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

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

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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

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

Last Answer : Solution :-

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

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

Last Answer : Solution :-

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

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

Description : Which of the following is a factor of (x+ y)3 – (x3 + y3) ? -Maths 9th

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Description : Which of the following is a factor of (x+ y)3 – (x3 + y3) ? -Maths 9th

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Description : If x-2 is a factor of x3-3x+5a then find the value of a -Maths 9th

Last Answer : -2/5 is value of a

Description : If x-2 is a factor of x3-3x+5a then find the value of a -Maths 9th

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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!

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

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

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

Description : The standard equation of a cubical spiral transition curve provided on roads, is (A) y = l²/6RL (B) y = x3 /6RL (C) y = x2 /6RL (D) y = x/6RL

Last Answer : Answer: Option B

Description : Big - O estimate for f(x) = (x + 1) log(x2 + 1) + 3x2 is given as (A) O(xlogx) (B) O(x2) (C) O(x3) (D) O(x2logx)

Last Answer : (B) O(x2)

Description : p(x)=x3+3x2+3x+1, g(x) = x+2 -Maths 9th

Last Answer : p(x) = x3+3x2+3x+1, g(x) = x+2 g(x) = 0 ⇒ x+2 = 0 ⇒ x = −2 ∴ Zero of g(x) is -2. Now, p(−2) = (−2)3+3(−2)2+3(−2)+1 = −8+12−6+1 = −1 ≠ 0 ∴By factor theorem, g(x) is not a factor of p(x

Description : For what value of m is x3 -2mx2 +16 divisible by x + 2 ? -Maths 9th

Last Answer : Let p(x) = x3 -2mx2 +16 Since, p(x) is divisible by (x+2), then remainder = 0 P(-2) = 0 ⇒ (-2)3 -2m(-2)2 + 16=0 ⇒ -8-8m+16=0 ⇒ 8 = 8 m m = 1 Hence, the value of m is 1 .

Description : Find the value of x3 +y3 -12xy + 64,when x+y = -4. -Maths 9th

Last Answer : Find the value of

Description : For what value of m is x3 -2mx2 +16 divisible by x + 2 ? -Maths 9th

Last Answer : Let p(x) = x3 -2mx2 +16 Since, p(x) is divisible by (x+2), then remainder = 0 P(-2) = 0 ⇒ (-2)3 -2m(-2)2 + 16=0 ⇒ -8-8m+16=0 ⇒ 8 = 8 m m = 1 Hence, the value of m is 1 .

Description : Find the value of x3 +y3 -12xy + 64,when x+y = -4. -Maths 9th

Last Answer : Find the value of

Description : p(x)=x3-2x2-4x-1, g(x)=x- -Maths 9th

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Description : p(x)=x3-2x2-4x-1, g(x)=x- -Maths 9th

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

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Description : factorize x3-2x2-x+2 -Maths 9th

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Description : Find the value of x3+y3+12xy-64 when x=4 -Maths 9th

Last Answer : On the value of x we put the value 4 When it is in power 64+y3+12x4y-64 64+y3+48y-64 64-64+y3+48y 0+y(y2+48) Y(y2+48)

Description : The position vectors of three particles are given by  X1 = (5i + 5j)m, X2 = (5ti + 5tj)m, and X3 = (5ti + 10t^2tj)

Last Answer : The position vectors of three particles are given by \(\overset\rightarrow{X}_1\) = (\(5\hat ... the velocity and acceleration for each, in SI units.

Description : The turns ratio of device 'B' shown in the illustration is two to one (total). If 220 volts were applied to terminals 'H1' & 'H2', what would be indicated across 'X1' & 'X4' with 'X2' & 'X3' connected and isolated? EL-0082 A. 55 volts B. 110 volts C. 220 volts D. 440 volts

Last Answer : Answer: B

Description : The turns ratio of device 'B' shown in the illustration is two to one (total). If 440 volts were applied to terminals 'H1' & 'H2', what would be indicated across 'X1' & 'X4' with 'X2' & 'X3' connected and isolated? EL-0082 A. 110 volts B. 220 volts C. 880 volts D. 1760 volts

Last Answer : Answer: B

Description : Consider the following LPP: Min. Z = x1+x2+x3 Subject to 3x1 + 4x3 ≤ 5 5x1 + x2 + 6x3 =7 8x1 + 9x3 ≥ 2 x1, x2, x3 ≥ 0 The standard form of this LPP shall be: (1) Min. Z = x1 + x2 + x3 + 0x4 + 0x5 Subject ... + 4x3 + x4 = 5; 5x1 + x2 + 6x3 + x6 = 7; 8x1 + 9x3 - x5 + x7 = 2; x1 to x7 ≥ 0

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Description : Factorise: x3 +13x2 + 32x + 20. -Maths 9th

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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

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

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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.

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

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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

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

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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.

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

Last Answer : Multiply this question

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

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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.

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

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

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

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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

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

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.

Description : The scores of an English test out of 100 of 20 students are given below : 75, 69, 88, 55, 95, 88, 73, 64, 75, 98, 88, 95, 90, 95, 88, 44, 59, 67, 88, 99. -Maths 9th

Last Answer : Median=n=even =n/2=20/2=10th observation =98 Mode =88