Prove that 2.4.6 ........... (2n) < (n + 1)^n. -Maths 9th

Prove that 2.4.6 ........... (2n) < (n + 1)^n. -Maths 9th
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If (log x)/(l + m - 2n) = (log y)/(m + n - 2l) = (log z)/(n + l - 2m), then xyz is equal to : -Maths 9th

Description : If (log x)/(l + m - 2n) = (log y)/(m + n - 2l) = (log z)/(n + l - 2m), then xyz is equal to : -Maths 9th

Last Answer : Let l+m−2nlogx​=m+n−2llogy​=n+l−2mlogz​=k(say) So, we get logx=k(l+m−2n) ....... (i) logy=k(m+n−2l) ....... (ii) logz=k(n+l−2m) ....... (iii) ∴logx+logy+logz=k(l+m−2n)+k(m+n−2l)+k(n+l−2m) ⇒logx+logy+logz=kl+km−2kn+km+kn−2kl+kn+kl−2km ⇒log(xyz)=0 ⇒logxyz=log1 ⇒xyz=1

If (log x)/(l + m - 2n) = (log y)/(m + n - 2l) = (log z)/(n + l - 2m), then xyz is equal to : -Maths 9th

Description : If (log x)/(l + m - 2n) = (log y)/(m + n - 2l) = (log z)/(n + l - 2m), then xyz is equal to : -Maths 9th

Last Answer : (b) 1Let \(rac{ ext{log}\,x}{l+m-2n}\) = \(rac{ ext{log}\,y}{m+n-2l}\) = \(rac{ ext{log}\,z}{n+l-2m}\) = k. Thenlog x = k(l + m – 2n), log y = k(m + n – 2l); log z = k(n + l – 2m) ⇒ log x + log y + log z = k(l + m – 2n) + k(m + n – 2l) + k(n + l – 2m)⇒ log(xyz) = 0 ⇒ log(xyz) = log 1 ⇒ xyz = 1.

A bag contains 2n + 1 coins. It is known that n of these coins have a head on both sides, whereas the remaining (n + 1) coins are fair. -Maths 9th

Description : A bag contains 2n + 1 coins. It is known that n of these coins have a head on both sides, whereas the remaining (n + 1) coins are fair. -Maths 9th

Last Answer : (a) 10As (n + 1) coins are fair P (Tossing a tail) = \(rac{rac{n+1}{2}}{2n+1}\) = \(rac{n+1}{2(2n+1)}\)∴ P (Tossing a head) = 1 - \(rac{n+1}{2(2n+1)}\) = \(rac{4n+2-n-1}{2(2n+1)}\) = \(rac{3n+1}{4n+2}\)Given, \(rac{3n+1}{4n+2}\) = \(rac{31}{42}\)⇒ 126n + 42 = 124n + 62 ⇒ 2n = 20 ⇒ n = 10.

If C(2n, 3) : C(n, 2) = 12 : 1 find n. -Maths 9th

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

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

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A group of 2n boys and 2n girls is divided at random into two equal batches. -Maths 9th

Description : A group of 2n boys and 2n girls is divided at random into two equal batches. -Maths 9th

Last Answer : (c) \(rac{(^{2n}C_n)^2}{^{4n}C_{2n}}\)Total number of boys and girls = 2n + 2n = 4n Since, there are two equal batches, each batch has 2n members ∴ Let S (Sample space) : Selecting one batch out of 2 ⇒ S : ... )2∴ Required probability = \(rac{n(E)}{n(S)}\) = \(rac{(^{2n}C_n)^2}{^{4n}C_{2n}}\)

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Description : 1.3.5.....(2n – 1) is -Maths 9th

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Without actual division, show that (x – 1)^2n – x^2n + 2x – 1 is divisible by 2x^3 – 3x^2 + x. -Maths 9th

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In the given figure, equal chords AB and CD of a circle with centre O cut at right angles at E. If M and N are the mid-points of AB and CD respectively, prove that OMEN is a square. -Maths 9th

