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Ivan
3 years ago
9

What is the difference between complementary and supplementary angles

Mathematics
2 answers:
Vsevolod [243]3 years ago
4 0
Supplementary angles are two angles whose sum is 180 degrees while complementary angles are two angles whose sum is 90 degrees. Supplementary and complementary angles do not have to be adjacent (sharing a vertex and side, or next to), but they can be. ... The difference is their sum.
777dan777 [17]3 years ago
4 0

Complementary angles are two angles whose sum is 90 degrees.

Supplementary angles are two angles whose sum is 180 degrees.

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45

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Manolo and Annalisa make handbags. Manolo made 16 handbags the first day and 12 each day after that. Annalisa made 8 the first d
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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.

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