Question

задати формулою лінійну функцію, якщо її кутовий коефіцієнт дорівнює 2/5 і графік проходить через точку В(-10;-4) БУДЬ ЛАСКА ПОМОЖІТЬ

Answer 1

Ответ:

уравнение прямой у=kx+b

k=$\frac{2}{5}$

B(-10;-4)

$-4 = -\frac{2}{5}*10+b\\ -4 = -4+b\\b=0$

тогда уравнение прямой

$y=\frac{2}{5}x$

Answer 2

Formulating the Linear Function

To formulate the linear function with a given slope and a point it passes through, we can use the point-slope form of a linear equation. The point-slope form is given by:

y - y₁ = m(x - x₁)

Where: - (x₁, y₁) is the given point through which the graph passes. - m is the slope of the line.

Applying the Given Information

In this case, the slope (kutoviy koeffitsiyent) of the linear function is given as 2/5, and the function passes through the point B(-10, -4).

Formulating the Linear Function

Using the point-slope form, we can substitute the given values into the equation to find the linear function:

y - (-4) = (2/5)(x - (-10))

Simplifying this equation gives us the linear function in the form of y = mx + b, where m is the slope and b is the y-intercept.

Calculating the Linear Function

Let's calculate the linear function using the given information: - y + 4 = (2/5)(x + 10)

Solving for y gives us the linear function: - y = (2/5)x + 8 - 4 - y = (2/5)x + 4

So, the linear function with a slope of 2/5 and passing through the point B(-10, -4) is y = (2/5)x + 4.