Description : In the given figure, equal chords AB and CD of a circle with centre O cut at right angles at E. If M and N are the mid-points of AB and CD respectively, prove that OMEN is a square. -Maths 9th

Last Answer : Join OE. In ΔOME and ΔONE, OM =ON [equal chords are equidistant from the centre] ∠OME = ∠ONE = 90° OE =OE [common sides] ∠OME ≅ ∠ONE [by SAS congruency] ⇒ ME = NE [by CPCT] In quadrilateral OMEN, ... =ON , ME = NE and ∠OME = ∠ONE = ∠MEN = ∠MON = 90° Hence, OMEN is a square. Hence proved.

If a is a positive rational number and n is a positive integer greater than 1, prove that an is a rational number . -Maths 9th

Description : If a is a positive rational number and n is a positive integer greater than 1, prove that an is a rational number . -Maths 9th

Last Answer : We know that product of two rational number is always a rational number. Hence if a is a rational number then a2 = a x a is a rational number, a3 = 4:2 x a is a rational number. ∴ an = an-1 x a is a rational number.

In the given figure, equal chords AB and CD of a circle with centre O cut at right angles at E. If M and N are the mid-points of AB and CD respectively, prove that OMEN is a square. -Maths 9th

Description : In the given figure, equal chords AB and CD of a circle with centre O cut at right angles at E. If M and N are the mid-points of AB and CD respectively, prove that OMEN is a square. -Maths 9th

Last Answer : Join OE. In ΔOME and ΔONE, OM =ON [equal chords are equidistant from the centre] ∠OME = ∠ONE = 90° OE =OE [common sides] ∠OME ≅ ∠ONE [by SAS congruency] ⇒ ME = NE [by CPCT] In quadrilateral OMEN, ... =ON , ME = NE and ∠OME = ∠ONE = ∠MEN = ∠MON = 90° Hence, OMEN is a square. Hence proved.

If a is a positive rational number and n is a positive integer greater than 1, prove that an is a rational number . -Maths 9th

Description : If a is a positive rational number and n is a positive integer greater than 1, prove that an is a rational number . -Maths 9th

Last Answer : We know that product of two rational number is always a rational number. Hence if a is a rational number then a2 = a x a is a rational number, a3 = 4:2 x a is a rational number. ∴ an = an-1 x a is a rational number.

PQ and RS are two equal and parallel line segments.Any points M not lying on PQ or RS is joined to Q and S and lines through P parallel to SM meet at N.Prove that line segments MN and PQ are equal and parallel to each other. -Maths 9th

Description : PQ and RS are two equal and parallel line segments.Any points M not lying on PQ or RS is joined to Q and S and lines through P parallel to SM meet at N.Prove that line segments MN and PQ are equal and parallel to each other. -Maths 9th

Last Answer : hope its clear

In Fig.5.6, if AC = BD, then prove that AB = CD. -Maths 9th

Description : In Fig.5.6, if AC = BD, then prove that AB = CD. -Maths 9th

Last Answer : if equals are subtracted to equals ,the remainders are equal subtract bc on both sides ab-bc=bd-bc ab=cd hence proved

In Fig. 6.14, if x+y = w+z,then then prove that AOB is a line. -Maths 9th

Description : In Fig. 6.14, if x+y = w+z,then then prove that AOB is a line. -Maths 9th

Last Answer : Solution :-

2. Prove that if chords of congruent circles subtend equal angles at their centres, then the chords are equal. -Maths 9th

Description : 2. Prove that if chords of congruent circles subtend equal angles at their centres, then the chords are equal. -Maths 9th

Last Answer : Consider the following diagram- Here, it is given that AOB = COD i.e. they are equal angles. Now, we will have to prove that the line segments AB and CD are equal i.e. AB = CD. Proof: In triangles AOB ... ) So, by SAS congruency, ΔAOB ΔCOD. ∴ By the rule of CPCT, we have AB = CD. (Hence proved).

