Question

2)x² + x - 6<0,x + x - 220.​

Answer 1

Ответ:

Ответ 12

Объяснение:

Answer 2

It looks like you have two inequalities:

1. $x^2 + x - 6 < 0$
2. $x + x - 220$

Let's solve each one separately:

1. $x^2 + x - 6 < 0$:

To solve this inequality, first factor the quadratic expression:

$x^2 + x - 6 = (x + 3)(x - 2)$

Now, you want to find the values of $x$ that make this expression less than zero. To do this, you can use a sign chart or test values within the intervals created by the roots of the equation:

The roots are $x = -3$ and $x = 2$. So, create three intervals: $(-∞, -3)$, $(-3, 2)$, and $(2, ∞)$.

1. Test a value in the first interval, for example, $x = -4$: $((-4 + 3)(-4 - 2)) < 0$ (negative)

2. Test a value in the second interval, for example, $x = 0$: $((0 + 3)(0 - 2)) < 0$ (positive)

3. Test a value in the third interval, for example, $x = 3$: $((3 + 3)(3 - 2)) < 0$ (negative)

So, the solution to $x^2 + x - 6 < 0$ is $x \in (-3, 2)$.

2. $x + x - 220$:

Simplify the expression:

$x + x - 220 = 2x - 220$

Now, you want to find the values of $x$ that satisfy this equation. To do that, set it equal to zero and solve for $x$:

$2x - 220 = 0$

Add 220 to both sides:

$2x = 220$

Divide both sides by 2:

$x = 110$

So, the solution to $x + x - 220$ is $x = 110$.

If you meant to have inequalities for the second expression as well, please provide them, and I can help you further.