1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
oksano4ka [1.4K]
4 years ago
10

Como se resuelve este problema 3{50-[6•2(9-5)+1]}

Mathematics
1 answer:
zmey [24]4 years ago
7 0
3{50-[6*2(9-5)+1]}=
3{50-[6*2(4)+1]}=
3[50-(6*8+1)]=
3[50-(48+1)]=
3(50-49)=
3(1)=
3
You might be interested in
Sid lives 3 miles north and 8 miles west of school. What is the direct distance from Sid's house to school?
Aliun [14]
The answer is D, about 8.5 miles. You can use the Pythagorean theorem
and plug in 3 and 8 as the length of the legs.
7 0
3 years ago
Read 2 more answers
some one plz help What is the difference? 1/6 - 5/8 a. -19/24 b - 11/24 c. 11/24 d. 2
Kobotan [32]


1/6-5/8

= (8-30)/48

= -22/48

= -11/24 which is b

6 0
3 years ago
Read 2 more answers
Please help me in this:
suter [353]

Answer:

;sjjdje

Step-by-step explanation:

4djdeje

3 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 <em>y(t)</em> be the mass of salt (in kg) that is in the tank at any time, <em>t</em> (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 <em>t</em> is just <em>y(t) </em>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 <em>t</em>

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
Solve for t<br> literal equation<img src="https://tex.z-dn.net/?f=y%20%3D%20%5Cfrac%7Bv-u%7D%7Bt%7D" id="TexFormula1" title="y =
balu736 [363]

Answer:

t = (v-u)/y

Step-by-step explanation:

yt = v - u

t = (v-u)/y

4 0
3 years ago
Other questions:
  • In the graph below, line "k" y = -k makes a 45° angle with the x- and y- axes. Rx:
    10·2 answers
  • Missing value 0.003 x what = 3
    15·1 answer
  • The oranges below are similar. Find the radius of the smaller orange. Correct answer will get brainliest.
    5·1 answer
  • I need help Answering
    12·2 answers
  • How to write a word problem
    13·1 answer
  • What is the inverse of the function f(x) = 1/9x + 2
    15·1 answer
  • What is the value of x in: 4x+3y=12 and 2x-3y=-30 *
    14·1 answer
  • Step 3: Displaying numerical data in dot plots, and describing and analyzing
    14·1 answer
  • A whole number is squared. The result is between 200 and 260. The number is between:
    14·1 answer
  • Luigi's Phone Shop charges $50.25 for its activation fee plus $27.07 for every gigabyte, g, of data you use over your limit. The
    5·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!