How to find absolute value - PSAT Math
Card 0 of 42
If 
, what is the value of 
?
If
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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.
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.
Let 
and 
both be negative numbers such that 
and 
. What is 
?
Let
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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.
For which of the following functions below does f(x) = |f(x)| for every value of x within its domain?
For which of the following functions below does f(x) = |f(x)| for every value of x within its domain?
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When we take the absolute value of a function, any negative values get changed into positive values. Essentially, |f(x)| will take all of the negative values of f(x) and reflect them across the x-axis. However, any values of f(x) that are positive or equal to zero will not be changed, because the absolute value of a positive number (or zero) is still the same number.
If we can show that f(x) has negative values, then |f(x)| will be different from f(x) at some points, because its negative values will be changed to positive values. In other words, our answer will consist of the function that never has negative values.
Let's look at f(x) = 2_x_ + 3. Obviously, this equation of a line will have negative values. For example, where x = –4, f(–4) = 2(–4) + 3 = –5, which is negative. Thus, f(x) has negative values, and if we were to graph |f(x)|, the result would be different from f(x). Therefore, f(x) = 2_x_ + 3 isn't the correct answer.
Next, let's look at f(x) = _x_2 – 9. If we let x = 1, then f(1) = 1 – 9 = –8, which is negative. Thus |f(x)| will not be the same as f(x), and we can eliminate this choice as well.
Now, let's examine f(x) = x_2 – 2_x. We know that _x_2 by itself can never be negative. However, if x_2 is really small, then adding –2_x could make it negative. Therefore, let's evaluate f(x) when x is a fractional value such as 1/2. f(1/2) = 1/4 – 1 = –3/4, which is negative. Thus, there are some values on f(x) that are negative, so we can eliminate this function.
Next, let's examine f(x) = _x_4 + x. In general, any number taken to an even-numbered power must always be non-negative. Therefore, _x_4 cannot be negative, because if we multiplied a negative number by itself four times, the result would be positive. However, the x term could be negative. If we let x be a small negative fraction, then _x_4 would be close to zero, and we would be left with x, which is negative. For example, let's find f(x) when x = –1/2. f(–1/2) = (–1/2)4 + (–1/2) = (1/16) – (1/2) = –7/16, which is negative. Thus, |f(x)| is not always the same as f(x).
By process of elimination, the answer is f(x) = _x_4 + (1 – x)2 . This makes sense because _x_4 can't be negative, and because (1 – x)2 can't be negative. No matter what we subtract from one, when we square the final result, we can't get a negative number. And if we add _x_4 and (1 – x)2, the result will also be non-negative, because adding two non-negative numbers always produces a non-negative result. Therefore, f(x) = _x_4 + (1 – x)2 will not have any negative values, and |f(x)| will be the same as f(x) for all values of x.
The answer is f(x) = _x_4 + (1 – x)2 .
When we take the absolute value of a function, any negative values get changed into positive values. Essentially, |f(x)| will take all of the negative values of f(x) and reflect them across the x-axis. However, any values of f(x) that are positive or equal to zero will not be changed, because the absolute value of a positive number (or zero) is still the same number.
If we can show that f(x) has negative values, then |f(x)| will be different from f(x) at some points, because its negative values will be changed to positive values. In other words, our answer will consist of the function that never has negative values.
Let's look at f(x) = 2_x_ + 3. Obviously, this equation of a line will have negative values. For example, where x = –4, f(–4) = 2(–4) + 3 = –5, which is negative. Thus, f(x) has negative values, and if we were to graph |f(x)|, the result would be different from f(x). Therefore, f(x) = 2_x_ + 3 isn't the correct answer.
Next, let's look at f(x) = _x_2 – 9. If we let x = 1, then f(1) = 1 – 9 = –8, which is negative. Thus |f(x)| will not be the same as f(x), and we can eliminate this choice as well.
