Begin by distributing each part:

2x(x² + 4ax – 3a²) = 2x * x² + 2x * 4ax – 2x * 3a² = 2x³ + 8ax² – 6a²x

The second:

–4a²(4x + 3a) = –16a²x – 12a³

Now, combine these:

2x³ + 8ax² – 6a²x – 16a²x – 12a³

The only common terms are those with a²x; therefore, this reduces to

2x³ + 8ax² – 22a²x – 12a³

This is the same as the given answer:

–12a³ – 22a²x + 8ax² + 2x³

In order to translate words into symbols and algebraic expressions, we must take each piece of the statement one at a time. First, the problem mentions the square of the sum of p and q. This means we must first find the sum of p and q, and then find the square of this quantity. The sum of p and q can be represented by p + q. The square of this expression would be written as (p + q)². 

Where the sentence says "is equal to," we must put an equal sign. This means we can put an equal sign after (p + q)². So our sentence is now as follows:

(p + q)² = 

Now, let's look at the last part of the sentence to see what should go on the right of the equal sign. 

We must write an expression to represent the product of n and the positive difference of j and k. A product requires multiplication. So we are going to multiply n by the positive difference of j and k.

The difference of j and k could be represented as either j – k or as k – j. However, the statement calls for the positive difference. We don't know whether j or k is larger, so we don't know if j – k is positive, or if k – j is positive. The only way to ensure that we get a positive result is to take the absolute value of the difference. In other words, we can write the positive difference as either |j – k| or as |k – j|, because the absolute value will ensure that we yield a positive value. Also, because, in general, |a – b| = |b – a|, it doesn't matter whether or not we represent the positive difference of j and k as |j – k| or as |k – j|.

Now, let us finish writing the product of n and the positive difference of j and k. Because a product requires multiplication, we must multiply n by the positive difference of j and k, which we just determined can be either |j – k| or |k – j|. Thus, symbolically, we can write this as n • |j – k| or n • |k – j|.

Finally, when we put all of the pieces together, the equation becomes:

(p + q)² = n • |k – j|

The answer is (p + q)² = n • |k – j|.

Since y = –100t² (in terms of t), we would like to replace the expression t with something in terms of x. Solving the first equation for t, we see:

x = 2t + 3

x – 3 = 2t

(x – 3)/2 = t

Therefore, y = –100((x – 3)/2)² = –100(x – 3)²/4 = –25(x – 3)²

"one half the value of three more than x" is the same as ¹/₂(x + 3)

"is" is the same as "="

"twice the value of three less than y" is the same as 2(y – 3)

So ¹/₂(x + 3) = 2(y – 3).

Multiply both sides by 2: x + 3 = 4(y – 3)

Distribute: x + 3 = 4y – 12

Add 12 to both sides: x + 15 = 4y

Divide both sides by 4: (x + 15)/4 = y

The problem states that x² – (y – z)² = 40, and x – y + z = 10. Because we are asked to find x, we need to find an equation for y and z in terms of x. This will be much easier to do in the second equation than in the first. Let's try to get the y and z terms by themselves, and then put everything else on the other side.

x – y + z = 10

Subtract x from both sides.

–y + z = 10 – x

Notice, that if we were to multiply –y + z by negative one, we would get y – z, which appears in our first equation. Let's multiply both sides by –1. 

–1(–y + z) = –1(10 – x)

Distribute on both sides.

y – z = –10 + x = x – 10

In short, we see that y – z = x – 10. Thus, we can substitute x – 10 into the first equation in place of y – z.

x² – (y – z)² = 40

x² – (x – 10)² = 40

Next, we will simplify.

x² – (x – 10)(x – 10) = 40

Use the FOIL technique to find (x – 10)(x – 10).

According to the FOIL method, we find the product of the first terms of the binomials, then the outer terms, then the inner binomials, then the last terms, and then we add these four products together.

Applying the FOIL technique, (x – 10)(x – 10) = x(x) + x(–10) + –10(x) + (–10)(–10) = x² – 10x – 10x + 100. Let's now replace (x – 10)(x – 10) with x² – 10x – 10x + 100.

x² – (x² – 10x – 10x + 100) = 40

Distribute the –1.

x² – x² + 10x + 10x – 100 = 40

Combine like terms.

20x – 100 = 40

Add 100 to both sides.

20x = 140

Divide both sides by 20.

x = 7.

The answer is 7.

