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Not sure how to tackle this question, any help would be appreciated. 

UpTheChels  Oct 11, 2017
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2+0 Answers

 #1
avatar+79894 
+1

 

[ 55t ] / [ t^2 + 25 ] ≥ 5     multiply both sides by  t^2 + 25

 

55t  ≥ 5 [ t^2 + 25]       simplify

 

55t ≥ 5t^2  + `125     divide  through by 5

 

11t ≥ t^2 + 25       subtract  11t  from both sides

 

0 ≥ t^2 - 11t + 25   and we can write

 

t^2 - 11t + 25 ≤ 0

 

Let's solve this

 

t^2 - 11t + 25 = 0

 

Putting this into the quadratic formula......we have the [approximate] solutions

 

t ≈ 3.2  hrs    and    t ≈  7.8  hrs

 

Choosing a test value between these two values  [ I'll choose 4 ]  and putting this into

 

t^2 - 11t + 25 ≤ 0  →  4^2 - 11(4) + 25  =  16 - 44 + 25  = -3  

 

So....since 4 makes the inequality true.....the interval between 3.2 hrs  and 7.8 hrs is the correct solution.....i.e, the concentration will be ≥ 5  between these two times

 

Here's the graph, substituting x for t : https://www.desmos.com/calculator/cjgoqwvrj5

 

Notice that  [ 55t ] / [ t^2 + 25 ]  is ≤ 5  between these two times

 

 

cool cool cool

CPhill  Oct 11, 2017
 #2
avatar+8842 
+1

Solve algebraically and graphically !

 

laugh

Omi67  Oct 11, 2017

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