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3 years ago

(0,-3) and(2,-4)whats the slope

Mathematics

2 answers:

Igoryamba3 years ago

8 0

The answer is m=-1/2

Usimov [2.4K]3 years ago

4 0

So slope, aka gradient = (y₂ - y₁) ÷ (x₂ - x₁)

Let y₂ = -4; y₁ = -3; x₂ = 2; x₁ = 0 [so practically you use the coordinates given
and you assign 1 and 2 to the x and y values ; just
ensuring that 1 and 1 are from the same pair

∴ slope = ((-4) - (-3)) ÷ (2 - 0)

= 1 ÷ 2

= $\frac{1}{2}$

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A line joins the point

Harrizon [31]

Intersection of the first two lines:

$\begin{cases}5x - 2y + 3 = 0\\4x - 3y + 1 = 0\end{cases}$

Multiply the first equation by 4 and the second by 5:

$\begin{cases}20x - 8y + 12 = 0\\20x - 15y + 5 = 0\end{cases}$

Subtract the two equations:

$(20x - 8y + 12)-(20x - 15y + 5)=0 \iff 7y+7=0 \iff y=-1$

Plug this value for y in one of the equation, for example the first:

$5x - 2\cdot (-1) + 3 = 0\iff 5x+5=0 \iff x=-1$

So, the first point of intersection is $(-1,-1)$

We can find the intersection of the other two lines in the same way: we start with

$\begin{cases}x=y\\x=3y+4\end{cases}$

Use the fact that x and y are the same to rewrite the second equation as

$x=3x+4 \iff 2x=-4 \iff x=-2$

And since x and y are the same, the second point is $(-2, -2)$

So, we're looking for a line passing through $(-1,-1)$ and $(-2, -2)$ . We may use the formula to find the equation of a line knowing two of its points, but in this case it is very clear that both points have the same coordinates, so the line must be $y=x$

In the attached figure, line $5x - 2y + 3 = 0$ is light green, line $4x - 3y + 1 = 0$ is dark green, and their intersection is point A.

Simiarly, line $x=y$ is red, line $x = 3y + 4$ is orange, and their intersection is B.

As you can see, the line connecting A and B is the red line itself.

5 0

3 years ago

<img src="https://tex.z-dn.net/?f=%28x%5E%7B2%7D%20%2Bx-3%29%3A%20%28x%5E%7B2%7D%20-4%29%5Cgeq%201" id="TexFormula1" title="(x^{

Jlenok [28]

Answer:

$x>2$

Step-by-step explanation:

When given the following inequality;

$(x^2+x-3):(x^2-4)\geq1$

Rewrite in a fractional form so that it is easier to work with. Remember, a ratio is another way of expressing a fraction where the first term is the numerator (value over the fraction) and the second is the denominator(value under the fraction);

$\frac{x^2+x-3}{x^2-4}\geq1$

Now bring all of the terms to one side so that the other side is just a zero, use the idea of inverse operations to achieve this:

$\frac{x^2+x-3}{x^2-4}-1\geq0$

Convert the (1) to have the like denominator as the other term on the left side. Keep in mind, any term over itself is equal to (1);

$\frac{x^2+x-3}{x^2-4}-\frac{x^2-4}{x^2-4}\geq0$

Perform the operation on the other side distribute the negative sign and combine like terms;

$\frac{(x^2+x-3)-(x^2-4)}{x^2-4}\geq0\\\\\frac{x^2+x-3-x^2+4}{x^2-4}\geq0\\\\\frac{x+1}{x^2-4}\geq0$

Factor the equation so that one can find the intervales where the inequality is true;

$\frac{x+1}{(x-2)(x+2)}\geq0$

Solve to find the intervales when the equation is true. These intervales are the spaces between the zeros. The zeros of the inequality can be found using the zero product property (which states that any number times zero equals zero), these zeros are as follows;

$-1, 2, -2$

Therefore the intervales are the following, remember, the denominator cannot be zero, therefore some zeros are not included in the domain

$x\leq-2\\-2$

Substitute a value in these intervales to find out if the inequality is positive or negative, if it is positive then the interval is a solution, if it is negative then it is not a solution. This is because the inequality is greater than or equal to zero;

$x\leq-2$ -> negative