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?
Please help. I'm in 7th grade and normally a very good math student but these are really hard. PLEASE RESPOND QUICKLY! I need this by tomarrow!!!
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.
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!
To solve this problem, you can use the information given about the rates at which the candles burn and the remaining length of each candle at a certain time.
1. Let's define Fb as the faster burning candle and Sb as the slower burning candle.
2. The rate at which Fb burns is L/6 inches per hour, where L is the original length of the candles.
3. The rate at which Sb burns is L/9 inches per hour.
4. In t hours, Fb burns a distance of tL/6 inches.
5. In the same t hours, Sb burns a distance of tL/9 inches.
6. The unburned lengths of the candles at t hours can be calculated as follows:
- For Fb: (L - tL/6), which can also be written as (6L - tL)/6
- For Sb: (L - tL/9), which can also be written as (9L - tL)/9
7. Since we are looking for the time when Sb's length is twice Fb's length, we can set up the equation (9L - tL)/9 = 2(6L - tL)/6.
8. Expanding the equation, we get 54L - 6tL = 108L - 18tL.
9. Simplifying, we eliminate the L's throughout and have 12t = 54.
10. Solving for t, we find t = 4.5 hours.
Therefore, after 4.5 hours, the slower burning candle will be twice the height of the faster burning candle.
To find the original height of the candles, you need to substitute t = 4.5 hours into the equation of either Fb or Sb. For example, you can use (6L - tL)/6 = (6L - 4.5L)/6 = L - 0.75L = 0.25L. So, the original height of the candles was 0.25L.