find domain
a/(a-6) - 3/(a+6)+(a^2)/36-a^2
To find the domain of the given expression, we need to identify any values of "a" that would make the expression undefined.
Let's take a look at each term separately:
1) a/(a-6): The expression is undefined when the denominator, (a-6), is equal to zero. Therefore, a cannot be equal to 6.
2) -3/(a+6): The expression is undefined when the denominator, (a+6), is equal to zero. Therefore, a cannot be equal to -6.
3) (a^2)/36: This expression is defined for all values of a.
4) -a^2: This expression is defined for all values of a.
Combining all these terms, the domain of the given expression is all real numbers except for a = 6 and a = -6.
x^2-1/5xy * x^y/1+x
find domain
To find the domain of the given expression, we need to identify any values of 'x' and 'y' that would make the expression undefined.
Let's break down the expression and analyze each term separately:
1) x^2 - 1/5xy: This expression is defined for all real values of 'x' and 'y' except for the values that would make the denominator equal to zero. Therefore, we need to check when 5xy = 0.
Since 5xy = 0 when either 'x' or 'y' is equal to 0, we need to exclude these values from the domain.
2) x^y: This expression is defined for all real values of 'x' and 'y'.
3) 1 + x: This expression is defined for all real values of 'x'.
Combining all these terms, the domain of the given expression is all real values of 'x' and 'y', except for 'x' = 0 and 'y' = 0.
no
Apologies for the confusion. Let's reevaluate the expression and find its domain.
The given expression is:
(x^2-1)/(5xy) * (x^y)/(1+x)
To find the domain, we need to identify any values of 'x' and 'y' that would make the expression undefined.
1) (x^2-1)/(5xy):
The expression is undefined when the denominator, 5xy, is equal to zero. Therefore, we need to check when 5xy = 0.
5xy = 0 when either 'x' or 'y' is equal to 0. Thus, the expression is undefined when 'x' or 'y' is equal to 0.
2) (x^y)/(1+x):
The expression is undefined when the denominator, 1+x, is equal to zero. Therefore, we need to check when 1+x = 0.
1+x = 0 when x = -1. Thus, the expression is undefined when 'x' is equal to -1.
Combining both expressions, the domain of the given expression is all real values of 'x' and 'y', except for 'x' = -1 and 'x' or 'y' = 0.
wrong
Apologies for the incorrect response. Let's re-evaluate the expression and find its domain.
The given expression is:
(x^2-1)/(5xy) * (x^y)/(1+x)
To find the domain, we need to identify any values of 'x' and 'y' that would make the expression undefined.
1) (x^2-1)/(5xy):
The expression is defined for all real values of 'x' and 'y' except for the values that would make the denominator, 5xy, equal to zero. Therefore, we need to check when 5xy = 0.
5xy = 0 when either 'x' or 'y' is equal to 0. Thus, the expression is undefined when 'x' or 'y' is equal to 0.
2) (x^y)/(1+x):
The expression is defined for all real values of 'x' and 'y' except for the values that would make the denominator, 1+x, equal to zero. Therefore, we need to check when 1+x = 0.
1+x = 0 when x = -1. Thus, the expression is undefined when 'x' is equal to -1.
Combining both expressions, the domain of the given expression is all real values of 'x' and 'y', except for 'x' = -1 and 'x' or 'y' = 0.
x^2-1/5xy * x^2y/1+x
Apologies for the confusion. Let's reevaluate the expression and find its domain.
The given expression is:
(x^2-1)/(5xy) * (x^2y)/(1+x)
To find the domain, we need to identify any values of 'x' and 'y' that would make the expression undefined.
1) (x^2-1)/(5xy):
The expression is defined for all real values of 'x' and 'y', except for the values that would make the denominator, 5xy, equal to zero. Therefore, we need to check when 5xy = 0.
5xy = 0 when either 'x' or 'y' is equal to 0. Thus, the expression is undefined when 'x' or 'y' is equal to 0.
2) (x^2y)/(1+x):
The expression is defined for all real values of 'x' and 'y' except for the values that would make the denominator, 1+x, equal to zero. Therefore, we need to check when 1+x = 0.
1+x = 0 when x = -1. Thus, the expression is undefined when 'x' is equal to -1.
Combining both expressions, the domain of the given expression is all real values of 'x' and 'y', except for 'x' = -1 and 'x' or 'y' = 0.
To find the domain of a given expression, we need to determine the values that the variable 'a' can take without causing any undefined or forbidden operations.
In this case, we have several terms involving 'a' in the expression:
1. The term 'a' in the denominator (a-6)
2. The term 'a' in the denominator (a+6)
3. The term 'a^2' in the denominator (36-a^2)
To find the domain, we need to consider each term separately and identify any values of 'a' that would make these terms equal to zero.
1. (a-6) in the denominator: The expression is undefined when the denominator is zero. Thus, we cannot have a-6 = 0. Solving this equation, we find that a = 6.
2. (a+6) in the denominator: Similar to the previous case, we cannot have a+6 = 0. Solving this equation, we find that a = -6.
3. (36-a^2) in the denominator: This expression is undefined when the denominator is zero. Therefore, we need to solve the equation 36-a^2 = 0. Rearranging the equation, we have a^2 = 36. Taking the square root of both sides, we get a = ±6.
Considering all the cases, we find that the values of 'a' that make the expression undefined are a = -6, a = 6, and a = -6. Hence, the domain of the given expression is all real numbers except for -6 and 6.