1. Recall that two circles are congruent if they have the same radii. Prove that equal chords of congruent circles subtend equal angles at their centres. -Maths 9th

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Last Answer : To recall, a circle is a collection of points whose every point is equidistant from its centre. So, two circles can be congruent only when the distance of every point of both the circles are equal from the centre ... ) So, by SSS congruency, ΔAOB ΔCOD ∴ By CPCT we have, AOB = COD. (Hence proved).

In the adjoining figure, ABCD is a parallelogram in which AB is produced to E so that BE = AB. Prove that ED bisects BC -Maths 9th

Description : In the adjoining figure, ABCD is a parallelogram in which AB is produced to E so that BE = AB. Prove that ED bisects BC -Maths 9th

Last Answer : Given, ABCD is a parallelogram. BE = AB To show, ED bisects BC Proof: AB = BE (Given) AB = CD (Opposite sides of ||gm) ∴ BE = CD Let DE intersect BC at F. Now, In ΔCDO and ΔBEO, ∠DCO = ... CD (Proved) ΔCDO ≅ ΔBEO by AAS congruence condition. Thus, BF = FC (by CPCT) Therefore, ED bisects BC. Proved

Prove that the quadrilateral formed by joining the mid points of quadrilateral forms parallelogram -Maths 9th

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Last Answer : Please see Exercise 8.2 - question 1 here in Quadrilaterals.

prove the following question -Maths 9th

Description : prove the following question -Maths 9th

Last Answer : Let ABCD is a quadrilateral in which P, Q, R, and S are the mid-points of sides AB, BC, CD, and DA respectively. Join PQ, QR, RS, SP, and BD. In ΔABD, S and P are the mid- ... SPQR is a parallelogram. We know that diagonals of a parallelogram bisect each other. Hence, PR and QS bisect each other.

PQRS is a square. A is a point on PS ,B is a point on PQ,C is a point on QR. ABC is a triangle inside square PQRS. Angle abc = 90° . If AP=BQ=CR then prove that angle BAC =45° -Maths 9th

Description : PQRS is a square. A is a point on PS ,B is a point on PQ,C is a point on QR. ABC is a triangle inside square PQRS. Angle abc = 90° . If AP=BQ=CR then prove that angle BAC =45° -Maths 9th

Last Answer : This is the sketch of the question but its hard to answer.

a squar ABCD in which AC =BE when BC produced .A is joined to E prove that FG=GE when AE intersect BD at F and CD at G -Maths 9th

Description : a squar ABCD in which AC =BE when BC produced .A is joined to E prove that FG=GE when AE intersect BD at F and CD at G -Maths 9th

Last Answer : Please give the figure to get your answer, as it is necessary to have figure to answer the question related to geometry.

An equation to prove -Maths 9th

Description : An equation to prove -Maths 9th

Last Answer : In mathematics, a proof is an inferential argument for a mathematical statement. In the argument, other previously established statements, such as theorems, can be used. In principle, a ... confirmatory cases. An unproved proposition that is believed to be true is known as a conjecture.

In a right angle triangle, prove that the hypotenuse is the longest side. -Maths 9th

Description : In a right angle triangle, prove that the hypotenuse is the longest side. -Maths 9th

Last Answer : The sum of angles of a triangle is180° If one aangke is of 90° then the sum of two angles is 90° It means that the angle forming 90° is biggest angle We know , Angle opposite to the longest side is largest. It means hypotenuse is the biggest side of right angled triangle

EF is the transversal to two parallel lines AB and CD. GM and HL are the bisector of the corresponding angles EGB and EHD.Prove that GL parallel to HL. -Maths 9th

Description : EF is the transversal to two parallel lines AB and CD. GM and HL are the bisector of the corresponding angles EGB and EHD.Prove that GL parallel to HL. -Maths 9th

Last Answer : AB || CD and a transversal EF intersects them ∴ ∠EGB = ∠GHD ( Corresponding Angles) ⇒ 2 ∠EGM = 2 ∠GHL ∵ GM and HL are the bisectors of ∠EGB and ∠EHD respectively. ⇒ ∠EGM = ∠GHL But these angles form a pair of equal corresponding angles for lines GM and HL and transversal EF. ∴ GM || HL.

