In order to find n, we need to figure out possible values for a, b, and c.

We are told that a, b, and c are integers. We are also told that 0 < a < b < c, which means that a, b, and c are all positive. It also means that a, b, and c are different numbers (they cannot be equal). It also means that a is the smallest, and c is the largest.

We are now told that a + b + c = 7.

Let's start with the smallest possible value that a could be. The smallest integer greater than 0 is 1.

Let us assume that a = 1.

If a = 1, then b > 1. The smallest integer that b can be is 2. So let's assume b = 2.

If a =1 and b = 2, then 1 + 2 + c = 7. This means that c = 4.

If a =1, b = 2, and c = 4, then all of the conditions for a, b, and c are met. Thus, we can use these values to find n.

What if we had assumed that a was 2? If a is 2, then the smallest value that b could be is 3. If a + b + c = 7, then c would have to be 2. But we are told that c is the largest number, so it can't be equal to 2.

Thus, the only value of a that works is 1.

Let's assume that if a = 1, then b = 3. If a + b + c  = 7, then c = 3. However, we are told that c > b. Thus, b cannot be 3.

Thus, b must equal 2, and c must equal 4, and a must be 1. This is the only possibility that satisfies all of the criteria given in the question.

Now we can use the values of a, b, and c to find n.

n =2^a3^b5^c

n = 2¹3²5⁴ = 2(9)(625) = 11250.

The answer is 11250.

Because we are dealing with an absolute value on the left side, we are going to have to consider two possible cases. Remember that, in general, |a| = a if a > 0, and |a|  = –a if a < 0. In other words, we need to consider two cases for the left side: x² – 2x – 8 and –(x² – 2x – 8).

Case 1: |x² – 2x – 8| = x² – 2x – 8 = 2x + 4

x² – 2x – 8 = 2x + 4

Subtract 2x from both sides.

x² – 4x – 8 = 4

Subtract 4 from both sides.

x² – 4x – 12 = 0

We must think of two numbers that multiply to give us –12 and add to give –4. These two numbers are –6 and 2. This means we can factor the left side as follows:

x² – 4x – 12 = (x – 6)(x + 2) = 0

Set each of the factors equal to 0.

x – 6 = 0

Add 6 to both sides.

x = 6

Next, x + 2 = 0

Subtract 2 from both sides.

x = –2

The values of x that solve the equation are –2 and 6.

However, before moving on to the next case, it's always important to check our work. Let's put x = –2 and x = 6 into our original equation and make sure that both sides are the same.

|x² – 2x – 8| = 2x + 4

Let x = –2:

|x² – 2x – 8| = |(–2)² – 2(–2) – 8| = |4 + 4 – 8| = 0

2x + 4 = 2(–2) + 4 = 0

x = –2 is a solution.

Let x = 6:

|6² – 2(6) – 8| = |36 – 12 – 8| = 16

2x + 4 = 2(6) + 4 = 16

x = 6 is also a solution.

Case 2:  |x² – 2x – 8| = –(x² – 2x – 8) = 2x + 4.

 –(x² – 2x – 8) = 2x + 4

Distribute the –1.

–x² + 2x + 8 = 2x + 4

Subtract 2x from both sides.

–x² + 8 = 4

Subtract 8 from both sides.

–x² = –4

Divide both sides by negative 1.

x² = 4

Take the square root.

x = 2 or –2

We have already established that x = –2 is a solution. Let's check to see if 2 is also a solution by going back to the original equation |x² – 2x – 8| = 2x + 4.

|x² – 2x – 8| = |2² – 2(2) – 8| = |–8| = 8

2x + 4 = 2(2) + 4 = 8

This means x = 2 is also a solution.

To summarize, the solutions to x are –2, 2, and 6.

We are asked to find their product, which is –2(2)(6) = –24.

The answer is therefore –24.

Think of the multiples of 10: 10, 20, 30, 40, 50, 60, 70, . . . 

I. Multiples of 2: 2, 4, 6, 8, 10, 12, 14, . . . 

Some of these already are not contained in S. 

II. Multiples of 5: 5, 10, 15, 20, 25, . . . 

Some of these already are not contained in S. 

III. Multiples of 20: 20, 40, 60, 80, 100, . . . 

All of these are also multiples of 10. Thus, our answer must be III only. 

When doing probability problems, we are looking for the number of successes over number of possible outcomes. There are 4 chances to successfully choose a green marble. The number of possible outcomes are 11, one for each of the 11 marbles in the bag. When we write the fraction, we get our answer. 

In mathematical words we get the following:

Looking at the second statement, isolate x on one side with all other constants and variables on the other side.

Looking at the third statement, isolate z on one side with all other constants and variables on the other side. 

Looking at the first statement, isolate y on one side with all other constants and variables on the other side. 

From here, use these equivalencies so solve for y.

Substituting twice:

Solve a quadratic inequality by first putting it in the standard form  (or similar); this is done by subtracting  from both sides:

Factor the quadratic trinomial on the left:

We are looking for two integers whose product is 32 and whose sum is . Through a little trial and error, we find  and , so

,

making the inequality 

The boundary points of the solution set are the points at which either factor is equal to 0:

, in which case , and

, in which case .

These points divide the real numbers into three intervals: 

We can choose one test point from each interval to determine the truth of each statement on the interval:

: choose  

True: include 

: choose 

False: exclude .

: choose 

True: include 

The boundaries are also included, since the inequality allows for inclusion ("greater than or equal to"). The solution set is .

The boundary points of a rational inequality are those values of the variable for which the numerator or the denominator are equal to 0:

If numerator

, then

.

Since equality is permitted in the statement (  ), include this value of .

If denominator 

,

then 

Since this value makes the denominator 0, exclude it, regardless of the inequality symbol.

There are three intervals which must be tested for inclusion or exclusion:

, , 

Test one value on each interval in the original inequality for truth or falsity in order to determine which intervals should be included:

: Set 

False; exclude .

: Set 

True; include 

: Set 

False; exclude 

Since  is excluded as a solution and 4 is included, the correct solution set is the interval .

Solve a quadratic inequality by first putting it in the standard form  (or similar); this is done by first applying the FOIL method to the product of the binomials on the right:

Subtract 21 from both sides:

The  method can be used to factor the trinomial; we are looking for two integers whose product is  and whose sum is ; by trial and error, we find that the numbers are 7 and . The trinomial can be rewritten:

Factoring by grouping:

The boundary points of the solution set are the points at which either factor is equal to 0:

and 

These points divide the real numbers into three intervals: 

Test one value on each interval in the original inequality for truth or falsity in order to determine which intervals should be included:

: Set 

True: include .

: Set 

False: exclude 

: Set 

True: include .

The boundaries are also included, since the inequality allows for inclusion ("greater than or equal to"). The solution set is .

Get all expressions in the rational inequality over to the same side by adding 4 to both sides:

Rewrite the expression on the left as a single rational expression, as follows:

The boundary points of a rational inequality are those values of the variable for which the numerator or the denominator are equal to 0:

If numerator

then

Since equality is permitted in the statement (  ), include this value of .

If denominator 

then

.

Since this value makes the denominator 0, exclude it, regardless of the inequality symbol.

There are three intervals which must be tested for inclusion or exclusion:

, , 

Test one value on each interval in the original inequality for truth or falsity in order to determine which intervals should be included:

: Set 

True; include 

: Set 

False: exclude 

: Set 

True: include .

Since  is excluded as a solution and  is included, the correct solution set is