Cuánto le falta a 3/5para llegar a 9/10

Answers

Answer 1
Answer:

Answer:

3/10

Step-by-step explanation:

3/5 se puede convertir en 6/10

(3x2)/(5x2).

9/10 - 6/10 = 3/10

le falta 3/10 para llegar a 9/10.


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Find the volume of a pyramid with a square base, where the perimeter of the base is 15.4 and the height of the pyramid is 13.1. Round your answer to the nearest tenth of a cubic meter.

2. What is the value of 6x – 3y if x = 5 and y = 1F.11.
G.33
H.6 with an exponent of 5
I.65

Answers

Step-by-step explanation:

x=5,y=-1

6(5)-3(-1)=30+3=33

answer G .33

A tank contains 300 liters of fluid in which 40 grams of salt is dissolved. Brine containing 1 gram of salt per liter is then pumped into the tank at a rate of 6 L/min; the well-mixed solution is pumped out at the same rate. Find the number A(t) of grams of salt in the tank at time t.

Answers

Answer:

A(t) = 300 -260e^(-t/50)

Step-by-step explanation:

The rate of change of A(t) is ...

A'(t) = 6 -6/300·A(t)

Rewriting, we have ...

A'(t) +(1/50)A(t) = 6

This has solution ...

A(t) = p + qe^-(t/50)

We need to find the values of p and q. Using the differential equation, we ahve ...

A'(t) = -q/50e^-(t/50) = 6 - (p +qe^-(t/50))/50

0 = 6 -p/50

p = 300

From the initial condition, ...

A(0) = 300 +q = 40

q = -260

So, the complete solution is ...

A(t) = 300 -260e^(-t/50)

___

The salt in the tank increases in exponentially decaying fashion from 40 grams to 300 grams with a time constant of 50 minutes.

The number of grams of salt in the tank at any time t is 40 grams. The inflow and outflow of brine do not affect the amount of salt in the tank because the solution is well-mixed, and the salt concentration remains constant.

To solve this problem, we need to set up a differential equation that describes the rate of change of salt in the tank over time. Let A(t) represent the number of grams of salt in the tank at time t.

Let's break down the components affecting the rate of change of salt in the tank:

Salt inflow rate: The brine is being pumped into the tank at a constant rate of 6 liters per minute, and it contains 1 gram of salt per liter. So, the rate of salt inflow is 6 grams per minute.

Salt outflow rate: The solution in the tank is being pumped out at the same rate of 6 liters per minute, which means the rate of salt outflow is also 6 grams per minute.

Mixing of the solution: Since the tank is well-mixed, the concentration of salt remains uniform throughout the tank.

Now, let's set up the differential equation for A(t):

dA/dt = Rate of salt inflow - Rate of salt outflow

dA/dt = 6 grams/min - 6 grams/min

dA/dt = 0

The above equation shows that the rate of change of salt in the tank is constant and equal to zero. This means the number of grams of salt in the tank remains constant over time.

Now, let's find the constant value of A(t) using the initial condition where the tank initially contains 40 grams of salt.

When t = 0, A(0) = 40 grams

Since the rate of change is zero, A(t) will be the same as the initial amount of salt in the tank at any time t:

A(t) = 40 grams

So, the number of grams of salt in the tank at any time t is 40 grams. The inflow and outflow of brine do not affect the amount of salt in the tank because the solution is well-mixed, and the salt concentration remains constant.

To know more about differential equation:

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Please helpTrying to help daughter do virtual. And I stink in this stuff!
w(t) = 3t – 1; t = 5

Answers

The answer is W =14/5 I believe

When you roll a pair of dice, there are several outcomes, as shown below:When rolling the dice, the outcome is expressed as a sum of the two dice. For
example, if after you rolled the dice, and one die was a 3 and the other was a 4, you
will have rolled a 7.
Select all of the TRUE statements below. There may be more than one.
The probability of rolling the same digit with each die is 1/4.
The odds in favour of rolling a 10 is 1:13.
The odds against rolling a number less than 7 is 7:5.

Answers

Answer:

M

Step-by-step explanation:

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M

M

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*4.8.21Question Help
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After the release of radioactive material into the atmosphere from a nuclear power plant in a country in 2000, the hay in that country was contaminated by a radioactive
isotope (half-life 7 days). If it is safe to feed the hay to cows when 14% of the radioactive isotope remains, how long did the farmers need to wait to use this hay?
The farmers needed to wait approximately days for it to be safe to feed the hay to the cows.
(Round to one decimal place as needed.)
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Answers

Answer:

  19.9 days

Step-by-step explanation:

The amount remaining after d days is ...

  a = (1/2)^(d/7)

We want to find d when a = 0.14

  log(a) = (d/7)log(1/2)

  d = 7·log(0.14)/log(1/2) ≈ 19.855 ≈ 19.9

The farmers need to wait about 19.9 days for it to be safe.

Help, ill mark brainliest

Answers

Answer: FIRST OPTION.

Step-by-step explanation:

First, it is important to remember that the Slope-Intercept form of the equation of a line is the shown below:

y=mx+b

Where "m" is the slope of the line and "b" is the y-intercept.

By definiton, given a System of Linear equations, if they are exactly the same line, then the System of equations have Infinely many solutions.

In this case you have the following System of Linear equations given in the exercise:

\left \{ {{y=-2x+5} \atop {y=ax+b}} \right.

So, since you need the system has Infinite solutions, you know that the slope and the y-intercept of both lines must be equal.

Therefore, you can identify that the value of "a" and "b" must be the following:

a=-2\n\nb=5

So the Linear System would be the shown below:

\left \{ {{y=-2x+5} \atop {y=-2x+5}} \right.