Question

(11-x)(1,5+x)<0 = help please

Answer 1

Объяснение:

)))))))))))))))))))))

Answer 2

Ответ:

как-то так. если что-то не понятно, пиши

Answer 3

To solve the inequality $(11 - x)(1.5 + x) < 0$, you can use the concept of interval notation and the properties of the factors involved.

First, you need to find the critical values of x where the expression $(11 - x)(1.5 + x)$ changes sign. These critical values are the values of x where the factors equal zero, because that's where the sign changes.

1. Set each factor equal to zero and solve for x: $(11 - x) = 0$ ⇒ $x = 11$ $(1.5 + x) = 0$ ⇒ $x = -1.5$

So, the critical values are $x = 11$ and $x = -1.5$.

Now, you can test the regions between and outside these critical values to determine where the inequality is satisfied:

• When $x < -1.5$, both factors $(11 - x)$ and $(1.5 + x)$ are negative, so the product $(11 - x)(1.5 + x)$ is positive. Thus, the inequality is not satisfied in this region.

• When $-1.5 < x < 11$, the factor $(11 - x)$ is positive, and the factor $(1.5 + x)$ is negative. A positive times a negative is negative, so the inequality is satisfied in this region.

• When $x > 11$, both factors $(11 - x)$ and $(1.5 + x)$ are positive, so the product $(11 - x)(1.5 + x)$ is positive. The inequality is not satisfied in this region.

To express this in interval notation, the solution to the inequality is: $x \in (-1.5, 11)$

So, the values of x that make the inequality $(11 - x)(1.5 + x) < 0$ true are in the open interval $(-1.5, 11)$.