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

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
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1 Answer

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} ⇒ R is reflexive • (3, 6) ∈ R but (6, 3) ∉R. Also (6, 12) ∈ R but (12, 6) ∉R. So (a, b) ∈R ⇒ (b, a) ∈R Hence R is not symmetric Now (3, 6) ∈ R, (6, 12) ∈ R and (3,12) ∈ R ⇒ (3, 12) ∈ R (x, y) ∈ R, (y, z) ∈ R ⇒ (x, z) ∈ R Hence R is transitive.
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Related questions

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 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 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.

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.

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 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 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 = {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.

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

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

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

Last Answer : The given relation stated in words is R = {(x, y) : x is the fourth power of y; x ∈ N, y ∈ {1, 2, 3, 4}}.

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.

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.

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.

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.

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.

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.

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 PS be the median of the triangle with vertices P(2, 2), Q(6, –1) and R(7, 3). -Maths 9th

Description : Let PS be the median of the triangle with vertices P(2, 2), Q(6, –1) and R(7, 3). -Maths 9th

Last Answer : (b) a = √2b Let D be the mid-point of BC. Then D ≡ \(\bigg(rac{a+0}{2},rac{0}{2}\bigg)\)i.e. \(\bigg(rac{a}{2},0\bigg)\)Let E be the mid-point of AC, thenE = \(\bigg(rac{a+0}{2},rac{0+b}{2}\bigg)\) = \(\bigg ... \(rac{b}{a}.\)∴ From (i), \(rac{-2b}{a}\) x \(rac{b}{a}\) = -1⇒ 2b2 = a2 ⇒ a = √2 .

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.

For any two real number a b and , we defined aRb if and only if sin^2a + cos^2b = 1. The relation R is -Maths 9th

Description : For any two real number a b and , we defined aRb if and only if sin^2a + cos^2b = 1. The relation R is -Maths 9th

Last Answer : (d) an equivalence relationGiven, a R b ⇒ sin2a + cos2b = 1 Reflexive: a R a ⇒ sin2 a + cos2 a = 1 ∀ a ∈ R (True) Symmetric: a R b ⇒ sin2 a + cos2 b = 1 ⇒ 1 - cos2 a + 1 - sin2 b = 1 ⇒ sin2 b + ... + cos2 b + sin2 b + cos2 c = 2 ⇒ sin2 a + cos2 c = 1 ⇒ a R c (True)∴ R is an equivalence relation.

Let u = (log2x)^2 – 6(log2x) + 12, where x is a real number. Then the equation x^u = 256 has : -Maths 9th

Description : Let u = (log2x)^2 – 6(log2x) + 12, where x is a real number. Then the equation x^u = 256 has : -Maths 9th

Last Answer : ⇒ u=(log2 x)2−log2 x+12 ⇒ We can take log2 x=y ⇒ Then equation becomes y2−6y+12=u ⇒ Given that xu=256 Applying log we get, ⇒ ulog2 x=8 ∴ u=log2 x8 =y8 So our equation becomes, ⇒ y2−6y+12=y8 ⇒ ... We get y=2 ⇒ So, log2 x=2 ⇒ x=22 ∴ x=4 ∴ The given equation has exactly one solution for x

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.

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.

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.

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.

The temperature of a liquid can be measured in Kelvin units as x K or in Fahrenheit units as y°F. The relation between the two system of measurement of temperature is given by the linear equation y = 9/5 ( x - 273 ) + 32 -Maths 9th

Description : The temperature of a liquid can be measured in Kelvin units as x K or in Fahrenheit units as y°F. The relation between the two system of measurement of temperature is given by the linear equation y = 9/5 ( x - 273 ) + 32 -Maths 9th

Last Answer : Solution :-

Let P(–3, 2), Q(–5, –5), R(2, –3) and S(4, 4) be four points in a plane. Then show that PQRS is a rhombus. Is it a square ? -Maths 9th

Description : Let P(–3, 2), Q(–5, –5), R(2, –3) and S(4, 4) be four points in a plane. Then show that PQRS is a rhombus. Is it a square ? -Maths 9th

