Identify the field of application of the following fraction:
Identify the field of application of the following fraction:
\( \frac{3}{x+2} \)
Identify the field of application of the following fraction:
\( \frac{8}{-2+x} \)
Identify the field of application of the following fraction:
\( \frac{7}{13+x} \)
Identify the field of application of the following fraction:
\( \frac{x+8}{3x} \)
\( \frac{x+y:3}{2x+6}=4 \)
What is the field of application of the equation?
Identify the field of application of the following fraction:
Let's examine the given expression:
As we know, the only restriction that applies to division is division by 0, since no number can be divided into 0 parts. Hence division by 0 is undefined.
Therefore, when we talk about a fraction, where the dividend (the number being divided) is in the numerator, and the divisor (the number we divide by) is in the denominator, the restriction applies only to the denominator, which must be different from 0,
In the given expression:
As stated, the restriction applies to the fraction's denominator only,
Therefore, in order for the given expression (the fraction - in this case) to be defined, we require that the expression in its denominator - does not equal zero, meaning we require that:
We will solve this inequality, which is a point inequality of first degree, in the same way we solve a first-degree equation, meaning - we isolate the variable on one side, by moving terms (and dividing both sides of the inequality by its coefficient if needed):
Therefore, the domain (definition domain) of the given expression is:
(This means that if we substitute for the variable x any number different fromthe expression will remain well-defined),
Therefore, the correct answer is answer D.
Note:
In general - solving an inequality of this form, meaning, a non-linear, but point inequality - that uses the sign and not the inequality signs: ,>,\hspace{2pt}<,\hspace{2pt}\geq,\hspace{2pt}\leq,\hspace{2pt} is identical in every aspect to an equation and therefore is solved in the same way and all rules used to solve an equation of any type are identical for it as well.
Identify the field of application of the following fraction:
Let's examine the following expression:
As we know, the only restriction that applies to division is division by 0, given that no number can be divided into 0 parts. Hence division by 0 is undefined.
Therefore, when we talk about a fraction, where the dividend (the number being divided) is in the numerator, and the divisor (the number we divide by) is in the denominator, the restriction applies only to the denominator, which must be different from 0,
In the given expression:
As stated, the restriction applies to the fraction's denominator only,
Therefore, in order for the given expression (the fraction - in this case) to be defined, we require that the expression in its denominator - does not equal zero, in other words:
We will solve this inequality, which is a point inequality of first degree, in the same way we solve a first-degree equation, meaning - we isolate the variable on one side, by moving terms (and dividing both sides of the inequality by its coefficient if needed):
Therefore, the domain (definition domain) of the given expression is:
(This means that if we substitute any number different from for x, the expression will remain well-defined),
Therefore, the correct answer is answer C.
Note:
In a general form - solving an inequality of this form, meaning, a non-graphical, but point inequality - that uses the sign and not the inequality signs: ,>,\hspace{2pt}<,\hspace{2pt}\geq,\hspace{2pt}\leq,\hspace{2pt} is identical in every way to an equation and therefore is solved in the same way and all rules used to solve an equation of any type are identical for it as well.
Identify the field of application of the following fraction:
Let's examine the given expression:
As we know, the only restriction that applies to division is division by 0. Given that no number can be divided into 0 parts, division by 0 is undefined.
Therefore, when we talk about a fraction, where the dividend (the number being divided) is in the numerator, and the divisor (the number we divide by) is in the denominator, the restriction applies only to the denominator, which must be different from 0,
In the given expression:
As stated, the restriction applies to the fraction's denominator only,
Therefore, in order for the given expression (the fraction - in this case) to be defined, we require that the expression in its denominator - does not equal zero, in other words:
We will solve this inequality, which is a point inequality of first degree, in the same way we solve a first-degree equation, meaning - we isolate the variable on one side, by moving terms (and dividing both sides of the inequality by its coefficient if needed):
Therefore, the domain (definition domain) of the given expression is:
(This means that if we substitute any number different from for the variable x, the expression will remain well-defined),
Therefore, the correct answer is answer D.
