Find the value of 27X3 + 8y3 if (i) 3x + 2y = 14 and xy = 8 (ii) 3x + 2y = 20 and xy = 149 -Maths 9th

Find the value of 27X3 + 8y3 if (i) 3x + 2y = 14 and xy = 8 (ii) 3x + 2y = 20 and xy = 149 -Maths 9th
by

1 Answer

Answer :

answer:
by

Related questions

If 3x – 2y= 11 and xy = 12, find the value of 27x3 – 8y3. -Maths 9th

Description : If 3x – 2y= 11 and xy = 12, find the value of 27x3 – 8y3. -Maths 9th

Last Answer : Given, (a−b)3=a3−3ab(a−b)−b3 (3x−2y)3=27x3−8y3−3(3x)(2y)(3x−2y) 113=27x3−8y3−18(12)(11) 27x3−8y3=3707

Compute the value of 9x2 + 4y2 if xy = 6 and 3x + 2y = 12. -Maths 9th

Description : Compute the value of 9x2 + 4y2 if xy = 6 and 3x + 2y = 12. -Maths 9th

Last Answer : Consider the equation 3x + 2y = 12 Now, square both sides: (3x + 2y)2 = 122 => 9x2 + 12xy + 4y2 = 144 =>9x2 + 4y2 = 144 – 12xy From the questions, xy = 6 So, 9x2 + 4y2 = 144 – 72 Thus, the value of 9x2 + 4y2 = 72

Find the value of x3-8y3-36xy-216 when x = 2y+6. -Maths 9th

Description : Find the value of x3-8y3-36xy-216 when x = 2y+6. -Maths 9th

Last Answer : Solution :-

Find the following products: (i) (3x + 2y + 2z) (9x2 + 4y2 + 4z2 – 6xy – 4yz – 6zx) (ii) (4x -3y + 2z) (16x2 + 9y2+ 4z2 + 12xy + 6yz – 8zx) (iii) (2a – 3b – 2c) (4a2 + 9b2 + 4c2 + 6ab – 6bc + 4ca) (iv) (3x -4y + 5z) (9x2 + 16y2 + 25z2 + 12xy- 15zx + 20yz) -Maths 9th

Description : Find the following products: (i) (3x + 2y + 2z) (9x2 + 4y2 + 4z2 – 6xy – 4yz – 6zx) (ii) (4x -3y + 2z) (16x2 + 9y2+ 4z2 + 12xy + 6yz – 8zx) (iii) (2a – 3b – 2c) (4a2 + 9b2 + 4c2 + 6ab – 6bc + 4ca) (iv) (3x -4y + 5z) (9x2 + 16y2 + 25z2 + 12xy- 15zx + 20yz) -Maths 9th

Last Answer : answer:

Find the following products: (i) (3x + 2y) (9X2 – 6xy + Ay2) (ii) (4x – 5y) (16x2 + 20xy + 25y2) (iii) (7p4 + q) (49p8 – 7p4q + q2) -Maths 9th

Description : Find the following products: (i) (3x + 2y) (9X2 – 6xy + Ay2) (ii) (4x – 5y) (16x2 + 20xy + 25y2) (iii) (7p4 + q) (49p8 – 7p4q + q2) -Maths 9th

Last Answer : answer:

Solve these following equations: (i) 3x + 3 = 15 (ii) 2y + 7 =19 -Maths 9th

Description : Solve these following equations: (i) 3x + 3 = 15 (ii) 2y + 7 =19 -Maths 9th

Last Answer : answer:

If x + y + z = 0, then x^2/(2x^2+yz)+y^2/(2y^2+zx)+z^2/(2z^2+xy) = -Maths 9th

Description : If x + y + z = 0, then x^2/(2x^2+yz)+y^2/(2y^2+zx)+z^2/(2z^2+xy) = -Maths 9th

Last Answer : answer:

