Let N be the set of natural numbers. Describe the following relation in words giving its domain -Maths 9th

Let N be the set of natural numbers. Describe the following relation in words giving its domain -Maths 9th
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The given relation stated in words is R = {(x, y) : x is the fourth power of y; x ∈ N, y ∈ {1, 2, 3, 4}}.
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I is the set of integers. Describe the following relations in words, giving its domain and range. -Maths 9th

Description : I is the set of integers. Describe the following relations in words, giving its domain and range. -Maths 9th

Last Answer : R = {(0, 0), (1, – 1), (2, – 2), (3, – 3) ...} = {(x, y) : y = – x, x ∈ W} Domain = {0, 1, 2, 3, ....} = W, Range = {...,– 3, – 2, – 1, 0}

On a set N of all natural numbers is defined the relation R by a R b iff the GCD of a and b is 2, then R is -Maths 9th

Description : On a set N of all natural numbers is defined the relation R by a R b iff the GCD of a and b is 2, then R is -Maths 9th

Last Answer : (c) Symmetric only Let a ∈N. Then (a, a) ∉R as the GCD of a' and a' is a' not 2. R is not reflexive Let a, b ∈N. Then, (a, b) ∉R ⇒ GCD of a' and b' is 2 ⇒ GCD of b' and a' is 2 ⇒ (b, a) ∈R ∴ R ... , let a = 4, b = 10, c = 12 GCD of (4, 10) = 2 GCD of (10, 12) = 2 But GCD of (4, 12) = 4.

If R is a relation defined on the set of natural numbers N such that (a, b) R (c, d) if and only if a + d = b + c, then R is -Maths 9th

Description : If R is a relation defined on the set of natural numbers N such that (a, b) R (c, d) if and only if a + d = b + c, then R is -Maths 9th

Last Answer : (d) An equivalence relationWe can check the given properties as follows: Reflexive: Let (a, b) ∈ N x N. Then (a, b) ∈ N ⇒ a + b = b + a (Communtative law of Addition) ⇒ (a, b) R (b, a) ⇒ (a, b) R (a, ... , f) ⇒ (a, b) R (e, f) on N x N so R is transitive.Hence R is an equivalence relation on N N.

Let R be a relation on the set N, defined by {(x, y) : 2x – y = 10} then R is -Maths 9th

Description : Let R be a relation on the set N, defined by {(x, y) : 2x – y = 10} then R is -Maths 9th

Last Answer : (a) ReflexiveGiven, {(\(x\), y) : 2\(x\) – y = 10} Reflexive, \(x\) R \(x\) = 2\(x\) – \(x\) = 10 ⇒ \(x\) = 10 ⇒ y = 10 ∴ Point (10, 10) ∈ N ⇒ R is reflexive.

The relation ‘is less than’ on a set of natural numbers is -Maths 9th

Description : The relation ‘is less than’ on a set of natural numbers is -Maths 9th

Last Answer : (c) Only transitiveLet N be the set of natural numbers. Then R = {(a, b) : a < b, a, b ∈N}A natural number is not less than itself ⇒ (a, a)∉R where a ∈N ⇒ R is not reflexive V a, b ∈N, (a, b) ∈R ⇒ a < b \( ot\ ... ∈N, (a, b) ∈R and (b, c) ∈R ⇒ a < b and b < c ⇒ a < c (a, c) ∈R ⇒ R is transitive.

Let R = {(3, 3), (6, 6), (9, 9), (12, 12), (6, 12), (3, 12), (3, 6)} be a relation on set A = {3, 6, 9, 12}. The relation is -Maths 9th

Description : Let R = {(3, 3), (6, 6), (9, 9), (12, 12), (6, 12), (3, 12), (3, 6)} be a relation on set A = {3, 6, 9, 12}. The relation is -Maths 9th

Last Answer : (c) Reflexive and transitive onlyHere A = {3, 6, 9, 12} and R = {(3, 3), (6, 6), (9, 9), (12, 12), (6, 12), (3, 12), (3, 6)} ∵ (3, 3), (6, 6), (9, 9), (12, 12) all belong to R ⇒ {(a, a): ∈R V a ∈A} ... 12) ∈ R and (3,12) ∈ R ⇒ (3, 12) ∈ R (x, y) ∈ R, (y, z) ∈ R ⇒ (x, z) ∈ R Hence R is transitive.

