Jovie is maintaining a camp fire. She has kept the fire steadily burning for 10 hours with 15 logs. She wants to know how many hours (h)left parenthesis, h, right parenthesis she could have kept the fire going with 9 logs. She assumes all logs are the same.

6 hours

Step-by-step explanation:

We can use ratios to solve

10 hours           x hours

=

15 logs            9 logs

Using cross products

90 = 15x

Divide by 15

90/15 = 15x/15

6 = x

6 hours

Below

Step-by-step explanation:

Jovie has maintained the fire burning with 15 logs for 10 hours.

● 15 => 10

Let's find how many hours do 9 logs keep the fire burning.

Let x be that time

● 9 => x

■■■■■■■■■■■■■■■■■■■■■■■■■■

● 15 => 10

● 9 => x

● 15 * x = 9×10

● 15x = 90

● x = 90/15

● x = 6

So the time that the fire will keep burning is 6 hours.

you would have to sell 24 bracelets.

step-by-step explanation:

84/3.5 = 24

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.