Question

16х²-8х+1≥0с рисунком, пожалуйстадаю 70 баллов​

Answer 1

Ответ:

объясню без рисунка

Объяснение:

16х²-8х+1≥0

(4x-1)² ≥0

xєR

Answer 2

Ответ:

x ∈ {0.25}

Объяснение:

Для начала решим неравенство (с помощью дискриминанта):

16х² - 8х + 1 ≤ 0

D = 64 - 4 * 16 * 1 = 64 - 64 = 0

Значит у неравенства только один корень (так как дискриминант равен нулю)

х = $\frac{8}{16*2} = 0.25$

x ∈ {0.25}

(Смотреть вложение). Обозначим точку 0.25 закрашенной на координатной прямой. Закрашенной так как знак (≤) меньше или равно. Если б было строго меньше или строго больше - точка была б не закрашенной (выколотой).

Answer 3

Quadratic Inequality: 16x² - 8x + 1 ≥ 0

To solve the quadratic inequality 16x² - 8x + 1 ≥ 0, we can use the method of graphing the related quadratic equation to determine the intervals where the inequality is satisfied.

First, let's find the solutions to the related quadratic equation 16x² - 8x + 1 = 0. We can use the quadratic formula to find the roots of the equation.

The quadratic formula states that for an equation of the form ax² + bx + c = 0, the solutions for x are given by:

x = (-b ± √(b² - 4ac)) / (2a)

In this case, for the equation 16x² - 8x + 1 = 0, the coefficients are: a = 16, b = -8, and c = 1.

Using the quadratic formula, we can find the solutions for x.

x = (-(-8) ± √((-8)² - 4*16*1)) / (2*16)

x = (8 ± √(64 - 64)) / 32

x = (8 ± √0) / 32

x = (8 ± 0) / 32

x = 8 / 32 = 1/4

So, the equation 16x² - 8x + 1 = 0 has a repeated root at x = 1/4.

Now, let's plot the graph of the related quadratic equation y = 16x² - 8x + 1 and analyze the intervals where the inequality 16x² - 8x + 1 ≥ 0 is satisfied.

The graph of the quadratic equation y = 16x² - 8x + 1 is a parabola. Since the coefficient of x² is positive (16 > 0), the parabola opens upwards.

The vertex of the parabola can be found using the formula: x = -b / (2a)

In this case, the x-coordinate of the vertex is: x = -(-8) / (2*16) = 8 / 32 = 1/4

So, the vertex of the parabola is at (1/4, f(1/4)), where f(1/4) is the value of the quadratic function at x = 1/4.

Since the coefficient of x² is positive, the parabola opens upwards, and the vertex represents the minimum point of the parabola.

Therefore, the inequality 16x² - 8x + 1 ≥ 0 is satisfied for all values of x in the interval around the vertex.

Conclusion: The solution to the inequality 16x² - 8x + 1 ≥ 0 is all real numbers x such that x belongs to the interval around the vertex of the parabola.

I hope this helps! If you have further questions or need additional assistance, feel free to ask!