# math (rate)

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I need help! Ok here's the problem.
Melissa lights 2 candles that are the exact same height at the exact same time. The blue candle takes 6 hours to burn completely. The purple candle takes 9 hours to burn completely. After how many hours will the slower burning candle be twice the height of the faster candle? How tall were the candles originally?

Bluerate=rb=1candle/6hrs
Purple rate=rp=1candle/9hrs.
Now, the length remaining is the original Length minus the burned part.
Length= Lorig - rburning*Time

Lengthblue= Lorig-Lorig/6*T
When the blue is one half the purple,
length blue= 1/2 Lp
Lorig (1- 1/6 T)= 1/2 Lorig(1-1/9 T)
1-T/6= 1/2 - T/18
solve for T.

Maria lights two candles of equal length at the same time. One candle takes 6
>hours to burn out and the other takes 9. How much time will pass until the
>slower-burning candle is exactly twice as long as the faster-burning one?
>Explain how you got the answer.

1--The faster burning candle, Fb, burns at the rate of L/6 inches per hour.
2--The slower burning candle, Sb, burns at the rate of L/9 inches per hour.
3--In t hours, Fb burns a distance of tL/6 inches.
4--In the same t hours, Fs burns a distance of tL/9 inches.
5--The unburned candle lengths at t hours are then (L - tL/6) or (6L - tL)/6 for Fb and (L - tL/9) or (6L - tL)/9 for Sb.
6--Since we are looking for the t where Sb's length is twice Fb's length, we can write (9L - tL)/9 = 2(6L - tL)/6.
7--Expanding, 54L - 6tL = 108L - 18tL.
8--Eliminating the L's throughout, we have 12t = 54 making t = 4.5 hours.

• math (rate) -

4.5 hours! It says it doesnt it?! stupid much! Unny way Im sorta a couple years of from when this was due! Im in 7th grade now Ive got a 99 in math1 No im not a nerd, quite the contrary! Im just smart, pretty, and i happen to have alot of freinds, unlike my BF David! JK! Unny way hope u got it rite! Byas!

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