basically, Barb pulled the plug in her bathtub and it started to drain. the amt of water in the bathtub as it drains is rep. by the equation L= -5t^2-8t+120, where L rep. the number of liters of water in the bathtub and "t" rep. the amt of time in min, since the plug was pulled.
1.How many liters of water were in the bathtub when barb pulled the plug?
2. To the nearest tenth of a min., the amt of tieme it takes for all the water in the bathtub to drain.
I used Axis of symmetry(-b/2a)
any more tips of formulas plz help
Thanks in Advance,
~*God-Bless*~
2007-12-09 08:32:14 · 3 answers · asked by modelgyal91 2 in Science & Mathematics ➔ Mathematics
what is 4.16 to the nearest ten of a min? is it 4min?
2007-12-09 09:00:01 · update #1
You are given the equation L= -5t²-8t+120 and asked to find the value of L at time zero. Substituting zero for t in the equation yields 120. You are then asked to find the value of t for which L = 0.
0 = -5t²-8t+120
This doesn't factor using integers so use the quadratic solution
t = ( 8 ± sqrt ( 8² - ( 4 ) ( -5 ) ( 120 ) ) ) / -10
t = ( 8 ± sqrt ( 2528 ) ) / -10
One solution is positive, the other negative; you want the positive solution.
2007-12-09 08:40:39 · answer #1 · answered by jgoulden 7 · 0⤊ 0⤋
Since at t=0 the tub is full. plug that into the eqution and you get 120 L
A. 120 L
Quadratic equation: set it equal to zero and solve for t will give you the time in
-5t^2-8t+120=0 or
5t^2+8t-120= 0 will give you two solutions (4.16m and -5.76m) (negative time has no meaning therefore 4.16m is the answer)
B. 4.16m
2007-12-09 08:42:49 · answer #2 · answered by Rector 2 · 0⤊ 0⤋
L= -5t²-8t+120
1. Barb pulled the plug at time t=0. Find L for t=0.
2. Find t for L=0, using quadratic formula.
2007-12-09 08:42:14 · answer #3 · answered by DWRead 7 · 0⤊ 0⤋