1. Martha can paint a room in 2 hours. Jamie can paint the same room in 6 hours. How long, to the nearest tenth of an hour, will it take them to paint
the room together?
2. Sally can paint a room in 4 hours. Joe can paint a room in 6 hours. How
long will it take if they paint the room together?
3. Nicole can paint a room in six hours, Alexandra can paint a room in 5 hours,
and Liana can paint a room in 4 If all three work together, how long
will it take them to paint the room?
4. Jake can paint a room in 7 hours, and Karen can paint the same room in 5
hours. How long will it take to paint the room if they work together?

3. 60/37 hours

4. 35/12 hours

Step-by-step explanation:

3.

Nicole: 1 room in 6 hours; in 1 hour: 1/6 room

Alexandra: 1 room in 5 hours; in 1 hour: 1/5 room

Liana: 1 room in 4 hours; in 1 hour: 1/4 room

Combined in 1 hour:

1/6 + 1/5 + 1/4

When they work together, they paint a room in x hours.

In 1 hour: 1/x room

1/6 + 1/5 + 1/4 = 1/x

10/60 + 12/60 + 15/60 = 1/x

37/60 = 1/x

x = 60/37

60/37 hours

4.

Jake: 1 room in 7 hours; in 1 hour: 1/7 room

Karen: 1 room in 5 hours; in 1 hour: 1/5 room

Combined in 1 hour:

1/7 + 1/5

When they work together, they paint a room in x hours.

In 1 hour: 1/x room

1/7 + 1/5 = 1/x

5/35 + 7/35 = 1/x

12/35 = 1/x

x = 35/12

35/12 hours

35/12 hours

abde or 1,2,4, 5 is the answer

step-by-step explanation:

tell you what. i'll find the length of the medians and you can fill in the blanks where they belong.

br

b = (2a,0)

r = (a/2,b/2) remember these things are medians. they go to the 1/2 way point of the line opposite the vertex they face.

br^2 =   (2a - a/2)^2 + (0 - b/2)^2

br^2 = (3/2 a) ^2 + b^2 / 4

br^2 = 9/4 a^2 + b^2 / 4   we need to find some relationship between a and b.

let's try ab = bc  

ab = 2a

bc = (2a - a)^2 + (b - 0)^2

bc = sqrt(a^2 + b^2)

ab = bc

2a = sqrt(a^2 + b^2) square both sides.

4a^2 = a^2 + b^2 subtract a^2 from both sides.

sqrt(3a^2) = sqrt(b^2)

sqrt(3)a = b

let's put b into br^2

br^2 = 9/4 a^2 + 3b^2 / 4

br^2 = 12 a^2 / 4

br^2 = 3a^2 

br = sqrt(3) * a

cp

c = (a,b) ; p = (a,0) this is another application of the distance formula, and it is a good one.

cp^2 = (a - a)^2 + (b - 0)^2

cp^2 = 0 + b^2

cp^2 = b^2 which you can see from the diagram. 

cp = sqrt(b^2)

cp = b but b = sqrt(3) * a 

cp = sqrt(3)*a

aq

a = (0,0)

q = (3/2) a, b/2

aq^2 = (3a/2 -0)^2 + (b/2 - 0)^2

aq^2 = 9a^2/4 + b^2/4

but b = sqrt(3) * a 

aq^2 = 9a^2 / 4 + (sqrt(3)*a)^2 /4

aq^2 = 9 a^2/ 4 + 3 a^2 / 4

aq^2 = 12 a^2/4

aq^2 = 3 a^2

aq = (sqrt  3) * a which agrees with the other two.