Let R be a relation defined on the set A of all triangles such that R = {(T1, T2) : T1 is similar to T2}. Then R is -Maths 9th

Description : Let R be a relation defined on the set A of all triangles such that R = {(T1, T2) : T1 is similar to T2}. Then R is -Maths 9th

Last Answer : (d) An equivalence relation.Every triangle is similar to itself, so (T1, T1) ∈ R ⇒ R is reflexive. (T1, T2) ∈ R ⇒ T1 ~ T2 ⇒T2 ~ T1, ⇒ (T2, T1) ∈ R ⇒ R is symmetrictransitive. ∴ R is an equivalence relation.

Let R be a relation on the set of integers given by a = 2^k .b for some integer k. Then R is -Maths 9th

Description : Let R be a relation on the set of integers given by a = 2^k .b for some integer k. Then R is -Maths 9th

Last Answer : (c) equivalence relationGiven, a R b = a = 2k .b for some integer. Reflexive: a R a ⇒ a = 20.a for k = 0 (an integer). True Symmetric: a R b ⇒ a = 2k b ⇒ b = 2-k . a ⇒ b R a as k, -k are both ... = 2k1 + k2 c, k1 + k2 is an integer. ∴ a R b, b R c ⇒ a R c True ∴ R is an equivalence relation.

Let A = {1, 2, 3, 4}, B = {5, 6, 7, 8}. Then R = {(1, 5), (1, 7), (2, 6)} is a relation from set A to B defined as : -Maths 9th

Description : Let A = {1, 2, 3, 4}, B = {5, 6, 7, 8}. Then R = {(1, 5), (1, 7), (2, 6)} is a relation from set A to B defined as : -Maths 9th

Last Answer : (d) R = {(a, b) : b/a is odd} Since (2, 6) ∈R, the relation a and b are odd does not exist. Since (1, 5) and (1, 7) ∈R, the relation a and b are even does not exist. None of the ... \(rac{7}{1}\) = 7, \(rac{6}{2}\) = 3, quotients being all odd numbers, the relation b/a is odd exists.

Let R be the relation in the set {1, 2, 3, 4} given by R = {(1, 2), (2, 2), (1, 1), (4, 4), (1, 3), (3, 3), (3, 2)}. Then R is -Maths 9th

Description : Let R be the relation in the set {1, 2, 3, 4} given by R = {(1, 2), (2, 2), (1, 1), (4, 4), (1, 3), (3, 3), (3, 2)}. Then R is -Maths 9th

Last Answer : (b) Reflexive and transitive but not symmetricLet A = {1, 2, 3, 4} • ∵ (1, 1), (2, 2), (3, 3) and (4, 4) ∈R ⇒ R is reflexive • ∵ (1, 2) ∈ R but (2, 1) ∉ R ; (1, 3) ∈ R and (3, 1) ∉ R ; (3, 2) ∈R and (2, 3) ∉ R ⇒ R is not symmetric • (1, 3) ∈ R and (3, 2) ∈ R and (1, 2) ∈ R ⇒ R is transitive.

