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3 years ago

- 2.5f + 0.4f - 13-5 – 2.5f +0.4f - 13-5=

Mathematics

1 answer:

Inessa05 [86]3 years ago

7 0

Answer:

=−4.2f−36

Step-by-step explanation:

combine like terms

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Write 6.089 repeating as a mixed number

Daniel [21]

$6.089=6\frac{89}{1000}$

⭐ Please consider brainliest! ⭐

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3 years ago

A 1000-liter (L) tank contains 500 L of water with a salt concentration of 10 g/L. Water with a salt concentration of 50 g/L flo

djverab [1.8K]

Answer:

a) $y(t)=50000-49990e^{\frac{-2t}{25}}$

b) $31690.7 g/L$

Step-by-step explanation:

By definition, we have that the change rate of salt in the tank is $\frac{dy}{dt}=R_{i}-R_{o}$ , where $R_{i}$ is the rate of salt entering and $R_{o}$ is the rate of salt going outside.

Then we have, $R_{i}=80\frac{L}{min}*50\frac{g}{L}=4000\frac{g}{min}$ , and

$R_{o}=40\frac{L}{min}*\frac{y}{500} \frac{g}{L}=\frac{2y}{25}\frac{g}{min}$

So we obtain. $\frac{dy}{dt}=4000-\frac{2y}{25}$ , then

$\frac{dy}{dt}+\frac{2y}{25}=4000$ , and using the integrating factor $e^{\int {\frac{2}{25}} \, dt=e^{\frac{2t}{25}$ , therefore $(\frac{dy }{dt}+\frac{2y}{25}}=4000)e^{\frac{2t}{25}$ , we get $\frac{d}{dt}(y*e^{\frac{2t}{25}})= 4000 e^{\frac{2t}{25}$ , after integrating both sides $y*e^{\frac{2t}{25}}= 50000 e^{\frac{2t}{25}}+C$ , therefore $y(t)= 50000 +Ce^{\frac{-2t}{25}}$ , to find $C$ we know that the tank initially contains a salt concentration of 10 g/L, that means the initial conditions $y(0)=10$ , so $10= 50000+Ce^{\frac{-0*2}{25}}$

$10=50000+C\\C=10-50000=-49990$

Finally we can write an expression for the amount of salt in the tank at any time t, it is $y(t)=50000-49990e^{\frac{-2t}{25}}$

b) The tank will overflow due Rin>Rout, at a rate of $80 L/min-40L/min=40L/min$ , due we have 500 L to overflow $\frac{500L}{40L/min} =\frac{25}{2} min=t$ , so we can evualuate the expression of a) $y(25/2)=50000-49990e^{\frac{-2}{25}\frac{25}{2}}=50000-49990e^{-1}=31690.7$ , is the salt concentration when the tank overflows

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4 years ago

WILL GIVE BRAINLIEST!!!!!!!!!!!!!

Phantasy [73]

Writing the domain in words: negative infinity to three halves and three halves to infinity

So, Option A is correct.

Step-by-step explanation:

We need to find domain of the function: $f(x)=\frac{x-5}{2x-3}$

Domain:

The domain of function is defined as all the input values for which the function is defined.

In other words we can say that any value of x that makes the denominator zero is not considered as domain of function.

In the function given: $f(x)=\frac{x-5}{2x-3}$

if x= 3/2 then the denominator will be zero and the function will not be defined.

So, value of x should be x>3/2 or x<3/2

We can write domain as: (-∞,3/2)(3/2,∞)

3/2 can be written as three halves.

Writing the domain in words: negative infinity to three halves and three halves to infinity

So, Option A is correct.

Keywords: Domain of Function

Learn more about Domain of function at:

#learnwithBrainly

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4 years ago

How do you do partial quationet

Orlov [11]

In partial-quotients division, it takes several steps to find the quotient. At each step, you find a partial answer (called a partial quotient); then you find the product of thepartial quotient and divisor and subtract it from the dividend. Finally, you add all thepartial quotients to find the final quotient.

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3 years ago

Unit Activity: Advanced Functions

Cloud [144]

See attached file for the solution/explanation to the 12 questions.

Download doc

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4 years ago