Last Answer : Let P(1, -1), Q \(\big(rac{-1}{2},rac{1}{2}\big)\) and R(1,2) be the vertices of the ΔPQR.Then, PQ = \(\sqrt{\big(rac{-1}{2}-1\big)^2+\big(rac{1}{2}+1\big)^2}\) = \(\sqrt{rac{9}{4}+rac{9}{4}} ... {3\sqrt2}{2}\)PR = \(\sqrt{(1-1)^2+(2+1)^2}\) = \(\sqrt9\) = 3∵ PQ = QR, the triangle PQR is isosceles.

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.

Let y varies directly as x. If y = 12 when x = 4, then write a linear equation. What is value of y when x = 5 ? -Maths 9th

Description : Let y varies directly as x. If y = 12 when x = 4, then write a linear equation. What is value of y when x = 5 ? -Maths 9th

Last Answer : Linear equation

Let y varies directly as x. If y = 12 when x = 4, then write a linear equation. What is value of y when x = 5 ? -Maths 9th

Description : Let y varies directly as x. If y = 12 when x = 4, then write a linear equation. What is value of y when x = 5 ? -Maths 9th

Last Answer : Linear equation

If AB = PQ, BC = QR and AC = PR, then write the congruence relation between the triangles. [Fig. 7.6] -Maths 9th

Description : If AB = PQ, BC = QR and AC = PR, then write the congruence relation between the triangles. [Fig. 7.6] -Maths 9th

Last Answer : Solution :- △ ABC ≅ △PQR

Let A = {2, 3, 5, 6}. Then, which of the following relations is transitive only? -Maths 9th

Description : Let A = {2, 3, 5, 6}. Then, which of the following relations is transitive only? -Maths 9th

Last Answer : (d) R = {(2, 3), (3, 5), (2, 5)}.

Let R1 and R2 be the remainders when the polynomials x^3 + 2x^2 – 5ax – 7 and x^2 + ax^2 – 12x + 6 are divided by (x + 1) and (x – 2) respectively. -Maths 9th

Description : Let R1 and R2 be the remainders when the polynomials x^3 + 2x^2 – 5ax – 7 and x^2 + ax^2 – 12x + 6 are divided by (x + 1) and (x – 2) respectively. -Maths 9th

Last Answer : answer:

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

Define the term of Relation with example. -Maths 9th

Description : Define the term of Relation with example. -Maths 9th

Last Answer : Relation: If A and B are any two non-empty sets, then any subset of A B is defined as a relation from A to B. For example, Suppose A = {1, 2, 3} and B = {1, 2, 3, 4}. Then {(2, 3), ... 3)} is a relation in A B. Many more relations (subsets) can be selected at random from our product set A B.

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}.

Define the term of Inverse of a relation. -Maths 9th

Description : Define the term of Inverse of a relation. -Maths 9th

Last Answer : For any binary relation R, a second relation can be constructed by merely interchanging first and second components in every ordered pair. The relation thus obtained is called the inverse of the first one and designated as R-1. Thus, R-1 = {(y, x ... {(1, 2), (2, 3), (3, 4), (5, 4)}. So (R-1)-1 = R.

Write ‘yes’ if each of the following relations is an equivalence relation. -Maths 9th

Description : Write ‘yes’ if each of the following relations is an equivalence relation. -Maths 9th

Last Answer : (i) is parallel to' is reflexive, because any line is parallel to itself. Symmetric, because if line l is parallel to line m, then line m is parallel to line l. Transitive, because if l || m ... y cannot be a factor of x. (v) No, because the relation is reflexive and transitive but not symmetric.