On the set R of all real numbers, a relation R is defined by R = {(a, b) : 1 + ab > 0}. Then R is -Maths 9th

Description : On the set R of all real numbers, a relation R is defined by R = {(a, b) : 1 + ab > 0}. Then R is -Maths 9th

Last Answer : (a) Reflexive and symmetric only(a, a) ∈ R ⇒ 1 + a . a = 1 + a2 > 0 V real numbers a ⇒ R is reflexive (a, b) ∈ R ⇒ 1 + ab > 0 ⇒ 1 + ba > 0 ⇒ (b, a) ∈ R ⇒ R is symmetricWe observe that \(\big(1,rac{1}{2}\big) ... }{2},-1\big)\) ∈ Rbut (1, - 1) ∉ R as 1 + 1 (-1) = 0 \( ot>\) 0 ⇒ R is not transitive.

Define the term of Domain, codomain and range of a relation: -Maths 9th

Description : Define the term of Domain, codomain and range of a relation: -Maths 9th

Last Answer : Let R be a relation from set A to set B. Then, the set of first element of the ordered pairs in R is called the domain and the set of second elements of the ordered pairs in R is called the range. The second set B is called ... 16, 25, 36}, Range of R = {4, 5, 6} and Codomain of R = {1, 4, 5, 6}.

If R is a relation on a finite set A having n elements, then the number of relations on A is -Maths 9th

Description : If R is a relation on a finite set A having n elements, then the number of relations on A is -Maths 9th

Last Answer : (d) \(2^{n^2}\)Set A has n elements ⇒ n(A) = n ⇒ A × A has n × n = n2 elements ∴ Number of relations on A = Number of subsets of A × A = \(2^{n^2}\)

Let R be a relation from A = {1, 2, 3, 4, 5, 6} to B = {1, 3, 5} which is defined as “x is less than y”. -Maths 9th

Description : Let R be a relation from A = {1, 2, 3, 4, 5, 6} to B = {1, 3, 5} which is defined as “x is less than y”. -Maths 9th

Last Answer : R = {a, b : a < b, a ∈ A, b ∈ B}, where A = {1, 2, 3, 4, 5, 6} and B = {1, 3, 5}. ∴ R = {(1, 3), (1, 5), (2, 3), (2, 5), (3, 5), (4, 5)} Domain of R = {1, 2, 3, 4} Range of R = {3, 5} Codomain of R = {1, 3, 5}.

Let R be a relation defined as a Rb if | a – b | > 0, then the relation is -Maths 9th

Description : Let R be a relation defined as a Rb if | a – b | > 0, then the relation is -Maths 9th

Last Answer : (d) Symmetric and transitive| a - a | = | 0 | = 0 so (a, a) ∉R ⇒ R is not reflexive(a, b) ∈ R ⇒ | a - b | > 0 ⇒ | b - a | > 0 ⇒ (b, a) ∈R (∵ | a - b | = | b - a |) ⇒ R is symmetric (a, b) ∈ R ... a, b, c. ∴ | a - b | > 0 and | b - c | > 0 ⇒ | a - c | > 0 ⇒ (a, c) ∈ R ⇒ R is transitive.

Let A = {(2, 5, 11)}, B = {3, 6, 10} and R be a relation from A to B defined by R = {(a, b) : a and b are co-prime}. Then R is -Maths 9th

Description : Let A = {(2, 5, 11)}, B = {3, 6, 10} and R be a relation from A to B defined by R = {(a, b) : a and b are co-prime}. Then R is -Maths 9th

Last Answer : (c) {(2, 3), (5, 3), (5, 6), (11, 3), (11, 6), (11, 10)}

Let A = {1, 2, 3} and R = {(1, 2), (1, 1), (2, 3)} be a relation on A. -Maths 9th

Description : Let A = {1, 2, 3} and R = {(1, 2), (1, 1), (2, 3)} be a relation on A. -Maths 9th

Last Answer : (a) 1For the relation R to become transitive: (1, 2) ∈R and (2, 3) ∈R should imply (1, 3) ∈R ∴ Minimum one ordered pair (1, 3) should be added to R.

