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

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
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 (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
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Related questions

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

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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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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

Description : 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 centroid of △ABC in which A(a,b), B(b,c) and C(c,a) is at the origin, then the value of a3 + b3 + c3 is: (a) abc (b) 2abc (c) 3abc (d) 0

Description : If the centroid of △ABC in which A(a,b), B(b,c) and C(c,a) is at the origin, then the value of a3 + b3 + c3 is: (a) abc (b) 2abc (c) 3abc (d) 0

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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

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Last Answer : Given, ax+by=a2−b2......(1) bx+ay=0.......(2). Now adding (1) and (2) we get, x(a+b)x+(a+b)y=a2−b2 or, (a+b)(x+y)=a2−b2 or, x+y=a−b.

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

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Last Answer : Find a2 + b2 +c2.

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

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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

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Last Answer : Given: Diameter of semicircle AEB= 2.8 cm RADIUS OF SEMICIRCLE(AEB)= 2.8/2= 1.4 cm Diameter of semicircle BFC=1.4 cm RADIUS OF SEMICIRCLE(BFC)= 1.4/2= 0.7 cm Diameter of semicircle ADC= 2.8+1.4 = ... ; .6 = 13.2 cm Hence, the perimeter of the shaded region is 13.2 cm HOPE THIS WILL HELP YOU...

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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)

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

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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

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

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Description : In the figure, arcs and drawn by taking vertices A, B and C of an equilateral triangle of side 10 cm to intersect the sides BC, CA and AB at their respective mid-points D, E and F. Find the area of teh shaded region. [use π = 3.14] -Maths 10th

Last Answer : Step-by-step explanation: We have been provided that, Triangle ABC is an Equilateral triangle. Side of triangle is = 10 cm The arcs are drawn from each vertices of the triangle. We get three sectors ... portion is, Remaining area = Area of triangle ABC - Area of all the sectors 39.25cm square

Check whether the following are quadratic equations: (i) (x+ 1)2=2(x-3) (ii) x – 2x = (- 2) (3-x) (iii) (x – 2) (x + 1) = (x – 1) (x + 3) (iv) (x – 3) (2x + 1) = x (x + 5) (v) (2x – 1) (x – 3) = (x + 5) (x – 1) (vi) x2 + 3x + 1 = (x – 2)2 (vii) (x + 2)3 = 2x(x2 – 1) (viii) x3 -4x2 -x + 1 = (x-2)3 -Maths 10th

Description : Check whether the following are quadratic equations: (i) (x+ 1)2=2(x-3) (ii) x - 2x = (- 2) (3-x) (iii) (x - 2) (x + 1) = (x - 1) (x + 3) (iv) (x - 3) (2x + 1) = x (x + 5) (v) (2x - 1) (x - 3) = (x ... vi) x2 + 3x + 1 = (x - 2)2 (vii) (x + 2)3 = 2x(x2 - 1) (viii) x3 -4x2 -x + 1 = (x-2)3 -Maths 10th

Last Answer : this is the correct answer!

Find the polynomial of least degree which should be subtracted from the polynomial x4 + 2x3 – 4x2 + 6x – 3 so that it is exactly divisible by x2 – x + 1. -Maths 10th

Description : Find the polynomial of least degree which should be subtracted from the polynomial x4 + 2x3 – 4x2 + 6x – 3 so that it is exactly divisible by x2 – x + 1. -Maths 10th

Last Answer : Here, p(x) = x4 + 2x3 - 4x2 + 6x - 3, g(x) = x2 - x +1 On dividing p(x) by g(x) Therefore (x-1) must be subtracted from the polynomial p(x) to make it divisible by g(x).

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

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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=5 and ab+bc+ca=10 find the value of a^3+b^3+c^3-3abc -Maths 9th

Description : If a+b+c=5 and ab+bc+ca=10 find the value of a^3+b^3+c^3-3abc -Maths 9th

Last Answer : We know , a³ + b³ + c³ -3abc = (a + b + c )(a² + b² + c² -ab -bc-ca) now , a + b + c = 5 ab + bc + ca = 10 (a + b + c)² = a² + b² + c² +2(ab + bc+ca) (5)² -2 10 = a² + b² + c² a² + b² + c² = ... )(a² + b² + c² -ab- bc-ca) =( 5)( 5 - 10) = 5 (-5) = -25 Hope this will help u..... by :- RAXTAR.....

Use Euclid’s Division Lemma to show that the cube of any positive integer is either of the form 9m, 9m + 1 or 9m + 8 -Maths 10th

Description : Use Euclid’s Division Lemma to show that the cube of any positive integer is either of the form 9m, 9m + 1 or 9m + 8 -Maths 10th

Last Answer : Let us consider a and b where a be any positive number and b is equal to 3. According to Euclid's Division Lemma a = bq + r where r is greater than or equal to zero and less than b (0 ≤ r < b) a = 3q + r so ... 8 Where m = (3q3 + 6q2 + 4q)therefore a can be any of the form 9m or 9m + 1 or, 9m + 8.

if A,Band c are three points on a line and B lies between A and C then prove that AB+BC=AC -Maths 9th

Description : if A,Band c are three points on a line and B lies between A and C then prove that AB+BC=AC -Maths 9th

Last Answer : Since complete line is AC and B is point on it. therefore, AC is divide into 2 parts AB&BC. therefore, AC=AB+BC

if A,Band c are three points on a line and B lies between A and C then prove that AB+BC=AC -Maths 9th

Description : if A,Band c are three points on a line and B lies between A and C then prove that AB+BC=AC -Maths 9th

Last Answer : AB=AC-BC BC =AC-AB AB+BC=AB HENCE PROVED

If AOB is a diameter of a circle [Fig. 10.8] and C is a point on the circle, then prove that AC* +BC*=AB*. -Maths 9th

Description : If AOB is a diameter of a circle [Fig. 10.8] and C is a point on the circle, then prove that AC* +BC*=AB*. -Maths 9th

Last Answer : Solution :- As, ∠ C = 90° (Angle in the semicircle) ∴ AC2 + BC2 = AB2 (By Pythagoras Theorem)

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,

In figure, AB || DE, AB = DE, AC|| DF and AC = OF. Prove that BC || EF and BC = EF. -Maths 9th

Description : In figure, AB || DE, AB = DE, AC|| DF and AC = OF. Prove that BC || EF and BC = EF. -Maths 9th

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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,

In figure, AB || DE, AB = DE, AC|| DF and AC = OF. Prove that BC || EF and BC = EF. -Maths 9th

Description : In figure, AB || DE, AB = DE, AC|| DF and AC = OF. Prove that BC || EF and BC = EF. -Maths 9th

Last Answer : Given In figure AB || DE and AC || DF, also AB = DE and AC = DF To prove BC ||EF and BC = EF Proof In quadrilateral ABED, AB||DE and AB = DE So, ABED is a parallelogram. AD || BE and AD = BE Now, ... = CF and BE||CF [from Eq. (iii)] So, BCFE is a parallelogram. BC = EF and BC|| EF . Hence proved.

The sum of two children is ‘a’. The age of the father is twice the ‘a’. After twenty years, his age will be equal to the addition of the ages of his children. Find the age of father. -Maths 10th

Description : The sum of two children is ‘a’. The age of the father is twice the ‘a’. After twenty years, his age will be equal to the addition of the ages of his children. Find the age of father. -Maths 10th

Last Answer : Let the sum of the ages of two children will be x and age of father will be y. A.T.Q. , y = 2x ----------(i) and , After 20 years, x+40=y+20 ==> ... i), x-2x = -20 ==> x = 20 Now, put x= 20 in (i), y = 2 x 20 = 40 Hence age of father will be 40 years.

Aftab tells his daughter, “Seven years ago, I was seven times as old as you were then. Also, three years from now, I shall be three times as old as you will be.” (Isn’t this interesting?) Represent this situation algebraically and graphically. -Maths 10th

Description : Aftab tells his daughter, “Seven years ago, I was seven times as old as you were then. Also, three years from now, I shall be three times as old as you will be.” (Isn’t this interesting?) Represent this situation algebraically and graphically. -Maths 10th

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If 2k + 1, 6, 3& + 1 are in AP, then find the value of k. -Maths 10th

Description : If 2k + 1, 6, 3& + 1 are in AP, then find the value of k. -Maths 10th

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Description : If the product of two zeroes of polynomial 2x3 + 3x2 – 5x – 6 is 3, then find its third zero. -Maths 10th

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If one zero of the polynomial 5z2 + 13z – p is reciprocal of the other, then find p. -Maths 10th

Description : If one zero of the polynomial 5z2 + 13z – p is reciprocal of the other, then find p. -Maths 10th

Last Answer : The value of p is -5 Step-by-step explanation: Given that if one zero of polynomial is reciprocal of the other then we have to find the value of p. a=5, b=13 and c=-p since the zeroes are reciprocal to each other. therefore their product must be 1 ⇒ ⇒ Hence, the value of p is -5

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Description : Find the zeroes of the of the polynomials p(x) = 4x2 – 12x + 9 -Maths 10th

Last Answer : P(x)= 4x² - 12x + 9 4x²- 12x + 9 = 0 4x²- 6x - 6x +9 = 0 2x(2x - 3) - 3(2x - 3) = 0 (2x - 3) (2x - 3) x = 3/2 x = 3/2 x = 3/2 ; 3/2 are the zeros of the given polynomial.

