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3 years ago

Write the point slope form of an equation of the line through the points

Mathematics

2 answers:

denis23 [38]3 years ago

7 0

Answer:

the answer is D

how to try the values of the x and y

(x,y)

MA_775_DIABLO [31]3 years ago

4 0

Equation of a line: $y = mx + c$ , where m is the slope and c is the y-intercept. To calculate the slope, we do so using the formula: $m = \frac{y2 - y1}{x2 - x1}$

Now plugging in our values, $m = \frac{4 - (-3)}{-7 - (-2)} = \frac{4 + 3}{-7 + 2} = \frac{7}{-5} = \frac{-7}{5}$

So our slope m, is = -7/5

We can now input the value of the slope. But in this question, we cannot use the above equation of a line since we are asked to use a point-slope equation expression, so the equation is $y - y1 = m(x - x1)$

Now we plug in our values,

$y - (-3) = \frac{-7}{5}(x - (-2))\\y + 3 = \frac{-7}{5}(x + 2)$

So the answer should be option D.

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densk [106]

Ummm what? I don’t get it

7 0

3 years ago

What is The equation of a vertical line passing through the point (-5,-1)

mihalych1998 [28]

The equation of a vertical line passing through the point (-5,-1) is x = -5

3 0

3 years ago

Read 2 more answers

Here’s a graph of a linear function. Write the equation that describes that function

inessss [21]

Answer:

The equation of the line is y = 1/4x - 4

Step-by-step explanation:

In order to find this, start with two points that are on the line. We'll use (0, -4) and (4, -3). Now we can use the slope formula to find the slope.

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

m = (-4 - -3)/(0 - 4)

m = -1/-4

m = 1/4

Now that we have this, we can use that slope and a point in point-slope form. Then we solve for y to get the equation.

y - y1 = m(x - x1)

y - -4 = 1/4(x - 0)

y + 4 = 1/4x

y = 1/4x - 4

8 0

4 years ago

How do I solve: 2 sin (2x) - 2 sin x + 2√3 cos x - √3 = 0

ziro4ka [17]

Answer:

$\displaystyle x = \frac{\pi}{3} +k\, \pi$ or $\displaystyle x =- \frac{\pi}{3} +2\,k\, \pi$ , where $k$ is an integer.

There are three such angles between $0$ and $2\pi$ : $\displaystyle \frac{\pi}{3}$ , $\displaystyle \frac{2\, \pi}{3}$ , and $\displaystyle \frac{4\,\pi}{3}$ .

Step-by-step explanation:

By the double angle identity of sines:

$\sin(2\, x) = 2\, \sin x \cdot \cos x$ .

Rewrite the original equation with this identity:

$2\, (2\, \sin x \cdot \cos x) - 2\, \sin x + 2\sqrt{3}\, \cos x - \sqrt{3} = 0$ .

Note, that $2\, (2\, \sin x \cdot \cos x)$ and $(-2\, \sin x)$ share the common factor $(2\, \sin x)$ . On the other hand, $2\sqrt{3}\, \cos x$ and $(-\sqrt{3})$ share the common factor $\sqrt[3}$ . Combine these terms pairwise using the two common factors:

$(2\, \sin x) \cdot (2\, \cos x - 1) + \left(\sqrt{3}\right)\, (2\, \cos x - 1) = 0$ .

Note the new common factor $(2\, \cos x - 1)$ . Therefore:

$\left(2\, \sin x + \sqrt{3}\right) \cdot (2\, \cos x - 1) = 0$ .

This equation holds as long as either $\left(2\, \sin x + \sqrt{3}\right)$ or $(2\, \cos x - 1)$ is zero. Let $k$ be an integer. Accordingly:

$\displaystyle \sin x = -\frac{\sqrt{3}}{2}$ , which corresponds to $\displaystyle x = -\frac{\pi}{3} + 2\, k\, \pi$ and $\displaystyle x = -\frac{2\, \pi}{3} + 2\, k\, \pi$ .
$\displaystyle \cos x = \frac{1}{2}$ , which corresponds to $\displaystyle x = \frac{\pi}{3} + 2\, k \, \pi$ and $\displaystyle x = -\frac{\pi}{3} + 2\, k \, \pi$ .

Any $x$ that fits into at least one of these patterns will satisfy the equation. These pattern can be further combined:

$\displaystyle x = \frac{\pi}{3} + k \, \pi$ (from $\displaystyle x = -\frac{2\,\pi}{3} + 2\, k\, \pi$ and $\displaystyle x = \frac{\pi}{3} + 2\, k \, \pi$ , combined,) as well as
$\displaystyle x =- \frac{\pi}{3} +2\,k\, \pi$ .

7 0

3 years ago

A system of equations with a linear and quadratic equation must have one solution.

Ierofanga [76]

That is false because, this type of system can have one solution, two solutions, or no solutions. Graph both equations on the same coordinate plane. Identify the point of intersection, if any.
Hope I Helped : )

8 0

3 years ago

Read 2 more answers