If a relation `R` on the set `N` of natural numbers is defined as `(x,y)hArrx^(2)-4xy+3y^(2)=0,Aax,yepsilonN`. Then the relation `R` is

Description : If a relation `R` on the set `N` of natural numbers is defined as `(x,y)hArrx^(2)-4xy+3y^(2)=0,Aax,yepsilonN`. Then the relation `R` is

Last Answer : If a relation `R` on the set `N` of natural numbers is defined as `(x,y)hArrx^(2) ... symmetric B. reflexive C. transitive D. an equivalence relation.

If R is a relation in N × N defined by (a, b) R (c, d) if and only if ad = bc, show that R is an equivalence relation. -Maths 9th

Description : If R is a relation in N × N defined by (a, b) R (c, d) if and only if ad = bc, show that R is an equivalence relation. -Maths 9th

Last Answer : (i) R is reflexive. For all (a, b) ∈ N N we have (a, b) R (a, b) because ab = ba ⇒ R is reflexive. (ii) R is symmetric. Suppose (a, b) R (c, d) Then (a, b) R (c, d) ⇒ ... ) R (e, f) ⇒ R is transitive. Since R is reflexive, symmetric and transitive, therefore, R is an equivalence relation on N N.

If the roots of the equation x^2 + x + 1 = 0 are in the ratio of m : n, then which one of the following relation holds ? -Maths 9th

Description : If the roots of the equation x^2 + x + 1 = 0 are in the ratio of m : n, then which one of the following relation holds ? -Maths 9th

Last Answer : answer:

Choose the correct option. The relation R = {(1, 1), (2, 2), (3, 3), (1, 2), (2, 3), (1, 3)} on a set A = {1, 2, 3} is -Maths 9th

Description : Choose the correct option. The relation R = {(1, 1), (2, 2), (3, 3), (1, 2), (2, 3), (1, 3)} on a set A = {1, 2, 3} is -Maths 9th

Last Answer : R = {(1, 1), (2, 2), (3, 3), (1, 2), (2, 3), (1, 3)} on set A = {1, 2, 3}. (i) Since (1, 1), (2, 2), (3, 3) ∈ R ⇒ R is reflexive. (ii) (1, 2) ∈ R but (2, 1) ∉ R ⇒ R is not symmetric. (iii) (1, 2) ∈ R, (2, 3) ∈ R and (1, 3) ∈R ⇒ R is transitive. ∴ Option (a) is the right answer.

Show that the relation R in the set A of all the books in a library of a school given by R = {(x, y): x and -Maths 9th

Description : Show that the relation R in the set A of all the books in a library of a school given by R = {(x, y): x and -Maths 9th

Last Answer : Given: A = {All books in a library of a school} R = {(x, y): x and y have the same number of pages} Reflexivity: (x, x) ∈ R ⇒ R is reflexive on A Symmetric: Since books x and y have the same number of pages ... and (y, z) ∈ R ⇒ (x, z) ∈ R ⇒ R is transitive on A. Hence, R is an equivalence relation.

Show that the relation ‘≅’ congruence on the set of all triangles in Euclidean plane geometry is an equivalence relation. -Maths 9th

Description : Show that the relation ‘≅’ congruence on the set of all triangles in Euclidean plane geometry is an equivalence relation. -Maths 9th

Last Answer : Reflexive : A ≅ A True Symmetric : if A ≅ B then B ≅ A True Transitive : if A ≅ B and B ≅ C, then A ≅ C True Therefore, the relation ‘≅’ is an equivalence relation.

