How to find excluded values - PSAT Math
Card 0 of 98
If the average (arithmetic mean) of 
, 
, and 
is 
, what is the average of 
, 
, and 
?
If the average (arithmetic mean) of
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If we can find the sum of x+2, y-6, and 10, we can determine their average. There is not enough information to solve for x or y individually, but we can find their sum, x+y.
Write out the average formula for the original three quantities. Remember, adding together and dividing by the number of quantities gives the average: $\frac{x + y + 9}{3}$ = 12
Isolate x+y:
x + y + 9 = 36
x + y = 27
Write out the average formula for the new three quantities:
$\frac{x + 2 + y - 6 + 10}{3}$ = ?
Combine the integers in the numerator:
$\frac{x + y + 6}{3}$ = ?
Replace x+y with 27:
$\frac{27+ 6}{3}$ = $\frac{33}{3}$ = 11
If we can find the sum of x+2, y-6, and 10, we can determine their average. There is not enough information to solve for x or y individually, but we can find their sum, x+y.
Write out the average formula for the original three quantities. Remember, adding together and dividing by the number of quantities gives the average: $\frac{x + y + 9}{3}$ = 12
Isolate x+y:
x + y + 9 = 36
x + y = 27
Write out the average formula for the new three quantities:
$\frac{x + 2 + y - 6 + 10}{3}$ = ?
Combine the integers in the numerator:
$\frac{x + y + 6}{3}$ = ?
Replace x+y with 27:
$\frac{27+ 6}{3}$ = $\frac{33}{3}$ = 11
Which of the following provides the complete solution set for 
?
Which of the following provides the complete solution set for
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The absolute value will always be positive or 0, therefore all values of z will create a true statement as long as 
. Thus all values except for 2 will work.
The absolute value will always be positive or 0, therefore all values of z will create a true statement as long as
Given the expression, 
, which value CANNOT be equal to 
?
Given the expression,
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cannot equal a value which would make the denominator equal to 0. In order to figure out what that value is, we must first simplify this fraction, then set each factor of the denominator equal to 0. As follows:
First, simplify by finding the original binomial multiples:
Now, set each factor of the denominator equal to 0
If 
OR 
, the denominator will equal 0. 
is the only choice provided by the answers.
Note: Even though you can cancel out 
from the numerator and denominator, 
still cannot be equal to two. The graph would have a hole at 
and an aymptote at 
.
First, simplify by finding the original binomial multiples:
Now, set each factor of the denominator equal to 0
If
Note: Even though you can cancel out
Define the function 
on the set of real numbers as follows:
Give the domain of 
.
Define the function
Give the domain of
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The domain of 
is restricted by two things.
First, the expression within the radical in the numerator must be nonnegative. We therefore solve for 
in the inequality
,
or, in interval notation, 
Second, the expression in the denominator must be nonzero. Therefore, we set the denominator equal to zero to determine the excluded value(s):
We exclude 4 from 
, so the correct response is
The domain of
First, the expression within the radical in the numerator must be nonnegative. We therefore solve for
or, in interval notation,
Second, the expression in the denominator must be nonzero. Therefore, we set the denominator equal to zero to determine the excluded value(s):
We exclude 4 from
Define the function 
as follows:
Give the domain of 
.
Define the function
Give the domain of
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The numerator, being a polynomial, is not restricting our domain. The domain is, however, restricted by the polynomial in the denominator, which must be nonzero. Therefore, we set the denominator equal to zero to determine the excluded values:
Therefore, the domain of 
is the set of all real numbers except 
- that is,
The numerator, being a polynomial, is not restricting our domain. The domain is, however, restricted by the polynomial in the denominator, which must be nonzero. Therefore, we set the denominator equal to zero to determine the excluded values:
Therefore, the domain of
Define the function 
on the set of real numbers as follows:
Give the domain of 
.
Define the function
Give the domain of
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There are two things restricting the domain of 
.
One is the radical symbol in the numerator. The expression inside the radical must be nonnegative, so solve the inequality:
,
or, in interval notation, 
The other is the denominator, which must be equal to 0, so set, and solve for 
in, the equation:
is already excluded from the domain; we exclude 4, so the domain is
.
There are two things restricting the domain of
One is the radical symbol in the numerator. The expression inside the radical must be nonnegative, so solve the inequality:
or, in interval notation,
The other is the denominator, which must be equal to 0, so set, and solve for
Define the function 
on the real numbers as follows:
Give the domain of 
.
