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dimaraw [331]
3 years ago
15

Put the following fractions in order from least to greatest, 1/8 2/5, 3/5, 1/6 1/4, 2/3, 1/3

Mathematics
2 answers:
Naily [24]3 years ago
7 0

Answer:

1/8, 1/6, 1/4, 1/3, 2/5, 3/5, 2/3

Brut [27]3 years ago
3 0

Answer:

Order from Least to Greatest

1/8 < 1/6 < 1/4 < 1/3 < 2/5 < 3/5 < 2/3

Step-by-step explanation:

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A tank contains 60 kg of salt and 1000 L of water. Pure water enters a tank at the rate 6 L/min. The solution is mixed and drain
MissTica

Answer:

(a) 60 kg; (b) 21.6 kg; (c) 0 kg/L

Step-by-step explanation:

(a) Initial amount of salt in tank

The tank initially contains 60 kg of salt.

(b) Amount of salt after 4.5 h

\text{Let A = mass of salt after t min}\\\text{and }r_{i} = \text{rate of salt coming into tank}\\\text{and }r_{0} =\text{rate of salt going out of tank}

(i) Set up an expression for the rate of change of salt concentration.

\dfrac{\text{d}A}{\text{d}t} = r_{i} - r_{o}\\\\\text{The fresh water is entering with no salt, so}\\ r_{i} = 0\\r_{o} = \dfrac{\text{3 L}}{\text{1 min}} \times \dfrac {A\text{ kg}}{\text{1000 L}} =\dfrac{3A}{1000}\text{ kg/min}\\\\\dfrac{\text{d}A}{\text{d}t} = -0.003A \text{ kg/min}

(ii) Integrate the expression

\dfrac{\text{d}A}{\text{d}t} = -0.003A\\\\\dfrac{\text{d}A}{A} = -0.003\text{d}t\\\\\int \dfrac{\text{d}A}{A} = -\int 0.003\text{d}t\\\\\ln A = -0.003t + C

(iii) Find the constant of integration

\ln A = -0.003t + C\\\text{At t = 0, A = 60 kg/1000 L = 0.060 kg/L} \\\ln (0.060) = -0.003\times0 + C\\C = \ln(0.060)

(iv) Solve for A as a function of time.

\text{The integrated rate expression is}\\\ln A = -0.003t +  \ln(0.060)\\\text{Solve for } A\\A = 0.060e^{-0.003t}

(v) Calculate the amount of salt after 4.5 h

a. Convert hours to minutes

\text{Time} = \text{4.5 h} \times \dfrac{\text{60 min}}{\text{1h}} = \text{270 min}

b.Calculate the concentration

A = 0.060e^{-0.003t} = 0.060e^{-0.003\times270} = 0.060e^{-0.81} = 0.060 \times 0.445 = \text{0.0267 kg/L}

c. Calculate the volume

The tank has been filling at 6 L/min and draining at 3 L/min, so it is filling at a net rate of 3 L/min.

The volume added in 4.5 h is  

\text{Volume added} = \text{270 min} \times \dfrac{\text{3 L}}{\text{1 min}} = \text{810 L}

Total volume in tank = 1000 L + 810 L = 1810 L

d. Calculate the mass of salt in the tank

\text{Mass of salt in tank } = \text{1810 L} \times \dfrac{\text{0.0267 kg}}{\text{1 L}} = \textbf{21.6 kg}

(c) Concentration at infinite time

\text{As t $\longrightarrow \, -\infty,\, e^{-\infty} \longrightarrow \, 0$, so A $\longrightarrow \, 0$.}

This makes sense, because the salt is continuously being flushed out by the fresh water coming in.

The graph below shows how the concentration of salt varies with time.

3 0
3 years ago
Dans school is planning a field trip to a museum. Bus company A charges a $40 rental fee plus $4 for each student. Bus company B
givi [52]

Answer:

30 students would have to go for the cost to be the same.

Step-by-step explanation:

Bus A: $40 + ($4 * 30) = $160

Bus B: $100 + ($2 * 30) = $160

7 0
3 years ago
If 0.00005893 is expressed in the form 5.893 × 10n, what is the value of n ?
yan [13]
In the scientific value of 5.893 x 10n. The standard value is 0.00005893 what is n? To better illustrate this phenomenon, we can explain it further under the rules of scientific notation.
For example. <span><span>
1. </span><span> 3 x 10^3 = 3 x 100 = 300</span></span>
<span><span>2. </span><span> 3 x 10^-3 = 3 x 0.001 = 0.003</span></span>

Solution:
0.00005893 = 5.893 x 0.00001 = 5.893 x 10^-5  
n= ^-5



7 0
3 years ago
Alright a real question I need help on
andreyandreev [35.5K]
B is your answer to this question
3 0
2 years ago
Solve the system of equations. -10y+9x=-9 <br> 10y+5x=-5
alexira [117]

by the use of elimination method

make all coefficients of subject to be eliminated similar..by multiplying the coefficients with one another

for eqn(i)

5(-10y+9x=-9)

-50y+45x=-45

for eqn(ii)

9(10y+5x=-5)

90y+45x=-45

-50y+45x=-45

90y+45x=-45

...subtract each set from the other...

we get

-140y+0=0

y=0

from eqn(i)

10y+5x=-5

0+5x=-5

x= -1

3 0
3 years ago
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