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3 years ago

write an equation in slope-intercept form for the line that jas a alope of 12 and passes through the point (3,20)

Mathematics

1 answer:

Ierofanga [76]3 years ago

3 0

Answer:

y = 12x - 16

Step-by-step explanation:

The equation of a line in slope- intercept form is

y = mx + c ( m is the slope and c the y- intercept )

here m = 12, hence

y = 12x + c ← is the partial equation

To find c substitute (3, 20) into the partial equation

20 = 36 + c ⇒ c = 20 - 36 = - 16

y = 12x - 16 ← equation of line in slope- intercept form

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5 pts

ruslelena [56]

Answer:

{8, 24, 72, 216, 648}

Step-by-step explanation:

Geometric sequence:

In a geometric sequence, the quotient of consecutive terms is always the same, that is, each term is the previous term multiplied by the common ratio.

In this question:

First element is 8, common ratio of 3. So

Second term: 8*3 = 24

Third term: 24*3 = 72

Fourth term: 72*3 = 216

Fifth term: 216*3 = 648

So the answer is {8, 24, 72, 216, 648}

4 0

3 years ago

Find the midpoint of the line segment joining the points (5,6) and (-1,-6).

kumpel [21]

Answer:

The midpoint is (2,0)

Step-by-step explanation:

To find the x coordinate of the midpoint, add the x coordinates of the endpoint and divide by 2

(5+-1)/2 =4/2 =2

To find the y coordinate of the midpoint, add the y coordinates of the endpoint and divide by 2

(6+-6)/2 =0/2 =0

The midpoint is (2,0)

8 0

3 years ago

Consider the three points ( 1 , 3 ) , ( 2 , 3 ) and ( 3 , 6 ) . Let ¯ x be the average x-coordinate of these points, and let ¯ y

loris [4]

Answer:

$m=\dfrac{3}{2}$

Step-by-step explanation:

Given points are: ( 1 , 3 ) , ( 2 , 3 ) and ( 3 , 6 )

The average of x-coordinate will be:

$\overline{x} = \dfrac{x_1+x_2+x_3}{\text{number of points}}$

1) Finding $(\overline{x},\overline{y})$

Average of the x coordinates:

$\overline{x} = \dfrac{1+2+3}{3}$

$\overline{x} = 2$

Average of the y coordinates:

similarly for y

$\overline{y} = \dfrac{3+3+6}{3}$

$\overline{y} = 4$

2) Finding the line through $(\overline{x},\overline{y})$ with slope m.

Given a point and a slope, the equation of a line can be found using:

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

in our case this will be

$(y-\overline{y})=m(x-\overline{x})$

$(y-4)=m(x-2)$

$y=mx-2m+4$

this is our equation of the line!

3) Find the squared vertical distances between this line and the three points.

So what we up till now is a line, and three points. We need to find how much further away (only in the y direction) each point is from the line.

Distance from point (1,3)

We know that when x=1, y=3 for the point. But we need to find what does y equal when x=1 for the line?

we'll go back to our equation of the line and use x=1.

$y=m(1)-2m+4$

$y=-m+4$

now we know the two points at x=1: (1,3) and (1,-m+4)

to find the vertical distance we'll subtract the y-coordinates of each point.

$d_1=3-(-m+4)$

$d_1=m-1$

finally, as asked, we'll square the distance

$(d_1)^2=(m-1)^2$

Distance from point (2,3)

we'll do the same as above here:

$y=m(2)-2m+4$

$y=4$

vertical distance between the two points: (2,3) and (2,4)

$d_2=3-4$

$d_2=-1$

squaring:

$(d_2)^2=1$

Distance from point (3,6)

$y=m(3)-2m+4$

$y=m+4$

vertical distance between the two points: (3,6) and (3,m+4)

$d_3=6-(m+4)$

$d_3=2-m$

squaring:

$(d_3)^2=(2-m)^2$

3) Add up all the squared distances, we'll call this value R.

$R=(d_1)^2+(d_2)^2+(d_3)^2$

$R=(m-1)^2+4+(2-m)^2$

4) Find the value of m that makes R minimum.

Looking at the equation above, we can tell that R is a function of m:

$R(m)=(m-1)^2+4+(2-m)^2$

you can simplify this if you want to. What we're most concerned with is to find the minimum value of R at some value of m. To do that we'll need to derivate R with respect to m. (this is similar to finding the stationary point of a curve)

$\dfrac{d}{dm}\left(R(m)\right)=\dfrac{d}{dm}\left((m-1)^2+4+(2-m)^2\right)$