A line segment has endpoints A(4, 8) and B(2, 10). The point M is the midpoint of AB. What is an equation of a line perpendicular to AB and passing through M?

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- anonymous

- chestercat

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- anonymous

ok you need to find the slope of the line

- anonymous

okay

- anonymous

use this formula
\[\frac{ y _{2}-y _{1} }{ x _{2}-x _{1} } = m\]

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## More answers

- anonymous

ok

- anonymous

I got y=-x+4

- anonymous

ok m is -1

- anonymous

okay

- anonymous

instead of 4 would it be

- anonymous

y=-x+c

- anonymous

plug in 1 of the point to the equation like this 8 = -1(4) +c
Solve for c

- anonymous

i got 3

- anonymous

i got -4

- anonymous

umm c should be 12
8 = -4 +12

- anonymous

okay

- anonymous

now we find the midpoint of this line with this formula
\[(\frac{ x _{1}+x _{2} }{ 2 } , \frac{ y _{1}+y _{2} }{ 2 })\]

- anonymous

okay

- anonymous

your midpoint should be (3,9)

- anonymous

okay

- anonymous

a. y = x + 4
b. y = âˆ’x + 4
c. y = âˆ’x + 6
d. y = x + 6
this is what I have to chose from

- anonymous

gradient of perpendicular line is
\[m _{1}m _{1} = -1\]
Since m1 is -1, we can assume m2 is 1

- anonymous

okay

- anonymous

so now we have y=x+c

- anonymous

we plug in the points of the midpoint into the equation we get
9 = 3 + c

- anonymous

it should be easy from here

- anonymous

12

- anonymous

umm no...look at it this way...3+ what number gives you 9?

- anonymous

6

- anonymous

yep so there you have it y=x+6

- anonymous

N is the midpoint of segment MP. O is the midpoint of segment NP.
Segment MP with midpoint N and O
M is located at (7, 8), and P is located at (14, -3). What are the coordinates of O?
Select one:
a. (-8.75, 4.25)
b. (24.50, -0.50)
c. (5.25, 1.25)
d. (12.25, -0.25)

- anonymous

ok looks confusing but this is what they want|dw:1438267893606:dw|

- anonymous

Find the midpoint of MP with the formula i gave you earlier

- anonymous

okay

- anonymous

would it b C?

- anonymous

I meant to say would it be C

- anonymous

It's D

- anonymous

you want the solution?

- anonymous

yes

- anonymous

OK midpoint of MP is
\[(\frac{ 7+14 }{ 2 },\frac{ 8+(-3) }{ 2 }) = (10.5, 2.5)\]

- anonymous

okay

- anonymous

N = (10.5,2.5), O is the midpoint of NP so
\[(\frac{ 10.5+14 }{ 2 }, \frac{ 2.5+(-3) }{ 2 }) = (12.25,-0.25)\]

- anonymous

okay

- anonymous

A triangle has vertices at (4, 5), (4, -4), and (-4, -3). What is the approximate perimeter of the triangle?
Select one:
a. 27
b. 29
c. 28
d. 30
I got c for this on I just wanted to make sure if it was right or wrong

- anonymous

ok you need to find distance between the points with this formula
\[\sqrt{(x _{2}-x _{1})+(y _{2}-y _{1})}\]

- anonymous

okay

- anonymous

so would I input which ever number to find the answer

- anonymous

no you need to get 2 points and put it into the formula to get the distance of 1 side of the triangle

- anonymous

okay

- anonymous

would be easier if you label it like A = (4,5) B = (4,-4) C =(-4,-3)

- anonymous

yes

- anonymous

Find the distance of AB and AC and BC then add them up

- anonymous

25?

- anonymous

lemme try

- anonymous

okay

- anonymous

I got 27

- anonymous

its C

- anonymous

okay

- anonymous

ill give you the distance you try and work it out
AB = 9
AC = 11.31
BC = 8.06

- anonymous

I got 28 as well :D

- anonymous

i mean use the formula and try to see if you get the same distance

- anonymous

okay

- anonymous

yes

- anonymous

The volume of a sphere is 63089.813626667 cubic inches. What is the approximate diameter of this sphere? (Volume of a sphere = \frac{4}{3}\pi {r^3})
Select one:
a. 5023.1 in
b. 203.4 in
c. 49.4 in
d. 24.7 in
I got b for this and I worked it out but I'm still not sure

- anonymous

\[63089.813626667 = \frac{ 4 }{ 3 } \pi r^3\]
\[63089.813626667(\frac{ 3 }{ 4 }) = \pi r^3\]
\[47317.36022 = \pi r^3\]

- anonymous

okay

- anonymous

\[\frac{ 47317.36022 }{ \pi } = 15061.58355 = r^3\]
\[\sqrt[3]{15061.58355} = 24.69582535 = r\]
\[dia. = 2r =24.69582535\times2 =49.39165069\]

- anonymous

so it's C

- anonymous

dayum sneaky teacher putting radius as an option

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