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3 years ago

5

the radius of the Earth is 6378.1 KM which is 2981 .1 km longer than the radius of Mars write and solve the equation to determin

e the radius of Mars show that your answer is reasonable

2 answers:

Bad White [126]3 years ago

7 0

$6,378.1 - 2,981.1

= 6,000 + 300 + 70 + 8 + 1/10 - 2,000 - 900 - 80 - 1 - 1/10

= (6,000 - 2,000) + (300 - 900) + (70 - 80) + (8-1) + (1/10-1/10)

= 4,000 - 600 - 10 + 7 + 0

= 3,400 - 3 

= 3,397$

Rasek [7]3 years ago

6 0

576.3
this is the answer to your question

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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:

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

<u>1) Finding $(\overline{x},\overline{y})$ </u>

Average of the x coordinates:

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

$\overline{x} = 2$

Average of the y coordinates:

similarly for y

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

$\overline{y} = 4$

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

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!

<u>3) Find the squared vertical distances between this line and the three points.</u>

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.

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

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

<u>4) Find the value of m that makes R minimum.</u>

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:

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

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

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2 years ago

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expeople1 [14]

Answer:

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Step-by-step explanation:

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$-0.8 + r = 12.6$ (given)

Add 0.8 to both sides of the equation (addition property of equality)

$-0.8 + r + 0.8 = 12.6 + 0.8$ (this is where the error occurred.)

$r = 13.4$

40. The correct steps to take in solving the equation, $-\frac{m}{3} = -4$ is as follows:

$-\frac{m}{3} = -4$ (given)

Multiply both sides by 3 (multiplication property of equality)

$3*-\frac{m}{3} = 3*(-4)$

$-m = -12$ (this is where the error occurred. This is what we should have at this line/step)

$m = 12$ (dividing both sides by -1)

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3 years ago