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leonid [27]
1 year ago
11

Solve using unit rates.

Mathematics
2 answers:
Mekhanik [1.2K]1 year ago
5 0

Answer:

59.99 minutes

Step-by-step explanation:

4 hours divided by 28 mins = 8.57142857 (rounded to 8.57)

7 x 8.57 = 59.99.

musickatia [10]1 year ago
5 0
Yes what they said on top
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Expreess 3.81 as a mixed number
Luba_88 [7]

Answer:

3.81 = 3 81/100

Step-by-step explanation:

This can’t be simplified so it stays as 3 \frac{81}{100}

8 0
2 years ago
Use PEMDAS to evaluate the expression 8 x 4 − 21 ÷ 3 + 5.<br><br> 30<br> 32<br> 34<br> 36
VikaD [51]
the answer should be 30
4 0
2 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
3 years ago
What are the answers to these thank you and please tell me how to get it
mote1985 [20]
5-3(1/2x+2)=-7
-3(1/2x+2)=-7-5
(1/2x+2)=4
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just to give u an idea
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I need help with these plzz
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Uhh my brains hurts looking at that
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