Now, let's examine f(x) = x_2 – 2_x. We know that _x_2 by itself can never be negative. However, if x_2 is really small, then adding –2_x could make it negative. Therefore, let's evaluate f(x) when x is a fractional value such as 1/2. f(1/2) = 1/4 – 1 = –3/4, which is negative. Thus, there are some values on f(x) that are negative, so we can eliminate this function.
Next, let's examine f(x) = _x_4 + x. In general, any number taken to an even-numbered power must always be non-negative. Therefore, _x_4 cannot be negative, because if we multiplied a negative number by itself four times, the result would be positive. However, the x term could be negative. If we let x be a small negative fraction, then _x_4 would be close to zero, and we would be left with x, which is negative. For example, let's find f(x) when x = –1/2. f(–1/2) = (–1/2)4 + (–1/2) = (1/16) – (1/2) = –7/16, which is negative. Thus, |f(x)| is not always the same as f(x).
By process of elimination, the answer is f(x) = _x_4 + (1 – x)2 . This makes sense because _x_4 can't be negative, and because (1 – x)2 can't be negative. No matter what we subtract from one, when we square the final result, we can't get a negative number. And if we add _x_4 and (1 – x)2, the result will also be non-negative, because adding two non-negative numbers always produces a non-negative result. Therefore, f(x) = _x_4 + (1 – x)2 will not have any negative values, and |f(x)| will be the same as f(x) for all values of x.
The answer is f(x) = _x_4 + (1 – x)2 .
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Evaluate 
if 
.
Evaluate
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Remember that the absolute value of a number is the positive value of that number. It is important to know what absolute value represents: the distance of any given number from the number 0. For that reason, absolute value cannot be negative, and we can eliminate 
and 
from our answer choices
To solve, substitute 
into the equation, taking extra care to correctly calculate the negative numbers:
The answer is 12.
Remember that the absolute value of a number is the positive value of that number. It is important to know what absolute value represents: the distance of any given number from the number 0. For that reason, absolute value cannot be negative, and we can eliminate
To solve, substitute
The answer is 12.
How many of these statements are correct?
I) 
II) 
III) 
IV) 
How many of these statements are correct?
I)
II)
III)
IV)
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The absolute value of a nonnegative number is the number itself; the absolute value of a negative number can be obtained by removing the negative symbol. Therefore,
.
The four statements can be rewritten as:
Both I and III) 
- This is true.
Both II and IV) 
- This is false.
The correct response is two.
The absolute value of a nonnegative number is the number itself; the absolute value of a negative number can be obtained by removing the negative symbol. Therefore,
The four statements can be rewritten as:
Both I and III)
Both II and IV)
The correct response is two.
If , what is the value of ?
If
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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.
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.
Let and both be negative numbers such that and . What is ?
Let
Tap to see back →
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.
For which of the following functions below does f(x) = |f(x)| for every value of x within its domain?
For which of the following functions below does f(x) = |f(x)| for every value of x within its domain?
Tap to see back →
When we take the absolute value of a function, any negative values get changed into positive values. Essentially, |f(x)| will take all of the negative values of f(x) and reflect them across the x-axis. However, any values of f(x) that are positive or equal to zero will not be changed, because the absolute value of a positive number (or zero) is still the same number.
If we can show that f(x) has negative values, then |f(x)| will be different from f(x) at some points, because its negative values will be changed to positive values. In other words, our answer will consist of the function that never has negative values.
Let's look at f(x) = 2_x_ + 3. Obviously, this equation of a line will have negative values. For example, where x = –4, f(–4) = 2(–4) + 3 = –5, which is negative. Thus, f(x) has negative values, and if we were to graph |f(x)|, the result would be different from f(x). Therefore, f(x) = 2_x_ + 3 isn't the correct answer.
Next, let's look at f(x) = _x_2 – 9. If we let x = 1, then f(1) = 1 – 9 = –8, which is negative. Thus |f(x)| will not be the same as f(x), and we can eliminate this choice as well.
