The product (a + b) (a – b) (a2 – ab + b2) (a2 + ab + b2) is equal to (a) a6 + b6 (b) a6 – b6 (c) a3 – b3 (d) a3 + b3 -Maths 9th

The product (a + b) (a – b) (a2 – ab + b2) (a2 + ab + b2) is equal to (a) a6 + b6 (b) a6 – b6 (c) a3 – b3 (d) a3 + b3 -Maths 9th
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If the roots ff the equation (c2 – ab)x2 – 2(a2 – bc)x + b2 – ac = 0 are equal, then prove that either a = 0 or a3 + b3 + c3 = 3abc -Maths 10th

Description : If the roots ff the equation (c2 – ab)x2 – 2(a2 – bc)x + b2 – ac = 0 are equal, then prove that either a = 0 or a3 + b3 + c3 = 3abc -Maths 10th

Last Answer : (c2 – ab) x2 + 2(bc - a2 ) x+ (b2 – ac) = 0 Comparing with Ax2 + Bx + C = 0 A = (c2 – ab), B = 2(bc - a2 ) and C = b2 – ac According to the question, B2 - 4AC = 0 Put the values in the above equation we get 4a(a3 + b3 + c3 -3abc) = 0 hence, a = 0 or a3 + b3 + c3 = 3ab

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

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

If a + b + c = 9 and ab + bc + ca = 23, then a3 + b3 + c3 – 3 abc = (a) 108 (b) 207 (c) 669 (d) 729 -Maths 9th

Description : If a + b + c = 9 and ab + bc + ca = 23, then a3 + b3 + c3 – 3 abc = (a) 108 (b) 207 (c) 669 (d) 729 -Maths 9th

Last Answer : a+b+c=9 and a2+b2+c2=35 Using formula, (a+b+c)2=a2+b2+c2+2(ab+bc+ca) 92=35+2(ab+bc+ca) 2(ab+bc+ca)=81−35=46 (ab+bc+ca)=23 using formula, (a3+b3+c3)−3abc=(a2+b2+c2−ab−bc−ca)(a+b+c) a3+b3+c3−3abc=(35−23)×9=9×12=108

If a – b = 4 and ab = 21, find the value of a3-b3. -Maths 9th

Description : If a – b = 4 and ab = 21, find the value of a3-b3. -Maths 9th

Last Answer : Given, a−b=4,ab=21 Then, ⇒(a−b)3=a3−3a2b+3ab2−b3=a3−3ab(a−b)−b3 ⇒43=a3−b3−3(21)(4) ⇒64+252=a3−b3 ∴a3−b3=316

If a + b = 10 and ab = 21, find the value of a3 + b3. -Maths 9th

Description : If a + b = 10 and ab = 21, find the value of a3 + b3. -Maths 9th

Last Answer : Given, a+b=10,ab=21 then, ⇒(a+b)3=a3+3ab(a+b)+b3 ⇒(10)3=a3+b3+3(21)(10) ⇒1000−630=a3+b3 ∴a3+b3=370

If a + b + c = 9 and ab + bc + ca = 26, find a2 + b2 +c2. -Maths 9th

Description : If a + b + c = 9 and ab + bc + ca = 26, find a2 + b2 +c2. -Maths 9th

Last Answer : Find a2 + b2 +c2.

If a + b + c = 9 and ab + bc + ca = 26, find a2 + b2 +c2. -Maths 9th

Description : If a + b + c = 9 and ab + bc + ca = 26, find a2 + b2 +c2. -Maths 9th

Last Answer : Find a2 + b2 +c2.

If a,b,c are all non-zero and a + b + c = 0, prove that a2/bc + b2/ca+ c2/ab = 3. -Maths 9th

Description : If a,b,c are all non-zero and a + b + c = 0, prove that a2/bc + b2/ca+ c2/ab = 3. -Maths 9th

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If a + b + c = 9 and ab + bc + ca = 23, find the value of a2 + b2 + c2 -Maths 9th

Description : If a + b + c = 9 and ab + bc + ca = 23, find the value of a2 + b2 + c2 -Maths 9th

Last Answer : (a+b+c)2=a2+b2+c2+2ab+2bc+2ca =a2+b2+c2+2(ab+bc+ca) Given, ⇒92=a2+b2+c2+2(23) ⇒81−46=a2+b2+c2 ∴a2+b2+c2=35

If a2 + b2 + c2 = 16 and ab + bc + ca = 10, find the value of a + b + c. -Maths 9th

Description : If a2 + b2 + c2 = 16 and ab + bc + ca = 10, find the value of a + b + c. -Maths 9th

Last Answer : ( a + b + c )^2 = a^2 + b^2 + c^2 + 2( ab + bc + ca ) => ( a + b + c )^2 = 16 + 2×10 => ( a + b + c )^2 = 36 => a + b + c = Root 36 = 6

Prove that a2 + b2 + c2 – ab – bc – ca is always non-negative for all values of a, b and c. -Maths 9th

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If a + b + c = 0 and a2 + b2 + c2 = 16, find the value of ab + be + ca. -Maths 9th

