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3 years ago

What would the equation be of the form y=mx

Mathematics

1 answer:

Ronch [10]3 years ago

4 0

Answer:

y = 1.5x

Step-by-step explanation:

y = mx

slope formula: $\frac{y2-y1}{x2-x1}$

Points: (x1, y1) = (-2, -3)

(x2, y2) = (4, 6)

To find the equation in y = mx form, we need to find the value of m. m is also known as the slope. To find the value of m, input the given points into the slope formula:

$\frac{6-(-3)}{4-(-2)}$

Simplify:

6 - (-3) = 6 + 3 = 9

4 - (-2) = 4 + 2 = 6

$\frac{9}{6} =\frac{3}{2}=1.5$

The value of m is 1.5. Now we can write the equation:

y = 1.5x

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50 POINTS SOLVE !! ::

satela [25.4K]

Answer:

Step-by-step explanation:

<u>k - 4</u> = 3

k/9 = 3(9) + 4

k = 31/9

k = 31/9

<u>8 + b </u> = -5

-4

8 + b = -5 (-4)

8 + b = 20

b = 20 - 8

b = 12

4 0

3 years ago

Find the area of the region that lies inside the first curve and outside the second curve.

marishachu [46]

Answer:

Step-by-step explanation:

From the given information:

r = 10 cos( θ)

r = 5

We are to find the the area of the region that lies inside the first curve and outside the second curve.

The first thing we need to do is to determine the intersection of the points in these two curves.

To do that :

let equate the two parameters together

So;

10 cos( θ) = 5

cos( θ) = $\dfrac{1}{2}$

$\theta = -\dfrac{\pi}{3}, \ \ \dfrac{\pi}{3}$

Now, the area of the region that lies inside the first curve and outside the second curve can be determined by finding the integral . i.e

$A = \dfrac{1}{2} \int \limits^{\dfrac{\pi}{3}}_{-\dfrac{\pi}{3}} (10 \ cos \ \theta)^2 d \theta - \dfrac{1}{2} \int \limits^{\dfrac{\pi}{3}}_{-\dfrac{\pi}{3}} \ \ 5^2 d \theta$

$A = \dfrac{1}{2} \int \limits^{\dfrac{\pi}{3}}_{-\dfrac{\pi}{3}} 100 \ cos^2 \ \theta d \theta - \dfrac{25}{2} \int \limits^{\dfrac{\pi}{3}}_{-\dfrac{\pi}{3}} \ \ d \theta$

$A = 50 \int \limits^{\dfrac{\pi}{3}}_{-\dfrac{\pi}{3}} \begin {pmatrix} \dfrac{cos \ 2 \theta +1}{2} \end {pmatrix} \ \ d \theta - \dfrac{25}{2} \begin {bmatrix} \theta \end {bmatrix}^{\dfrac{\pi}{3}}_{-\dfrac{\pi}{3}}$

$A =\dfrac{ 50}{2} \int \limits^{\dfrac{\pi}{3}}_{-\dfrac{\pi}{3}} \begin {pmatrix} {cos \ 2 \theta +1} \end {pmatrix} \ \ d \theta - \dfrac{25}{2} \begin {bmatrix} \dfrac{\pi}{3} - (- \dfrac{\pi}{3} )\end {bmatrix}$

$A =25 \begin {bmatrix} \dfrac{sin2 \theta }{2} + \theta \end {bmatrix}^{\dfrac{\pi}{3}}_{\dfrac{\pi}{3}} \ \ - \dfrac{25}{2} \begin {bmatrix} \dfrac{2 \pi}{3} \end {bmatrix}$

$A =25 \begin {bmatrix} \dfrac{sin (\dfrac{2 \pi}{3} )}{2}+\dfrac{\pi}{3} - \dfrac{ sin (\dfrac{-2\pi}{3}) }{2}-(-\dfrac{\pi}{3}) \end {bmatrix} - \dfrac{25 \pi}{3}$

$A = 25 \begin{bmatrix} \dfrac{\dfrac{\sqrt{3}}{2} }{2} +\dfrac{\pi}{3} + \dfrac{\dfrac{\sqrt{3}}{2} }{2} + \dfrac{\pi}{3} \end {bmatrix}- \dfrac{ 25 \pi}{3}$

$A = 25 \begin{bmatrix} \dfrac{\sqrt{3}}{2 } +\dfrac{2 \pi}{3} \end {bmatrix}- \dfrac{ 25 \pi}{3}$

$A = \dfrac{25 \sqrt{3}}{2 } +\dfrac{25 \pi}{3}$

The diagrammatic expression showing the area of the region that lies inside the first curve and outside the second curve can be seen in the attached file below.

Download docx

7 0

3 years ago

Quadrilateral BCDE is a square. What is mZDCF?

o-na [289]

Answer:

4 squares make up a triangle so divide the degrees if the square by the triangle.

7 0

3 years ago

Why is every prime number greater than 2 an odd number?

Arlecino [84]

Because if it was even, then it would be divisble by 2 which makes it not prime

4 0

3 years ago

Read 2 more answers

(NEED DONE SOON) The measure of two supplementary angles are 4x - 24 and 4(2x - 3). <br> What is x

Zarrin [17]

Answer:

x = 12

Step-by-step explanation:

A pair of supplementary angles added together MUST equal 180. Therefore, as you are given two angles with the measures of (4x - 24) and 4(2x - 3), they must add together to equal 180. So, we get this equation:

(4x - 24) + 4(2x - 3) = 180

4x - 24 + 8x - 12 = 180 (we distributed 4(2x-3))

12x - 36 = 180

12x = 144

x = 12, -12

Therefore, x is 12 because -12 would mean that (4x-24) and 4(2x-3) are both negative, which is impossible for an angle measure.

5 0

2 years ago

What would the equation be of the form y=mx​

What would the equation be of the form y=mx