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Let R = {(1, 3), (4, 2), (2, 4), (2, 3), (3, 1)} be a relation on the set A = {1, 2, 3, 4}. Then relation R is
We have the relation R = {(1, 3), (4, 2), (2, 4), (2, 3), (3, 1)} on the set A = {1, 2, 3, 4}. Definitions: 1. Function: A relation R is a function if each element of A occurs as the first component in at most one pair of R. 2. Transitive: A relation R is transitive if for all a, b, c ∈ A, wheneverRead more
We have the relation R = {(1, 3), (4, 2), (2, 4), (2, 3), (3, 1)} on the set A = {1, 2, 3, 4}.
Definitions:
1. Function: A relation R is a function if each element of A occurs as the first component in at most one pair of R.
2. Transitive: A relation R is transitive if for all a, b, c ∈ A, whenever (a, b) ∈ R and (b, c) ∈ R, then (a, c) ∈ R.
3. Symmetric: A relation R is symmetric if for all a, b ∈ A, whenever (a, b) ∈ R, then (b, a) ∈ R.
4. Reflexive: A relation R is reflexive if for all a ∈ A, (a, a) ∈ R.
Analysis:
1. Is R a function?
– The element 2 occurs as the first element in both (2, 4) and (2, 3), which goes against the definition of a function. Thus, R is not a function.
2. Is R transitive?
– Since (1, 3) ∈ R and (3, 1) ∈ R, we would need (1, 1) ∈ R for it to be transitive, but it isn’t there.
– For (2, 4) ∈ R and (4, 2) ∈ R, we would need (2, 2) ∈ R, which is absent too.
– Thus, R is not transitive.
3. Is R symmetric?
– Do we have (a, b) ∈ R implies (b, a) ∈ R?
– (1, 3) ∈ R, and (3, 1) ∈ R (symmetric pair).
– (4, 2) ∈ R, but (2, 4) ∈ R (symmetric pair).
– All needed symmetric pairs are there. So, R is symmetric.
4. Is R reflexive?
– See if (a, a) ∈ R for every a ∈ A:
– (1, 1), (2, 2), (3, 3), and (4, 4) are not in R.
– Hence, R is not reflexive.
Conclusion:
The relation R is:
– Not a function
– Not transitive
– Symmetric
– Not reflexive
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Let A = {1, 2, 3}, B = {4, 5, 6, 7} and let f = {(1, 4), (2, 5), (3, 6)} be a function from A to B. Based on the given information, f is best defined as:
Let A = {1, 2, 3} and B = {4, 5, 6, 7}. Let f = {(1, 4), (2, 5), (3, 6)} be a function from A to B. 1. Injective (One-to-one): Each element of A goes to a distinct element in B. There are no two elements of A going to the same element in B. Therefore, f is injective. 2. Surjective (Onto): Not all elRead more
Let A = {1, 2, 3} and B = {4, 5, 6, 7}.
Let f = {(1, 4), (2, 5), (3, 6)} be a function from A to B.
1. Injective (One-to-one): Each element of A goes to a distinct element in B. There are no two elements of A going to the same element in B.
Therefore, f is injective.
2. Surjective (Onto):
Not all elements of B are covered by f. For instance, 7 is not mapped by any element of A.
Therefore, f is not surjective.
3. Bijective (One-to-one and Onto):
Since f is not surjective, it cannot be bijective.
4. Function:
f is a function since every element of A is related to exactly one element of B.
Therefore, the correct answer is: Injective function.
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Let R be the relation in the set N given by R = {(a, b) : a = b – 2, b – 2, b > 6}, then
We are given the relation R on the set N, defined as: R = {(a, b) : a = b - 2, b > 6} This is to say, for every pair (a, b), a should be b - 2, and b must be greater than 6. (6, 8) ∈ R - a = 6 and b = 8. - a = b - 2 gives 6 = 8 - 2, which is true. - b = 8 > 6, so the condition is satisfied.Read more
We are given the relation R on the set N, defined as:
R = {(a, b) : a = b – 2, b > 6}
This is to say, for every pair (a, b), a should be b – 2, and b must be greater than 6.
(6, 8) ∈ R
– a = 6 and b = 8.
– a = b – 2 gives 6 = 8 – 2, which is true.
– b = 8 > 6, so the condition is satisfied.
– Therefore, (6, 8) ∈ R.
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Mud houses are cooler in summer and warmer in winter because
Mud houses do not get warmer in summer because mud is a bad conductor of heat. They resist the movement of heat that would make its interior warm with hot weather and cold in winter by either preventing the movement of heat within or outside in. Click here for more: https://www.tiwariacademy.com/nceRead more
Mud houses do not get warmer in summer because mud is a bad conductor of heat. They resist the movement of heat that would make its interior warm with hot weather and cold in winter by either preventing the movement of heat within or outside in.
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A body cools from 80 °C to 64 °C in 5 minutes and same body cools from 80 °C to 52 °C in 10 minutes. What is the temperature of surrounding?
We can solve this using Newton's Law of Cooling, which states, (dT/dt) = -k(T - Tₛ) (dT/dt) is the rate of change of temperature k is a constant T is the temperature of the body Tₛ is the temperature of the surroundings. We are given that the body cools from 80°C to 64°C in 5 minutes and again fromRead more
We can solve this using Newton’s Law of Cooling, which states,
(dT/dt) = -k(T – Tₛ)
(dT/dt) is the rate of change of temperature
k is a constant
T is the temperature of the body
Tₛ is the temperature of the surroundings.
We are given that the body cools from 80°C to 64°C in 5 minutes and again from 80°C to 52°C in 10 minutes. Applying Newton’s Law, we can derive the temperature of the surroundings.
We can use this formula to find the temperatures:
ln[(T₁ – Tₛ) / (T₂ – Tₛ)] = k(t₂ – t₁)
For the first cooling process, where it cools from 80 °C to 64 °C in 5 minutes:
ln[(80 – Tₛ) / (64 – Tₛ)] = k × 5
For the second cooling (from 80 °C to 52 °C in 10 minutes):
ln[(80 – Tₛ) / (52 – Tₛ)] = k × 10
This system of equations solves for the value of Tₛ.
After solving, we find that the temperature of the surroundings is 25 °C.
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