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3 years ago

Write the equation of the line that passes through the points (-3, - 5) and (-9,3). Express your answer in

Mathematics

1 answer:

Leokris [45]3 years ago

7 0

that's the answer

Step-by-step explanation:

$slope = \frac{y2 - y1}{x2 - x1} \\ = \frac{3 - - 5}{ - 9 - - 3} = \frac{8}{ - 6} \\ \frac{y - 3}{x - - 9} = \frac{8}{ - 6} \\ - 6(y - 3) = 8(x - - 9) \\ - 6y + 18 = 8x + 72 \\ - 6y = 8x + 72 -18 \\ - 6y = 8x + 54 \\ \frac{ - 6y}{ - 6} = \frac{8x}{ - 6} + \frac{54}{ - 6} \\ y = \frac{8x}{ - 6} - 9$

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4 years ago

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A circle has its center at the origin, and (5, -12) is a point on the circle. How long is the radius of the circle?

Amanda [17]

Option C

The radius of circle is 13 units

Solution:

The equation of circle is given by formula:

$(x-h)^2+(y-k)^2 = r^2$

Where, the center being at the point (h, k) and the radius being "r"

Given that circle has its center at the origin

(h, k) = (0, 0)

(5, -12) is a point on the circle

(x, y) = (5, 12)

Substituting in equation we get,

$(5-0)^2+(12-0)^2 = r^2\\\\5^2+12^2 = r^2\\\\r^2 = 5^2+12^2\\\\r^2 = 25 + 144\\\\r^2 = 169\\\\Taking\ square\ root\ on\ both\ sides\\\\r = \sqrt{169}\\\\r = \pm 13\\\\Since\ radius\ cannot\ be\ negative\ ignore\ r = -13\\\\Thus\ the\ solution\ is:\\\\ \r = 13$

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4 0

4 years ago

BRAINLIEST ASAP! PLEASE HELP ME :)

jenyasd209 [6]

<h3>Answers:</h3><h3>There are four solutions and they are</h3>

$\theta = 0$
$\theta = \pi$
$\theta = \frac{7\pi}{6}$ ... this says "7pi over 6"
$\theta = \frac{11\pi}{6}$ ... this says "11pi over 6"

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Work Shown:

$\sin(\theta)+1 = \cos(2\theta)\\\\\sin(\theta)+1 = 1-2\sin^2(\theta)\\\\\sin(\theta)+1-1+2\sin^2(\theta)=0\\\\2\sin^2(\theta)+\sin(\theta)=0\\\\\sin(\theta)(2\sin(\theta)+1)=0\\\\\sin(\theta)=0 \ \text{ or } \ 2\sin(\theta)+1=0\\\\$

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Solving $\sin(\theta)=0$ leads to $\theta=n\pi$ for any integer n.

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Solving $2\sin(\theta)+1=0$ leads to...

$2\sin(\theta)+1=0\\\\2\sin(\theta)=-1\\\\\sin(\theta)=-\frac{1}{2}\\\\\theta=\arcsin\left(-\frac{1}{2}\right) \ \text{ or } \ \theta=\pi-\arcsin\left(-\frac{1}{2}\right)\\\\\theta=\frac{11}{6}\pi+2\pi n \ \text{ or } \ \theta=\pi-\frac{11}{6}\pi+2\pi n\\\\\theta=\frac{11}{6}\pi+2\pi n \ \text{ or } \ \theta=\frac{-5}{6}\pi+2\pi n\\\\\theta=\frac{11}{6}\pi+2\pi n \ \text{ or } \ \theta=\frac{7}{6}\pi+2\pi n\\\\$

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The general solution set is

$\{\theta=n*\pi, \theta=\frac{11}{6}\pi+2\pi n, \theta=\frac{7}{6}\pi+2\pi n\}$

Again, n is any integer.

Let's look at a table of values where we plug in various integers for n. See the attached image below. Note the stuff in the highlighted yellow cells represents expressions that are between 0 and 2pi = 6.28

Therefore the four solutions are $\{0, \pi, \frac{7\pi}{6}, \frac{11\pi}{6}\}$ after we plug the proper values of n into the expressions, to have things match what the table shows.

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3 years ago