SOLUTION: The function Q (t) = 0.003t^2 – 0.625t + 25 represents the amount of energy in a battery after t minutes of use. (a) State the amount of energy held by the battery immediately b

Algebra ->  Quadratic Equations and Parabolas  -> Quadratic Equations Lessons  -> Quadratic Equation Lesson -> SOLUTION: The function Q (t) = 0.003t^2 – 0.625t + 25 represents the amount of energy in a battery after t minutes of use. (a) State the amount of energy held by the battery immediately b      Log On


   

Question 1069403: The function Q (t) = 0.003t^2 – 0.625t + 25 represents the amount of energy in a battery after t minutes of use.
(a) State the amount of energy held by the battery immediately before it was used.
(c) Given that Q (10) = 19.05, find the average amount of energy produced per minute for the interval 10 ≤ t ≤ 20.
(d) Calculate the number of minutes it takes for the energy to reach zero.
Answer by josmiceli(19441) About Me  (Show Source):
You can put this solution on YOUR website!
(a)
At +t+=+0+
+Q%28t%29+=+.003t%5E2+-+.625t+%2B+25+
+Q%28t%29+=+.003%2A0%5E2+-+.625%2A0+%2B+25+
+Q%280%29+=+25+
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(b)
+Q%2810%29+=+19.05+
+Q%2820%29+=+.003%2A20%5E2+-+.625%2A20+%2B+25+
+Q%2820%29+=+.003%2A400+-+12.5+%2B+25+
+Q%2820%29+=+1.2+-+12.5+%2B+25+
+Q%2820%29+=+13.7+
+13.7+-+19.05+=+-5.35+
units of energy used in +20+-+10+=+10+ min
+5.35%2F10+=+.535+ average energy/min
produced for +10+%3C=+t+%3C=+20+
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(c)
+Q%28t%29+=+.003t%5E2+-+.625t+%2B+25+
When does +Q%28t%29+=+0+ ?
+0+=+.003t%5E2+-+.625t+%2B+25+
Use quadratic formula
+t+=+%28+-b+%2B-+sqrt%28+b%5E2+-+4%2Aa%2Ac+%29%29%2F%282%2Aa%29+
+a+=+.003+
+b+=+-.625+
+c+=+25+
------------------
+t+=+%28+-%28-.625%29+%2B-+sqrt%28+%28-.625%29%5E2+-+4%2A.003%2A25+%29%29%2F%282%2A.003%29+
+t+=+%28+.625+%2B-+sqrt%28+.3906+-+.3+%29%29%2F.006+
+t+=+%28+.625+%2B-+sqrt%28+.0906+%29%29%2F.006+
+t+=+%28+.625+%2B+.301+%29+%2F+.006+
+t+=+.926%2F.006+
+t+=+154.33+
and
+t+=+%28+.625+-+.301+%29+%2F+.006+
+t+=+.324%2F.006+
+t+=+54+
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Here's the plot:
+graph%28+400%2C+400%2C+-5%2C+70%2C+-5%2C+50%2C+.003x%5E2+-+.625x+%2B+25+%29+
This tells me that at +t+=+54+ min the energy is zero
I can ignore the root +t+=+154.33+