This is a question with double inequality.

First solve the left side which will be which will give you   and then solve the right side which is and solution is  which is really equal to

This question deals with absolute value inequalities and as a result you get a set of solutions.

The first solution involves solving which gives you .

Next solve for which will give you .

Since and overlap, the correct solution is [-1,4]

The repeating pattern after the decimal is 63475, so the 22nd number would be 3.

This problem deals with absolute value inequalities.

An absolute value expression can never be less than 0.  So the correct answer is "No Solution"

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: