+0

0
435
2
+79

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

Oct 11, 2017

#1
+99661
+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

Oct 11, 2017
#2
+10254
+1

Solve algebraically and graphically !

Oct 11, 2017