((a^2-b^2)^3 + (b^2-c^2)^3 + (c^2-a^2)^3)/((a-b)^3 + (b-c)^3 + (c-a)^3) on simplification is equal to : -Maths 9th

((a^2-b^2)^3 + (b^2-c^2)^3 + (c^2-a^2)^3)/((a-b)^3 + (b-c)^3 + (c-a)^3) on simplification is equal to : -Maths 9th
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Three equal cubes are placed adjacently in a row. Find the ratio of total surface area of the new cuboid to that of the sum of the surface areas of the three cubes. -Maths 9th

Description : Three equal cubes are placed adjacently in a row. Find the ratio of total surface area of the new cuboid to that of the sum of the surface areas of the three cubes. -Maths 9th

Last Answer : Let each side of a cube = a cm Then surface area = 6a² cm² and surface area of 3 such cubes = 3 x 6a² = 18a² cm² By placing three cubes side by side we get a cuboid whose ... + 3a²] = 14 a² ∴ Ratio between their surface areas = 14a² : 18a² = 7 : 9

The paint in a certain container is sufficient to paint an area equal to 9.375 m2. How many bricks of dimensions 22.5 cm×10 cm×7.5 cm can be painted out of this container? -Maths 9th

Description : The paint in a certain container is sufficient to paint an area equal to 9.375 m2. How many bricks of dimensions 22.5 cm×10 cm×7.5 cm can be painted out of this container? -Maths 9th

Last Answer : Total surface area of one brick = 2(lb +bh+lb) = [2(22.5 10+10 7.5+22.5 7.5)] cm2 = 2(225+75+168.75) cm2 = (2 468.75) cm2 = 937.5 cm2 Let n bricks can be painted out by the ... 93750 cm2 So, we have, 93750 = 937.5n n = 100 Therefore, 100 bricks can be painted out by the paint of the container.

Find the area of a triangle having perimeter 32cm. One side of its side is equal to 11cm and difference of the other two is 5cm. -Maths 9th

Description : Find the area of a triangle having perimeter 32cm. One side of its side is equal to 11cm and difference of the other two is 5cm. -Maths 9th

Last Answer : Solutions :- We have, Perimeter of triangle = 32 cm One of its side = 11 cm Let the second side be x And third side be x + 5 Perimeter of triangle = sum of three sides A/q => 11 + x + x + 5 ... 13 cm Now, By using heron's formula, Find the area of a triangle :- Answer : Area of triangle = 43.81 cm²

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

Description : 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

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).

5. Show that if the diagonals of a quadrilateral are equal and bisect each other at right angles, then it is a square. -Maths 9th

Description : 5. Show that if the diagonals of a quadrilateral are equal and bisect each other at right angles, then it is a square. -Maths 9th

Last Answer : Solution: Given that, Let ABCD be a quadrilateral and its diagonals AC and BD bisect each other at right angle at O. To prove that, The Quadrilateral ABCD is a square. Proof, In ΔAOB and ΔCOD, AO = ... right angle. Thus, from (i), (ii) and (iii) given quadrilateral ABCD is a square. Hence Proved.

4. Show that the diagonals of a square are equal and bisect each other at right angles. -Maths 9th

Description : 4. Show that the diagonals of a square are equal and bisect each other at right angles. -Maths 9th

Last Answer : Solution: Let ABCD be a square and its diagonals AC and BD intersect each other at O. To show that, AC = BD AO = OC and ∠AOB = 90° Proof, In ΔABC and ΔBAD, AB = BA (Common) ∠ABC = ∠BAD = ... = ∠COB ∠AOB+∠COB = 180° (Linear pair) Thus, ∠AOB = ∠COB = 90° , Diagonals bisect each other at right angles

find two solution of 4 x + Y is equal to 2 -Maths 9th

Description : find two solution of 4 x + Y is equal to 2 -Maths 9th

Last Answer : When solving for y in this problem(4x+y=2) we need to get y by itself in the equation. To do this we need to move 4x over to the right side of the equation. To do this we subtract, 4x from ... we do to one side of the equation we have to do the same to the other side of the equation. So,..

An exterior angle of a triangle is 110° and the two interior opposite angles are equal find the interior opposite angels -Maths 9th

Description : An exterior angle of a triangle is 110° and the two interior opposite angles are equal find the interior opposite angels -Maths 9th

Last Answer : each interior opposite angles are 55

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.

