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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.

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

$0.2P > 0$

=> $P > 0$

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

$P < 135$

Now when the first value of P i.e $P< 0$ for $\frac{dP}{dt} > 0$

$P_2 > 135$

So when population increasing the values of P are

$P > 0$ and $P < 135$

OR

$P > 0$ and $P < 135$

So to obtain initial values of P where the population converge to the carrying capacity as $t \to [\infty]$

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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