Now, let's examine f(x) = x_2 – 2_x. We know that _x_2 by itself can never be negative. However, if x_2 is really small, then adding –2_x could make it negative. Therefore, let's evaluate f(x) when x is a fractional value such as 1/2. f(1/2) = 1/4 – 1 = –3/4, which is negative. Thus, there are some values on f(x) that are negative, so we can eliminate this function.
Next, let's examine f(x) = _x_4 + x. In general, any number taken to an even-numbered power must always be non-negative. Therefore, _x_4 cannot be negative, because if we multiplied a negative number by itself four times, the result would be positive. However, the x term could be negative. If we let x be a small negative fraction, then _x_4 would be close to zero, and we would be left with x, which is negative. For example, let's find f(x) when x = –1/2. f(–1/2) = (–1/2)4 + (–1/2) = (1/16) – (1/2) = –7/16, which is negative. Thus, |f(x)| is not always the same as f(x).
By process of elimination, the answer is f(x) = _x_4 + (1 – x)2 . This makes sense because _x_4 can't be negative, and because (1 – x)2 can't be negative. No matter what we subtract from one, when we square the final result, we can't get a negative number. And if we add _x_4 and (1 – x)2, the result will also be non-negative, because adding two non-negative numbers always produces a non-negative result. Therefore, f(x) = _x_4 + (1 – x)2 will not have any negative values, and |f(x)| will be the same as f(x) for all values of x.
The answer is f(x) = _x_4 + (1 – x)2 .
When we take the absolute value of a function, any negative values get changed into positive values. Essentially, |f(x)| will take all of the negative values of f(x) and reflect them across the x-axis. However, any values of f(x) that are positive or equal to zero will not be changed, because the absolute value of a positive number (or zero) is still the same number.
If we can show that f(x) has negative values, then |f(x)| will be different from f(x) at some points, because its negative values will be changed to positive values. In other words, our answer will consist of the function that never has negative values.
Let's look at f(x) = 2_x_ + 3. Obviously, this equation of a line will have negative values. For example, where x = –4, f(–4) = 2(–4) + 3 = –5, which is negative. Thus, f(x) has negative values, and if we were to graph |f(x)|, the result would be different from f(x). Therefore, f(x) = 2_x_ + 3 isn't the correct answer.
Next, let's look at f(x) = _x_2 – 9. If we let x = 1, then f(1) = 1 – 9 = –8, which is negative. Thus |f(x)| will not be the same as f(x), and we can eliminate this choice as well.
Now, let's examine f(x) = x_2 – 2_x. We know that _x_2 by itself can never be negative. However, if x_2 is really small, then adding –2_x could make it negative. Therefore, let's evaluate f(x) when x is a fractional value such as 1/2. f(1/2) = 1/4 – 1 = –3/4, which is negative. Thus, there are some values on f(x) that are negative, so we can eliminate this function.
Next, let's examine f(x) = _x_4 + x. In general, any number taken to an even-numbered power must always be non-negative. Therefore, _x_4 cannot be negative, because if we multiplied a negative number by itself four times, the result would be positive. However, the x term could be negative. If we let x be a small negative fraction, then _x_4 would be close to zero, and we would be left with x, which is negative. For example, let's find f(x) when x = –1/2. f(–1/2) = (–1/2)4 + (–1/2) = (1/16) – (1/2) = –7/16, which is negative. Thus, |f(x)| is not always the same as f(x).
By process of elimination, the answer is f(x) = _x_4 + (1 – x)2 . This makes sense because _x_4 can't be negative, and because (1 – x)2 can't be negative. No matter what we subtract from one, when we square the final result, we can't get a negative number. And if we add _x_4 and (1 – x)2, the result will also be non-negative, because adding two non-negative numbers always produces a non-negative result. Therefore, f(x) = _x_4 + (1 – x)2 will not have any negative values, and |f(x)| will be the same as f(x) for all values of x.
