Write the coefficient of x2 in each of the following -Maths 9th

Write the coefficient of x2 in each of the following -Maths 9th
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The coefficient of x2 in each of the following .
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Write the coefficient of x2 in each of the following -Maths 9th

Description : Write the coefficient of x2 in each of the following -Maths 9th

Last Answer : The coefficient of x2 in each of the following .

Use the Factor Theorem to determine whether g(x) is a factor of p(x) in each of the following cases: (i) p(x) = 2x3+x2–2x–1, g(x) = x+1 -Maths 9th

Description : Use the Factor Theorem to determine whether g(x) is a factor of p(x) in each of the following cases: (i) p(x) = 2x3+x2–2x–1, g(x) = x+1 -Maths 9th

Last Answer : Solution: p(x) = 2x3+x2–2x–1, g(x) = x+1 g(x) = 0 ⇒ x+1 = 0 ⇒ x = −1 ∴Zero of g(x) is -1. Now, p(−1) = 2(−1)3+(−1)2–2(−1)–1 = −2+1+2−1 = 0 ∴By factor theorem, g(x) is a factor of p(x

Evaluate each of the following using identities: (i) (2x –1x)2 (ii) (2x + y) (2x – y) (iii) (a2b – b2a)2 (iv) (a – 0.1) (a + 0.1) (v) (1.5.x2 – 0.3y2) (1.5x2 + 0.3y2) -Maths 9th

Description : Evaluate each of the following using identities: (i) (2x –1x)2 (ii) (2x + y) (2x – y) (iii) (a2b – b2a)2 (iv) (a – 0.1) (a + 0.1) (v) (1.5.x2 – 0.3y2) (1.5x2 + 0.3y2) -Maths 9th

Last Answer : (i) (2x - 1/x)2 [Use identity: (a - b)2 = a2 + b2 - 2ab ] (2x - 1/x)2 = (2x) 2 + (1/x)2 - 2 (2x)(1/x) = 4x2 + 1/x2 - 4 (ii) (2x + y) (2x - y) [Use identity: (a - b)(a + b) = a2 - b 2 ] (2x + y) (2x - ... ) = a2 - b 2 ](1.5 x 2 - 0.3y2 ) (1.5x2 + 0.3y2 ) = (1.5 x 2 ) 2 - (0.3y2 ) 2 = 2.25 x4 - 0.09y4

Determine which of the following polynomials has (x + 1) a factor: (i) x3+x2+x+1 -Maths 9th

Description : Determine which of the following polynomials has (x + 1) a factor: (i) x3+x2+x+1 -Maths 9th

Last Answer : Solution: Let p(x) = x3+x2+x+1 The zero of x+1 is -1. [x+1 = 0 means x = -1] p(−1) = (−1)3+(−1)2+(−1)+1 = −1+1−1+1 = 0 ∴By factor theorem, x+1 is a factor of x3+x2+x+1

If both (x+1) and (x -1) are factors of ax3 + x2 - 2x + b , find a and b. -Maths 9th

Description : If both (x+1) and (x -1) are factors of ax3 + x2 - 2x + b , find a and b. -Maths 9th

Last Answer : Let p(x) = ax3 + x2 - 2x + b Since (x+1) and (x-1) are the factors of p(x), ∴ p(-1) = 0 and p(1) = 0 ∴ p(-1) = a(-1)3 + (-1)2 - 2 (-1) + b = 0 ⇒ - a + 1 + 2 + b = 0 ⇒ a - b = 3 ---- (i) ... 0 ⇒ a + 1 - 2 + b = 0 ⇒ a + b = 1 ----- (ii) solving equations (i) and (ii) we get a = 2 and b = -1

If x2 - 1 is a factor of ax4 + bx3 + cx2 + dx + e , show that a + c + e = b + d = 0. -Maths 9th

Description : If x2 - 1 is a factor of ax4 + bx3 + cx2 + dx + e , show that a + c + e = b + d = 0. -Maths 9th

Last Answer : Since x2 - 1 = (x - 1) is a factor of p(x) = ax4 + bx3 + cx2 + dx + e ∴ p(x) is divisible by (x+1) and (x-1) separately ⇒ p(1) = 0 and p(-1) = 0 p(1) = a(1)4 + b(1)3 + c(1)2 + d(1) + e = 0 ... (b+d) = 0 ⇒ b + d = 0 ---- (iii) comparing equations (ii) and (iii) , we get a + c + e = b + d = 0

