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adoni [48]
3 years ago
14

19

Mathematics
2 answers:
const2013 [10]3 years ago
8 0

Answer:

10:2:3

Step-by-step explanation:

PLEASE GIVE BRAINLIEST

Viktor [21]3 years ago
4 0

Answer:

10 : 2 : 3

Step-by-step explanation:

Water used = 1 litre = 1000ml

Sucrose used = 200ml

Saline used = 300ml

Ratio:

1000 : 200 : 300

10 : 2 : 3

The amount of water, sucrose and saline solution as a ratio = 10 : 2 : 3

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The set of the ordered pairs is a {(1, 6), (2, 12), (3, 18), (4, 24)}.

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5 0
2 years ago
6. A tank contains 100-gallon of pure water. At time t = 0, a solution containing 2 lb of salt per gallon flows into the tank at
SOVA2 [1]

Answer:

The answer is below

Step-by-step explanation:

Let Q represent the amount of salt in the tank at time t.

\frac{dQ}{dt}= flow\ in - flow\ out\\ \\flow\ in=3\ gal/min*2\ lb/gal=6\ lb/min\\\\net\ gain\ in\ tank\ volume=3-4=-1, henceflow\ out= \frac{4Q}{100-t} \\\\\frac{dQ}{dt}= 6-\frac{4Q}{100-t} \\\\\frac{dQ}{dt}+ \frac{4Q}{100-t}=6\\\\The \ integrating\ factor\ is:\\\\IF=e^{\int\limits {\frac{4}{100-t} } \, dt }=e^{-4\int\limits {\frac{-1}{100-t}}=e^{-4ln(100-t)}=(100-t)^{-4}}\\\\Multiplying\ through  \ by\ IF: \\\\(100-t)^{-4}\frac{dQ}{dt}+ (100-t)^{-4}\frac{4Q}{100-t}=6(100-t)^{-4}\\\\

Integrating:\\\\A(100-t)^{-4}=-2(100-t)^{-3}+c\\\\A=-2(100-t)+\frac{c}{(100-t)^{-4}} \\\\at, t=0,A=0\\\\0=-2(100-0)+\frac{c}{(100-0)^{-4}}\\\\c=0.02\\\\A=-2(100-t)+\frac{0.02}{(100-t)^{-4}}

7 0
3 years ago
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The sum of three consecutive integers is 108. Which equation would be used to solve this word problem? (HINT: The equation has b
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