How do you Evaluate the expression X2 and y4?

How do you Evaluate the expression X2 and y4?
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It is x*x + y*y*y*y
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Last Answer : this is the correct answer!

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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

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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

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

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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

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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.

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

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

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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

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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

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

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

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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

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Last Answer : Factorisation

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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.

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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

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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.

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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

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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

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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.

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

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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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