The answer is f(x) = _x_4 + (1 – x)2 .
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Evaluate if .
Evaluate
Tap to see back →
Remember that the absolute value of a number is the positive value of that number. It is important to know what absolute value represents: the distance of any given number from the number 0. For that reason, absolute value cannot be negative, and we can eliminate and from our answer choices
To solve, substitute into the equation, taking extra care to correctly calculate the negative numbers:
The answer is 12.
Remember that the absolute value of a number is the positive value of that number. It is important to know what absolute value represents: the distance of any given number from the number 0. For that reason, absolute value cannot be negative, and we can eliminate
To solve, substitute
The answer is 12.
How many of these statements are correct?
I)
II)
III)
IV)
How many of these statements are correct?
I)
II)
III)
IV)
Tap to see back →
The absolute value of a nonnegative number is the number itself; the absolute value of a negative number can be obtained by removing the negative symbol. Therefore,
.
The four statements can be rewritten as:
Both I and III) - This is true.
Both II and IV) - This is false.
The correct response is two.
The absolute value of a nonnegative number is the number itself; the absolute value of a negative number can be obtained by removing the negative symbol. Therefore,
The four statements can be rewritten as:
Both I and III)
Both II and IV)
The correct response is two.
If , what is the value of ?
If
Tap to see back →
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.
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.
Let and both be negative numbers such that and . What is ?
Let
Tap to see back →
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.
For which of the following functions below does f(x) = |f(x)| for every value of x within its domain?
For which of the following functions below does f(x) = |f(x)| for every value of x within its domain?
Tap to see back →
When we take the absolute value of a function, any negative values get changed into positive values. Essentially, |f(x)| will take all of the negative values of f(x) and reflect them across the x-axis. However, any values of f(x) that are positive or equal to zero will not be changed, because the absolute value of a positive number (or zero) is still the same number.
If we can show that f(x) has negative values, then |f(x)| will be different from f(x) at some points, because its negative values will be changed to positive values. In other words, our answer will consist of the function that never has negative values.
Let's look at f(x) = 2_x_ + 3. Obviously, this equation of a line will have negative values. For example, where x = –4, f(–4) = 2(–4) + 3 = –5, which is negative. Thus, f(x) has negative values, and if we were to graph |f(x)|, the result would be different from f(x). Therefore, f(x) = 2_x_ + 3 isn't the correct answer.
Next, let's look at f(x) = _x_2 – 9. If we let x = 1, then f(1) = 1 – 9 = –8, which is negative. Thus |f(x)| will not be the same as f(x), and we can eliminate this choice as well.
Now, let's examine f(x) = x_2 – 2_x. We know that _x_2 by itself can never be negative. However, if x_2 is really small, then adding –2_x could make it negative. Therefore, let's evaluate f(x) when x is a fractional value such as 1/2. f(1/2) = 1/4 – 1 = –3/4, which is negative. Thus, there are some values on f(x) that are negative, so we can eliminate this function.
Next, let's examine f(x) = _x_4 + x. In general, any number taken to an even-numbered power must always be non-negative. Therefore, _x_4 cannot be negative, because if we multiplied a negative number by itself four times, the result would be positive. However, the x term could be negative. If we let x be a small negative fraction, then _x_4 would be close to zero, and we would be left with x, which is negative. For example, let's find f(x) when x = –1/2. f(–1/2) = (–1/2)4 + (–1/2) = (1/16) – (1/2) = –7/16, which is negative. Thus, |f(x)| is not always the same as f(x).
By process of elimination, the answer is f(x) = _x_4 + (1 – x)2 . This makes sense because _x_4 can't be negative, and because (1 – x)2 can't be negative. No matter what we subtract from one, when we square the final result, we can't get a negative number. And if we add _x_4 and (1 – x)2, the result will also be non-negative, because adding two non-negative numbers always produces a non-negative result. Therefore, f(x) = _x_4 + (1 – x)2 will not have any negative values, and |f(x)| will be the same as f(x) for all values of x.
