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3 years ago

17/39 + 11/12 −( 5/12 + 4 /39 )

Mathematics

2 answers:

Nana76 [90]3 years ago

8 0

Answer:

.83333333 or 5/6

Step-by-step explanation:

Make common denominator

204/468 + 429/468 = 633/468 - (195/468 + 48/468)

Simplify: 633/468 - 243/468

390/468 can be further broken down to 5/6 if you divide by a common factor which is 78.

statuscvo [17]3 years ago

8 0

Answer:

≈ 1.35 i hope this helps! :)

Step-by-step explanation:

$\frac{17}{39}+\frac{11}{12}=1.35256410256$

$-(\frac{5}{12}+\frac{4}{39})$ = -0.51923076923

17/39+11/12 = 1.35256410256

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3 years ago

(1 point) A fish tank initially contains 15 liters of pure water. Brine of constant, but unknown, concentration of salt is flowi

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Answer:

a. $\dfrac{dx}{dt} = 6\frac{2}{3} \cdot c$

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c. $c = \dfrac{3}{8} \ g/L$

Step-by-step explanation:

a. The volume of water initially in the fish tank = 15 liters

The volume of brine added per minute = 5 liters per minute

The rate at which the mixture is drained = 5 liters per minute

The amount of salt in the fish tank after t minutes = x

Where the volume of water with x grams of salt = 15 liters

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$\dfrac{dx}{dt} = 6\frac{2}{3} \cdot c$

b. The amount of salt, x after t minutes is given by the relation

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$dx = 6\frac{2}{3} \cdot c \cdot dt$

$x(t) = \int\limits \, dx = \int\limits \left ( 6\frac{2}{3} \cdot c \right) \cdot dt$

$x(t) = 6\frac{2}{3} \cdot c \cdot t$

c. Given that in 10 minutes, the amount of salt in the tank = 25 grams, and the volume is 15 liters, we have;

$x(10) = 25 \ grams(15 \ in \ liters) = 6\frac{2}{3} \times c \times 10$

$6\frac{2}{3} \times c =\dfrac{25 \ grams }{10}$

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$c = \dfrac{3}{8} \ g/L$

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