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No. This is not an inequality, because you need something > something_else, or less than or 'not equal' or 'greater than or equal', etc. Since it has an x cubed term, it is not a quadratic.
In a linear inequality the variable is only present raised to the first power (which is usually not explicitly shown). In a quadratic the square of the variable is present (or implied). The square can be implied in an inequality such as x + 1/x < 6 (x not 0) This is equivalent to x2 - 6x + 1 < 0
The answer depends on the nature of the inequality: whether it is linear, quadratic or has some other functional form.
Use the quadratic formula for the equality. Then, depending on the coefficient of x2 and the nature of the inequality [>, ≥, ≤, <], determine whether you need the open or closed intervals between the roots or beyond the roots.
A linear inequality is all of one side of a plane. A quadratic inequality is either the inside of a parabola or the outside.
Substitute the values of the variables into the inequality. If the inequality is true then they are a solution, if not, they are not.Substitute the values of the variables into the inequality. If the inequality is true then they are a solution, if not, they are not.Substitute the values of the variables into the inequality. If the inequality is true then they are a solution, if not, they are not.Substitute the values of the variables into the inequality. If the inequality is true then they are a solution, if not, they are not.
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A bivariate linear inequality.
No. This is not an inequality, because you need something > something_else, or less than or 'not equal' or 'greater than or equal', etc. Since it has an x cubed term, it is not a quadratic.
In a linear inequality the variable is only present raised to the first power (which is usually not explicitly shown). In a quadratic the square of the variable is present (or implied). The square can be implied in an inequality such as x + 1/x < 6 (x not 0) This is equivalent to x2 - 6x + 1 < 0
If the equal sign in a linear equation in two variables is replaced with an inequality symbol, the result is a linear inequality in two variables. 3x-2y>7 x<-5
The answer depends on the nature of the inequality: whether it is linear, quadratic or has some other functional form.
Use the quadratic formula for the equality. Then, depending on the coefficient of x2 and the nature of the inequality [>, ≥, ≤, <], determine whether you need the open or closed intervals between the roots or beyond the roots.
If you use a variable, or variables, with an equation, or with an inequality, it is neither true nor false until you replace the variables with specific values.
a2b2 - 2ab - 25 is a quadratic expression in the variables ab. There is no equation or inequality in the question so there is nothing that can be solved. Because of the nature of the expression a and b cannot be separated in any meaningful way.
Yes, and no. The solution set to an inequality are those points which satisfy the inequality. A linear inequality is one in which no variable has a power greater than 1. Only if there are two variables will the solution be points in a plane; if there are more than two variables then the solution set will be points in a higher space, for example the solution set to the linear inequality x + y + z < 1 is a set of points in three dimensional space.