NO LINKS PLEASE help

Mathematics

1 answer:

liubo4ka [24]2 years ago

7 0

Answer:

Midpoint for A is (1,5) and for B it is (2.5,4.5)

Step-by-step explanation:

To get the midpoint you add the x coordinates and divide by 2 and then the y coordinates and divide by 2.

X = (3 + -1)/2 y = (8+ 2)/2

=2/2 = 1 10/2=5

You might be interested in

What is the factoring quadratic expression for 8x^2-18=0

mash [69]

$8x^2-18=0\ /:2\\\\4x^2-9=0\ \ \ \Leftrightarrow\ \ \ (2x)^2-3^2=0\ \ \ \Leftrightarrow\ \ \ (2x-3)(2x+3)=0\\\\2x-3=0\ \ \ \ \ \ \ or\ \ \ \ 2x+3=0\\\\2x=3\ /:2\ \ \ \ \ \ or\ \ \ \ 2x=-3\ /:2\\\\x=1.5\ \ \ \ \ \ \ \ \ \ \ \ or\ \ \ \ x=-1.5$

3 0

2 years ago

Read 2 more answers

Who know how to do this

koban [17]

A(-6,4) and E(6,0)
slope m = (4 - 0) / ( -6-6) = 4/-12 = -1/3

perpendicular, slope is opposite and reciprocal so slope = 3

A perpendicular bisector of a line segment is a line segment perpendicular to and passing through the midpoint

Midpoint:
so x = ((-6+6)/2 = 0 and y = (4+0)/ 2 = 2
so M = (0,2)

equation of perpendicular line: y = 3x + 2

answer:
equation: y = 3x + 2
point lies on perpendicular bisector: (0,2)

5 0

3 years ago

Find the sum \[\sum_{k = 1}^{2004} \frac{1}{1 + \tan^2 (\frac{k \pi}{2 \cdot 2005})}.\]

vampirchik [111]

Answer:1002

Step-by-step explanation:

$\Rightarrow \sum_{k=1}^{2004}\left ( \frac{1}{1+\tan ^2\left ( \frac{k\pi }{2\cdot 2005}\right )}\right )$

and $1+\tan ^2\theta =\sec^2\theta$

and $\cos \theta =\frac{1}{\sec \theta }$

$\Rightarrow \sum_{k=1}^{2004}\left ( \cos^2\frac{k\pi }{2\cdot 2005}\right )$

as $\cos ^2\theta =\cos ^2(\pi -\theta )$

Applying this we get

$\Rightarrow \sum_{1}^{1002}\left [ \cos^2\left ( \frac{k\pi }{2\cdot 2005}\right )+\cos^2\left ( \frac{(2005-k)\pi }{2\cdot 2005}\right )\right ]$

every $\theta$ there exist $\pi -\theta$

such that $\sin^2\theta +\cos^2\theta =1$

therefore

$\Rightarrow \sum_{1}^{1002}\left [ \cos^2\left ( \frac{k\pi }{2\cdot 2005}\right )+\cos^2\left ( \frac{(2005-k)\pi }{2\cdot 2005}\right )\right ]=1002$