If p (x) = x2 – 4x + 3, then evaluate p(2) – p (-1) + p (1/2). -Maths 9th

Description : If p (x) = x2 – 4x + 3, then evaluate p(2) – p (-1) + p (1/2). -Maths 9th

Last Answer : Solution of this question

Show that, x + 3 is a factor of 69 + 11c – x2 + x3 -Maths 9th

Description : Show that, x + 3 is a factor of 69 + 11c – x2 + x3 -Maths 9th

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

If x +1 is a factor of ax3 +x2 -2x + 4a - 9, then find the value of a. -Maths 9th

Description : If x +1 is a factor of ax3 +x2 -2x + 4a - 9, then find the value of a. -Maths 9th

Last Answer : The value of a

Factorise : x2 + 9x +18 -Maths 9th

Description : Factorise : x2 + 9x +18 -Maths 9th

Last Answer : Factorisation

Without actual division, prove that 2x4 – 5x3 + 2x2 – x+ 2 is divisible by x2-3x+2. -Maths 9th

Description : Without actual division, prove that 2x4 – 5x3 + 2x2 – x+ 2 is divisible by x2-3x+2. -Maths 9th

Last Answer : Let p(x) = 2x4 - 5x3 + 2x2 - x+ 2 firstly, factorise x2-3x+2. Now, x2-3x+2 = x2-2x-x+2 [by splitting middle term] = x(x-2)-1 (x-2)= (x-1)(x-2) Hence, 0 of x2-3x+2 are land 2. We have to prove that, 2x4 ... )2 - 2 + 2 = 2x16-5x8+2x4+ 0 = 32 - 40 + 8 = 40 - 40 =0 Hence, p(x) is divisible by x2-3x+2.

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

If both (x+1) and (x -1) are factors of ax3 + x2 - 2x + b , find a and b. -Maths 9th

Description : If both (x+1) and (x -1) are factors of ax3 + x2 - 2x + b , find a and b. -Maths 9th

Last Answer : Let p(x) = ax3 + x2 - 2x + b Since (x+1) and (x-1) are the factors of p(x), ∴ p(-1) = 0 and p(1) = 0 ∴ p(-1) = a(-1)3 + (-1)2 - 2 (-1) + b = 0 ⇒ - a + 1 + 2 + b = 0 ⇒ a - b = 3 ---- (i) ... 0 ⇒ a + 1 - 2 + b = 0 ⇒ a + b = 1 ----- (ii) solving equations (i) and (ii) we get a = 2 and b = -1

If x2 - 1 is a factor of ax4 + bx3 + cx2 + dx + e , show that a + c + e = b + d = 0. -Maths 9th

Description : If x2 - 1 is a factor of ax4 + bx3 + cx2 + dx + e , show that a + c + e = b + d = 0. -Maths 9th

Last Answer : Since x2 - 1 = (x - 1) is a factor of p(x) = ax4 + bx3 + cx2 + dx + e ∴ p(x) is divisible by (x+1) and (x-1) separately ⇒ p(1) = 0 and p(-1) = 0 p(1) = a(1)4 + b(1)3 + c(1)2 + d(1) + e = 0 ... (b+d) = 0 ⇒ b + d = 0 ---- (iii) comparing equations (ii) and (iii) , we get a + c + e = b + d = 0

If p (x) = x2 – 4x + 3, then evaluate p(2) – p (-1) + p (1/2). -Maths 9th

Description : If p (x) = x2 – 4x + 3, then evaluate p(2) – p (-1) + p (1/2). -Maths 9th

Last Answer : Solution of this question

Show that, x + 3 is a factor of 69 + 11c – x2 + x3 -Maths 9th

Description : Show that, x + 3 is a factor of 69 + 11c – x2 + x3 -Maths 9th

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

If x +1 is a factor of ax3 +x2 -2x + 4a - 9, then find the value of a. -Maths 9th

Description : If x +1 is a factor of ax3 +x2 -2x + 4a - 9, then find the value of a. -Maths 9th

Last Answer : The value of a

Factorise : x2 + 9x +18 -Maths 9th

Description : Factorise : x2 + 9x +18 -Maths 9th

Last Answer : Factorisation

Without actual division, prove that 2x4 – 5x3 + 2x2 – x+ 2 is divisible by x2-3x+2. -Maths 9th

Description : Without actual division, prove that 2x4 – 5x3 + 2x2 – x+ 2 is divisible by x2-3x+2. -Maths 9th

Last Answer : Let p(x) = 2x4 - 5x3 + 2x2 - x+ 2 firstly, factorise x2-3x+2. Now, x2-3x+2 = x2-2x-x+2 [by splitting middle term] = x(x-2)-1 (x-2)= (x-1)(x-2) Hence, 0 of x2-3x+2 are land 2. We have to prove that, 2x4 ... )2 - 2 + 2 = 2x16-5x8+2x4+ 0 = 32 - 40 + 8 = 40 - 40 =0 Hence, p(x) is divisible by x2-3x+2.