If 3x -7y = 10 and xy = -1, find the value of 9x2 + 49y2 -Maths 9th

Description : If 3x -7y = 10 and xy = -1, find the value of 9x2 + 49y2 -Maths 9th

Last Answer : answer:

If 9x2 + 25y2 = 181 and xy = -6, find the value of 3x + 5y. -Maths 9th

Description : If 9x2 + 25y2 = 181 and xy = -6, find the value of 3x + 5y. -Maths 9th

Last Answer : answer:

Write any two solutions of the linear equation 3x + 2y =9. -Maths 9th

Description : Write any two solutions of the linear equation 3x + 2y =9. -Maths 9th

Last Answer : Solution :-

Draw a graph of the equation x + Y = 5 & 3x - 2y =0 on the same graph paper. Find the coordinates of the point whose two lines intersect. -Maths 9th

Description : Draw a graph of the equation x + Y = 5 & 3x - 2y =0 on the same graph paper. Find the coordinates of the point whose two lines intersect. -Maths 9th

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

Draw a graph of the equation x+ y=5 & 3x -2y=0 in the same graph paper find the coordinates of the point whose two two lines intersect. -Maths 9th

Description : Draw a graph of the equation x+ y=5 & 3x -2y=0 in the same graph paper find the coordinates of the point whose two two lines intersect. -Maths 9th

Last Answer : From x + y = 5, If x = 0 0 + y = 5 y = 5 Therefore (0,5) If x = 1 1 + y = 5 y =5 - 1 y = 4 Therefore (1,4) Draw a graph for this And From 3x - 2y = 0 If x = 0 3 (0) - 2y = 0 0 - ... 2y = 0 -2y = -6 y = -6/-2 y = 3 Therefore (2,3) Draw a graph for these points And the point of intersection is (2,3)

Find the equation of the straight line passing through the point (4, 5) and perpendicular to 3x – 2y + 5 = 0. -Maths 9th

Description : Find the equation of the straight line passing through the point (4, 5) and perpendicular to 3x – 2y + 5 = 0. -Maths 9th

Last Answer : There are two ways to prove it. 1st way: Area of triangle formed by the given points = 0 if they are collinear.∴ \(rac{1}{2}\) [\(x\)(2 - (y + 1) ) + 1((y + 1) - 1) + 0(1 - 2)] = 0⇒ \(rac{1}{2}\) [2\(x\) - \(x\)y - \( ... y - \(x\)y - 1 + \(x\) ⇒ x + y = \(x\)y ⇒ \(rac{1}{x}\) + \(rac{1}{y}\) = 1.

What is the angle between the lines whose equations are: 3x + y – 7 = 0 and x + 2y + 9 = 0. -Maths 9th

Description : What is the angle between the lines whose equations are: 3x + y – 7 = 0 and x + 2y + 9 = 0. -Maths 9th

Last Answer : (c) (8, 6)Let AB be the given line 4x + 3y = 25 Let O′(a, b) be the image of O in the given line AB. Let O O′ cut AB in point P. Also OP ⊥ AB and P is the mid-point of OO′. ∴ Co-ordinates of P are \(\bigg( ... 4 imes6}{3}\) = 8∴ The image of the point O(0, 0) in the line 4x + 3y - 25 = 0 is (8, 6).

If x+y=9 and xy=20,then find the value of x(square) + y(square). -Maths 9th

Description : If x+y=9 and xy=20,then find the value of x(square) + y(square). -Maths 9th

Last Answer : Solution :-

if 2x + 3y = 8 and xy = 2, find the value of 4X2 + 9y2. -Maths 9th

Description : if 2x + 3y = 8 and xy = 2, find the value of 4X2 + 9y2. -Maths 9th

Last Answer : Given 2x+3y=8 and xy=2, formula, (a+b)2=a2+b2+2ab ∴(2x+3y)2=4x2+9y2+2(2x)(3y) (2x+3y)2=4x2+9y2+12xy 82=4x2+9y2+12(2) ∴4x2+9y2=64−24=40