Let's look at the function g(x) = (4x – 1)(4x + 1). We want to find g(–x) and see if that equals g(x).

g(–x) = (4(–x) – 1)(4(–x) + 1)

= (–4x – 1)(–4x + 1)

We can factor out a –1 from both binomials.

(–4x – 1)(–4x + 1) = (–1)(4x + 1)(–1)(4x – 1)

Notice that we could arrange this as (–1)(–1)(4x + 1)(4x – 1). And because (–1)(–1) = 1, we would be left with (4x + 1)(4x – 1), which is our original g(x).

Therefore, g(x) = (4x – 1)(4x + 1) is indeed an even function.

Next, we will examine g(x) = 4x⁴ – x⁶

g(–x) = 4(–x)⁴ – (–x)⁶

Because of the property of exponents which states that (ab)^c = a^cb^c, we can rewrite the above expressions as:

4(–1)⁴x⁴ – (–1)⁶(x⁶)

Whenever -1 is raised to an even-valued exponent, the result is positive one.  For example, (–1)⁴ = (–1)(–1)(–1)(–1) = 1. Let's go back to simplifying f(–x).

g(–x) = 4(–1)⁴x⁴ – (–1)⁶(x⁶) = 4(1)x⁴ – (1)x⁶ = 4x⁴ – x⁶ = g(x).

Thus, 4x⁴ – x⁶ is an even function.

Next, let's examine g(x) = 4 + 3x² + x^–2.

g(–x) = 4 + 3(–x)² + (–x)^–2

= 4 + 3(–1)²x² + (–1)^–2x^–2 = 4 + 3x² + x^–2 = g(x). This function is even as well.

Now, let's look at g(x) = x²(x⁴ – 1)^–1.

g(–x) = (–x)²((–x)⁴ – 1)^–1

= x²(x⁴ – 1) = g(x)

Finally, we will analyze g(x) = 2x(x – 1)².

g(–x) = 2(–x)(–x – 1)² = –2x(–1)²(x + 1)² = –2x(x + 1)², which doesn't equal 2x(x – 1)² .

In this case, g(–x) is not the same as g(x), so this function cannot be even.

The answer is g(x) = 2x(x – 1)².

Traveling 20 mph faster than Gary, it will take the officers 30 minutes to catch up to Gary. The answer is 3:30 PM.

In order to solve this problem, we need to look at each function separately and then derive expressions for f(a + b), f(a), and f(b). Then, we need to see whether f(a + b) = f(a) + f(b).

Let's start with the function 2^x.

f(a + b) = 2^(a+b)

Using our property of exponents that x^yx^z = x^y+z, we can rewrite 2^a+b.

f(a + b) = 2^(a+b) = 2^a2^b

Now, let's derive an expression for f(a) + f(b).

f(a) = 2^a, and f(b) = 2^b. Thus, f(a) + f(b) = 2^a + 2^b

Let's compare f(a + b) and f(a) + f(b).

Does 2^a2^b = 2^a + 2^b? It might, but not for every value of a and b. For example, let a and b both equal 0.

2⁰2⁰ = 1(1) = 1

2⁰ + 2⁰ = 1 + 1 = 2

If a and b are both zero, then it isn't true that f(a + b) = f(a) + f(b). Thus, f(x) = 2^x isn't special.

Next, let's examine the function f(x) = |x|.

f(a + b) = |a + b|

f(a) + f(b) = |a| + |b|

|a + b| doesn't always equal |a| + |b|. For example, if a = –1 and b = 1, then |a + b| = |–1 + 1| = 0, while f(a) + f(b) = |–1| + |1| = 1 + 1 = 2. Because f(a + b) is not always going to equal f(a) + f(b), the function f(x) = |x| isn't special.

The next function we can analyze is f(x) = x².

f(a + b) = (a + b)² = a² +2ab + b²

f(a) + f(b) = a² + b²

a² + 2ab + b² doesn't always equal a² + b² . Thus, f(x) = x² isn't special.

The next function is f(x) = (x + 1)² – x². We can simplify f(x) a little bit to make it easier to work with.

f(x) = (x + 1)² – x² = (x² + 2x + 1) – x² = 2x + 1.

f(x) = 2x + 1.

f(a + b) = 2(a + b) + 1 = 2a + 2b + 1

f(a) + f(b) = (2a + 1) + (2b + 1) = 2a + 2b + 2

2a + 2b + 1 doesn't equal 2a + 2b + 2, so this isn't a special function either.

