Mark draw two lines that form a right angle what words describe the lines mark Drew

plz help will pick brainlest Douglas is placing strings of lights around the edge of a circular patio with a diameter of 7.5 met

andrew11 [14]

I think the answer is 5, if you divide 7.5 by 1.45 you get 5.2 or 5.17 so approximately 5.

7 0

3 years ago

Read 2 more answers

Center: (10,-10)

vova2212 [387]

This is the standard form equation $\frac{(x - 10)^2}{100} + \frac{(y + 10)^2}{-100} = 1$

What is the ellipse?

The equation for an ellipse is typically written as x² a² + y² b² = 1. x² a² + y² b² = 1. An ellipse with its origin at the center is defined by this equation. The ellipse is stretched further in both the horizontal and vertical directions if a > b, a > b, and if b > a, b > a, respectively.

The standard form of the equation of an ellipse with center (h, k)and major axis parallel to the x-axis is:

$\frac{(x - h)^2}{a^2} + \frac{(y - k)^2}{b^2} = 1$

where,

a > b

the length of the major axis is 2a

the coordinates of the vertices are (h±a,k)

the length of the minor axis is 2b

the coordinates of the co-vertices are (h,k±b)

the coordinates of the foci are (h±c,k),

where c² = a² − b².

so,

$\frac{(x - 10)^2}{100} + \frac{(y + 10)^2}{-100} = 1$

Hence, this is the standard form equation $\frac{(x - 10)^2}{100} + \frac{(y + 10)^2}{-100} = 1$ .

To learn more about ellipse, visit

brainly.com/question/16904744

#SPJ1

7 0

1 year ago

What's 221 rounded to the nearest 10

ycow [4]

It would be 220 rounded to the nearest ten because the number in the ones place does not equal or exceed 5.

So 220 is your answer :)

4 0

4 years ago

erica earned 30,000 bonus points on her computer assignment this is ten times as many points as she earned last week how many bo

alisha [4.7K]

If last week she made x points, and this week she made 30, 000
then 30 000 = 10x (solve for x)
30 000/10 = x
so she made 3000 points last week

3 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)

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