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3 years ago

I couod use some help plz

Mathematics

1 answer:

Anni [7]3 years ago

7 0

Answer:

Yes, you definitely couod! ;) I think it is B and C. (Not an expert tho)

Step-by-step explanation:

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Help me someone plz I need help

Tema [17]

Your answer would be -17 I think

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4 years ago

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Solve: (1 1/3) 24 = 32 b: 875 + 61/2= 1/2x + 31/2y= 91/5x +91/5= Please help me

In-s [12.5K]

Answer:

875 + 61/2= 1811/2

1/2x + 31/2y= y + 31x/2xy

91/5x +91/5= 91 + 91x/5

Step-by-step explanation:

I did not understood the question but I hope thet was the Answer

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3 years ago

PLEASE HELP ASAP!!! WILL MARK BRAINLIEST!!! (Solving Equations with Distributive Property)

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-1/2*12p = -6p

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3 years ago

A. B. C. D. Help me please!!

Alecsey [184]

It's B!!! ik cuz i ggogled it lol

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3 years ago

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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

4 years ago

I couod use some help plz​

I couod use some help plz