A pair of linear equations which has a unique solution x = 2, y = -3 is (a) x + y = -1 ; 2x – 3y = -5 (b) 2x + 5y = -11 ; 4x + 10y = -22(c) 2x – y = 1 ; 3x + 2y = 0 (d) x – 4y – 14 = 0 ;5x – y – 13 = 0

Description : A pair of linear equations which has a unique solution x = 2, y = -3 is (a) x + y = -1 ; 2x – 3y = -5 (b) 2x + 5y = -11 ; 4x + 10y = -22(c) 2x – y = 1 ; 3x + 2y = 0 (d) x – 4y – 14 = 0 ;5x – y – 13 = 0

Last Answer : (d) x – 4y – 14 = 0 ;5x – y – 13 = 0

If 2x + 3y = 13 and xy = 6, find the value of 8x3 + 21y3. -Maths 9th

Description : If 2x + 3y = 13 and xy = 6, find the value of 8x3 + 21y3. -Maths 9th

Last Answer : 2x +3y = 13-----(1) xy =6-----(2) 8x³ +27y³ = (2x)³ +(3y)³ = (2x+3y)³ - 3*2x*3y(2x+3y) [using (a+b)³ = a³+b³+3ab(a+b)] = (13)³- 18 * 6 *13 [ using (1) and (2)] = 2197 - 1404 =793

If (x^4 – 2x^2y^2 + y^2)^(a –1) = (x – y)^2a (x + y) ^–2, then the value of a is -Maths 9th

Description : If (x^4 – 2x^2y^2 + y^2)^(a –1) = (x – y)^2a (x + y) ^–2, then the value of a is -Maths 9th

Last Answer : answer:

Find the solution of the linear equation x+2y = 8 which represents a point on -Maths 9th

Description : Find the solution of the linear equation x+2y = 8 which represents a point on -Maths 9th

Last Answer : We have, x + 2y = 8 ,..(i) (i) When the point is on the X-axis, then put y = 0 in Eq. (i), we get x+2 (0)=8 ⇒ x = 8 Hence, the required point is (8, 0). (ii) When the point is on the Y-axis, then put x = 0 in Eq. (i), we get 0 + 2y = 8 ⇒ y = 8/2 = 4 Hence, the required point is (0, 4).

Find the solution of the linear equation x+2y = 8 which represents a point on -Maths 9th

Description : Find the solution of the linear equation x+2y = 8 which represents a point on -Maths 9th

Last Answer : We have, x + 2y = 8 ,..(i) (i) When the point is on the X-axis, then put y = 0 in Eq. (i), we get x+2 (0)=8 ⇒ x = 8 Hence, the required point is (8, 0). (ii) When the point is on the Y-axis, then put x = 0 in Eq. (i), we get 0 + 2y = 8 ⇒ y = 8/2 = 4 Hence, the required point is (0, 4).

If (x, y) is a solution of the following pair of linear equations in two variables, then the value of expression (√ is: x + 2y = 4 and 3x - y = 5 (a) 2 (b) 3 (c) 4 (d) √

Description : If (x, y) is a solution of the following pair of linear equations in two variables, then the value of expression (√ is: x + 2y = 4 and 3x - y = 5 (a) 2 (b) 3 (c) 4 (d) √2

Last Answer : (a) 2

Give the geometric interpretations of 5x + 3 = 3x – 7 as an equation (i) in one variable (ii) in two variables. -Maths 9th

Description : Give the geometric interpretations of 5x + 3 = 3x – 7 as an equation (i) in one variable (ii) in two variables. -Maths 9th

Last Answer : Solution :-

If x + y = 5 and xy = 4 , find x - y , using identities. -Maths 9th

Description : If x + y = 5 and xy = 4 , find x - y , using identities. -Maths 9th

Last Answer : (x+y)2 = x2 - 2xy + y2 = x2 + 2xy + y2 - 4xy = (x+y)2 - 4xy = 52 - 4 × 4 = 25 - 16 = 9 ∴ x - y = √9 = 3

