Question

(14.18. 1) y = (x - 2)2 – 4; 3) y = 0,1(x - 1)2 - 1. 2​

Answer 1

Ответ:

1)4 2)24

Объяснение:

1)0=(x-2)×2-4

0=22x-4-4

0=2x-8

x=4

2)0=0,1(x-12)-1,2

0=0.1x-1,2-1,2

0=0,11x-2,4

-0,1x=-2,4

x=24

Answer 2

Ответ:

Система линейных уравнений с двумя неизвестными

x + y = 5

2x - 3y = 1

Система линейных ур-ний с тремя переменными

2*x = 2

5*y = 10

x + y + z = 3

Система дробно-рациональных уравнений

x + y = 3

1/x + 1/y = 2/5

Система четырёх уравнений

x1 + 2x2 + 3x3 - 2x4 = 1

2x1 - x2 - 2x3 - 3x4 = 2

3x1 + 2x2 - x3 + 2x4 = -5

2x1 - 3x2 + 2x3 + x4 = 11

Система линейных уравнений с четырьмя неизвестными

2x + 4y + 6z + 8v = 100

3x + 5y + 7z + 9v = 116

3x - 5y + 7z - 9v = -40

-2x + 4y - 6z + 8v = 36

Система трёх нелинейных ур-ний, содержащая квадрат и дробь

2/x = 11

x - 3*z^2 = 0

2/7*x + y - z = -3

Система двух ур-ний, содержащая куб (3-ю степень)

x = y^3

x*y = -5

Система ур-ний c квадратным корнем

x + y - sqrt(x*y) = 5

2*x*y = 3

Система тригонометрических ур-ний

x + y = 5*pi/2

sin(x) + cos(2y) = -1

Система показательных и логарифмических уравнений

y - log(x)/log(3) = 1

x^y = 3^12

Объяснение:

Answer 3

It seems like you have provided two quadratic equations in different forms. I'll help you analyze and work with each of them:

1. $y = (x - 2)^2 - 4$:

   This equation represents a parabola with its vertex at the point $(2, -4)$. The vertex form of a quadratic equation is given by $y = a(x - h)^2 + k$, where $(h, k)$ is the vertex of the parabola.

   In this case:

   • $a = 1$
   • $h = 2$
   • $k = -4$

   So, the vertex of this parabola is $(2, -4)$, and it opens upward because the coefficient of $x^2$ is positive (1). If you want to find additional information, such as the x-intercepts (where the graph crosses the x-axis), you can set $y = 0$ and solve for $x$: $0 = (x - 2)^2 - 4$

   Now, solve for $x$:

   $(x - 2)^2 = 4$

   Taking the square root of both sides:

   $x - 2 = ±2$

   Solve for $x$:

   $x = 2 + 2 = 4$ or $x = 2 - 2 = 0$

   So, the x-intercepts are $x = 4$ and $x = 0$.

2. $y = 0.1(x - 1)^2 - 1$:

   This equation also represents a parabola, but it's a bit different from the previous one. This equation has its vertex at $(1, -1)$.

   In this case:

   • $a = 0.1$
   • $h = 1$
   • $k = -1$

   So, the vertex of this parabola is $(1, -1)$, and it opens upward because the coefficient of $x^2$ is positive (0.1).

   To find the x-intercepts, set $y = 0$ and solve for $x$:

   $0 = 0.1(x - 1)^2 - 1$

   Add 1 to both sides:

   $1 = 0.1(x - 1)^2$

   Divide by 0.1:

   $10 = (x - 1)^2$

   Taking the square root of both sides:

   $±√10 = x - 1$

   Solve for $x$:

   $x = 1 ± √10$

   So, the x-intercepts are $x = 1 + √10$ and $x = 1 - √10$.

These are the key characteristics and solutions for the two quadratic equations you provided. If you have any further questions or need more information, please feel free to ask!