ABC and ADC are two right triangles with common hypotenuse AC. Prove that angle CAD = angle CAB -Maths 9th

Description : ABC and ADC are two right triangles with common hypotenuse AC. Prove that angle CAD = angle CAB -Maths 9th

Last Answer : Given, AC is the common hypotenuse. ∠B = ∠D = 90°. To prove, ∠CAD = ∠CBD Proof: Since, ∠ABC and ∠ADC are 90°. These angles are in the semi circle. Thus, both the triangles are lying in the semi ... D are concyclic. Thus, CD is the chord. ⇒ ∠CAD = ∠CBD (Angles in the same segment of the circle)

Two circles intersect at A and B. AC and AD are respectively the diameters of the circles. Prove that C, B and D are collinear. -Maths 9th

Description : Two circles intersect at A and B. AC and AD are respectively the diameters of the circles. Prove that C, B and D are collinear. -Maths 9th

Last Answer : Join CB, BD and AB, Since, AC is a diameter of the circle with centre O. ∴ ∠ABC = 90° [angle in semi circle] ---- (i) Also, AD is a diameter of the circle with center O . ∴ ∠ABD = 90° [angle in ... ⇒ ∠ABC + ∠ABD = 180° So. CBD is a straight line. Hence C, B and D are collinear . Hence proved.

Prove that the line joining the mid-points of two parallel chords of a circle passes through the centre. -Maths 9th

Description : Prove that the line joining the mid-points of two parallel chords of a circle passes through the centre. -Maths 9th

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If two chords of a circle with a common end-point are inclined equally to the diameter through this common end-point, then prove that chords are equal. -Maths 9th

Description : If two chords of a circle with a common end-point are inclined equally to the diameter through this common end-point, then prove that chords are equal. -Maths 9th

Last Answer : Let AB and AC be two chords and AD be the diameter of the circle . Draw OL ⊥ AB and OM ⊥ AC . In ΔOLA and ΔOMA ∠ OLA = ∠ OMA = 90° OA = OA [common sides] ∠ OAL = ∠ OAM [given] . OLA ≅ OMA [by ASA congruency] OL =OM (by CPCT) Chords AM and AC are equidistant from O. ∴ AB = AC Hence proved.

State and prove-line joining the midpoint of any two sides of a triangle is parallel to throw side and is equal to 1/2 of it -Maths 9th

Description : State and prove-line joining the midpoint of any two sides of a triangle is parallel to throw side and is equal to 1/2 of it -Maths 9th

Last Answer : Here, In △△ ABC, D and E are the midpoints of sides AB and AC respectively. D and E are joined. Given: AD = DB and AE = EC. To Prove: DE ∥∥ BC and DE = 1212 BC. Construction: Extend line segment DE to ... we have DF ∥∥ BC and DF = BC DE ∥∥ BC and DE = 1212BC (DE = EF by construction) Hence proved.

ABCD is a parallelogram x and y are midpoints of BC and CD respectively.Prove that- Area of triangle axy =3/8 area of parallelogram ABCD -Maths 9th

Description : ABCD is a parallelogram x and y are midpoints of BC and CD respectively.Prove that- Area of triangle axy =3/8 area of parallelogram ABCD -Maths 9th

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

ABC is an isosceles triangle in which AB=AC.AD bisects exterior angles PAC and CD parallel AB.Prove that-i)angle DAC=angle BAC ii)∆BCD is a parallelogram -Maths 9th

