Expand using suitable identity (-2x + 5y - 3z) to the whole square. -Maths 9th

Expand using suitable identity (-2x + 5y - 3z) to the whole square. -Maths 9th
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Find (2x – y + 3z) (4x2 + y2 + 9z2 + 2xy + 3yz – 6xz). -Maths 9th

Description : Find (2x – y + 3z) (4x2 + y2 + 9z2 + 2xy + 3yz – 6xz). -Maths 9th

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Find (2x – y + 3z) (4x2 + y2 + 9z2 + 2xy + 3yz – 6xz). -Maths 9th

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Simplify (2x- 5y)3 – (2x+ 5y)3. -Maths 9th

Description : Simplify (2x- 5y)3 – (2x+ 5y)3. -Maths 9th

Last Answer : (2x -5y)3 - (2x + 5y)3 = [(2x)3 - (5y)3 - 3(2x)(5y)(2x - 5y)] -[(2x)3 + (5y)3 + 3(2x)(5y)(2x+5y)] [using identity, (a - b)3 = a3 -b3 - 3ab and (a + b)3 =a3 +b3 + 3ab] = (2x)3 - (5y)3 - ... 2x)3- (5y)3 - 30xy (2x + 5y) = -2 (5y)3 - 30xy(2x - 5y + 2x + 5y) = -2 x 125y3 - 30xy(4x) = -250y3 -120x2y

The linear equation 2x – 5y = 7 has -Maths 9th

Description : The linear equation 2x – 5y = 7 has -Maths 9th

Last Answer : (c) In the given equation 2x – 5y = 7, for every value of x, we get a corresponding value of y and vice-versa. Therefore, the linear equation has infinitely many solutions.

The equation 2x+ 5y = 7 has a unique solution, if x and y are -Maths 9th

Description : The equation 2x+ 5y = 7 has a unique solution, if x and y are -Maths 9th

Last Answer : (a) In natural numbers, there is only one pair i.e., (1, 1) which satisfy the given equation but in positive real numbers, real numbers and rational numbers there are many pairs to satisfy the given linear equation.

Simplify (2x- 5y)3 – (2x+ 5y)3. -Maths 9th

Description : Simplify (2x- 5y)3 – (2x+ 5y)3. -Maths 9th

Last Answer : (2x -5y)3 - (2x + 5y)3 = [(2x)3 - (5y)3 - 3(2x)(5y)(2x - 5y)] -[(2x)3 + (5y)3 + 3(2x)(5y)(2x+5y)] [using identity, (a - b)3 = a3 -b3 - 3ab and (a + b)3 =a3 +b3 + 3ab] = (2x)3 - (5y)3 - ... 2x)3- (5y)3 - 30xy (2x + 5y) = -2 (5y)3 - 30xy(2x - 5y + 2x + 5y) = -2 x 125y3 - 30xy(4x) = -250y3 -120x2y

The linear equation 2x – 5y = 7 has -Maths 9th

Description : The linear equation 2x – 5y = 7 has -Maths 9th

Last Answer : (c) In the given equation 2x – 5y = 7, for every value of x, we get a corresponding value of y and vice-versa. Therefore, the linear equation has infinitely many solutions.

The equation 2x+ 5y = 7 has a unique solution, if x and y are -Maths 9th

Description : The equation 2x+ 5y = 7 has a unique solution, if x and y are -Maths 9th

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Simplify: (2x–5y)3 – (2x + 5y)3 -Maths 9th

Description : Simplify: (2x–5y)3 – (2x + 5y)3 -Maths 9th

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How many solutions does the equation 2x +5y=8 has? -Maths 9th

Description : How many solutions does the equation 2x +5y=8 has? -Maths 9th

Last Answer : Solution :- Infinitely many solutions.

Using suitable identity, evaluate the following. -Maths 9th

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Using suitable identity, evaluate the following -Maths 9th

Description : Using suitable identity, evaluate the following -Maths 9th

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Using suitable identity, evaluate the following. -Maths 9th

Description : Using suitable identity, evaluate the following. -Maths 9th

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Using suitable identity, evaluate the following -Maths 9th

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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

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If the polynomials az3 +4z2 + 3z-4 and z3-4z + 0 leave the same remainder when divided by z – 3, -Maths 9th

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6x^+17xy+5y^ -Maths 9th

Description : 6x^+17xy+5y^ -Maths 9th

Last Answer : what we have to do to this question??? we have to factorise or do something else?

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The coordinates of the points where 2x + 5y – 10 =0 meets y axis is a) (0 , 2) b) (0 , -2) c) (2 , 2) d) (-2 , -2)

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