I need help with these pre- calculus questions. Please help it is due tonight.

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

Drupady [299]

Answer:

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

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

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

dx = (5·c - 5·c/3)×dt = 20/3·c = $6\frac{2}{3} \cdot c \cdot dt$

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

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

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

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

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

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

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

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

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

$c =\dfrac{25 \ g/L }{10 \times 6\frac{2}{3} } = \dfrac{25 \ g/L}{10 \times \dfrac{20}{3} } =\dfrac{3}{200} \times 25 \ g/L= \dfrac{75}{200} \ g/L = \dfrac{3}{8} \ g/L$

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

4 0

3 years ago

Solve the inequality <br>1/2x - 4 < 3 - 2/3 x

34kurt

Answer:

X<6

Step-by-step explanation:

3x-24<18-4x

7x<42

5 0

3 years ago

A region where there is a large Amusement park is experiencing a heatwave officials at the park at a scatter plot showing the nu

Andreas93 [3]

Answer:

A

Step-by-step explanation:

Vertex appear to be at (5,115)

A) h = -b/2a

= -(20)/2(-2) = 5

k = -2(5²) + 20(5) + 60 = 110

6 0

3 years ago

the weight of an object on mars varies directly as the weight of the object on earth a 90 pound object on earth weighs 34 pounds

Mariulka [41]

We have been given that the weight of an object on the mars varies directly as the weight of the object on earth.

Let y be weight of an object on Mars and x be weight of an object on Earth.

$\\
y=kx$

We have been given that the weight of an object on Mars is 34 pounds and weight of the same object on Earth is 90 pounds.

Putting the given values in equation we get

$\\
34=k(90)\\
\\
k=\frac{34}{90}=\frac{17}{45}$

Therefore, our equation now becomes

$\\
y= \frac{17}{45}x$

Now we have been given that the weight of an object on Earth is 135 pounds and we have to find the weight of the same object on Mars,

We have been given that $\\
x=135$

Putting the given values in our equation we get

$\\
y= \frac{17}{45}(135)=\frac{17\cdot 135}{45}=17\cdot 3 = 51$

Therefore weight of the object on Mars will be 51 pounds.

7 0

3 years ago

Read 2 more answers

HELP ASAP!!!!! Explaiiin

zloy xaker [14]

Answer:

$x = \sqrt{y}$

Step-by-step explanation:

$x^{3/4 } = y^{3/8 }$

Note that

$x^{3/4 } = \sqrt[4]{x^3}$

$y^{3/8 } = \sqrt[8]{y^3}$

So,

$\sqrt[4]{x^3} = \sqrt[8]{y^3}$

$( \sqrt[4]{x^3} ) ^8= ( \sqrt[8]{y^3} )^8$

$( \sqrt[4]{x^3} ) ^8= y^3$

Once

$( \sqrt[4]{x^3} )^8 = (( \sqrt[4]{x^3} )^4)^2= (x^3)^2 = x^6$

$x^6 = y^3$

$x = \sqrt[6]{y^3}$

$x = y^{3\cdot 1/6}$

$x = y^{1/2}$

$x = \sqrt{y}$

6 0

3 years ago