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2 years ago

Does anyone understand how to solve this?

Mathematics

1 answer:

yanalaym [24]2 years ago

3 0

You solve it by filling in the function value and simplifying the result.

$\displaystyle \frac{f(x)-f(a)}{x-a}=\frac{\left(3x^{2}+x+3\right)-\left(3a^{2}+a+3\right)}{x-a}\\\\=\frac{3\left(x^{2}-a^{2}\right)+\left(x-a\right)}{x-a}\\\\=\frac{3(x+a)(x-a)+(x-a)}{x-a}=3(x+a)+1$

$\displaystyle \frac{f(x+h)-f(x)}{h}=\frac{\left(3(x+h)^{2}+(x+h)+3\right)-\left(3x^{2}+x+3\right)}{h}\\\\=\frac{\left(3\left(x^{2}+2hx+h^{2}\right)+(x+h)+3\right)-\left(3x^{2}+x+3\right)}{h}\\\\=\frac{3\left(2hx+h^{2}\right)+h}{h}=6x+1+3h$

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Initially a tank contains 10 liters of pure water. Brine of unknown (but constant) concentration of salt is flowing in at 1 lite

zhenek [66]

Answer:

Therefore the concentration of salt in the incoming brine is 1.73 g/L.

Step-by-step explanation:

Here the amount of incoming and outgoing of water are equal. Then the amount of water in the tank remain same = 10 liters.

Let the concentration of salt be a gram/L

Let the amount salt in the tank at any time t be Q(t).

$\frac{dQ}{dt} =\textrm {incoming rate - outgoing rate}$

Incoming rate = (a g/L)×(1 L/min)

=a g/min

The concentration of salt in the tank at any time t is = $\frac{Q(t)}{10}$ g/L

Outgoing rate = $(\frac{Q(t)}{10} g/L)(1 L/ min)$ $\frac{Q(t)}{10} g/min$

$\frac{dQ}{dt} = a- \frac{Q(t)}{10}$

$\Rightarrow \frac{dQ}{10a-Q(t)} =\frac{1}{10} dt$

Integrating both sides

$\Rightarrow \int \frac{dQ}{10a-Q(t)} =\int\frac{1}{10} dt$

$\Rightarrow -log|10a-Q(t)|=\frac{1}{10} t +c$ [ where c arbitrary constant]

Initial condition when t= 20 , Q(t)= 15 gram

$\Rightarrow -log|10a-15|=\frac{1}{10}\times 20 +c$

$\Rightarrow -log|10a-15|-2=c$

Therefore ,

$-log|10a-Q(t)|=\frac{1}{10} t -log|10a-15|-2$ .......(1)

In the starting time t=0 and Q(t)=0

Putting t=0 and Q(t)=0 in equation (1) we get

$- log|10a|= -log|10a-15| -2$

$\Rightarrow- log|10a|+log|10a-15|= -2$

$\Rightarrow log|\frac{10a-15}{10a}|= -2$

$\Rightarrow |\frac{10a-15}{10a}|=e ^{-2}$

$\Rightarrow 1-\frac{15}{10a} =e^{-2}$

$\Rightarrow \frac{15}{10a} =1-e^{-2}$

$\Rightarrow \frac{3}{2a} =1-e^{-2}$

$\Rightarrow2a= \frac{3}{1-e^{-2}}$

$\Rightarrow a = 1.73$

Therefore the concentration of salt in the incoming brine is 1.73 g/L

8 0

3 years ago

Find the product.<br> (3b + 6at)(b - at)

adoni [48]

Answer:

3bat + 3b² - 6a²t²

Step-by-step explanation:

First you have to expand it to get;

3b(b - at) + 6at(b - at)

Then you can now multiply.

3b² - 3bat + 6bat - 6at²

Group like terms

6bat - 3bat + 3b² -6at²

3bat + 3b² - 6a²t²

4 0

3 years ago

Read 2 more answers

Which of the following is a coefficient in the expression 822 + 6?

ioda

Answer:

There are no coefficients.

Step-by-step explanation:

6 0

2 years ago

One year on Venus is equivalent to 224.7 days on Earth. How many days on Earth, in decimal from, are equivalent to 9 ½ years on

Ivan

By using simple rule of three, a period of 9.5 years on Venus is equivalent to 2134.65 terrestrial days.

<h3>How many terrestrial days are equivalent time in Venus?</h3>

One year on Earth is equivalent to 365.3 days and one year on Venus is equivalent to 224.7 days, the equivalent terrestrial time of 9.5 years on Venus is found by simple rule of three:

x = 9.5 yr × (224.7 days / 1 yr)

x = 2134.65 days

A period of 9.5 years on Venus is equivalent to 2134.65 terrestrial days.

To learn more on simple rule of three: brainly.com/question/15209325

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6 0

1 year ago

2) Which choice best represents:<br> Divide 8 by N<br> A. 8-N<br> B. N -8

Nataly_w [17]

Answer: A

Step-by-step explanation:

4 0

3 years ago