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.

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.

If the diagonals of a parallelogram are equal, then show that it is a rectangle. -Maths 9th

Description : If the diagonals of a parallelogram are equal, then show that it is a rectangle. -Maths 9th

Last Answer : Given : A parallelogram ABCD , in which AC = BD TO Prove : ABCD is a rectangle . Proof : In △ABC and △ABD AB = AB [common] AC = BD [given] BC = AD [opp . sides of a | | gm] ⇒ △ABC ≅ △BAD [ ... ∵ ∠ABC = ∠BAD] ⇒ 2∠ABC = 180° ⇒ ∠ABC = 1 /2 180° = 90° Hence, parallelogram ABCD is a rectangle.

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

The line segment joining the mid-points of any two sides of a triangle is parallel to the third side and equal to half of it. -Maths 9th

Description : The line segment joining the mid-points of any two sides of a triangle is parallel to the third side and equal to half of it. -Maths 9th

Last Answer : Given = A △ABC in which D and E are the mid-points of side AB and AC respectively. DE is joined . To Prove : DE || BC and DE = 1 / 2 BC. Const. : Produce the line segment DE to F , such that DE = ... of ||gm are equal and parallel] Also, DE = EF [by construction] Hence, DE || BC and DE = 1 / 2 BC

Two parallelograms are on equal bases and between the same parallels. -Maths 9th

Description : Two parallelograms are on equal bases and between the same parallels. -Maths 9th

Last Answer : Two Parallelograms on the equal based and between the same parallels are equal in area.

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)

Equal chords of a circle subtend equal angles at the centre. -Maths 9th

Description : Equal chords of a circle subtend equal angles at the centre. -Maths 9th

Last Answer : Given : In a circle C(O,r), chord AB = chord CD. To Prove : ∠AOB = ∠COD. Proof : In△AOB and △COD AO = CO [radii of same circle] BO = DO [radii of same circle] Chord AB = Chord CD [given] ⇒ △AOB ≅ △COD [by SSS congruence axiom] ⇒ ∠AOB = ∠COD. [c.p.c.t.]

If the angles subtended by the chords of a circle at the centre are equal, then chords are equal. -Maths 9th

Description : If the angles subtended by the chords of a circle at the centre are equal, then chords are equal. -Maths 9th

Last Answer : Given : In a circle C(O,r) , ∠AOB = ∠COD To Prove : Chord AB = Chord CD . Proof : In △AOB and △COD AO = CO [radii of same circle] BO = DO [radii of same circle] ∠AOB = ∠COD [given] ⇒ △AOB ≅ △COD [by SAS congruence axiom] ⇒ Chord AB = Chord CD [c.p.c.t]

If the volume of a sphere is numerically equal to its surface area, then find the diameter of the sphere. -Maths 9th

Description : If the volume of a sphere is numerically equal to its surface area, then find the diameter of the sphere. -Maths 9th

Last Answer : Let r be the radius of the sphere. and Volume of a sphere = surface area of the sphere ⇒ 4 / 3πr3 = 4πr2 ⇒ r = 3 cm ∴ Diameter of the sphere = 2r = 2 × 3 = 6 cm

A spherical ball is divided into two equal halves. If the curved surface area of each half is 56.57 cm?, find the volume of the spherical ball.11531/cylinder-radius-halved-and-height-doubled-then-find-volume-with-respect-original-volume -Maths 9th

Description : A spherical ball is divided into two equal halves. If the curved surface area of each half is 56.57 cm?, find the volume of the spherical ball.11531/cylinder-radius-halved-and-height-doubled-then-find-volume-with-respect-original-volume -Maths 9th

Last Answer : since curved surface of half of the spherical ball = 56.57 cm2 ∴ 2πr2 = 56.57 ⇒ r2 = 56.57 / 2 × 3.14 = 9 ⇒ r = 3 cm Now, volume of spherical ball = 4 / 3 πr3 = 4 / 3 × 3.14 × 3 × 3 × 3 = 113.04 cm3

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

Which of the following is equal to X ? -Maths 9th

Description : Which of the following is equal to X ? -Maths 9th

Last Answer : x is equal to

If p (x) = x + 3, then p(x)+ p (- x) is equal to -Maths 9th

Description : If p (x) = x + 3, then p(x)+ p (- x) is equal to -Maths 9th

Last Answer : (d) Given p(x) = x+3, put x = -x in the given equation, we get p(-x) = -x+3 Now, p(x)+ p(-x) = x+ 3+ (-x)+ 3=6

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.

