Herbivore class does not contain any attributes or methods other than those found in the Animal class.

The constructor to the Herbivore class accepts two parameters for the species and name of the herbivore. The constructor passes those parameters along with the string literal "herbivore" to the Animal class to construct the object.

The following table shows the intended behavior of the Herbivore class.

Statement Result
Herbivore gary = new Herbivore("giraffe", "Gary"); A new Herbivore object is created.
gary. toString(); The string "Gary the giraffe is a
herbivore" is returned.
(b) Write the complete Herbivore class. Your implementation must meet all specifications and conform to the behavior shown in the table.

Question 3
The Elephant class is a subclass of Herbivore. The Elephant class contains one additional attribute not found in Herbivore: a double variable representing the length of the elephant’s tusks, in meters.

The constructor to the Elephant class accepts two parameters for the name and tusk length of the elephant. The constructor passes those parameters along with the string literal "elephant" to the Herbivore class to construct the object.

The following table shows the intended behavior of the Elephant class.

Statement Result
Elephant percy = new Elephant("Percy", 2.0); A new Elephant object is created.
percy. toString(); The string "Percy the elephant is a
herbivore with tusks 2.0 meters long" is returned.
(c) Write the complete Elephant class. Your implementation must meet all specifications and conform to the behavior shown in the table.

CAN ANYONE PROVIDE THE COLLEGEBOARD FILE FOR THE SOLUTION TO THIS FRQ? THIS IS AP COMPUTER SCIENCE A UNIT 9 FRQ.

Answers: 2

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Answer from: Quest

tthe answer to your question is c. extrusive. hope this

Answer from: Quest

Answer: the interval on which the curve is concave up is (-∞, -1/2)====== $\displaystyle y = \int_{0}^x \frac{1}{1 + t + t^2} \; dt$ we find the interval where this curve is concave up by finding the interval in which the 2nd derivative is positive.differentiate the curve equation with respect to x $\begin{aligned} \frac{dy}{dx} & = \frac{d}{dx} \int_{0}^x \frac{1}{1 + t + t^2} \; dt \end{aligned}$ let the function f denote be the anti-derivative of the inside expression, i.e. a function such that $(d/dx)f(x) = 1/(1 + t + t^2)$ .then use the fundamental theorem of calculus: $\begin{aligned} \frac{dy}{dx} & = \frac{d}{dx} \int_{0}^x \frac{1}{1 + t + t^2} \; dt \\& = \frac{d}{dx} \big[ f(x) - f(0) \big] & & \text{\footnotesize apply ftc} \\ & = \frac{d}{dx}f(x) - 0 \\ & = \frac{1}{x + x + x^2} & & \text{\footnotesize since $\frac{d}{dx}f(x) = \frac{1}{1 + x + x^2}$ } \end{aligned}$ differentiate again to get second derivative. we can use chain rule with 1/x as the outside function and 1 + x + x² as the inside function $\begin{aligned} \frac{d}{dx} \left( \frac{dy}{dx} \right) & = \frac{d}{dx}\left(\frac{1}{1 + x + x^2} \right) \\ \frac{d^2y}{dx^2} & = \frac{-1}{\left(1 + x + x^2\right)^2} \cdot \tfrac{d}{dx}(1 + x + x^2) \\ & = \frac{-1}{\left(1 + x + x^2\right)^2} \cdot (1 + 2x) \\ & = \frac{-(1 + 2x)}{\left(1 + x + x^2\right)^2} \end{aligned}$ find critical numbers of the second derivative. these numbers are where the 2nd deriv. is undefined or zero.where the 2nd derivative is zero: set (d²y/dx²) to equal 0 and solve for x $\begin{aligned} 0 & = \frac{-(1 + 2x)}{\left(1 + x + x^2\right)^2} \\ 0 & = -(1+2x) \\ -1 & = 2x \\ x & = -1/2 \end{aligned}$ where the 2nd derivative is undefined: if the denominator is equal to zero, the derivative will be undefeined. attempting to solve the equation 1 + x + x² = 0results in a discriminant of $b^2 - 4ac = 1^2 - 4(1)(1) = -3$ meaning that the equal 1 + x + x² = 0 has no real solutions and therefore will never equal zero.======our only critical number is x = -1/2.this forms two intervals to test the sign for: (-∞, -1/2) and (-1/2, ∞)test values into the 2nd derivative for (-∞, -1/2)if we try x = -1, we get $\begin{aligned} \left.\frac{d^2y}{dx^2}\right|_{x = -1} & = \frac{-(1+2(-1))}{\left(1+(-1)+(-1)^2\right)^2} \\ & = \frac{-(1-2)}{1} \\ & = 1 \end{aligned}$ so the since the 2nd derivative of the curve is positive on the interval (-∞, -1/2), the curve is concave up on (-∞, -1/2)test values into the 2nd derivative for (-1/2, ∞)if we try x = 0, we get $\begin{aligned} \left.\frac{d^2y}{dx^2}\right|_{x = 0} & = \frac{-(1+2(0))}{\left(1+(0)+(0)^2\right)^2} \\ & = \frac{-(1)}{1} \\ & = -1 \end{aligned}$ so the since the 2nd derivative of the curve is positive on the interval (-∞, -1/2), the curve is concave down on (-∞, -1/2).the interval on which the curve is concave up is (-∞, -1/2)

Answer from: Quest

answer: no

explanation:

i thought it would be hard. yeah it took a little while to get used to but in all you will be fine. just do your work and stay out of drama.

Answer from: Quest

You should ease the seat belt back slightly and slowly

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