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trasher [3.6K]
2 years ago
14

A model rocket is launched into the air from ground level.

Mathematics
1 answer:
Anon25 [30]2 years ago
8 0

p(x)=-16x^2+32x

p(x)=0


-16x^2+32x = 0

-16x(x-2)=0

x_1=0, x_2=2

Rocket will be in the air for 2 seconds.


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(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
2 years ago
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