I keep getting this wrong and I only have one chance to get it right. Please help! It's timed aswell.

Can I get some help, it is due in 15 minutes and I need help. Thanks, any help appreciated

koban [17]

Well, I'm way past the 15 min mark, but here's how to do the question.

With this, you will need to use the distance formula, $\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$ , on XY, YZ, and ZX.

XY: $\sqrt{(3-1)^2+(1-6)^2}$

Firstly, solve inside the parentheses: $\sqrt{(2)^2+(-5)^2}$

Next, solve the exponents: $\sqrt{4+25}$

Next, solve the addition, and XY's distance will be √29

(The process is the same with the other 2 sides, so I'll go through them real quickly)

YZ:

$\sqrt{(6-3)^2+(3-1)^2}\\ \sqrt{(3)^2+(2)^2}\\ \sqrt{9+4}\\ \sqrt{13}$

ZX:

$\sqrt{(1-6)^2+(6-3)^2}\\ \sqrt{(-5)^2+(3)^2}\\ \sqrt{25+9}\\ \sqrt{34}$

Now that we got the 3 sides, we can add them up: $\sqrt{29}+\sqrt{13} +\sqrt{34} =14.8$

In short, your answer is 14.8, or the second option.

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

70*6000 associative and commutative properties

stealth61 [152]

The commutative property of multiplication is 6000*70

8 0

4 years ago

How much tax is withheld from $45,580 if the tax rate is 4%?

mel-nik [20]

Answer:

1,823.20

Step-by-step explanation:

I'm smart!

8 0

4 years ago

Adam tried to compute the average of his 7 test scores. He mistakenly divided the correct sum of all of his test scores by 6, wh

Harman [31]

Adam got the sum of his grades right, but he divided that correct figure by 6, getting 84 Before that division, the sum of his grades must have been 84 * 6, or 504. He should have divided 504 by 7, getting 72. Answer: His correct average test score is 72.

3 0

3 years ago