Note:
In a general way - solving an inequality of this form, meaning, a non-linear, but point inequality - that uses the sign and not the inequality signs: ,>,\hspace{2pt}<,\hspace{2pt}\geq,\hspace{2pt}\leq,\hspace{2pt} is identical in every way to an equation and therefore is solved in the same way and all rules used to solve an equation of any type are identical for it as well.
Identify the field of application of the following fraction:
Let's examine the given expression:
As we know, the only restriction that applies to division is division by 0. Given that no number can be divided into 0 parts, division by 0 is undefined.
Therefore, when we talk about a fraction, where the dividend (the number being divided) is in the numerator, and the divisor (the number we divide by) is in the denominator, the restriction applies only to the denominator, which must be different from 0,
In the given expression:
As stated, the restriction applies to the fraction's denominator only,
Therefore, in order for the given expression (the fraction - in this case) to be defined, we require that the expression in its denominator - does not equal zero, in other words:
We will solve this inequality, which is a point inequality of first degree, in the same way we solve a first-degree equation, meaning - we isolate the variable on one side, by moving terms (and dividing both sides of the inequality by its coefficient if needed):
Therefore, the domain (definition domain) of the given expression is:
(This means that if we substitute any number different from for x, the expression will remain well-defined),
Therefore, the correct answer is answer A.
Note:
In a general form - solving an inequality of this form, meaning, a non-linear, but point inequality - which uses the sign and not the inequality signs: ,>,\hspace{2pt}<,\hspace{2pt}\geq,\hspace{2pt}\leq,\hspace{2pt} is identical in every aspect to an equation and therefore is solved in the same way and all rules used to solve an equation of any type are identical for it as well.
What is the field of application of the equation?
To solve this problem, we'll follow these steps to find the domain:
Thus, the domain of the given expression is all real numbers except . This translates to:
Select the field of application of the following fraction:
\( \frac{x}{x+3} \)
Identify the field of application of the following fraction:
\( \frac{-8-x}{-x} \)
Choose the field of application of the following fraction:
\( \frac{-8-x}{-3x+2} \)
\( \frac{3x:4}{y+6}=6 \)
What is the field of application of the equation?
Solve the following equation:
\( \frac{(2x+1)^2}{x+2}+\frac{(x+2)^2}{2x+1}=4.5x \)
Select the field of application of the following fraction:
Let's examine the given expression:
As we know, the only restriction that applies to division is division by 0. Given that no number can be divided into 0 parts, division by 0 is undefined.
Therefore, when we talk about a fraction, where the dividend (the number being divided) is in the numerator, and the divisor (the number we divide by) is in the denominator, the restriction applies only to the denominator, which must be different from 0,
In the given expression:
As stated, the restriction applies to the fraction's denominator only,
Therefore, in order for the given expression (the fraction - in this case) to be defined, we require that the expression in its denominator - does not equal zero, meaning we require that:
We will solve this inequality, which is a point inequality of first degree, in the same way we solve a first-degree equation, meaning - we isolate the variable on one side, by moving terms (and dividing both sides of the inequality by its coefficient if needed):
Therefore, the domain (definition domain) of the given expression is:
(This means that if we substitute for the variable x any number different fromthe expression will remain well-defined),
Therefore, the correct answer is answer D.
Note:
In a general form - solving an inequality of this form, meaning, a non-linear, but point inequality - that uses the sign and not the inequality signs: ,>,\hspace{2pt}<,\hspace{2pt}\geq,\hspace{2pt}\leq,\hspace{2pt} is identical in every way to an equation and therefore is solved in the same way and all rules used to solve an equation of any type are identical for it as well.
Identify the field of application of the following fraction:
Let's examine the given expression:
As we know, the only restriction that applies to division is division by 0. Given that no number can be divided into 0 parts, division by 0 is undefined.