Description : ABC is an isosceles triangle in which AB=AC.AD bisects exterior angles PAC and CD parallel AB.Prove that-i)angle DAC=angle BAC ii)∆BCD is a parallelogram -Maths 9th

Last Answer : AB =AC(given) Angle ABC =angle ACB (angle opposite to equal sides) Angle PAC=Angle ABC +angle ACB (Exterior angle property) Angle PAC =2 angle ACB - - - - - - (1) AD BISECTS ANGLE PAC. ANGLE ... AND AC IS TRANSVERSAL BC||AD BA||CD (GIVEN ) THEREFORE ABCD IS A PARALLEGRAM. HENCE PROVED........

If a and b are two rational numbers, prove that a + b, a - b, ab are rational numbers. -Maths 9th

Description : If a and b are two rational numbers, prove that a + b, a - b, ab are rational numbers. -Maths 9th

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Prove that the area of a parallelogram is the product of its side and the corresponding height. -Maths 9th

Description : Prove that the area of a parallelogram is the product of its side and the corresponding height. -Maths 9th

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Prove that a diagonal of a parallelogram divide it into two congruent triangles. -Maths 9th

Description : Prove that a diagonal of a parallelogram divide it into two congruent triangles. -Maths 9th

Last Answer : Given: A parallelogram ABCD and AC is its diagonal . To prove : △ABC ≅ △CDA Proof : In △ABC and △CDA, we have ∠DAC = ∠BCA [alt. int. angles, since AD | | BC] AC = AC [common side] and ∠BAC = ∠DAC [alt. int. angles, since AB | | DC] ∴ By ASA congruence axiom, we have △ABC ≅ △CDA

Prove that parallelogram on equal bases and between the same parallels are equal in area. -Maths 9th

Description : Prove that parallelogram on equal bases and between the same parallels are equal in area. -Maths 9th

Last Answer : Suppose AL and PM are the altitudes corresponding to equal bases AB and PQ of ||gm ABCD and PQRS respectively . Since the ||gm are between the same parallels PB and SC. ∴ AL = PM Now, ar(||gm ABCD) = AB AL ar(|| ... PM But, AB = PQ [given] AL = PM [proved] ∴ ar(||gm ABCD) = ar(||gm PQRS)

ABCD is a parallelogram whose diagonals intersect at O. If P is any point on BO, prove that : -Maths 9th

Description : ABCD is a parallelogram whose diagonals intersect at O. If P is any point on BO, prove that : -Maths 9th

Last Answer : (i) Since diagonals of a parallelogram bisect each other. ∴ O is the mid - point AC as well as BD. In △ADC, OD is a median. ∴ ar(△ADO) = ar(△CDO) [∵ A median of a triangle divide it into two triangles of equal ... and (i) , we have ar(△AOB) - ar(△AOP) = ar(△BOC) - ar(△COP) ⇒ ar(△ABP) = (△CBP)

In the given figure, ABCD is a square. Side AB is produced to points P and Q in such a way that PA = AB = BQ. Prove that DQ = CP. -Maths 9th

Description : In the given figure, ABCD is a square. Side AB is produced to points P and Q in such a way that PA = AB = BQ. Prove that DQ = CP. -Maths 9th

Last Answer : In △PAD, ∠A = 90° and DA = PA = PB ⇒ ∠ADP = ∠APD = 90° / 2 = 45° Similarly, in △QBC, ∠B = 90° and BQ = BC = AB ⇒∠BCQ = ∠BQC = 90° / 2 = 45° In △PAD and △QBC , we have PA = QB [given] ∠A = ... [each = 90° + 45° = 135°] ⇒ △PDC = △QCD [by SAS congruence rule] ⇒ PC = QD or DQ = CP

If two opposite sides of a cyclic quadrilateral are parallel , then prove that - (a) remaining two sides are equal (b) both the diagonals are equal -Maths 9th

Description : If two opposite sides of a cyclic quadrilateral are parallel , then prove that - (a) remaining two sides are equal (b) both the diagonals are equal -Maths 9th

