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2 years ago

What is the slope of the line represented by 9x - 14y = 28?

Mathematics

1 answer:

Soloha48 [4]2 years ago

4 0

Answer:

The slope of the line is $\displaystyle \frac{9}{14}$ .

Step-by-step explanation:

We are given the line in standard form:

$9x - 14y = 28$

We need to get this into slope-intercept form in order to determine the slope of the line.

Slope-intercept form is $y = mx + b$ , where m is our slope and b is the y-intercept.

Therefore, we need to get:

$9x - 14y = 28$

converted to:

$y = mx +b$ .

$\displaystyle 9x - 14y = 28\\\\-14y = -9x + 28\\\\\frac{-14y}{-14}=\frac{-9x+28}{-14}\\\\y = \frac{9}{14}x-2$

Therefore, by solving the equation in slope-intercept form, we see that m equals 9/14, so our slope is $\displaystyle \frac{9}{14}$ .

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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

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Answer:

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

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

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

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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.

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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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