Someone help me - Quadratics Help
I'm struggling on quadratics, can anyone help me out? Thanks in advance!
Water is pumped into an empty tank at a steady rate of 0.2l/min. after 1 hr the depth of water in the tank is 5cm, after 5 hrs the depth is 10cm
a. if the volume of the water in the tank is V litres when the depth is x and there is a quadratic realtionship between v and x, write down a rule
b. it is known that the maximum possible depth of water in the tank is 20 cm. For how long, from the beginning can water be pumped into the tank at the same rate without overflowing?
Answer:
A) V = 0.72x^2 -1.2x
B) 22 hours
THANKS
May 8, 2012
4 Comments • Newest first
@Watermelon: @LowWillpower:
omg thank you bless your soul
^ damn too fast
Okay, there are a few things we know here.
First, we know that the volume will always be 0.2*t where t is in minutes. We also know 3 times and respective heights. We have 3 unknowns in the quadratic equation, so this should work.
The first point we know is t=0, x=0, which in the formula V = ax^2 + bx + c means V = c, where V = 0.2 * t = 0 so c = 0.
We then know the points t = 60 (change hours to minutes), x = 5 and t = 300, x = 10. Putting the values of t and x into the quadratic, 0.2*t = ax^2 + bx will give you 2 equations with a and b, which you can use substitution or elimination to solve.
For the second question, just use the quadratic from before, except now knowing a and b, knowing x, and solving for t.
http://au.answers.yahoo.com/question/index?qid=20110402013738AAaEml8
http://au.answers.yahoo.com/question/index?qid=20100425235138AA3URq5
http://au.answers.yahoo.com/question/index?qid=20120425060635AACxfkh
http://au.answers.yahoo.com/question/index?qid=20120309233657AAJTou5 <-- This one was pretty easy to understand.