I am clueless about classifying polynomials by degree and number of terms. What kind of help can you give me?
1 answer:
Whenever there's an interruption of say a sign like + or -, what's before it and after it is a "term", so for example,
a + b <---- two terms
a + b + c <---- three terms
25x⁻¹ - 17y² + 3 - 21x² <---- four terms.
the degree of a polynomial, is based on the exponents of the variables in each term, you add the exponents of the variables for that term, and the term with the highest degree, is the degree of the polynomial,
notice the polynomial here, it has 4 terms... what's its degree? let's check the degree of each term,
the term with the highest degree is the first one, with 12, and therefore, the degree of that polynomial is 12 then.
a + b <---- two terms
a + b + c <---- three terms
25x⁻¹ - 17y² + 3 - 21x² <---- four terms.
the degree of a polynomial, is based on the exponents of the variables in each term, you add the exponents of the variables for that term, and the term with the highest degree, is the degree of the polynomial,
notice the polynomial here, it has 4 terms... what's its degree? let's check the degree of each term,
the term with the highest degree is the first one, with 12, and therefore, the degree of that polynomial is 12 then.
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Answer:
a) (dy/dt) = 2 - [3y/(100 + 2t)]
b) The solved differential equation gives
y(t) = 0.4 (100 + 2t) - 40000 (100 + 2t)⁻¹•⁵
c) Concentration of salt in the tank after 20 minutes = 0.2275 kg/L
Step-by-step explanation:
First of, we take the overall balance for the system,
Let V = volume of solution in the tank at any time
The rate of change of the volume of solution in the tank = (Rate of flow into the tank) - (Rate of flow out of the tank)
The rate of change of the volume of solution = dV/dt
Rate of flow into the tank = Fᵢ = 5 L/min
Rate of flow out of the tank = F = 3 L/min
(dV/dt) = Fᵢ - F
(dV/dt) = (Fᵢ - F)
dV = (Fᵢ - F) dt
∫ dV = ∫ (Fᵢ - F) dt
Integrating the left hand side from 100 litres (initial volume) to V and the right hand side from 0 to t
V - 100 = (Fᵢ - F)t
V = 100 + (5 - 3)t
V = 100 + (2) t
V = (100 + 2t) L
Component balance for the amount of salt in the tank.
Let the initial amount of salt in the tank be y₀ = 0 kg
Let the rate of flow of the amount of salt coming into the tank = yᵢ = 0.4 kg/L × 5 L/min = 2 kg/min
Amount of salt in the tank, at any time = y kg
Concentration of salt in the tank at any time = (y/V) kg/L
Recall that V is the volume of water in the tank. V = 100 + 2t
Rate at which that amount of salt is leaving the tank = 3 L/min × (y/V) kg/L = (3y/V) kg/min
Rate of Change in the amount of salt in the tank = (Rate of flow of salt into the tank) - (Rate of flow of salt out of the tank)
(dy/dt) = 2 - (3y/V)
(dy/dt) = 2 - [3y/(100 + 2t)]
To solve this differential equation, it is done in the attached image to this question.
The solution of the differential equation is
y(t) = 0.4 (100 + 2t) - 40000 (100 + 2t)⁻¹•⁵
c) Concentration after 20 minutes.
After 20 minutes, volume of water in tank will be
V(t) = 100 + 2t
V(20) = 100 + 2(20) = 140 L
Amount of salt in the tank after 20 minutes gives
y(t) = 0.4 (100 + 2t) - 40000 (100 + 2t)⁻¹•⁵
y(20) = 0.4 [100 + 2(20)] - 40000 [100 + 2(20)]⁻¹•⁵
y(20) = 0.4 [100 + 40] - 40000 [100 + 40]⁻¹•⁵
y(20) = 0.4 [140] - 40000 [140]⁻¹•⁵
y(20) = 56 - 24.15 = 31.85 kg
Amount of salt in the tank after 20 minutes = 31.85 kg
Volume of water in the tank after 20 minutes = 140 L
Concentration of salt in the tank after 20 minutes = (31.85/140) = 0.2275 kg/L
Hope this Helps!!!
