If ((n+1)/2)^n ((2n+1)/3)^n > (n!)^k, then k = -Maths 9th

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Description : If x + 1 is a factor of the polynomial 2x2 + kx, then the value of k is -Maths 9th

Last Answer : (c) Let p(x) = 2x2 + kx Since, (x + 1) is a factor of p(x), then p(-1)=0 2(-1)2 + k(-1) = 0 ⇒ 2-k = 0 ⇒ k= 2 Hence, the value of k is 2.

Description : If (2, 0) is a solution of the linear equation 2x + 3y = k, then the value of k is -Maths 9th

Last Answer : (a) Since, (2, 0) is a solution of the given linear equation 2x + 3y = k, then put x =2 and y= 0 in the equation. ⇒ 2 (2) + 3 (0) = k ⇒ k = 4 Hence, the value of k is 4.

Description : If x + 1 is a factor of the polynomial 2x2 + kx, then the value of k is -Maths 9th

Last Answer : (c) Let p(x) = 2x2 + kx Since, (x + 1) is a factor of p(x), then p(-1)=0 2(-1)2 + k(-1) = 0 ⇒ 2-k = 0 ⇒ k= 2 Hence, the value of k is 2.

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Last Answer : (a) Since, (2, 0) is a solution of the given linear equation 2x + 3y = k, then put x =2 and y= 0 in the equation. ⇒ 2 (2) + 3 (0) = k ⇒ k = 4 Hence, the value of k is 4.

Description : If x+1 is a factor of the polynomial 3x(square) - kx,then find the value of k. -Maths 9th

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Description : If y = (1/(a^(1-loga x))), z = (1/(a^(1-loga y))) and x = a^k, then k = -Maths 9th

Last Answer : (b) \(rac{1}{{1-log_az}}\) y = \(rac{1}{a^{1-log_ax}}\) = \(a^{-(1-log_ax)}\)⇒ logay = \(rac{1}{{1-log_ax}}\) and loga z = \(rac{1}{{1-log_ay}}\)∴ logaz = \(rac{1}{1-\bigg(rac{1}{1-log_ax}\bigg)}\) = \( ... x = \(rac{1}{{1-log_az}}\)⇒ x = \(a^{rac{1}{1-log_az}}\) = ak ⇒ k = \(rac{1}{{1-log_az}}\).

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

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Description : If (x + k) is a common factor of x^2 + px + q and x^2 + lx + m, then the value of k is -Maths 9th

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Description : If the three points (k, 2k), (2k, 3k) and (3, 1) are collinear then k is equal to -Maths 9th

Last Answer : (d) 3Let (x, y) be the co-ordinates of the third vertex of the triangle. Then\(rac{0+2+x}{3}\) = 1 and \(rac{0+0+y}{3}\) = 1⇒ 2 + \(x\) = 3 and y = 3 ⇒ \(x\) = 1, y = 3. ∴ Co-ordinates of vertices of the triangle ... - y3) + x2 (y2 - y3) + x3(y1 - y2)]= \(rac{1}{2}\) [0+6+0] = \(rac{6}{2}\) = 3.

Description : If the sum of the zeroes of the polynomial p(x) = (k2 – 14) x2 – 2x – 12 is 1, then find the value of k. -Maths 9th

Last Answer : p(x) = (k2 – 14) x2 – 2x – 12 Here a = k2 – 14, b = -2, c = -12 Sum of the zeroes, (α + β) = 1 …[Given] ⇒ − = 1 ⇒ −(−2)2−14 = 1 ⇒ k2 – 14 = 2 ⇒ k2 = 16 ⇒ k = ±4

Description : If the sum of zeroes of the quadratic polynomial 3x2 – kx + 6 is 3, then find the value of k. -Maths 9th

Last Answer : Here a = 3, b = -k, c = 6 Sum of the zeroes, (α + β) = − = 3 …..(given) ⇒ −(−)3 = 3 ⇒ k = 9

Description : If vertices of a triangles are (1, k), (4, -3) and (-9, 7) and its area is 15 sq. units then find then the value of k. -Maths 9th

Last Answer : hope it helps if the vertices of a triangle are (1,k),(4,−3)(−9,7) area = 15 sq.units. find the value of k. Area of △ 21 [x1 (y2 −y3 )+x2 (y3 −y1 )+x3 (y1 −y2 )]=15 21 [1(−3−7)+ ... k+3)]=15 21 [(−10+28−4k−9k−27)]=15 −10+28−4k−9k−27=30 −10+28−13k−27=30 −13k=30+10+27−28 −13k=39 k=1339 k=−3 thank u

Description : In Fig. 6.10, if m|n, then find the value of x. -Maths 9th

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Description : If 1/3 log3 M + 3 log3 N = 1 + log 0.008 5, then -Maths 9th

Last Answer : (b) \(N^9=rac{9}{M}\)\(rac{1}{3}\) log3 M + 3 log3 N = 1 + log0.008 5⇒ log3 M1/3 + log3N3 = 1 + log0.0085 ⇒ log3 M1/3 N3 = 1 + log0.0085 ⇒ M1/3 N3 = 3(1 + log0.0085) ⇒ M1/3 N3 = 31 . 3log0.0085⇒ \(N^9=rac{27 ... ^9=rac{27}{M}\big(3^{log_{rac{1}{5}}5}\big)\) = \(rac{1}{M}\) (27)(3-1) = \(rac{9}{M}.\)

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Last Answer : (b) 235n(A) = 5 and n(B) = 7 ∴ n(A × B) = 5 × 7 = 35 Total number of relations from A to B = Total number of subsets of (A × B) = 235.

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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.

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

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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 = \(2^{n^2}\)

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