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3 years ago

8 minus the quotient of 15 and y

Mathematics

1 answer:

STALIN [3.7K]3 years ago

6 0

8 minus the quotient of 15 and y

8 - 15 : y = 8 - 15/y

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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

A container weighs 23 2/5 pounds holds sports equipment. Of the equipment in the container 1/3 of the weight is baseball equipme

hram777 [196]

What we need to know is how much is 1/3 of 25 2/5.

We will be multiplying fraction by a fraction and for this, it's good to change mixed numbers into improper fractions:

25 2/5= $\frac{23*5+2}{5}= \frac{117}{5}$

and now we multiply it by 1/3:

$\frac{1}{3}\frac{117}{5}=\frac{39}{5}=7\frac{4}{5}$

and that's the answer: the basketball equipment weights 7 4/5 pounds.

8 0

3 years ago

Please answer the question from the attachment.

ryzh [129]

-x - 7y = 18

4x + 7y = -30

The -7y and 7y cancel by addition.

Thus, 3x = -12

Divide both sides by 3

x = -4

Plug in -4 for x

Thus, -(-4) - 7y = 18

4 - 7y = 18

Subtract 4 to both sides; 4 - 4 - 7y = 18 - 4

Simplify, -7y = 14

Divide both sides by -7; -7y/-7 = 14 / -7

y = -2

5 0

4 years ago

Five students are seated in a circle. Each student has either a red disc or a green disc painted on

lesantik [10]

Answer:

Students 3 and 4 have green discs. Students 1, 2, and 5 have red discs.

Step-by-step explanation:

If Student 1 is telling the truth, Students 2 and 3 must be lying. If that were true, Student 1 would see at least 2 red discs, not 1. So Student 1 is lying, and therefore has a red disc.

Therefore, Student 5 is also lying, and must also have a red disc.

If Student 2 is telling the truth, the other four students must be lying. That means Student 2 would have a green disc, and the other four would have red discs. But if that were true, then Student 3 would be telling the truth. Therefore, Student 2 is lying.

So far, we know Students 1, 2, and 5 are lying. If Student 3 is telling the truth, then she and Student 4 each have a green disc. If Student 3 is lying, then all five students would have red discs, which would mean Student 2 was telling the truth.

Therefore, Students 3 and 4 have green discs. Students 1, 2, and 5 have red discs.

3 0

3 years ago

You pick two marbles from a bag containing 30 marbles: 3 green, 6 white, 4 red, 7 yellow, and 10 blue. Find the given probabilit

VARVARA [1.3K]

Total balls in bag(sample space)=30
Red balls=4
Green=3
White=6
Yellow=7
Blue=10
event P(E) = (Number of favorable outcomes) ÷ (Sample space).
So p(red)=no red balls/total balls in the marble
P(red)=4/30=2/15
P(blue)=10/30=1/3
P(white)=6/30=1/5
P(yellow)=7/30
P(green)=3/30=1/10

4 0

2 years ago