Therefore, when we talk about a fraction, where the dividend (the number being divided) is in the numerator, and the divisor (the number we divide by) is in the denominator, the restriction applies only to the denominator, which must be different from 0,
In the given expression:
As stated, the restriction applies to the fraction's denominator only,
Therefore, in order for the given expression (the fraction - in this case) to be defined, we require that the expression in its denominator - does not equal zero, in other words:
We will solve this inequality, which is a point inequality of first degree, in the same way we solve a first-degree equation, meaning - we isolate the variable on one side, by moving terms (and dividing both sides of the inequality by its coefficient if needed):
Therefore, the domain (definition domain) of the given expression is:
(This means that if we substitute for the variable x any number different fromthe expression will remain well-defined),
Therefore, the correct answer is answer A.
Note:
In a general form - solving an inequality of this form, meaning, a non-linear, but point inequality - that uses the sign and not the inequality signs: ,>,\hspace{2pt}<,\hspace{2pt}\geq,\hspace{2pt}\leq,\hspace{2pt} is identical in every way to an equation and therefore is solved in the same way and all rules used to solve an equation of any type are identical for it as well.
Choose the field of application of the following fraction:
Let's examine the given expression:
As we know, the only restriction that applies to division is division by 0, since no number can be divided into 0 parts, therefore, division by 0 is undefined.
Therefore, when we talk about a fraction, where the dividend (the number being divided) is in the numerator, and the divisor (the number we divide by) is in the denominator, the restriction applies only to the denominator, which must be different from 0,
In the given expression:
As stated, the restriction applies to the fraction's denominator only,
Therefore, in order for the given expression (the fraction - in this case) to be defined, we require that the expression in its denominator - does not equal zero, meaning we require that:
We will solve this inequality, which is a point inequality of first degree, in the same way we solve a first-degree equation, meaning - we isolate the variable on one side, by moving terms (and dividing both sides of the inequality by its coefficient if needed):
Therefore, the domain (definition domain) of the given expression is:
(This means that if we substitute any number different from for x, the expression will remain well-defined),
Therefore, the correct answer is answer C.
Note:
In a general form - solving an inequality of this form, meaning, a non-graphical, but point inequality - that uses the sign and not the slope signs: ,>,\hspace{2pt}<,\hspace{2pt}\geq,\hspace{2pt}\leq,\hspace{2pt} is identical in every way to an equation and therefore is solved in an identical way and all rules used to solve an equation of any type are identical for it as well.
What is the field of application of the equation?
To determine the field of application of the equation , we must identify values of for which the equation is defined.
Therefore, the field of application, or the domain of the equation, is all real numbers except .
We must conclude that .
Comparing with the provided choices, the correct answer is choice 3: .
Solve the following equation:
In order to solve the equation, start by removing the denominators.
To do this, we'll multiply the denominators:
Open the parentheses on the left side, making use of the distributive property:
Continue to open the parentheses on the right side of the equation:
Simplify further:
Go back and simplify the parentheses on the left side of the equation:
Combine like terms:
Notice that all terms can be divided by 9 as shown below:
Move all numbers to one side:
We obtain the following:
In order to remove the one-half coefficient, multiply the entire equation by 2
Apply the square root formula, as shown below-
Apply the properties of square roots in order to simplify the square root of 12:
Divide both the numerator and denominator by 2 as follows:
\( 2x+\frac{6}{x}=18 \)
What is the domain of the above equation?
\( 2x-3=\frac{4}{x} \)
What is the domain of the exercise?
\( 22(\frac{2}{x}-1)=30 \)
What is the domain of the equation above?
What is the domain of the exercise?
\( \frac{5x+8}{2x-6}=30 \)
Select the field of application of the following fraction:
\( 16+\frac{8}{2x} \)
What is the domain of the above equation?
x≠0
What is the domain of the exercise?
x≠0
What is the domain of the equation above?
x≠0
What is the domain of the exercise?
x≠3
Select the field of application of the following fraction:
Select the field of application of the following fraction:
\( -8+\frac{3}{x+2} \)
Solve the following equation:
\( \frac{x^3+1}{(x+1)^2}=x \)
Select the field of application of the following fraction:
All numbers except (-2)
Solve the following equation: