The symbols express the idea that if a number is less than a second number, which is less than a third, then the first number is less than the third. This is the transitive property of inequality.

The symbols express the idea that if a number is less than a second number, which is less than a third, then the first number is less than the third. This is the transitive property of inequality.

We need to solve the two equations |2a – 3| = 5 and |3 – 4b| = 11, in order to determine the possible values of a and b. When solving equations involving absolute values, we must remember to consider both the positive and negative cases. For example, if |x| = 4, then x can be either 4 or –4.

Let's look at |2a – 3| = 5. The two equations we need to solve are 2a – 3 = 5 and 2a – 3 = –5.

2a – 3 = 5 or 2a – 3 = –5

Add 3 to both sides.

2a = 8 or 2a = –2

Divide by 2.

a = 4 or a = –1

Therefore, the two possible values for a are 4 and –1. However, the problem states that both a and b are negative. Thus, a must equal –1.

Let's now find the values of b.

3 – 4b = 11 or 3 – 4b = –11

Subtract 3 from both sides.

–4b = 8 or –4b = –14

Divide by –4.

b = –2 or b = 7/2

Since b must also be negative, b must equal –2.

We have determined that a is –1 and b is –2. The original question asks us to find |b – a|.

|b – a| = |–2 – (–1)| = | –2 + 1 | = |–1| = 1.

The answer is 1.

We need to solve the two equations |2a – 3| = 5 and |3 – 4b| = 11, in order to determine the possible values of a and b. When solving equations involving absolute values, we must remember to consider both the positive and negative cases. For example, if |x| = 4, then x can be either 4 or –4.

Let's look at |2a – 3| = 5. The two equations we need to solve are 2a – 3 = 5 and 2a – 3 = –5.

2a – 3 = 5 or 2a – 3 = –5

Add 3 to both sides.

2a = 8 or 2a = –2

Divide by 2.

a = 4 or a = –1

Therefore, the two possible values for a are 4 and –1. However, the problem states that both a and b are negative. Thus, a must equal –1.

Let's now find the values of b.

3 – 4b = 11 or 3 – 4b = –11

Subtract 3 from both sides.

–4b = 8 or –4b = –14

Divide by –4.

b = –2 or b = 7/2

Since b must also be negative, b must equal –2.

We have determined that a is –1 and b is –2. The original question asks us to find |b – a|.

|b – a| = |–2 – (–1)| = | –2 + 1 | = |–1| = 1.

The answer is 1.

First, we must find out by how much they are spaced by. It cannot be 1, since 4(4) = 16, which is too great of a step in the positive direction and exceeds the equal-spacing limit. 2 works perfectly, however, as 4(2) equals 8 and fits in line with the equal spacing.

Next, we can find the values of x and y since we are given a value of 6 for the third tick mark. As such, x (6 – 4) and y (6 – 2) are 2 and 4, respectively.

Finally, z is 4 steps away from y, and since each step has a value of 2, 2(4) = 8, plus the value that y is already at, 8 + 4 = 12 (or can simply count).

Finding the average of all 3 values, we get (2 + 4 + 12)/3 = 18/3 = 6.

First, we must find out by how much they are spaced by. It cannot be 1, since 4(4) = 16, which is too great of a step in the positive direction and exceeds the equal-spacing limit. 2 works perfectly, however, as 4(2) equals 8 and fits in line with the equal spacing.

Next, we can find the values of x and y since we are given a value of 6 for the third tick mark. As such, x (6 – 4) and y (6 – 2) are 2 and 4, respectively.

Finally, z is 4 steps away from y, and since each step has a value of 2, 2(4) = 8, plus the value that y is already at, 8 + 4 = 12 (or can simply count).

Finding the average of all 3 values, we get (2 + 4 + 12)/3 = 18/3 = 6.

Since the inequality represents one range of values between two end points (both of which are included, given the sign being "less than or equal"), you know that whatever you answer, it must be convertible to the form:

Now, you know that it is impossible to get this out of the choices that have no absolute values involved in them. Therefore, the only options that make sense are the two having absolute values; however, here you should choose only the ones that have a , for only that will yield a range like this. Thus, we can try both of our options.

The wrong answer is simplified in this manner:

And you can stop right here, for you know you will never have for the left terminus.

The other option is simplified in this manner:

This is just what you need!

Substitute – 4 in for x. Remember that when a negative number is raised to the third power, it is negative. - = – 64. – 64 – 36 = – 100. Since you are asked to take the absolute value of – 100 the final value of f(-4) = 100. The absolute value of any number is positive.

Since the inequality represents one range of values between two end points (both of which are included, given the sign being "less than or equal"), you know that whatever you answer, it must be convertible to the form:

Now, you know that it is impossible to get this out of the choices that have no absolute values involved in them. Therefore, the only options that make sense are the two having absolute values; however, here you should choose only the ones that have a , for only that will yield a range like this. Thus, we can try both of our options.

The wrong answer is simplified in this manner:

And you can stop right here, for you know you will never have for the left terminus.

The other option is simplified in this manner:

This is just what you need!

First, it will help us to determine which of the answer choices are positive and which are negative.

Because n is positive, we know that n2 is positive, because any number squared is positive. Similarly, 1/n2 is also positive.

Let's look at the answer choice –1/n. This must be negative, because a negative number divided by a positive one will give us a negative number. Similarly, –1/n3 will also be negative.

The last choice is n3 – 1. We are told that 0 < n < 1. Because n is a positive value less than one, we know that 0 < n3 < n2 < n < 1. In other words, n3 will be a small positive value, but it will still be less than one. Thus, because n3 < 1, if we subtract 1 from both sides, we see that n3 – 1 < 0. Therefore, n3 - 1 is a negative value.

All negative numbers are less than positive numbers. Thus, we can eliminate n2 and 1/n2, which are both positive. We are left with –1/n, –1/n3, and n3 – 1.

Let us compare –1/n and –1/n3. First, let us assume that –1/n < –1/n3.

–1/n < –1/n3

Multiply both sides by n3. We don't need to switch the signs because n3 is positive.

–n2 < –1

Multiply both sides by –1.

n2 > 1.

We know that n2 will only be bigger than 1 when n > 1 or if n < –1. But we know that 0 < n < 1, so –1/n is not less than –1/n3. Therefore, –1/n3 must be smaller.

Finally, let's compare –1/n3 and n3 – 1. Let us assume that –1/n3 < n3 – 1.

–1/n3 < n3 – 1

Multiply both sides by n3.

–1 < n6 – n3

Add one to both sides.

n6 – n3 + 1 > 0. If this inequality is true, then it will be true that –1/n3 is the smallest number.

Here it will be helpful to try some values for n. Let's pick n = 1/2 and see what happens. It will help to use our calculator.

(1/2)6 – (1/2)3 + 1 = 0.891 > 0.

Therefore, we suspect that because n6 – n3 + 1 > 0, –1/n3 is indeed the smallest number. We can verify this by trying more values of n.

The answer is –1/n3.