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Slav-nsk [51]
3 years ago
10

Bill makes leather shoes. He charges $45 for a pair of women's shoes and $50 for a pair of men's shoes. He sells x pairs of wome

n’s shoes and y pairs of men’s shoes each week and earns a total income of $2,500.
This situation can be represented by the equation
. If Bill has no female customers during a particular week, he needs to sell
pairs of men’s shoes to earn his weekly income of $2,500.
Mathematics
1 answer:
Kruka [31]3 years ago
7 0

Answer:

Bill needs to sell 50 pairs of men’s clothes.

Step-by-step explanation:

50y = 2500

Divide by 50 on both sides

2500/50 = 50

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1. Consider the population model dP dt = 0.2P 1 − P 135 , where P(t) is the population at time t. (a) For what values of P is th
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a

  P = 0  OR  P = 135

b

P > 0 and P < 135

OR

P > 0 and P < 135

c

Generally the carrying capacity is can be defined as the highest amount of population and environment can support for an unlimited duration or time period

d

P = 67.5

Step-by-step explanation:

From the question we are told that

The population model is \frac{dP}{dt}  =  0.2P(1 - \frac{P}{135} )

Generally at equilibrium

\frac{dP}{dt} = 0

So

0.2P = 0

=> P = 0

Or

(1 - \frac{P}{135} ) = 0

=> P = 135

Thus at equilibrium P = 0 or P = 135

Generally when the population is increasing we have that

\frac{dP}{dt} > 0

So

0.2P > 0

=> P > 0

and

(1 - \frac{P}{135} ) > 0

P < 135

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P_2 > 135

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The rate equation can be represented as

\frac{dP}{dt}  =  \frac{1}{5}P (1 - \frac{P}{135} )

So we will differentiate the equation again we have that

\frac{d^2 P}{dt^2} = \frac{(1 - \frac{P}{135} )}{5}  - \frac{P}{675}

Now as  t \to [\infty]

\frac{d^2 P}{dt^2} \to  0

So

   \frac{(1 - \frac{P}{135} )}{5}  - \frac{P}{675} = 0      

=>    \frac{(1 - \frac{P}{135} )}{5}   =  \frac{P}{675}

=> P = 67.5

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3 years ago
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