Check whether the relation R defined in the set {1, 2, 3, 4, 5, 6} as R = {(a, b): b = a + 1} is reflexive, symmetric or transitive. -Maths 9th

Description : Check whether the relation R defined in the set {1, 2, 3, 4, 5, 6} as R = {(a, b): b = a + 1} is reflexive, symmetric or transitive. -Maths 9th

Last Answer : Reflexive: R = {(a, b) : b = a +1} = {(a, a + l) : a, a + 1∈{l, 2, 3, 4, 5, 6}} = {(1, 2), (2, 3), (3, 4), (4, 5), (5, 6)} ⇒ R is not reflexive since (a, a) ∉R for all a. Symmetric: R is not symmetric as (a ... as (a, b) ∈ R and (b, c) ∈ R but (a, c) ∉ R e.g., (1, 2) ∈ R (2, 3) ∈ R but (1, 3) ∉R

The relation R in the set {1, 2, 3} given by R = {(1, 2), (2, 1)} is: -Maths 9th

Description : The relation R in the set {1, 2, 3} given by R = {(1, 2), (2, 1)} is: -Maths 9th

Last Answer : (c) Symmetric onlyAs (1, 1), (2, 2) ∉ R so R is not reflexive. A relation a R b is said to symmetric if (a, b) ∈ R ⇒ (b, a) ∈ R. Here (1, 2) ∈ R and (2, 1) ∈ R so it is symmetric. Also, as (1, 2) ∈ R, but (2, 3), (1, 3) ∉ R, so R is not transitive.

Which of the following is an equivalence relation defined on set A = {1, 2, 3} -Maths 9th

Description : Which of the following is an equivalence relation defined on set A = {1, 2, 3} -Maths 9th

Last Answer : (c) {(1, 1), (2, 2), (3, 1), (1, 3), (3, 3)} Option (c) satisfies all the conditions of an equivalence relation. (1, 1), (2, 2), (3, 3) ∈ R ⇒ (a, a) ∈ R V a ∈ A ⇒ R is reflexive (3, 1) ∈ R and (1, 3) ∈ R ... R and (1, 1) ∈ R ⇒ (a, b) ∈ R, (b, c) ∈ R ⇒ (a, c) ∈ R V a, b, c ∈ A ⇒ R is transitive.

Given the relation R = {(1, 2), (2, 3)} on the set A = {1, 2, 3}, -Maths 9th

Description : Given the relation R = {(1, 2), (2, 3)} on the set A = {1, 2, 3}, -Maths 9th

Last Answer : (d) 7A = {1, 2, 3}. R = {(1, 2), (2, 3)} To make R an equivalence relation, it should be: (i) Reflexive: So three more ordered pairs (1, 1), (2, 2), (3, 3) should be added to R to make it ... 2, 1), (3, 2), (1, 3), (3, 1)} is an equivalence relation. So minimum 7 ordered pairs are to be added.

The relation “is parallel to” on a set S of all straight lines in a plane is : -Maths 9th

Description : The relation “is parallel to” on a set S of all straight lines in a plane is : -Maths 9th

Last Answer : (d) An equivalence relationLet R = {(x, y) : line x is parallel to line y, x y ∈ set of coplanar straight lines}. Every line is parallel to itself. So, if x ∈S, then (x, x) ∈R ⇒ R is ... | z ⇒ (x, z) ∈R ⇒ R is transitive ∴ R being reflexive, symmetric and transitive, it is an equivalence relation.

Let x and y be rational and irrational numbers, respectively. -Maths 9th

Description : Let x and y be rational and irrational numbers, respectively. -Maths 9th

Last Answer : Yes, (x + y) is necessarily an irrational number.

Let x and y be rational and irrational numbers, respectively. -Maths 9th

Description : Let x and y be rational and irrational numbers, respectively. -Maths 9th

Last Answer : Yes, (x + y) is necessarily an irrational number.

