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4 years ago

Using the slope formula, find the slope of the line through the given points. (8,-6) and (1, -6)

Mathematics

1 answer:

Darina [25.2K]4 years ago

8 0

Answer:

m = 0 or slope = 0

Step-by-step explanation:

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Use absolute value to express the distance between −10 and 16 on the number line.

Ede4ka [16]

I think 12,13,14 that's what I think

4 0

4 years ago

Read 2 more answers

Directions: write a mixed number in simplest form to represent the number of pounds equivalent to each number of ounces. ( I wou

I am Lyosha [343]

16 oz to 1 pound so:

1. 40 oz = 2 $\frac{1}{2}$
2. 50 oz = 3 $\frac{1}{8}$
3. 44 oz = 2 $\frac{3}{4}$
4. 68 oz = 2 $\frac{1}{4}$
5. 22 oz = 1 $\frac{3}{8}$
6. 94 oz = 5 $\frac{7}{8}$

Hope that helps

3 0

3 years ago

** if u can answer this quickly it will be very appreciated!**

REY [17]

Answer:

Step-by-step explanation:

Answers

8 x - 1/2 = -4 - middle line

-3/4 x 8 = - 6 - side line

-3/4 - - 1/2 = -1/4 - top line

-1/2 - 1/2 = -1 - vertical line

7 0

3 years ago

A newspaper charges $3 per line for the first 4 lines plus a $20 to advertise.

sineoko [7]

Answer:

it is 32

and it is continuous

Step-by-step explanation:

And for the last one is the domain refers to the set of possible input values, the domain of a graph consists of all the input values shown on the x-axis. The range is the set of possible output values, which are shown on the y-axis.

i hope it helps :)

6 0

3 years ago

A brine solution of salt flows at a constant rate of 8L/min into a large tank that initially held 100L of brine solution in whic

GaryK [48]

Answer:

The mass of salt in the tank after t minutes is

$y(t) = 5-4.5e^{-\frac{2t}{25}}$

The concentration of salt in the tank reach 0.02 kg/L when $t=-\frac{25\ln \left(\frac{83}{75}\right)}{2} \approx -1.2669$

Step-by-step explanation:

Let y(t) be the mass of salt (in kg) that is in the tank at any time, t (in minutes).

The main equation that we will be using to model this mixing process is:

Rate of change of $\frac{dy}{dt}$ = Rate of salt in - Rate of salt out

We need to determine the rate at which salts enters the tank. From the information given we know:

The brine flows into the tank at a rate of $8\:\frac{L}{min}$
The concentration of salt in the brine entering the tank is $0.05\:\frac{kg}{L}$

The Rate of salt in = (flow rate of liquid entering) x (concentration of salt in liquid entering)

$(8\:\frac{L}{min}) \cdot (0.05\:\frac{kg}{L})=0.4 \:\frac{kg}{min}$

Next, we need to determine the output rate of salt from the tank.

The Rate of salt out = (flow rate of liquid exiting) x (concentration of salt in liquid exiting)

The concentration of salt in any part of the tank at time t is just y(t) divided by the volume. From the information given we know:

The tank initially contains 100 L and the rate of flow into the tank is the same as the rate of flow out.

$(8\:\frac{L}{min}) \cdot (\frac{y(t)}{100} \:\frac{kg}{L})= \frac{2y(t)}{25} \:\frac{kg}{min}$

At time t = 0, there is 0.5 kg of salt, so the initial condition is y(0) = 0.5. And the mathematical model for the mixing process is

$\frac{dy}{dt}=0.4-\frac{2y(t)}{25}, \quad{y(0)=0.5}$

$\frac{dy}{dt}=0.4-\frac{2y(t)}{25}\\\\\mathrm{Rewrite\:in\:the\:form\:of\:a\:first\:order\:separable\:ODE}\\\\ \frac{1}{2}\frac{dy}{5-y}=\frac{1}{25}dt \\\\\frac{1}{2}\int \frac{dy}{5-y}=\int \frac{1}{25}dt\\\\\frac{1}{2}\left(-\ln \left|5-y\right|+C\right)=\frac{1}{25}t+C\\\\-\frac{1}{2}\ln \left|5-y\right|+C_1\right)=\frac{1}{25}t+C\\\\-\frac{1}{2}\ln \left|5-y\right|=\frac{1}{25}t+C_2\\\\5-y=C_3e^{-\frac{2t}{25} }\\\\y(t) =5-C_3e^{-\frac{2t}{25} }$

Using the initial condition y(0)=0.5

$y(t) =5-C_3e^{-\frac{2t}{25} }\\y(0)=0.5=5-C_3e^{-\frac{2(0)}{25}} \\C_3=4.5$

The mass of salt in the tank after t minutes is

$y(t) = 5-4.5e^{-\frac{2t}{25}}$

To determine when the concentration of salt is 0.02 kg/L, we solve for t

$y(t) = 5-4.5e^{-\frac{2t}{25}}\\\\0.02=5-4.5e^{-\frac{2t}{25}}\\\\5\cdot \:100-4.5e^{-\frac{2t}{25}}\cdot \:100=0.02\cdot \:100\\\\500-450e^{-\frac{2t}{25}}=2\\\\500-450e^{-\frac{2t}{25}}-500=2-500\\\\-450e^{-\frac{2t}{25}}=-498\\\\e^{-\frac{2t}{25}}=\frac{83}{75}\\\\\ln \left(e^{-\frac{2t}{25}}\right)=\ln \left(\frac{83}{75}\right)\\\\\frac{2t}{25}=\ln \left(\frac{83}{75}\right)\\\\t=-\frac{25\ln \left(\frac{83}{75}\right)}{2} \approx -1.2669$

5 0

3 years ago