## anonymous one year ago 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?

1. anonymous

ok you need to find the slope of the line

2. anonymous

okay

3. anonymous

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

4. anonymous

ok

5. anonymous

I got y=-x+4

6. anonymous

ok m is -1

7. anonymous

okay

8. anonymous

instead of 4 would it be

9. anonymous

y=-x+c

10. anonymous

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

11. anonymous

i got 3

12. anonymous

i got -4

13. anonymous

umm c should be 12 8 = -4 +12

14. anonymous

okay

15. anonymous

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

16. anonymous

okay

17. anonymous

18. anonymous

okay

19. anonymous

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

20. anonymous

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

21. anonymous

okay

22. anonymous

so now we have y=x+c

23. anonymous

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

24. anonymous

it should be easy from here

25. anonymous

12

26. anonymous

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

27. anonymous

6

28. anonymous

yep so there you have it y=x+6

29. 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)

30. anonymous

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

31. anonymous

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

32. anonymous

okay

33. anonymous

would it b C?

34. anonymous

I meant to say would it be C

35. anonymous

It's D

36. anonymous

you want the solution?

37. anonymous

yes

38. anonymous

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

39. anonymous

okay

40. anonymous

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

41. anonymous

okay

42. 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

43. anonymous

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

44. anonymous

okay

45. anonymous

so would I input which ever number to find the answer

46. anonymous

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

47. anonymous

okay

48. anonymous

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

49. anonymous

yes

50. anonymous

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

51. anonymous

25?

52. anonymous

lemme try

53. anonymous

okay

54. anonymous

I got 27

55. anonymous

its C

56. anonymous

okay

57. anonymous

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

58. anonymous

I got 28 as well :D

59. anonymous

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

60. anonymous

okay

61. anonymous

yes

62. 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

63. anonymous

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

64. anonymous

okay

65. anonymous

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

66. anonymous

so it's C

67. anonymous

dayum sneaky teacher putting radius as an option