I'm sort of the person who hates these types of questions since I only like math problems with diagrams, and formulas. Anyways....here is the question for all you math lovers. (This isn't my homework, it's just practice I was doing, and no don't get the answer from Google) Oh, explain to me your answer too, I'm dumb.
Chantelle has two candles, one of which was 32cm longer than the other.
She lit the longer one at 3p.m. and lit the shorter one at 7.pm.
At 9 p.m., they were both the same length.
The longer one was completely burned out at 10p.m. and the shorter ones was completely burned at midnight (12 p.m.)
The two candles burned at different, but constant rates.
What was the sum of the original lengths of the two candles?
b42cm (B)48cm (C)60cm (D)80cm (E)52cm
(F)Wow, you actually spent time typing this? (G)°0° I'm not going to answer it. (H) ]:[/b]
Choose an answer from above.
G, lol.
I do more math than i need to.
The answer is (e) 52 cm
The taller candle is 42 cm tall, burning at a rate of 6 cm/hr The shorter candle is 10 cm tall, burning at a rate of 2 cm/hr
Does that help?
H
You don't want my help. I have a D- in Algebra. ):
The answer is M, for whatever answer my mom gives me when I ask her to do the problem. ;D
let x = length of candle 1 let y = length of candle 2 let x0 = original length of candle 1 let y0 = original length of candle 2 let dx/dt = rate of burning candle 1 let dy/dt = rate of burning candle 2
x0 - y0 = 32 x0 = 32 + y0
x = x0 - t*dx/dt
y = y0 - t*dy/dt
x0 - 6dx/dt = y0 - 2dy/dt
0 = x0 - 7dx/dt 7dx/dt = x0 7*dx/dt = 32 + y0
0 = y0 - 5dy/dt 5dy/dt = y0
x0 - 7dx/dt = y0 - 5dy/dt
x0 + y0 = ?
therefore:
7dx/dt - 6dx/dt = 5dy/dt - 2dy/dt dx/dt = 3*dy/dt
73dy/dt = 32 + y0 21dy/dt - 32 = y0 21 dy/dt - 32 = 5dy/dt -32 = 5dy/dt - 21dy/dt -32 = -16*dy/dt 2 = dy/dt
dx/dt = 3dy/dt dx/dt = 32 dx/dt = 6
y0 = 5dy/dt y0 = 52 y0 = 10 cm
x0 = 7dx/dt x0 = 76 x0 = 42 cm
y0 + x0 = ? 10 + 42 = 52 cm
God am I a nerd.
Sorry. ):
It probably wasn't necessary to use that Leibniz notation crap.