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

Let x be the mean of x1, x2,….,xn and y be the mean of y1, y2, ……,yn the mean of z is x1, x2,….,xn , y1, y2, ……,yn then z is equal to -Maths 9th

Description : Let x be the mean of x1, x2,….,xn and y be the mean of y1, y2, ……,yn the mean of z is x1, x2,….,xn , y1, y2, ……,yn then z is equal to -Maths 9th

Last Answer : NEED ANSWER

FOR WHAT VALUE OF K, THE POLYNOMIAL X2+(4-K)X+2 IS DIVISIBLE BY X-2 -Maths 9th

Description : FOR WHAT VALUE OF K, THE POLYNOMIAL X2+(4-K)X+2 IS DIVISIBLE BY X-2 -Maths 9th

Last Answer : As the given polynomial divisible by x-2 means the polynomial satisfies for the value x=2 So putting x=2 in x²+(4-k)x+2 yields 0 ⇒2²+(4-k)2+2=0 ⇒4+8-2k+2=0 ⇒ 2k=14 ⇒ k= ... ;-3x+2 if factorized yields (x-1)(x-2). Thus is divisible by x-2 as well as divisible by x-1.

Let x be the mean of x1, x2,….,xn and y be the mean of y1, y2, ……,yn the mean of z is x1, x2,….,xn , y1, y2, ……,yn then z is equal to -Maths 9th

Description : Let x be the mean of x1, x2,….,xn and y be the mean of y1, y2, ……,yn the mean of z is x1, x2,….,xn , y1, y2, ……,yn then z is equal to -Maths 9th

Last Answer : According to question find the value of z

FOR WHAT VALUE OF K, THE POLYNOMIAL X2+(4-K)X+2 IS DIVISIBLE BY X-2 -Maths 9th

Description : FOR WHAT VALUE OF K, THE POLYNOMIAL X2+(4-K)X+2 IS DIVISIBLE BY X-2 -Maths 9th

Last Answer : The value of 'k' is 4

Using factor theorem, factorise the polynomial x3 + x2 - 4x - 4. -Maths 9th

Description : Using factor theorem, factorise the polynomial x3 + x2 - 4x - 4. -Maths 9th

Last Answer : Solution :-

If x2 + 1/x2 = 34, find x3 + 1/x3 - 9. -Maths 9th

Description : If x2 + 1/x2 = 34, find x3 + 1/x3 - 9. -Maths 9th

Last Answer : Solution :-

If one of the roots of the equation x^2 + ax + 3 = 0 is 3 and one of the roots of the equation x2 + ax + b = 0 is three -Maths 9th

Description : If one of the roots of the equation x^2 + ax + 3 = 0 is 3 and one of the roots of the equation x2 + ax + b = 0 is three -Maths 9th

Last Answer : answer:

If the sum of the zeroes of the polynomial p(x) = (k2 – 14) x2 – 2x – 12 is 1, then find the value of k. -Maths 9th

Description : If the sum of the zeroes of the polynomial p(x) = (k2 – 14) x2 – 2x – 12 is 1, then find the value of k. -Maths 9th

Last Answer : p(x) = (k2 – 14) x2 – 2x – 12 Here a = k2 – 14, b = -2, c = -12 Sum of the zeroes, (α + β) = 1 …[Given] ⇒ − = 1 ⇒ −(−2)2−14 = 1 ⇒ k2 – 14 = 2 ⇒ k2 = 16 ⇒ k = ±4

the product (x2 – 1) (x4 + x2 + 1) is equal to -Maths 9th

Description : the product (x2 – 1) (x4 + x2 + 1) is equal to -Maths 9th

Last Answer : (x^2-1)(x^4+x^2+1)=x^6+x^4+x^2-x^4-x^2-1. =x^6+1.

