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3 years ago

Please help I will make BRAINLIEST!!!!!

Mathematics

1 answer:

Anon25 [30]3 years ago

3 0

Answer:

variable= x

x ≤ 100

Step-by-step explanation:

it would be x ≤ 100 because it says it can hold as many AS 100, making it less than or equal to 100

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3 years ago

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3 years ago

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3 years ago

Read 2 more answers

A tank initially contains 60 gallons of brine, with 30 pounds of salt in solution. Pure water runs into the tank at 3 gallons pe

adoni [48]

Answer:

the amount of time until 23 pounds of salt remain in the tank is 0.088 minutes.

Step-by-step explanation:

The variation of the concentration of salt can be expressed as:

$\frac{dC}{dt}=Ci*Qi-Co*Qo$

being

C1: the concentration of salt in the inflow

Qi: the flow entering the tank

C2: the concentration leaving the tank (the same concentration that is in every part of the tank at that moment)

Qo: the flow going out of the tank.

With no salt in the inflow (C1=0), the equation can be reduced to

$\frac{dC}{dt}=-Co*Qo$

Rearranging the equation, it becomes

$\frac{dC}{C}=-Qo*dt$

Integrating both sides

$\int\frac{dC}{C}=\int-Qo*dt\\ln(\abs{C})+x1=-Qo*t+x2\\ln(\abs{C})=-Qo*t+x\\C=exp^{-Qo*t+x}$

It is known that the concentration at t=0 is 30 pounds in 60 gallons, so C(0) is 0.5 pounds/gallon.

$C(0)=exp^{-Qo*0+x}=0.5\\exp^{x} =0.5\\x=ln(0.5)=-0.693\\$

The final equation for the concentration of salt at any given time is

$C=exp^{-3*t-0.693}$

To answer how long it will be until there are 23 pounds of salt in the tank, we can use the last equation:

$C=exp^{-3*t-0.693}\\(23/60)=exp^{-3*t-0.693}\\ln(23/60)=-3*t-0.693\\t=-\frac{ln(23/60)+0.693}{3}=-\frac{-0.959+0.693}{3}= -\frac{-0.266}{3}=0.088$

5 0

3 years ago

I really need help on this question

jekas [21]

P(t) = [ 64/ (1 + 11.e⁽⁻⁰ ⁰⁸t⁾]
In 1991, t = 1, hence:

P(t) = [ 64/ (1 + 11.e⁽⁻⁰ ⁰⁸ˣ¹⁾] = 5.7377 billion or rounded 5.74 billion
(Answer A)

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3 years ago