This means that f(x) = (x + 1)² – (x – 1)² must be special. Let's see why.

First, let's simplify f(x).

f(x) = (x + 1)² – (x – 1)² = (x² + 2x + 1) – (x² – 2x + 1) = x² + 2x + 1 – x² + 2x – 1 = 4x.

f(x) = 4x.

f(a + b) = 4(a + b) = 4a + 4b

f(a) + f(b) = 4a + 4b

4a + 4b is always equal to 4a + 4b. This means that the function is special.

The answer is f(x) = (x + 1)² – (x – 1)² .

First, let's rewrite x & y in a form that is much easier to analyze.

x & y = |(y – x)(x – y)|. First, let's factor out –1 from the y – x term.

|(y – x)(x – y)| = |–(–y + x)(x – y)|

Notice that we can rewrite (–y + x) as (x – y)

|–(–y + x)(x – y)| = |–(x – y)(x – y)| = |–(x – y)²|.

Now, we can make use of the property of absolute values which states that |–a| = |a|.

|–(x – y)²| = |(x – y)²|

Lastly, notice that because (x – y)² is never negative, we can remove the absolute value sign. This is because, in general, |a²| = a².

|(x – y)²| = (x – y)²

To summarize, x & y = (x – y)².

Next, let's go through each of the answer choices and determine if the left side equals the right side.

First, let's examine x & y = y & x.

We have already established that x & y = (x – y)². This means that y & x = (y – x)². We want to see if (x – y)² = (y – x)². Let's factor out a –1 from y – x.

(y – x)² = ((–1)(–y + x))² = ((–1)(x – y))².

Remember that, according to an important property of exponents, (ab)^c = a^cb^c.

((–1)(x – y))² = (–1)²(x – y)² = (1)(x – y)² = (x – y)².

Thus, because y & x = (x – y)², we can see that y & x = x & y. This answer choice can be eliminated.

Next, we will analyze x & y = (–x) & (–y). Again, our strategy will be to show whether or not (–x) & (–y) = (x – y)²

(–x) & (–y) = (–x – (–y))² = (–x + y)² = ((–1)(x – y))² = (–1)²(x – y)² = (x – y)²

Therefore, x & y = (–x) & (–y). We can eliminate this answer choice.

Next, we will look at x & y = (–y) & (–x).

(–y) & (–x) = (–y – (–x))² = (–y + x)² = (x – y)²

This means that x & y = (–y) & (–x), and we can eliminate this answer choice.

Let's now see if a²(x & y) = (ax) & (ay). We want to see if the left side equals the right. Remember that x & y = (x – y)².

a²(x&y) = a²(x – y)²

Now, we want to see if (ax) & (ay) = a²(x – y)².

(ax) & (ay) = (ax – ay)² = (a(x – y))² = a²(x – y)².

Therefore, a²(x & y) = a²(x – y)² .

By process of elimination, the answer must be –(x & y) = (–x) & y. Let's see why.

–(x & y) = –(x – y)²

We need to see if (–x) & y = –(x – y)².

(–x) & y = (–x – y)² = (–1(x + y))² = (–1)²(x + y)² = (x + y)², which doesn't have to equal –(x – y)².

For example, let's let x = 1 and y = 2 and see if –(1 & 2) = (–1) & 2

–(1 & 2) = –1(1 – 2)² = –1(–1)² = –1

(–1) & 2 = (–1 – 2)² = (–3)² = 9, which doesn't equal –1.

The answer is –(x & y) = (–x) & y.

We are told that 2a = 4b and 12b = 8c. We are asked to find the value of a. All of the answer choices are in terms of c, so we will need to find a in terms of c. This means that we need to eliminate b somehow and create an equation with only a and c.

We know that 12b = 8c. If we can figure out how many a = 12b, then we can create an equation that contains only a and c by replacing 12b with some value of a.

The first equation tells us that 2a = 4b. If we multiply both sides by 3, then we will know how many a are equal to 12b.

3(2a) = 3(4b)

6a = 12b

This means that we can replace 12b with 6a in the equation 12b = 8c.

12b = 6a = 8c

6a = 8c

Now we have an equation with only a and c. However, we are ultimately asked to find 3a. If we were to multiply both sides of the equation by one-half, we would have 3a on the left side.

(1/2)6a = (1/2)8c

3a = 4c

The answer is 4c.