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Description : How many solutions does the pair of equations y = 0 and y = -5 have? -Maths 10th

Last Answer : y = 0 and y = -5 are Parallel lines, hence no solution.

Solve graphically 4x-3y+4=0, 4x+3y-20=0 -Maths 10th

Description : Solve graphically 4x-3y+4=0, 4x+3y-20=0 -Maths 10th

Last Answer : 4x−3y+4=0 ⟹x=4−4+3y Let, y=4, then x=4−4+3(4) =2 Similarly, when y=0,x=−1 4x+3y−20=0 ⟹x=420−3y Let, y=4, then x=420−3(4) =2 Similarly, when y=0,x=5 Plotting the above points on the graph, ... point i.e(2,4) Area of the region bounded by these lines and x-axis =21 6units 4 units =12 sq. units.

If the equation x2 minus ax plus 2b equals 0 has prime roots where a and b are positive integers then a minus b is equal to?

Description : If the equation x2 minus ax plus 2b equals 0 has prime roots where a and b are positive integers then a minus b is equal to?

Last Answer : I think you mean the roots are prime numbers.Let the two roots be primes p and qThen the equation factorises to (x - p)(x - q) = 0 which can beexpanded to give:x² - (p + q)x + pq = 0Which comparing ... = 2(It doesn't matter if the other prime is even (2) or not as itcancels out from a - b.)

In a spreadsheet software, the formula =A1 +$2 was entered in cell A3 and then copied into cell B3. What is the formula copied into B3? -Technology

Description : In a spreadsheet software, the formula =A1 +$2 was entered in cell A3 and then copied into cell B3. What is the formula copied into B3? -Technology

Last Answer : Need Answer

If a1, a2, .... an are positive numbers such that a1.a2.a3 .... an = 1, then their sum is -Maths 9th

Description : If a1, a2, .... an are positive numbers such that a1.a2.a3 .... an = 1, then their sum is -Maths 9th

Last Answer : answer:

If a1, a2, a3 ....... an are positive real numbers whose product is a fixed number ‘c’, then the minimum value of a1 + a2 ..... + an–1 + 2an is -Maths 9th

Description : If a1, a2, a3 ....... an are positive real numbers whose product is a fixed number ‘c’, then the minimum value of a1 + a2 ..... + an–1 + 2an is -Maths 9th

Last Answer : answer:

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Description : Find the distance between the following pairs of points: (i) (2, 3), (4, 1) (ii) (-5, 7), (-1, 3) (iii) (a, b), (- a, – b) -Maths 10th

Last Answer : Distance between two points (x1 ,y1 ) and (x2 ,y2 ) is (x1 −x2 )2+(y1 −y2 )2 (i) Distance between(2,3) and (4,1) is =(2−4)2+(3−1)2 =(−2)2+(2)2 =4+4 =8 =22 (ii) Distance between (−5,7) and (−1,3) ... 42 (iii )Distance between (a,b) and(−a,−b) is =(a−(−a))2+(b−(−b))2 =(2a)2+(2b)2 =4a2+4b2 =2a2+b2

The coach of a cricket team buys 3 bats and 6 balls for Rs 3900. Later, she buys another bat and 2 more balls of the same kind for Rs 1300. Represent this situation algebraically and geometrically. -Maths 10th

Description : The coach of a cricket team buys 3 bats and 6 balls for Rs 3900. Later, she buys another bat and 2 more balls of the same kind for Rs 1300. Represent this situation algebraically and geometrically. -Maths 10th

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The cost of 2 kg of apples and 1 kg of grapes on a day was found to be Rs 160. After a month, the cost of 4 kg of apples and 2 kg of grapes is Rs 300. Represent the situation algebraically and geometrically. -Maths 10th

Description : The cost of 2 kg of apples and 1 kg of grapes on a day was found to be Rs 160. After a month, the cost of 4 kg of apples and 2 kg of grapes is Rs 300. Represent the situation algebraically and geometrically. -Maths 10th

Last Answer : y 20 40 60 80Let the cost of 1 kg of apples be x and that of 1 kg be y So the algebraic representation can be as follows: 2x+y=160 4x+2y=300⇒2x+y=150 The situation can be represented ... that the lines do not intersect anywhere, i.e. they are parallel. Hence we can not arrive at a solution.

What are trigonometric ratios? -Maths 10th

Description : What are trigonometric ratios? -Maths 10th

Last Answer : The trigonometric ratios of the angle A in the right triangle ABC are defined as follows: sine of ∠A = side opposite to angle Ahypotenuseside opposite to angle Ahypotenuse = BCACBCAC cosine of ∠A = side ... of angle A = ACABACAB cotangent of ∠A = 1tangent of angle A1tangent of angle A = ABBCABBC