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Kaylis [27]
3 years ago
11

Define a variable, write an inequality, and solve each problem. Check your solution.

Mathematics
1 answer:
grandymaker [24]3 years ago
7 0

Answer:

Step-by-step explanation:

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What is the area of this trapezoid?<br><br> 26 cm²<br><br> 32 cm²<br><br> 40 cm²<br><br> 72 cm²
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2 years ago
The most important component of a physical structure is a solid foundation on which the rest of the structure sits. a strong fou
eduard

The foundational concept of algebra is to use the alphabetical letters to find the unknown number.

Algebra is the concept that uses the alphabetical letters such as x, y, z, etc. to calculate the unknow numbers. here the letters are also called as variables. The values that are known in the expression as termed as constants.

The general operations performed on the algebra is

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Here the alphabets x and y are called variables.

Order of operations:

The order of operation in algebra is given as follows

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To know mire about Algebra here

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7 0
1 year ago
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
Kryger [21]

Answer:

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

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

5 0
3 years ago
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