Let a, b, c be positive numbers, then a/(b+c) + b/(c+a) + c/(a+b) is -Maths 9th

Description : Let a, b, c be positive numbers, then a/(b+c) + b/(c+a) + c/(a+b) is -Maths 9th

Last Answer : answer:

Let a, b, c be positive numbers lying in the interval (0, 1], then a/(1+b+ca)+b/(a+c+ab)+c/(1+a+bc) is -Maths 9th

Description : Let a, b, c be positive numbers lying in the interval (0, 1], then a/(1+b+ca)+b/(a+c+ab)+c/(1+a+bc) is -Maths 9th

Last Answer : answer:

Let a1, a2, ..... an be positive real numbers such that a1a2a3 ...... an = 1. Then (1 + a1) (1 + a2) ..... (1 + an) is -Maths 9th

Description : Let a1, a2, ..... an be positive real numbers such that a1a2a3 ...... an = 1. Then (1 + a1) (1 + a2) ..... (1 + an) is -Maths 9th

Last Answer : answer:

Let O be any point inside a triangle ABC. Let L, M and N be the points on AB, BC and CA respectively, -Maths 9th

Description : Let O be any point inside a triangle ABC. Let L, M and N be the points on AB, BC and CA respectively, -Maths 9th

Last Answer : answer:

Let ABCD be a parallellogram. Let m and n be positive integers such that n < m < 2n. Let AC = 2 mn -Maths 9th

Description : Let ABCD be a parallellogram. Let m and n be positive integers such that n < m < 2n. Let AC = 2 mn -Maths 9th

Last Answer : answer:

How To Solve A Function F(x) Has These Properties The Domain Of F Is The Set Of Natural Numbers F (1)1 F (x plus 1)f(x) plus 3x(x plus 1) plus 1 A. Determine F (2) F(3) F (4) F (5) F(6) THANKS!!!!!!?

Description : How To Solve A Function F(x) Has These Properties The Domain Of F Is The Set Of Natural Numbers F (1)1 F (x plus 1)f(x) plus 3x(x plus 1) plus 1 A. Determine F (2) F(3) F (4) F (5) F(6) THANKS!!!!!!?

Last Answer : The function (sequence generator) f(x) with x∈ℕ has been defined as a recursive function (sequence), with the initial value defined for some x, ie starting form some some natural number (in this case 1), the value of the function (sequence) is ... (4 + 1) = f(4) + ...f(6) = f(5 + 1) = f(5) + ...

Let X be the set of all graduates in India. Elements x and y in X are said to be related, if they are graduates of the same university. -Maths 9th

Description : Let X be the set of all graduates in India. Elements x and y in X are said to be related, if they are graduates of the same university. -Maths 9th

Last Answer : R = {(x, y)}: x and y are graduates of same university, x, y ∈ {All graduates of India}. R is reflexive as (x, x) ∈ R, since x and x are graduates from the same university. R ... graduates from the same university ⇒ (x, z) ∈ R. R being reflexive, symmetric and transitive is an equivalence relation.

Determine the mean of first 10 natural numbers. -Maths 9th

Description : Determine the mean of first 10 natural numbers. -Maths 9th

Last Answer : Mean = 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10 / 10 = 55 / 10 = 5.5

Determine the mean of first 10 natural numbers. -Maths 9th

Description : Determine the mean of first 10 natural numbers. -Maths 9th

Last Answer : Mean = 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10 / 10 = 55 / 10 = 5.5

Find the arithmetic mean of first-five natural numbers. -Maths 9th

Description : Find the arithmetic mean of first-five natural numbers. -Maths 9th

Last Answer : Mean = 1 + 2 + 3 + 4 + 5/5 = 15/5 = 3

Out of 3n consecutive natural numbers 3 natural numbers are chosen at random without replacement. -Maths 9th

Description : Out of 3n consecutive natural numbers 3 natural numbers are chosen at random without replacement. -Maths 9th

Last Answer : (c) \(rac{3n^2-3n+2}{(3n-1)(3n-2)}\)In 3n consecutive natural numbers, (i) n numbers are of the form 3p (ii) n numbers are of the form 3p + 1 (iii) n numbers are of the form 3p + 2 For the ... ) We can select one number from each set.∴ Favourable number of cases = nC3 + nC3 + nC3 + (nC1 nC1 nC1)

