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3 years ago

Please Hurry!!!

Mathematics

1 answer:

Valentin [98]3 years ago

5 0

The point-slope form:

$y-y_1=m(x-x_1)$

The formula of a slope:

$m=\dfrac{y_2-y_1}{x_2-x_1}$

We have the points A(-2, 6) and B(2, -2). Substitute:

$m=\dfrac{-2-6}{2-(-2)}=\dfrac{-8}{4}=-2\\\\y-6=-2(x-(-2))\\\\\boxed{y-6=-2(x+2)}\to\boxed{(B)}$

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3 years ago

To solve a system of inequalities so you can graph it how do you change these two equations into something like the two that are

Zigmanuir [339]

Explanation

Problem #2

We must find the solution to the following system of inequalities:

$\begin{gathered} 3x-2y\leq4, \\ x+3y\leq6. \end{gathered}$

(1) We solve for y the first inequality:

$-2y\leq4-3x.$

Now, we multiply both sides of the inequality by (-1), this changes the signs on both sides and inverts the inequality symbol:

$\begin{gathered} 2y\ge-4+3x, \\ y\ge\frac{3}{2}x-2. \end{gathered}$

The solution to this inequality is the set of all the points (x, y) over the line:

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• slope m = 3/2,

• y-intercept b = -2.

(2) We solve for y the second inequality:

$\begin{gathered} x+3y\leq6, \\ 3y\leq6-x, \\ y\leq-\frac{1}{3}x+2. \end{gathered}$

The solution to this inequality is the set of all the points (x, y) below the line:

$y=-\frac{1}{3}x+2.$

This line has:

• slope m = -1/3,

• y-intercept b = 2.

(3) Plotting the lines of points (1) and (2), and painting the region:

• over the line from point (1),

• and below the line from point (2),

we get the following graph:

Answer

The points that satisfy both inequalities are given by the intersection of the blue and red regions:

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1 year ago