Question

3.График функции у = kx+в пересекает оси координат в точках А(0;-6) и В(3;0).Найдите значения кив. = 4.Решите систему уравнений 2х + y = 3 3х – 5 y = 37 5.Решите задачу: Найдите четыре последовательных натуральных числа таких, что произведение третьего и четвертого из этих чисел на 22 больше произведения первого и второго.​

Answer 1

Ответ:

y=kx+b

{-6=k*0+b                     {b=-6                {b=-6

{0=k*3+b                      {3k=6                {k=2

y=2x-6    

Пошаговое объяснение:

Answer 2

3. График функции y = kx пересекает оси координат в точках А(0;-6) и В(3;0). Найдите значения k.

To find the value of k, we can use the information given in the problem.

First, let's find the slope of the line passing through points A(0,-6) and B(3,0). The slope of a line is given by the formula:

m = (y2 - y1) / (x2 - x1)

Using the values from the points A(0,-6) and B(3,0), we have:

m = (0 - (-6)) / (3 - 0) m = 6 / 3 m = 2

The slope of the line is 2, which means k (the coefficient of x) is equal to 2.

Therefore, the value of k is 2.

4. Решите систему уравнений 2x + y = 3 и 3x – 5y = 37.

To solve the system of equations, we can use the method of substitution or elimination.

Let's solve the system of equations using the method of substitution:

Step 1: Solve the first equation for y: 2x + y = 3 y = 3 - 2x

Step 2: Substitute the value of y in the second equation: 3x - 5(3 - 2x) = 37

Step 3: Simplify and solve for x: 3x - 15 + 10x = 37 13x - 15 = 37 13x = 37 + 15 13x = 52 x = 52 / 13 x = 4

Step 4: Substitute the value of x back into the first equation to find y: 2(4) + y = 3 8 + y = 3 y = 3 - 8 y = -5

Therefore, the solution to the system of equations is x = 4 and y = -5.

5. Решите задачу: Найдите четыре последовательных натуральных числа таких, что произведение третьего и четвертого из этих чисел на 22 больше произведения первого и второго.

To solve the problem, let's assume the first natural number is n.

The four consecutive natural numbers would be n, n+1, n+2, and n+3.

According to the problem, the product of the third and fourth numbers, multiplied by 22, is greater than the product of the first and second numbers.

This can be expressed as: (n+2)(n+3) * 22 > n(n+1)

Expanding and simplifying the inequality: 22n^2 + 88n + 66 > n^2 + n

Subtracting n^2 + n from both sides: 21n^2 + 87n + 66 > 0

Now, we can solve this quadratic inequality by factoring or using the quadratic formula.

The factors of 21n^2 + 87n + 66 are (3n + 11)(7n + 6).

Setting each factor greater than zero and solving for n: 3n + 11 > 0 n > -11/3

7n + 6 > 0 n > -6/7

Since we are looking for natural numbers, the smallest possible value for n is 1.

Therefore, the four consecutive natural numbers that satisfy the condition are 1, 2, 3, and 4.