How many natural numbers not exceeding 4321 can be formed with the digits 1, 2, 3, 4 if repetition is allowed? -Maths 9th

Description : How many natural numbers not exceeding 4321 can be formed with the digits 1, 2, 3, 4 if repetition is allowed? -Maths 9th

Last Answer : answer:

Show that If m > 1, then the sum of the mth powers of underline (n)even numbers is greater than n (n + 1)^m -Maths 9th

Description : Show that If m > 1, then the sum of the mth powers of underline (n)even numbers is greater than n (n + 1)^m -Maths 9th

Last Answer : answer:

What is the probability that a number selected at random from the set of numbers {1, 2, 3, …, 100} is a perfect cube? -Maths 9th

Description : What is the probability that a number selected at random from the set of numbers {1, 2, 3, …, 100} is a perfect cube? -Maths 9th

Last Answer : (a) \(rac{1}{25}\) Let us assume S as the sample space in all questions. S means the set denoting the total number of outcomes possible. Let S = {1, 2, 3, , 100} be the sample space. Then, n(S) = 100 Let A : ... ∴Required probability P(A) = \(rac{n(A)}{n(S)}\) = \(rac{4}{100}\) = \(rac{1}{25}\)

The solution set for the inequality 2x – 10 < 3x – 15 over the set of real numbers is -Maths 9th

Description : The solution set for the inequality 2x – 10 < 3x – 15 over the set of real numbers is -Maths 9th

Last Answer : answer:

The set of all real numbers x, for which x^2 – |x + 2| + x > 0, is -Maths 9th

Description : The set of all real numbers x, for which x^2 – |x + 2| + x > 0, is -Maths 9th

Last Answer : answer:

………..defines the structure of a relation which consists of a fixed set of attribute-domain pairs. A) Instance B) Schema c) Program D) Super Key

Description : ………..defines the structure of a relation which consists of a fixed set of attribute-domain pairs. A) Instance B) Schema c) Program D) Super Key

Last Answer : B) Schema

Determine the domain and range of the following relations: (i) {(–3, 1), (–1, 1), (1, 0), (3, 0)} -Maths 9th

Description : Determine the domain and range of the following relations: (i) {(–3, 1), (–1, 1), (1, 0), (3, 0)} -Maths 9th

Last Answer : (i) Domain = {-3, -1, 1, 3}, Range = {0, 1} (ii) Domain = {x : x is a multiple of 3} = {3n : n ∈ Z} Range = {y : y is a multiple of 5} = {5n : n ∈ Z} (iii) Relation = {(x, x2) : x is a prime number ... (7, 49), (11, 121), (13, 169)} Domain = {2, 3, 5, 7, 11, 13}, Range = {4, 9, 25, 49, 121, 169}

Given A = {–2, –1, 0, 1, 2}, which of the following relations on A have both domain and range equal to A? -Maths 9th

Description : Given A = {–2, –1, 0, 1, 2}, which of the following relations on A have both domain and range equal to A? -Maths 9th

Last Answer : (d) (i) and (iii)Let us examine the domain and range of the each relation individually: (i) R : is equal to means R = {(a, b) : a = b, a ∈ A, b ∈A} ∴ R = {(-2, -2), (-1, -1), (0, 0), (1, ... = {-2, -1, 0, 1} and range = {-1, 0, 1, 2}. ∴ Relations given in (i) and (iii) satisfy the given condition.

While discussing the properties of a parallelogram teacher asked about the relation between two angles x and y of a parallelogram as shown ... -Maths 9th

Description : While discussing the properties of a parallelogram teacher asked about the relation between two angles x and y of a parallelogram as shown ... -Maths 9th

Last Answer : (a) Yes , x < y is correct (b) Ð ADB =Ð DBC = y (alternate int. angles) since BC < CD (angle opp. to smaller side is smaller) there for, x < y (c) Truth value