Description : If a + b + c = 0 and a2 + b2 + c2 = 16, find the value of ab + be + ca. -Maths 9th

Last Answer : (a + b + c)2 = a2 + b2 + c2 + 2ab + 2bc + 2ca = a2 + b2 + c2 + 2(ab + bc + ca) 0 = 16 + 2(ab + bc + ca) 2(ab + bc + ca) = - 16 (ab + bc + ca) = -8

If a + b + c =0, then a3+b3 + c3 is equal to -Maths 9th

Description : If a + b + c =0, then a3+b3 + c3 is equal to -Maths 9th

Last Answer : (d) Now, a3+b3 + c3 = (a+ b + c) (a2 + b2 + c2 – ab – be – ca) + 3abc [using identity, a3+b3 + c3 – 3 abc = (a + b + c)(a2+b2+c2 –ab–bc-ca)] = 0 + 3abc [∴ a + b + c = 0] a3+b3 + c3 = 3abc.

If a + b + c =0, then a3+b3 + c3 is equal to -Maths 9th

Description : If a + b + c =0, then a3+b3 + c3 is equal to -Maths 9th

Last Answer : (d) Now, a3+b3 + c3 = (a+ b + c) (a2 + b2 + c2 – ab – be – ca) + 3abc [using identity, a3+b3 + c3 – 3 abc = (a + b + c)(a2+b2+c2 –ab–bc-ca)] = 0 + 3abc [∴ a + b + c = 0] a3+b3 + c3 = 3abc.

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Description : From the following figure, determine the resultant of four forces A1, A2, A3, A4.

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Description : The total extension in a bar, consists of 3 bars of same material, of varying sections is a. P/E(L1/A1+L2/A2+L3/A3) b. P/E(L1A1+L2A2+L3A3) c. PE(L1/A1+L2/A2+L3/A3) d. PE(L1/A1+L2/A2+L3/A3)

Last Answer : a. P/E(L1/A1+L2/A2+L3/A3)

Factorise: a3 -8b3 -64c3 -24abc -Maths 9th

Description : Factorise: a3 -8b3 -64c3 -24abc -Maths 9th

Last Answer : Factorisation

Factorise: a3 -8b3 -64c3 -24abc -Maths 9th

Description : Factorise: a3 -8b3 -64c3 -24abc -Maths 9th

Last Answer : Factorisation

If a1, a2, ....., an are distinct positive real numbers such that a1 + a2 + ..... + an = 1, then -Maths 9th

Description : If a1, a2, ....., an are distinct positive real numbers such that a1 + a2 + ..... + an = 1, then -Maths 9th

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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

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A circle with centre O and diameter COB is given. If AB and CD are parallel, then show that chord AC is equal to chord BD. -Maths 9th

Description : A circle with centre O and diameter COB is given. If AB and CD are parallel, then show that chord AC is equal to chord BD. -Maths 9th

Last Answer : O Join AC and BD. Given, COB is the diameter of circle. ∠CAB = ∠BDC = 90° [angle in a semi-circle] Also, AB II CD ∠ABC = ∠DCB (alternate angles] Now, ∠ACB = 90° - ∠ABC and ∠DBC = 90° - ∠DCB = ... = ∠DBC BC = BC [common sides] ΔABC = ΔDCB [by ASA congruency] ∴ AC = BD [by CPCT] Hence Proved.

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.

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

Opposite angles of a quadrilateral ABCD are equal. If AB = 4 cm, determine CD. -Maths 9th

Description : Opposite angles of a quadrilateral ABCD are equal. If AB = 4 cm, determine CD. -Maths 9th

Last Answer : Given, opposite angles of a quadrilateral are equal. So, ABCD is a parallelogram and we know that, in a parallelogram opposite sides are also equal. ∴ CD = AB = 4cm

In figure, if OA = 5 cm, AB = 8 cm and OD is perpendicular to AB, then CD is equal to -Maths 9th

Description : In figure, if OA = 5 cm, AB = 8 cm and OD is perpendicular to AB, then CD is equal to -Maths 9th

Last Answer : (a) We know that, the perpendicular from the centre of a circle to a chord bisects the chord. AC = CB = 1/2 AB = 1/2 x 8 = 4 cm given OA = 5 cm AO2 = AC2 + OC2 (5)2 = (4)2 + OC2 25 = 16 + OC2 ... length is always positive] OA = OD [same radius of a circle] OD = 5 cm CD = OD - OC = 5 - 3 = 2 cm

AB and AC are two equal chords of a circle. -Maths 9th

Description : AB and AC are two equal chords of a circle. -Maths 9th

Last Answer : Given AS and AC are two equal chords whose centre is O. To prove Centre O lies on the bisector of ∠BAC. Construction Join SC, draw bisector AD of ∠BAC. Proof In ΔSAM and ΔCAM, AS = AC [given] ∠BAM = ∠CAM [given]

Two equal chords AB and CD of a circle when produced intersect at a point P. Prove that PB = PD. -Maths 9th

Description : Two equal chords AB and CD of a circle when produced intersect at a point P. Prove that PB = PD. -Maths 9th

Last Answer : According to question prove that PB = PD.