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.

The difference between the price to repair the car or buy a new one is $11,000. You divide this by the monthly savings, $220, giving you 50 months. 

To simplify this, we will want to use the correct order of operations. The mnemonic device PEMDAS is usually very helpful.

Parenthesis (1st)

Exponents (2nd)

Multiply, Divide (3rd)

Add, Subtract (4th)

PEMDAS tells us to evaluate parentheses first, then exponents. After exponents, we evaluate multiplication and division from left to right, and then we evaluate addition and subtraction from left to right.

Let's look at (xy)² – x((4x)(y)² – (4x)²) – 4²x²

We want to start with parentheses first. We will simplify the (xy)² by using the general rule of exponents, which states that (ab)^c = a^cb^c. Thus we can replace (xy)² with x²y².

x²y² – x((4x)(y)² – (4x)²) – 4²x²

When we have parentheses within parentheses, we want to move from the innermost parentheses to the outermost. This means we will want to simplify the expression (4x)(y)² – (4x)² first, which becomes 4xy² – 16x². We can now replace (4x)(y)² – (4x)² with 4xy² – 16x² .

x²y² – x(4xy² – 16x²) – 4²x²

In order to remove the last set of parentheses, we will need to distribute the x to  4xy² – 16x² . We will also make use of the property of exponents which states that a^ba^c = a^b+c.

x²y² – x(4xy²) – x(16x² ) – 4²x²
= x²y² – 4x²y² – 16x³ – 4²x²

We now have the parentheses out of the way. We must now move on to the exponents. Really, the only exponent we need to simplify is –4², which is equal to –16. Remember that –4² = –(4²), which is not the same as (–4)².

x²y² – 4x²y² – 16x³ – 16x²

Now, we want to use addition and subtraction. We need to add or subtract any like terms. The only like terms we have are x²y² and –4x²y². When we combine those, we get –3x²y²

–3x²y² – 16x³ – 16x²

The answer is –3x²y² – 16x³ – 16x² .

It's much easier to use factoring and canceling than it is to use long division for this problem. 9x² – 1 is a difference of squares. The difference of squares formula is a² – b² = (a + b)(a – b). So 9x² – 1 = (3x + 1)(3x – 1). Putting the numerator and denominator together, (9x² – 1) / (3x – 1) = (3x + 1)(3x – 1) / (3x – 1) = 3x + 1.

First, let's expand (a – b)² using the FOIL method. 

(a – b)(a – b) = a² – ab – ba + b² = a² – 2ab + b²

The value of a² + b² is already given to us. We can manipulate the equation ab = –8 to determine the value of –2ab. Multiply both sides of the equation by –2.

–2ab = 16

We will now substitute the values of –2ab and a² + b² to find the value of (a – b)².

(a – b)(a – b) = a² – 2ab + b² = 16 + 20 = 36

The answer is 36.

We are told that 4x – 8y = 18, but we want to know the value of 4y – 2x. In order to do this, we want to manipulate the expression 4x – 8y until it looks like 4y – 2x. We want to go from 4x to –2x. In order to do this, we could divide both sides of the equation by –2. If we did this, then we would also go from –8y to 4y. In short, we can find the value of 4y – 2x if we divide both sides of the equation 4x – 8y = 18 by negative two.

4x – 8y = 18

We can use the algebraic rule which states that , as long as c is nonzero. We will also use the fact that –a + b = b – a.

The answer is –9.

Using the multiplication property of logarithms we get. 

Knowing  we get

Assume there are x pennies.  Hence the number of nickels is . 

Now you have to set up an equation for the dollar value of the pennies and nickels which will be . 

Now solving for  results in (number of pennies).  Hence the number of nickels will be .