Define the function
Give the domain of
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The numerator, being a polynomial, is not restricting our domain. The domain is, however, restricted by the expression in the denominator, which must be nonzero. Furthermore, the radicand must be nonnegative. Combined, these facts mean that the radicand must be positive, and that the following inequality be solved:
or, equivalently, 
.
In interval form, this is 
The numerator, being a polynomial, is not restricting our domain. The domain is, however, restricted by the expression in the denominator, which must be nonzero. Furthermore, the radicand must be nonnegative. Combined, these facts mean that the radicand must be positive, and that the following inequality be solved:
In interval form, this is
Define the function 
as follows:
Give the domain of 
.
Define the function
Give the domain of
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The domain of 
is restricted by two different denominators, neither of which can be equal to 0, so the excluded values are:
The correct response is therefore 
.
The domain of
The correct response is therefore
Define the function 
as follows:
Give the domain of 
.
Define the function
Give the domain of
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The numerator, being a polynomial, does not restrict our domain. The denominator, however, does restrict it to the values for which it is not equal to 0. We set the denominator equal to 0 to find the excluded values:
The domain, in interval notation, is therefore
.
The numerator, being a polynomial, does not restrict our domain. The denominator, however, does restrict it to the values for which it is not equal to 0. We set the denominator equal to 0 to find the excluded values:
The domain, in interval notation, is therefore
Define the functions 
and 
as follows:
and 
Give the domain of the function 
.
Define the functions
Give the domain of the function
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The domain of the product of two functions is the intersection of the domains of the individual functions.
The domain of 
is restricted to all values of 
that yield a nonzero denominator. Since this means that
,
then, subsequently,
,
so the domain is the set of all real numbers except 4.
Similarly, the domain of 
is restricted to all values of 
that yield a nonzero denominator. This set is found to be the set of all real numbers except 7.
The intersection of these sets is the set of all real numbers except 4 and 7, or
.
The domain of the product of two functions is the intersection of the domains of the individual functions.
The domain of
then, subsequently,
so the domain is the set of all real numbers except 4.
Similarly, the domain of
The intersection of these sets is the set of all real numbers except 4 and 7, or
Define the function 
as follows:
Give the domain of 
.
Define the function
Give the domain of
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The definition of 
has two denominators of two fractions (one within the other), so we must exclude the values of 
that make either denominator equal to zero.
One denominator is 
. Since it cannot be zero, we have
,
and, subsequently,
.
The value 2 is excluded from the domain.
The other denominator is 
. If it is equal to zero, then
Therefore, this value is also excluded from the domain.
The correct domain is the set 
.
The definition of
One denominator is
and, subsequently,
The value 2 is excluded from the domain.
The other denominator is
Therefore, this value is also excluded from the domain.
The correct domain is the set
Define the function 
as follows:
Give the domain of 
.
Define the function
Give the domain of
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The numerator 
, being a polynomial, does not restrict the domain.
appears as a radicand of a square root in the definition of 
, so 
must be restricted to nonnegative numbers. Also, 
is a denominator, which means 0 must also be excluded. Therefore, we are restricted so far to positive numbers.
There is one more denominator, which is 
; it must be nonzero. We set this equal to zero to determine any additional value(s) that must be excluded:
Therefore, the domain is the set of all positive numbers except 9 - or
.
The numerator
There is one more denominator, which is
Therefore, the domain is the set of all positive numbers except 9 - or
Which of the following are answers to the equation below?
I. -3
II. -2
III. 2
Which of the following are answers to the equation below?
I. -3
II. -2
III. 2
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Given a fractional algebraic equation with variables in the numerator and denominator of one side and the other side equal to zero, we rely on a simple concept. Zero divided by anything equals zero. That means we can focus in on what values make the numerator (the top part of the fraction) zero, or in other words,
The expression 
is a difference of squares that can be factored as
Solving this for 
gives either 
or 
. That means either of these values will make our numerator equal zero. We might be tempted to conclude that both are valid answers. However, our statement earlier that zero divided by anything is zero has one caveat. We can never divide by zero itself. That means that any values that make our denominator zero must be rejected. Therefore we must also look at the denominator.
The left side factors as follows
This means that if 
is 
or 
, we end up dividing by zero. That means that 
cannot be a valid solution, leaving 
as the only valid answer. Therefore only #3 is correct.