Simplify: (i) (a + b + c)2 + (a – b + c)2 (ii) (a + b + c)2 – (a – b + c)2 (iii) (a + b + c)2 + (a – b + c)2 + (a + b – c)2 (iv) (2x + p – c)2 – (2x – p + c)2 (v) (x2 + y2 – z2)2 – (x2 – y2 + z2)2 -Maths 9th

Description : Simplify: (i) (a + b + c)2 + (a – b + c)2 (ii) (a + b + c)2 – (a – b + c)2 (iii) (a + b + c)2 + (a – b + c)2 + (a + b – c)2 (iv) (2x + p – c)2 – (2x – p + c)2 (v) (x2 + y2 – z2)2 – (x2 – y2 + z2)2 -Maths 9th

Last Answer : answer:

Find the value of x3 + y3 + z3 – 3xyz if x2 + y2 + z2 = 83 and x + y + z = 15 -Maths 9th

Description : Find the value of x3 + y3 + z3 – 3xyz if x2 + y2 + z2 = 83 and x + y + z = 15 -Maths 9th

Last Answer : Consider the equation x + y + z = 15 From algebraic identities, we know that (a + b + c)2 = a2 + b2 + c2 + 2(ab + bc + ca) So, (x + y + z)2 = x2 + y2 + z2 + 2(xy + yz + xz) From the question, x2 + y2 + z2 ... y3 + z3 - 3xyz = 15(83 - 71) => x3 + y3 + z3 - 3xyz = 15 12 Or, x3 + y3 + z3 - 3xyz = 180

If cosec |=3x and cot |=3/x. then find the value of (x2-1/x2) -Maths 9th

Description : If cosec |=3x and cot |=3/x. then find the value of (x2-1/x2) -Maths 9th

Last Answer : cosecβ = = 3X and cotβ = 3/X , to find value ( X2 - 1/X2 ) . we know , cosec2β = cot2β +1 putting value , (3X )2 = ( 3/X )2 +1 , OR 9X2 = 9 /X2 + 1 , OR 9X2 – 9/X2 = 1, OR 9( X2 - 1/X2 ) = 1, SO (X2 – 1/X2 ) = 1/9

If x sin3|+ y cos3|=sin|cos| and xsin|=ycos|, prove x2+y2=1. -Maths 9th

Description : If x sin3|+ y cos3|=sin|cos| and xsin|=ycos|, prove x2+y2=1. -Maths 9th

Last Answer : xsin3θ+ycos3θ=sinθcosθ (xsinθ)sin2θ+(ycosθ)cos2θ=sinθcosθ (xsinθ)sin2θ+(xsinθ)cos2θ=sinθcosθ xsinθ=sinθcosθ x=cosθ Again, ycosθ=xsinθ ycosθ=cosθsinθ y=sinθ Therefore, x2+y2=sin2θ+cos2θ=1.

Write the coefficient of y in the expansion of (5-y) to the whole square. -Maths 9th

Description : Write the coefficient of y in the expansion of (5-y) to the whole square. -Maths 9th

Last Answer : Solution :-

what is the coefficient -Maths 9th

Description : what is the coefficient -Maths 9th

Last Answer : 2xy = coefficient is 2. the numerical value of a variable is called coefficient.

The coefficient of x in the expansion of (x + 3)3 is -Maths 9th

Description : The coefficient of x in the expansion of (x + 3)3 is -Maths 9th

Last Answer : (d) Now, (x + 3)3 = x3 + 33 + 3x (3)(x + 3) [using identity, (a + b)3 = a3 + b3 + 3ab (a + b)] = x3 + 27 + 9x (x + 3) = x3 +27 + 9x2+27x Hence, the coefficient of x in (x + 3)3 is 27.

what is the coefficient -Maths 9th

Description : what is the coefficient -Maths 9th

Last Answer : 2xy = coefficient is 2. the numerical value of a variable is called coefficient.

The coefficient of x in the expansion of (x + 3)3 is -Maths 9th

Description : The coefficient of x in the expansion of (x + 3)3 is -Maths 9th

Last Answer : (d) Now, (x + 3)3 = x3 + 33 + 3x (3)(x + 3) [using identity, (a + b)3 = a3 + b3 + 3ab (a + b)] = x3 + 27 + 9x (x + 3) = x3 +27 + 9x2+27x Hence, the coefficient of x in (x + 3)3 is 27.

