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3 years ago

18.2 + 45.4 + x < 100

2 answers:

8 0

<span>Simplifying 18.2 + 45.4 + x = 100
Combine like terms: 18.2 + 45.4 = 63.6 63.6 + x = 100
Solving 63.6 + x = 100
Solving for variable 'x'.
Move all terms containing x to the left, all other terms to the right.
Add '-63.6' to each side of the equation. 63.6 + -63.6 + x = 100 + -63.6 Combine like terms: 63.6 + -63.6 = 0.0 0.0 + x = 100 + -63.6 x = 100 + -63.6 Combine like terms: 100 + -63.6 = 36.4 x = 36.4
Simplifying x = 36.4</span>

NemiM [27]3 years ago

4 0

18.2+45.4=63.6
100-63.6=36.4

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Katie is getting her laundry done it cost $2.25 per square foot of flowers added to the yard if she wants 12.4 ft.² of flowers h

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3 years ago

Which expressions have a quotient of 4/5

goblinko [34]

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b and e

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2 years ago

Read 2 more answers

A swimming pool with a volume of 30,000 liters originally contains water that is 0.01% chlorine (i.e. it contains 0.1 mL of chlo

SpyIntel [72]

Answer:

$R_{in}=0.2\dfrac{mL}{min}$

$C(t)=\dfrac{A(t)}{30000}$

$R_{out}= \dfrac{A(t)}{1500} \dfrac{mL}{min}$

$A(t)=300+2700e^{-\dfrac{t}{1500}},$ A(0)=3000$

Step-by-step explanation:

The volume of the swimming pool = 30,000 liters

(a) Amount of chlorine initially in the tank.

It originally contains water that is 0.01% chlorine.

0.01% of 30000=3000 mL of chlorine per liter

A(0)= 3000 mL of chlorine per liter

(b) Rate at which the chlorine is entering the pool.

City water containing 0.001%(0.01 mL of chlorine per liter) chlorine is pumped into the pool at a rate of 20 liters/min.

$R_{in}=$ (concentration of chlorine in inflow)(input rate of the water)

$=(0.01\dfrac{mL}{liter}) (20\dfrac{liter}{min})\\R_{in}=0.2\dfrac{mL}{min}$

Volume of the pool =30,000 Liter

$Concentration, C(t)= \dfrac{Amount}{Volume}\\C(t)=\dfrac{A(t)}{30000}$

(d) Rate at which the chlorine is leaving the pool

$R_{out}=$ (concentration of chlorine in outflow)(output rate of the water)

$= (\dfrac{A(t)}{30000})(20\dfrac{liter}{min})\\R_{out}= \dfrac{A(t)}{1500} \dfrac{mL}{min}$

(e) Differential equation representing the rate at which the amount of sugar in the tank is changing at time t.

$\dfrac{dA}{dt}=R_{in}-R_{out}\\\dfrac{dA}{dt}=0.2- \dfrac{A(t)}{1500}$

We then solve the resulting differential equation by separation of variables.

$\dfrac{dA}{dt}+\dfrac{A}{1500}=0.2\\$The integrating factor: e^{\int \frac{1}{1500}dt} =e^{\frac{t}{1500}}\\$Multiplying by the integrating factor all through\\\dfrac{dA}{dt}e^{\frac{t}{1500}}+\dfrac{A}{1500}e^{\frac{t}{1500}}=0.2e^{\frac{t}{1500}}\\(Ae^{\frac{t}{1500}})'=0.2e^{\frac{t}{1500}}$

Taking the integral of both sides

$\int(Ae^{\frac{t}{1500}})'=\int 0.2e^{\frac{t}{1500}} dt\\Ae^{\frac{t}{1500}}=0.2*1500e^{\frac{t}{1500}}+C, $(C a constant of integration)\\Ae^{\frac{t}{1500}}=300e^{\frac{t}{1500}}+C\\$Divide all through by e^{\frac{t}{1500}}\\A(t)=300+Ce^{-\frac{t}{1500}}$

Recall that when t=0, A(t)=3000 (our initial condition)

$3000=300+Ce^{0}\\C=2700\\$Therefore:\\A(t)=300+2700e^{-\dfrac{t}{1500}}$

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3 years ago

Explain how you can find the constant of proportionality from a graph representing a proportional relationship when it shows a p

Korvikt [17]

For a proportional relationship, the constant is found dividing all values of y by each respective value of x.

<h3>What is a proportional relationship?</h3>

A proportional relationship is a function in which the output variable is given by the input variable multiplied by a constant of proportionality, that is:

y = kx

In which k is the constant of proportionality.

The constant can be represented as follows:

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Hence the constant is found dividing all values of y by each respective value of x.

More can be learned about proportional relationships at brainly.com/question/10424180

#SPJ1

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2 years ago