For what value of c, the linear equation 2x + cy = 8 has equal values of x and y for its solution? -Maths 9th

Description : For what value of c, the linear equation 2x + cy = 8 has equal values of x and y for its solution? -Maths 9th

Last Answer : The given linear equation is 2x + cy= 8. …(i) Now, by condition, x and y-coordinate of given linear equation are same, i.e., x = y. Put y = x in Eq. (i), we get

If one angle of a triangle is equal to the sum of the other two angles, then the triangle is -Maths 9th

Description : If one angle of a triangle is equal to the sum of the other two angles, then the triangle is -Maths 9th

Last Answer : (d) Let the angles of a AABC be ∠A, ∠B and ∠C. Given, ∠A = ∠B+∠C …(i) InMBC, ∠A+ ∠B+ ∠C-180° [sum of all angles of a triangle is 180°]…(ii) From Eqs. (i) and (ii), ∠A+∠A = 180° ⇒ 2 ∠A = 180° ⇒ 180° /2 ∠A = 90° Hence, the triangle is a right triangle.

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.

‘If two sides and an angle of one triangle are equal to two sides and an angle of another triangle , then the two triangles must be congruent’. -Maths 9th

Description : ‘If two sides and an angle of one triangle are equal to two sides and an angle of another triangle , then the two triangles must be congruent’. -Maths 9th

Last Answer : No, because in the congruent rule, the two sides and the included angle of one triangle are equal to the two sides and the included angle of the other triangle i.e., SAS rule.

All the angles of a quadrilateral are equal. What special name is given to this quadrilateral ? -Maths 9th

Description : All the angles of a quadrilateral are equal. What special name is given to this quadrilateral ? -Maths 9th

Last Answer : We know that, sum of all angles in a quadrilateral is 360°. If ABCD is a quadrilateral, ∠A+ ∠B+ ∠C + ∠D = 360° (i) But it is given all angles are equal. ∠A = ∠B = ∠C = ∠D From Eq. (i ... ⇒ 4 ∠A = 360° ∠A = 90° So, all angles of a quadrilateral are 90°. Hence, given quadrilateral is a rectangle.

Diagonals of a rectangle are equal and perpendicular. Is this statement true ? -Maths 9th

Description : Diagonals of a rectangle are equal and perpendicular. Is this statement true ? -Maths 9th

Last Answer : No, diagonals of a rectangle are equal but need not be perpendicular.

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 parallelogram ABCD and rectangle ABEM are of equal area, then -Maths 9th

Description : In figure, if parallelogram ABCD and rectangle ABEM are of equal area, then -Maths 9th

Last Answer : (c) In rectangle ABEM, AB = EM [sides of rectangle] and in parallelogram ABCD, CD = AB On adding, both equations, we get AB + CD = EM + AB (i) We know that, the perpendicular distance between two ... AB+BE + EM+ AM [∴ CD = AB = EM] Perimeter of parallelogram ABCD > perimeter of rectangle ABEM

The mid-point of the sides of a triangle along with any of the vertices as the fourth point make a parallelogram of area equal to -Maths 9th

Description : The mid-point of the sides of a triangle along with any of the vertices as the fourth point make a parallelogram of area equal to -Maths 9th

Last Answer : Solution of this question

Two parallelograms are on equal bases and between the same parallels. -Maths 9th

Description : Two parallelograms are on equal bases and between the same parallels. -Maths 9th

Last Answer : (b) We know that, parallelogram on the equal bases and between the same parallels are equal in area. So, ratio of their areas is 1 :1.

ABCD is a quadrilateral whose diagonal AC divides it into two parts, equal in area, then ABCD -Maths 9th

Description : ABCD is a quadrilateral whose diagonal AC divides it into two parts, equal in area, then ABCD -Maths 9th

Last Answer : (d) Here, ABCD need not be any of rectangle, rhombus and parallelogram because if ABCD is a square, then its diagonal AC also divides it into two parts which are equal in area.

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]