X and y are points on the side LN of the triangle LMN , such that LX = XY = YN . Through X, a line is drawn parallel to LM to meet MN at Z. -Maths 9th

Description : X and y are points on the side LN of the triangle LMN , such that LX = XY = YN . Through X, a line is drawn parallel to LM to meet MN at Z. -Maths 9th

Last Answer : Here, △XZM and △XZL are on the same base (XZ) and lie between the same parallels (XZ || LM). ∴ ar(△XZL) = ar( △XZM) Adding ar(△XZY) on both sides , we have ar(△XZL) + ar(△XZY) = ar(△XZM) + ar(△XZY) ⇒ ar(△LZY) = ar(quad.MZYX)

Let x be rational and y be irrational. Is xy necessarily irrational? Justify your answer by an example. -Maths 9th

Description : Let x be rational and y be irrational. Is xy necessarily irrational? Justify your answer by an example. -Maths 9th

Last Answer : No, (xy) is necessarily an irrational only when x ≠0. Let x be a non-zero rational and y be an irrational. Then, we have to show that xy be an irrational. If possible, let xy be a rational number. Since ... wrong. Hence, xy is an irrational number. But, when x = 0, then xy = 0, a rational number.

X and Y are points on the side LN of the triangle LMN such that LX = XY = YN. -Maths 9th

Description : X and Y are points on the side LN of the triangle LMN such that LX = XY = YN. -Maths 9th

Last Answer : Given X and Y are points on the side LN such that LX = XY = YN and XZ || LM To prove ar (ΔLZY) = ar (MZYX) Proof Since, ΔXMZ and ΔXLZ are on the same base XZ and between the same parallel lines LM and XZ. ... get ar (ΔXMZ) + ar (ΔXXZ) = ar (ΔXLZ) + ar (ΔXYZ) => ar (MZYX) = ar (ΔLZY) Hence proved.

If x + y = 5 and xy = 4 , find x - y , using identities. -Maths 9th

Description : If x + y = 5 and xy = 4 , find x - y , using identities. -Maths 9th

Last Answer : (x+y)2 = x2 - 2xy + y2 = x2 + 2xy + y2 - 4xy = (x+y)2 - 4xy = 52 - 4 × 4 = 25 - 16 = 9 ∴ x - y = √9 = 3

X and y are points on the side LN of the triangle LMN , such that LX = XY = YN . Through X, a line is drawn parallel to LM to meet MN at Z. -Maths 9th

Description : X and y are points on the side LN of the triangle LMN , such that LX = XY = YN . Through X, a line is drawn parallel to LM to meet MN at Z. -Maths 9th

Last Answer : Here, △XZM and △XZL are on the same base (XZ) and lie between the same parallels (XZ || LM). ∴ ar(△XZL) = ar( △XZM) Adding ar(△XZY) on both sides , we have ar(△XZL) + ar(△XZY) = ar(△XZM) + ar(△XZY) ⇒ ar(△LZY) = ar(quad.MZYX)

Let x be rational and y be irrational. Is xy necessarily irrational? Justify your answer by an example. -Maths 9th

Description : Let x be rational and y be irrational. Is xy necessarily irrational? Justify your answer by an example. -Maths 9th

Last Answer : No, (xy) is necessarily an irrational only when x ≠0. Let x be a non-zero rational and y be an irrational. Then, we have to show that xy be an irrational. If possible, let xy be a rational number. Since ... wrong. Hence, xy is an irrational number. But, when x = 0, then xy = 0, a rational number.

X and Y are points on the side LN of the triangle LMN such that LX = XY = YN. -Maths 9th

Description : X and Y are points on the side LN of the triangle LMN such that LX = XY = YN. -Maths 9th

Last Answer : Given X and Y are points on the side LN such that LX = XY = YN and XZ || LM To prove ar (ΔLZY) = ar (MZYX) Proof Since, ΔXMZ and ΔXLZ are on the same base XZ and between the same parallel lines LM and XZ. ... get ar (ΔXMZ) + ar (ΔXXZ) = ar (ΔXLZ) + ar (ΔXYZ) => ar (MZYX) = ar (ΔLZY) Hence proved.

