How did you get this. Can you show your solution and explain how it become like this? KarrStep2b)leFor the population tothelongterm. Weensurewithin thestable50,50,= 9000-m +0.00006 (1000 + m) (9000 m)= 9000 -mv + 6×10-5 (9x106 +8000m - m²)= 9540 - (1-48) m-m= 9540 -0 ≤ 9540 -Ofor610⁰ < 100m².thetheTheexpressiongives positive vahretothatweКли52108Hencelongm< 100m², 52m +weed52021m2survive inweed to+52m - 0.954X106 <0.46000 < 106.an+1range.mm < 10000100m²⇒ 100m² + 52m -954000 ≤0⇒ (m-97.4) (m + 97.9) 10.97.9and100m² +52m + 46000 always.for myo.find the range of m0≤m ≤ 97maximum value ofpopulationformfalls+52m + 46000 ≤106is 97,m ≤ 97.4.msurvives infor whichthe

For the population to the term we long ensure within the 0 ≤ 9540- => O that 106< - stable 52 160 < 100m². surovive weed to an+1 range m - * falls my < 10000 +52m - 0.954X106 ≤0. < 100m, 52m + 46000 ≤ 10⁰.

For the population to the term we long ensure within the 0 ≤ 9540- => O that 106< - stable 52 160 < 100m². surovive weed to an+1 range m - * falls my < 10000 +52m - 0.954X106 ≤0. < 100m, 52m + 46000 ≤ 10⁰.

Algebra & Trigonometry with Analytic Geometry

13th Edition

ISBN:9781133382119

Author:Swokowski

Publisher:Swokowski

Chapter2: Equations And Inequalities

Section2.1: Equations

Problem 78E

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Question

How did you get this. Can you show your solution and explain how it become like this?

Karr
Step2
b)
le
For the population to
the
long
term. We
ensure
within the
stable
50,
50,
= 9000-m +0.00006 (1000 + m) (9000 m)
= 9000 -mv + 6×10-5 (9x106 +8000m - m²)
= 9540 - (1-48) m
-m
= 9540 -
0 ≤ 9540 -
O
for
6
10⁰ < 100m².
the
the
The
expression
gives positive vahre
to
that
we
Кли
52
108
Hence
long
m
< 100m², 52m +
weed
52
021
m
2
survive in
weed to
+52m - 0.954X106 <0.
46000 < 106.
an+1
range.
mm < 10000
100m²
⇒ 100m² + 52m -
954000 ≤0
⇒ (m-97.4) (m + 97.9) 10.
97.9
and
100m² +52m + 46000 always.
for myo.
find the range of m
0≤m ≤ 97
maximum value of
population
form
falls
+52m + 46000 ≤106
is 97,
m ≤ 97.4.
m
survives in
for which
the

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