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3 years ago

14

WILL AWARD A MEDAL AND FIVE STARS!!!!

1 answer:

ycow [4]3 years ago

6 0

Hello,

f(x)-2x-7

g(x)=-4x+3

(fog)(x)=f(g(x))=f(-4x+3)=-2(-4x+3)-7=8x-6-7=8x-13

(fog)(-5)=8*(-5)-13=-53

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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

On a scale drawing of Princeton, two landmarks are shown 2 centimeters apart. If the scale of the map is 1 centimeter : 35 kilom

tankabanditka [31]

Answer:

70 km

Step-by-step explanation:

2 cm is twice as much as 1 cm. So, the distance between landmarks will be twice as much as 35 km.

The actual distance is 70 km.

3 0

3 years ago

Solve this system of linear equation the x- and y-value with a comma. 18x+13y=60 6x+2y=6

OLEGan [10]

Answer:

18x + 13y = 60

6x + 2y = 6 ---> 6x = -2y +6 ---> 18x = -6y+18

Substitution

(-6y +18)+13y =60

-18 -18

--------------------------

7y = 42 so y=6

Then, 6x + 2(6) =6, which you will get x = -1

(-1, 6)

8 0

3 years ago

Read 2 more answers

How many solutions would there be to make an equation true?

mariarad [96]

Answer:

Step-by-step explanation:

Only one

5 0

3 years ago

Please help me with this question

jek_recluse [69]

A is the correct answer

E is 90 degrees

ABE is 52 so EBD is 64
$(180 - 52) \div 2 = 128 \div 2 = 64$

so D is 26:
$180 - (90 + 64) = 180 - 154 \\ = 126$

good luck

3 0

3 years ago