Prove that x^3+y^3=x+y(x^2+y^2-xy) -Maths 9th

Description : Prove that x^3+y^3=x+y(x^2+y^2-xy) -Maths 9th

Last Answer : NEED ANSWER

Prove that x^3+y^3=x+y(x^2+y^2-xy) -Maths 9th

Description : Prove that x^3+y^3=x+y(x^2+y^2-xy) -Maths 9th

Last Answer : x3 + y3 = x+y (x2 + y2 -xy) RHS = x+y (x2 + y2 -xy) On expanding, = x3 + xy2 - x2y + x2y + y3 - xy2 = x3 + y3 LHS = RHS

In Fig.5.7, AC = XD, c is the mid-point of AB and D is the mid-point of XY. Using a Euclid's axiom,show that AB=XY. -Maths 9th

Description : In Fig.5.7, AC = XD, c is the mid-point of AB and D is the mid-point of XY. Using a Euclid's axiom,show that AB=XY. -Maths 9th

Last Answer : Solution :-

In the given figure, ABC is an equilateral triangle of side length 30 cm. XY is parallel to BC, XP is parallel to AC and YQ is parallel to AB. -Maths 9th

Description : In the given figure, ABC is an equilateral triangle of side length 30 cm. XY is parallel to BC, XP is parallel to AC and YQ is parallel to AB. -Maths 9th

Last Answer : answer:

Let ABC be a triangle of area 16 cm^2 . XY is drawn parallel to BC dividing AB in the ratio 3 : 5. If BY is joined, then the area of triangle BXY is -Maths 9th

Description : Let ABC be a triangle of area 16 cm^2 . XY is drawn parallel to BC dividing AB in the ratio 3 : 5. If BY is joined, then the area of triangle BXY is -Maths 9th

Last Answer : answer:

Which one of the following is one of the factors of x^2 (y – z) + y^2 (z – x) – z (xy – yz – zx) ? -Maths 9th

Description : Which one of the following is one of the factors of x^2 (y – z) + y^2 (z – x) – z (xy – yz – zx) ? -Maths 9th

Last Answer : answer:

x^4 + xy^3 + x^3y + xz^3 + y^4 + yz^3 is divisible by : -Maths 9th

Description : x^4 + xy^3 + x^3y + xz^3 + y^4 + yz^3 is divisible by : -Maths 9th

Last Answer : answer:

If a = (xy)/(x+y), b = (xz)/(x+z), and c = (yz)/(y+z), where a, b and c are non-zero, then what is x equal to ? -Maths 9th

Description : If a = (xy)/(x+y), b = (xz)/(x+z), and c = (yz)/(y+z), where a, b and c are non-zero, then what is x equal to ? -Maths 9th

Last Answer : answer:

WXYZ is a square of side length 30. V is a point on XY and P is a point inside the square with PV perpendicular to XY. PW = PZ = PV – 5. Find PV. -Maths 9th

Description : WXYZ is a square of side length 30. V is a point on XY and P is a point inside the square with PV perpendicular to XY. PW = PZ = PV – 5. Find PV. -Maths 9th

Last Answer : answer:

In the given figure, YZ is parallel to MN, XY is parallel is LM and XZ is parallel to LN. Then MY is -Maths 9th

Description : In the given figure, YZ is parallel to MN, XY is parallel is LM and XZ is parallel to LN. Then MY is -Maths 9th

Last Answer : answer:

Find the remainder when y3 + y2 - 2y + 5 is divided by y - 5. -Maths 9th

Description : Find the remainder when y3 + y2 - 2y + 5 is divided by y - 5. -Maths 9th

Last Answer : Remainder = 145 Again, we should evaluate p(5) Let p(y) = y3 + y2 - 2y + 5 ∴ p(5) = 53 + 52 - 2 x 5 + 5 = 125 + 25 - 10 + 5 = 145 Thus , we find that p(5) is the remainder when p(y) is divided by y - 5 .