For what value of p is the coefficient of x^2 in the product (2x – 1) (x – k) (px + 1) equal to 0 and the constant term equal to 2 ? -Maths 9th

Description : For what value of p is the coefficient of x^2 in the product (2x – 1) (x – k) (px + 1) equal to 0 and the constant term equal to 2 ? -Maths 9th

Last Answer : answer:

Write the coordinates of each of the points P, Q, R, S, T and 0 from the figure . -Maths 9th

Description : Write the coordinates of each of the points P, Q, R, S, T and 0 from the figure . -Maths 9th

Last Answer : Here, points P and S lie in I quadrant so their both coordinates will be positive. Now, perpendicular distance of P from both axes is 1, so coordinates of P are (1, 1). Also, perpendicular distance of S ... 0 is the intersection of both axes, so it is the origin and its coordinates are O (0,0).

Write the linear equation such that each point on its graph has an ordinate 3 times its abscissa. -Maths 9th

Description : Write the linear equation such that each point on its graph has an ordinate 3 times its abscissa. -Maths 9th

Last Answer : Let the abscissa of the point be x, According to the question, Ordinate (y) = 3 x Abscissa ⇒ y=3x When x = 1, then y = 3 x 1 = 3 and when x = 2, then y = 3 x 2 = 6. Here, ... the line AB. Hence, y = 3x is the required equation such that each point on its graph has an ordinate 3 times its abscissa.

Write the coordinates of each of the points P, Q, R, S, T and 0 from the figure . -Maths 9th

Description : Write the coordinates of each of the points P, Q, R, S, T and 0 from the figure . -Maths 9th

Last Answer : Here, points P and S lie in I quadrant so their both coordinates will be positive. Now, perpendicular distance of P from both axes is 1, so coordinates of P are (1, 1). Also, perpendicular distance of S ... 0 is the intersection of both axes, so it is the origin and its coordinates are O (0,0).

Write the linear equation such that each point on its graph has an ordinate 3 times its abscissa. -Maths 9th

Description : Write the linear equation such that each point on its graph has an ordinate 3 times its abscissa. -Maths 9th

Last Answer : Let the abscissa of the point be x, According to the question, Ordinate (y) = 3 x Abscissa ⇒ y=3x When x = 1, then y = 3 x 1 = 3 and when x = 2, then y = 3 x 2 = 6. Here, ... the line AB. Hence, y = 3x is the required equation such that each point on its graph has an ordinate 3 times its abscissa.

Write ‘yes’ if each of the following relations is an equivalence relation. -Maths 9th

Description : Write ‘yes’ if each of the following relations is an equivalence relation. -Maths 9th

Last Answer : (i) is parallel to' is reflexive, because any line is parallel to itself. Symmetric, because if line l is parallel to line m, then line m is parallel to line l. Transitive, because if l || m ... y cannot be a factor of x. (v) No, because the relation is reflexive and transitive but not symmetric.

Write each of the following as decimals. (a) 20 + 9 + 4 / 10 + 1 / 100 (b) 137 + 5 / 100 (c) 7 / 10 + 6 / 100 + 4 / 1000 (d) 23 + 2 / 10 + 6 / 1000 (e) 700 + 20 + 5 + 9 / 100 -Maths 9th

Description : Write each of the following as decimals. (a) 20 + 9 + 4 / 10 + 1 / 100 (b) 137 + 5 / 100 (c) 7 / 10 + 6 / 100 + 4 / 1000 (d) 23 + 2 / 10 + 6 / 1000 (e) 700 + 20 + 5 + 9 / 100 -Maths 9th

Last Answer : (a) 20 + 9 + 4 / 10 + 1 / 100 = 29 + 0.4 + 0.01 = 29.41 (b) 137 + 5 / 100 = 137 + 0.05 = 137.05 (c) 7 / 10 + 6 / 100 + 4 / 1000 = 0.7 + 0.06 + 0.004 =0.764 (d) 23 + 2 / 10 + 6 / 1000 = 23 + 0.2 + 0.006 = 23.206 (e) 700 + 20 + 5 + 9 / 100 = 725 + 0.09 = 725.09

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 the roots of the equation (b – c)x2 + (c – a)x + (a – b) = 0 are equal, then prove that 2b = a + c. -Maths 10th

Description : If the roots of the equation (b – c)x2 + (c – a)x + (a – b) = 0 are equal, then prove that 2b = a + c. -Maths 10th

Last Answer : following is the equation of 2b = a+c =

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 the mean of five observations x, x + 2, x + 4, x + 6, x + 8 is 11, then write the value of x. -Maths 9th

Description : If the mean of five observations x, x + 2, x + 4, x + 6, x + 8 is 11, then write the value of x. -Maths 9th

Last Answer : x + x + 2 + x + 4 + x + 6 + x + 8 / 5 = 11 5x + 20 = 55 5x = 35 ⇒ x = 7