The scale on a map is 1:200000. What is the distance on the map in cm for a real distance of:

Hatshy [7]

The answer to your equation is 1008

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)

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

$\dfrac{dR}{dm}=2(m-1)+0+2(2-m)(-1)$

now to find the minimum value we'll just use a condition that $\dfrac{dR}{dm}=0$

$0=2(m-1)+2(2-m)(-1)$

now solve for m:

$0=2m-2-4+2m$

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

This is the value of m for which the sum of the squared vertical distances from the points and the line is small as possible!

5 0

3 years ago

Which pair shows equivalent expressions?

maria [59]

The equivalent expression is A

7 0

2 years ago

90% of what number is 57

aliina [53]

To solve this, set up the equation, 90% of x being .9*x=57. Then, solve for x, x=57/.9= $6 \frac{1}{3}$ =6.333

4 0

3 years ago

The length of a rectangular fenced enclosure is 12 feet more than the width. If Farmer Dan has 100 feet of fencing, write an ine

Kitty [74]

Answer:
length = 31 ft
width = 19 ft

Explanation:
Assume that the width of the rectangle is w.
We are given that the length is 12 ft more than the width. This means that the length of the rectangle is w + 12

The perimeter of the rectangle in this case would be:
perimeter = 2 (length + width)
perimeter = 2 (12 + w + w)
perimeter = 24 + 4w

Assume that Dan would use all 100 ft of fencing to surround the yard. This would mean that the largest perimeter is 100 ft.

Therefore:
perimeter = 24 + 4w
100 = 24 + 4w
4w = 100 - 24
4w = 76
w = 19 ft

Since we have calculated that the width of the yard is 19 ft, we can substitute to get the length as follows:
length = 12 + w
length = 12 + 19
length = 31 ft

Hope this helps :)

7 0

4 years ago

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