Given a fractional algebraic equation with variables in the numerator and denominator of one side and the other side equal to zero, we rely on a simple concept. Zero divided by anything equals zero. That means we can focus in on what values make the numerator (the top part of the fraction) zero, or in other words,
The expression
Solving this for
The left side factors as follows
This means that if
Find the excluded values of the following algebraic fraction
Find the excluded values of the following algebraic fraction
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To find the excluded values of a algebraic fraction you need to find when the denominator is zero. To find when the denominator is zero you need to factor it. This denominator factors into
so this is zero when x=4,7 so our answer is
To find the excluded values of a algebraic fraction you need to find when the denominator is zero. To find when the denominator is zero you need to factor it. This denominator factors into
so this is zero when x=4,7 so our answer is
If the average (arithmetic mean) of , , and is , what is the average of , , and ?
If the average (arithmetic mean) of
Tap to see back →
If we can find the sum of x+2, y-6, and 10, we can determine their average. There is not enough information to solve for x or y individually, but we can find their sum, x+y.
Write out the average formula for the original three quantities. Remember, adding together and dividing by the number of quantities gives the average: $\frac{x + y + 9}{3}$ = 12
Isolate x+y:
x + y + 9 = 36
x + y = 27
Write out the average formula for the new three quantities:
$\frac{x + 2 + y - 6 + 10}{3}$ = ?
Combine the integers in the numerator:
$\frac{x + y + 6}{3}$ = ?
Replace x+y with 27:
$\frac{27+ 6}{3}$ = $\frac{33}{3}$ = 11
If we can find the sum of x+2, y-6, and 10, we can determine their average. There is not enough information to solve for x or y individually, but we can find their sum, x+y.
Write out the average formula for the original three quantities. Remember, adding together and dividing by the number of quantities gives the average: $\frac{x + y + 9}{3}$ = 12
Isolate x+y:
x + y + 9 = 36
x + y = 27
Write out the average formula for the new three quantities:
$\frac{x + 2 + y - 6 + 10}{3}$ = ?
Combine the integers in the numerator:
$\frac{x + y + 6}{3}$ = ?
Replace x+y with 27:
$\frac{27+ 6}{3}$ = $\frac{33}{3}$ = 11
Which of the following provides the complete solution set for ?
Which of the following provides the complete solution set for
Tap to see back →
The absolute value will always be positive or 0, therefore all values of z will create a true statement as long as . Thus all values except for 2 will work.
The absolute value will always be positive or 0, therefore all values of z will create a true statement as long as
Define the function as follows:
Give the domain of .
Define the function
Give the domain of
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The domain of is restricted by two different denominators, neither of which can be equal to 0, so the excluded values are:
The correct response is therefore .
The domain of
The correct response is therefore
Given the expression, , which value CANNOT be equal to ?
Given the expression,
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cannot equal a value which would make the denominator equal to 0. In order to figure out what that value is, we must first simplify this fraction, then set each factor of the denominator equal to 0. As follows:
First, simplify by finding the original binomial multiples:
Now, set each factor of the denominator equal to 0
If OR , the denominator will equal 0. is the only choice provided by the answers.
Note: Even though you can cancel out from the numerator and denominator, still cannot be equal to two. The graph would have a hole at and an aymptote at .
First, simplify by finding the original binomial multiples:
Now, set each factor of the denominator equal to 0
If
Note: Even though you can cancel out
Define the function on the set of real numbers as follows:
Give the domain of .
Define the function
Give the domain of
Tap to see back →
The domain of is restricted by two things.
First, the expression within the radical in the numerator must be nonnegative. We therefore solve for in the inequality
,
or, in interval notation,
Second, the expression in the denominator must be nonzero. Therefore, we set the denominator equal to zero to determine the excluded value(s):
We exclude 4 from , so the correct response is
The domain of
First, the expression within the radical in the numerator must be nonnegative. We therefore solve for
or, in interval notation,
Second, the expression in the denominator must be nonzero. Therefore, we set the denominator equal to zero to determine the excluded value(s):
We exclude 4 from
Define the function as follows:
Give the domain of .
Define the function
Give the domain of
Tap to see back →
The numerator, being a polynomial, is not restricting our domain. The domain is, however, restricted by the polynomial in the denominator, which must be nonzero. Therefore, we set the denominator equal to zero to determine the excluded values:
Therefore, the domain of is the set of all real numbers except - that is,
The numerator, being a polynomial, is not restricting our domain. The domain is, however, restricted by the polynomial in the denominator, which must be nonzero. Therefore, we set the denominator equal to zero to determine the excluded values:
Therefore, the domain of