Last Answer : Let ABCD be quadrilateral with ab||cd Join be. In triangle abd and CBD, Angle abd=angle cdb(alternate angles) Anglecbd=angle adb(alternate angles) Bd=bd(common) Abd=~CBD by asa test Ad=BC by cpct Since ad ... c(from 1) Ad =bc(proved above) Triangle adc=~bcd by sas test Ac=bd by cpct Hence proved

Without actual division, prove that 2x4 – 5x3 + 2x2 – x+ 2 is divisible by x2-3x+2. -Maths 9th

Description : Without actual division, prove that 2x4 – 5x3 + 2x2 – x+ 2 is divisible by x2-3x+2. -Maths 9th

Last Answer : Let p(x) = 2x4 - 5x3 + 2x2 - x+ 2 firstly, factorise x2-3x+2. Now, x2-3x+2 = x2-2x-x+2 [by splitting middle term] = x(x-2)-1 (x-2)= (x-1)(x-2) Hence, 0 of x2-3x+2 are land 2. We have to prove that, 2x4 ... )2 - 2 + 2 = 2x16-5x8+2x4+ 0 = 32 - 40 + 8 = 40 - 40 =0 Hence, p(x) is divisible by x2-3x+2.

If a+b+c= 5 and ab+bc+ca =10, then prove that a3 +b3 +c3 – 3abc = -25. -Maths 9th

Description : If a+b+c= 5 and ab+bc+ca =10, then prove that a3 +b3 +c3 – 3abc = -25. -Maths 9th

Last Answer : Prove that a3 +b3 +c3 – 3abc = -25

Prove that (a +b +c)3 -a3 -b3 – c3 =3(a +b)(b +c)(c +a). -Maths 9th

Description : Prove that (a +b +c)3 -a3 -b3 – c3 =3(a +b)(b +c)(c +a). -Maths 9th

Last Answer : Solution to (a +b +c) 3 -a 3 -b 3 – c 3 =3(a +b)(b +c)(c +a)

If two lines intersect prove that the vertically opposite angles are equal. -Maths 9th

Description : If two lines intersect prove that the vertically opposite angles are equal. -Maths 9th

Last Answer : Given Two lines AB and CD intersect at point O.

Prove that through a given point, we can draw only one perpendicular to a given line. -Maths 9th

Description : Prove that through a given point, we can draw only one perpendicular to a given line. -Maths 9th

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Prove that two lines that are respectively perpendicular to two intersecting lines intersect each other. -Maths 9th

Description : Prove that two lines that are respectively perpendicular to two intersecting lines intersect each other. -Maths 9th

Last Answer : Given Let lines l and m are two intersecting lines. Again, let n and p be another two lines which are perpendicular to the intersecting lines meet at point D. To prove Two lines n and p ... is a contradiction. Thus, our assumption is wrong. Therefore, lines n and p intersect at a point.

Prove that a triangle must have at least two acute angles. -Maths 9th

Description : Prove that a triangle must have at least two acute angles. -Maths 9th

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state and prove angle sum property of triangle -Maths 9th

Description : state and prove angle sum property of triangle -Maths 9th

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

P, Q, R and S are respectively the mid-points of sides AB, BC, CD and DA of quadrilateral ABCD in which AC = BD and AC ⊥ BD. Prove that PQRS is a square. -Maths 9th

Description : P, Q, R and S are respectively the mid-points of sides AB, BC, CD and DA of quadrilateral ABCD in which AC = BD and AC ⊥ BD. Prove that PQRS is a square. -Maths 9th

Last Answer : Given In quadrilateral ABCD, P, Q, R and S are the mid-points of the sides AB, BC, CD and DA, respectively. Also, AC = BD and AC ⊥ BD. To prove PQRS is a square. Proof Now, in ΔADC, S and R are the mid-points of the sides AD and DC respectively, then by mid-point theorem,