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

If C(2n, 3) : C(n, 2) = 12 : 1 find 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

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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

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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.

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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.

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Last Answer : Let f(x) = xn + an x + a will be the factor of xn + an if f(-a) = 0 Now f(-a) = (-a)n + an = 0 (since n is a odd +ve integer) Thus (x +a) is a factor of xn + an .

l, m and n are three parallel lines intersected by transversals p and q such that l, m and n cut off equal intercepts AB and BC on p (see figure). -Maths 9th

Description : l, m and n are three parallel lines intersected by transversals p and q such that l, m and n cut off equal intercepts AB and BC on p (see figure). -Maths 9th

Last Answer : Though E, draw a line parallel to p intersecting L at G and n at H respectively. Since l | | m ⇒ AG | | BE and AB | | GE [by construction] ∴ Opposite sides of quadrilateral AGEB are ... ∠DGE = ∠FHE [alternate interior angles] By ASA congruence axiom, we have △DEG ≅ △FEH Hence, DE = EF

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.

Show that x + a is a factor of xn + an for any odd +ve integer n. -Maths 9th

Description : Show that x + a is a factor of xn + an for any odd +ve integer n. -Maths 9th

Last Answer : Let f(x) = xn + an x + a will be the factor of xn + an if f(-a) = 0 Now f(-a) = (-a)n + an = 0 (since n is a odd +ve integer) Thus (x +a) is a factor of xn + an .

l, m and n are three parallel lines intersected by transversals p and q such that l, m and n cut off equal intercepts AB and BC on p (see figure). -Maths 9th

Description : l, m and n are three parallel lines intersected by transversals p and q such that l, m and n cut off equal intercepts AB and BC on p (see figure). -Maths 9th

Last Answer : Though E, draw a line parallel to p intersecting L at G and n at H respectively. Since l | | m ⇒ AG | | BE and AB | | GE [by construction] ∴ Opposite sides of quadrilateral AGEB are ... ∠DGE = ∠FHE [alternate interior angles] By ASA congruence axiom, we have △DEG ≅ △FEH Hence, DE = EF

A force of 50 N acts on a body for 1/10 s. Find the change in the momentum of the body. -Maths 9th

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Show that x+a is a factor of x^n+a^n for any odd positive n -Maths 9th

Description : Show that x+a is a factor of x^n+a^n for any odd positive n -Maths 9th

Last Answer : Let f(x)=xn+an. In order to prove that x+a is a factor of f(x) for any odd positive integer n, it is sufficient to show that f(−a)=0. f(−a)=(−a)n+an=(−1)nan+an f(−a)=(−1+1)an [ n is odd positive integer ] f(−a)=0×an=0 Hence, x+a is a factor of xn+an, when n is an odd positive integer.

A force of 50 N acts on a body for 1/10 s. Find the change in the momentum of the body. -Maths 9th

Description : A force of 50 N acts on a body for 1/10 s. Find the change in the momentum of the body. -Maths 9th

Last Answer : F = 50 N t = 1 / 10 mv - mu = ? F = m ( v - u ) / t ( newton ' s 2 nd law of motion ) 50 N = mv - mu / 1/10 sec 500 kg m / s = v - u

Show that x+a is a factor of x^n+a^n for any odd positive n -Maths 9th

Description : Show that x+a is a factor of x^n+a^n for any odd positive n -Maths 9th

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

Two lines l and m are perpendicular to the same line n.Are l and m perpendicular to each other ? -Maths 9th

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Last Answer : Solution :- No, they are parallel.

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

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In Fig. 8.40, points M and N are taken on opposite sides AB and CD, respectively of a parallelogram ABCD such that AM = CN. Show that AC and MN bisect each other. -Maths 9th

Description : In Fig. 8.40, points M and N are taken on opposite sides AB and CD, respectively of a parallelogram ABCD such that AM = CN. Show that AC and MN bisect each other. -Maths 9th

Last Answer : Solution :-

l,m and n are three parallel lines intersected by transversal p and q such that l,m and n cut-off equal intersepts AB and BC on p (Fig.8.55). Show that l,m and n cut - off equal intercepts DE and EF on q also. -Maths 9th

Description : l,m and n are three parallel lines intersected by transversal p and q such that l,m and n cut-off equal intersepts AB and BC on p (Fig.8.55). Show that l,m and n cut - off equal intercepts DE and EF on q also. -Maths 9th

Last Answer : Given:l∥m∥n l,m and n cut off equal intercepts AB and BC on p So,AB=BC To prove:l,m and n cut off equal intercepts DE and EF on q i.e.,DE=EF Proof:In △ACF, B is the mid-point of ... a triangle, parallel to another side, bisects the third side. Since E is the mid-point of DF DE=EF Hence proved.