The answer is f(x) = _x_4 + (1 – x)2 .
When we take the absolute value of a function, any negative values get changed into positive values. Essentially, |f(x)| will take all of the negative values of f(x) and reflect them across the x-axis. However, any values of f(x) that are positive or equal to zero will not be changed, because the absolute value of a positive number (or zero) is still the same number.
If we can show that f(x) has negative values, then |f(x)| will be different from f(x) at some points, because its negative values will be changed to positive values. In other words, our answer will consist of the function that never has negative values.
Let's look at f(x) = 2_x_ + 3. Obviously, this equation of a line will have negative values. For example, where x = –4, f(–4) = 2(–4) + 3 = –5, which is negative. Thus, f(x) has negative values, and if we were to graph |f(x)|, the result would be different from f(x). Therefore, f(x) = 2_x_ + 3 isn't the correct answer.
Next, let's look at f(x) = _x_2 – 9. If we let x = 1, then f(1) = 1 – 9 = –8, which is negative. Thus |f(x)| will not be the same as f(x), and we can eliminate this choice as well.
Now, let's examine f(x) = x_2 – 2_x. We know that _x_2 by itself can never be negative. However, if x_2 is really small, then adding –2_x could make it negative. Therefore, let's evaluate f(x) when x is a fractional value such as 1/2. f(1/2) = 1/4 – 1 = –3/4, which is negative. Thus, there are some values on f(x) that are negative, so we can eliminate this function.
Next, let's examine f(x) = _x_4 + x. In general, any number taken to an even-numbered power must always be non-negative. Therefore, _x_4 cannot be negative, because if we multiplied a negative number by itself four times, the result would be positive. However, the x term could be negative. If we let x be a small negative fraction, then _x_4 would be close to zero, and we would be left with x, which is negative. For example, let's find f(x) when x = –1/2. f(–1/2) = (–1/2)4 + (–1/2) = (1/16) – (1/2) = –7/16, which is negative. Thus, |f(x)| is not always the same as f(x).
By process of elimination, the answer is f(x) = _x_4 + (1 – x)2 . This makes sense because _x_4 can't be negative, and because (1 – x)2 can't be negative. No matter what we subtract from one, when we square the final result, we can't get a negative number. And if we add _x_4 and (1 – x)2, the result will also be non-negative, because adding two non-negative numbers always produces a non-negative result. Therefore, f(x) = _x_4 + (1 – x)2 will not have any negative values, and |f(x)| will be the same as f(x) for all values of x.
The answer is f(x) = _x_4 + (1 – x)2 .
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Evaluate if .
Evaluate
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Remember that the absolute value of a number is the positive value of that number. It is important to know what absolute value represents: the distance of any given number from the number 0. For that reason, absolute value cannot be negative, and we can eliminate and from our answer choices
To solve, substitute into the equation, taking extra care to correctly calculate the negative numbers:
The answer is 12.
Remember that the absolute value of a number is the positive value of that number. It is important to know what absolute value represents: the distance of any given number from the number 0. For that reason, absolute value cannot be negative, and we can eliminate
To solve, substitute
The answer is 12.
How many of these statements are correct?
I)
II)
III)
IV)
How many of these statements are correct?
I)
II)
III)
IV)
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The absolute value of a nonnegative number is the number itself; the absolute value of a negative number can be obtained by removing the negative symbol. Therefore,
.
The four statements can be rewritten as:
Both I and III) - This is true.
Both II and IV) - This is false.
The correct response is two.
The absolute value of a nonnegative number is the number itself; the absolute value of a negative number can be obtained by removing the negative symbol. Therefore,
The four statements can be rewritten as:
Both I and III)
Both II and IV)
The correct response is two.
If , what is the value of ?
If
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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.
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.
Let and both be negative numbers such that and . What is ?
Let
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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.