If a, b, c, d are positive reals such that a + b + c + d = 2, then M = (a + b) (c + d) satisfies the relation -Maths 9th

Description : If a, b, c, d are positive reals such that a + b + c + d = 2, then M = (a + b) (c + d) satisfies the relation -Maths 9th

Last Answer : answer:

The lengths of the sides a, b, c of a ΔABC are connected by the relation a^2 + b^2 = 5c^2. The angle between medians drawn to the sides 'a' and 'b' is -Maths 9th

Description : The lengths of the sides a, b, c of a ΔABC are connected by the relation a^2 + b^2 = 5c^2. The angle between medians drawn to the sides 'a' and 'b' is -Maths 9th

Last Answer : Let median through C be CF. AF=FB=c2 CF=122(a2+b2)−c2−−−−−−−−−−−−√=3c2 CG=c where G is the centroid and GF=c2 34( ... +M2b+9c24 c2=(23M2a)+(23M2b) BC2=AG2+BG2 So medians through A and B are perpendicular.

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:

Find the relation between x and y if points (2, 1), (x, y) and (7, 5) are collinear. -Maths 9th

Description : Find the relation between x and y if points (2, 1), (x, y) and (7, 5) are collinear. -Maths 9th

Last Answer : answer:

Let R and S be two fuzzy relations defined as: Image Then, the resulting relation, T, which relates elements of universe x to elements of universe z using max-min composition is given by

Description : Let R and S be two fuzzy relations defined as: Then, the resulting relation, T, which relates elements of universe x to elements of universe z using max-min composition is given by

Last Answer : Answer: C

Let pk(R) denotes primary key of relation R. A many-to-one relationship that exists between two relations R1 and R2 can be expressed as follows: (1) pk(R2)→pk(R1) (2) pk(R1)→pk(R2) (3) pk(R2)→R1∩R2 (4) pk(R1)→R1∩R2

Description : Let pk(R) denotes primary key of relation R. A many-to-one relationship that exists between two relations R1 and R2 can be expressed as follows: (1) pk(R2)→pk(R1) (2) pk(R1)→pk(R2) (3) pk(R2)→R1∩R2 (4) pk(R1)→R1∩R2

Last Answer : Answer: 2

A closed iron tank 12 m long 9 m wide and 4 m deep is to be made . Determine the cost of iron sheet used at the rate of rs 5 per meter , sheet being 2 m wide. -Maths 9th

Description : A closed iron tank 12 m long 9 m wide and 4 m deep is to be made . Determine the cost of iron sheet used at the rate of rs 5 per meter , sheet being 2 m wide. -Maths 9th

Last Answer : This answer was deleted by our moderators...

NCERT Solutions for class 9 Maths Chapter 12 Heron’s Formula Exercise 12.1 -Maths 9th

Description : NCERT Solutions for class 9 Maths Chapter 12 Heron’s Formula Exercise 12.1 -Maths 9th

Last Answer : 1. A traffic signal board, indicating SCHOOL AHEAD', is an equilateral triangle with side a'. Find the area of the signal board, using Heron's formula. If its perimeter is 180 cm, what will be the area of the signal board? 2 ... x = 30 - 12 - 12 cm ⇒ x = 6 cm Now, semi-perimeter = (30/2)cm = 15 cm

NCERT Solutions for class 9 Maths Chapter 12 Heron’s Formula Exercise 12.2 -Maths 9th

Description : NCERT Solutions for class 9 Maths Chapter 12 Heron’s Formula Exercise 12.2 -Maths 9th

Last Answer : 1. A park, in the shape of a quadrilateral ABCD, has - C = 90°, AB = 9 m, BC = 12 m, CD = 5 m and AD = 8 m. How much area does it occupy? Solution: Let us join B and D, such that ΔBCD is ... The given field is in the form of a trapezium ABCD such that parallel sides AB = 10 m and DC = 25 m

MCQ Questions for Class 9 Maths Chapter 12 Heron's Formula with answers Heron`s Formula -Maths 9th

Description : MCQ Questions for Class 9 Maths Chapter 12 Heron's Formula with answers Heron`s Formula -Maths 9th

Last Answer : Below you will find MCQ Questions of Chapter 12 Heron's Formula Class 9 Maths Free PDF Download that will help you in gaining good marks in the examinations and also cracking competitive exams. ... Heron's Formula MCQ Questions will help you in practising more and more questions in less time.