Without finding the cubes, factorise (x- 2y)3 + (2y – 3z)3 + (3z – x)3. -Maths 9th

Description : Without finding the cubes, factorise (x- 2y)3 + (2y – 3z)3 + (3z – x)3. -Maths 9th

Last Answer : We know that, a3 + b3 + c3 – 3 abc = (a + b + c)(a2 + b2 + c2 -ab-bc-ca) Also, if a + b + c = 0, then a3 + b3 + c3 = 3abc Here, we see that (x-2y) +(2y-3z)+ (3z-x) = 0 Therefore, (x-2y)3 + (2y-3z)3 + (3z-x)3 = 3(x-2y)(2y-3z)(3z-x).

Multiply x2 + 4y2 + z2 + 2xy + xz – 2yz by (-z + x-2y). -Maths 9th

Description : Multiply x2 + 4y2 + z2 + 2xy + xz – 2yz by (-z + x-2y). -Maths 9th

Last Answer : Multiply this question

Find the remainder when y3 + y2 - 2y + 5 is divided by y - 5. -Maths 9th

Description : Find the remainder when y3 + y2 - 2y + 5 is divided by y - 5. -Maths 9th

Last Answer : Remainder = 145 Again, we should evaluate p(5) Let p(y) = y3 + y2 - 2y + 5 ∴ p(5) = 53 + 52 - 2 x 5 + 5 = 125 + 25 - 10 + 5 = 145 Thus , we find that p(5) is the remainder when p(y) is divided by y - 5 .

Without finding the cubes, factorise (x- 2y)3 + (2y – 3z)3 + (3z – x)3. -Maths 9th

Description : Without finding the cubes, factorise (x- 2y)3 + (2y – 3z)3 + (3z – x)3. -Maths 9th

Last Answer : We know that, a3 + b3 + c3 – 3 abc = (a + b + c)(a2 + b2 + c2 -ab-bc-ca) Also, if a + b + c = 0, then a3 + b3 + c3 = 3abc Here, we see that (x-2y) +(2y-3z)+ (3z-x) = 0 Therefore, (x-2y)3 + (2y-3z)3 + (3z-x)3 = 3(x-2y)(2y-3z)(3z-x).

Multiply x2 + 4y2 + z2 + 2xy + xz – 2yz by (-z + x-2y). -Maths 9th

Description : Multiply x2 + 4y2 + z2 + 2xy + xz – 2yz by (-z + x-2y). -Maths 9th

Last Answer : Multiply this question

factorise (x-2y)^3-(x+2y)^3 -Maths 9th

Description : factorise (x-2y)^3-(x+2y)^3 -Maths 9th

Last Answer : NEED ANSWER

factorise (x-2y)^3-(x+2y)^3 -Maths 9th

Description : factorise (x-2y)^3-(x+2y)^3 -Maths 9th

Last Answer : Given ( x - 2y) 3 + (2y - 3z) 3 + ( 3z - x) 3 Let a = ( x - 2y), b =(2y - 3z), c= ( 3z - x) a + b + c = ( x - 2y)+(2y - 3z)+( 3z - x) = 0 Recall that if (a + b + c) = 0 then a 3 + b 3 + c 3 = 3abc Thus, ( x - 2y) 3 + (2y - 3z) 3 + ( 3z - x) 3 = 3( x - 2y)(2y - 3z)( 3z - x)

Draw a graph of the equation x - Y = 4 & 2x+ 2y =4 on the same graph paper find the coordinates of the point whose two lines intersect. -Maths 9th

Description : Draw a graph of the equation x - Y = 4 & 2x+ 2y =4 on the same graph paper find the coordinates of the point whose two lines intersect. -Maths 9th

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