Solve x^2 – 5x + 4 > 0. -Maths 9th

Solve x^2 – 5x + 4 > 0. -Maths 9th
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If x = 0 and y = k is a solution of the equation 5x - 3 y = 0, find the value of k. -Maths 9th

Description : If x = 0 and y = k is a solution of the equation 5x - 3 y = 0, find the value of k. -Maths 9th

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Find the solution set of x^2 – 5x + 6 > 0. -Maths 9th

Description : Find the solution set of x^2 – 5x + 6 > 0. -Maths 9th

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Find the value of k for 5x +2ky =3k, if x =1 and y =1 is its solution. -Maths 9th

Description : Find the value of k for 5x +2ky =3k, if x =1 and y =1 is its solution. -Maths 9th

Last Answer : Given, equation is 5x + 2ky = 3k. On putting x =1 and y =1 in this equation, we get 5(1) + 2k(1) =3k ⇒ 5 + 2k =3k ⇒ 5 = 3k - 2k ⇒ k = 5 Hence, required value of k is 5.

The value of the polynomial 5x – 4x2 + 3, when x = -1 is -Maths 9th

Description : The value of the polynomial 5x – 4x2 + 3, when x = -1 is -Maths 9th

Last Answer : (a) Let p (x) = 5x – 4x2 + 3 …(i) On putting x = -1 in Eq. (i), we get p(-1) = 5(-1) -4(-1)2 + 3= - 5 - 4 + 3 = -6

If both x – 2 and x -(1/2) are factors of px2+ 5x+r, then show that p = r. -Maths 9th

Description : If both x – 2 and x -(1/2) are factors of px2+ 5x+r, then show that p = r. -Maths 9th

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Find the value of k for 5x +2ky =3k, if x =1 and y =1 is its solution. -Maths 9th

Description : Find the value of k for 5x +2ky =3k, if x =1 and y =1 is its solution. -Maths 9th

Last Answer : Given, equation is 5x + 2ky = 3k. On putting x =1 and y =1 in this equation, we get 5(1) + 2k(1) =3k ⇒ 5 + 2k =3k ⇒ 5 = 3k - 2k ⇒ k = 5 Hence, required value of k is 5.

The value of the polynomial 5x – 4x2 + 3, when x = -1 is -Maths 9th

Description : The value of the polynomial 5x – 4x2 + 3, when x = -1 is -Maths 9th

Last Answer : (a) Let p (x) = 5x – 4x2 + 3 …(i) On putting x = -1 in Eq. (i), we get p(-1) = 5(-1) -4(-1)2 + 3= - 5 - 4 + 3 = -6

If both x – 2 and x -(1/2) are factors of px2+ 5x+r, then show that p = r. -Maths 9th

Description : If both x – 2 and x -(1/2) are factors of px2+ 5x+r, then show that p = r. -Maths 9th

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Find the value of k if (x-2)is a factor of polynomial p(x) = 2x(cube) - 6x(square) + 5x + k. -Maths 9th

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What must be added to 2x(square) - 5x + 6 to get x(cube) - 3x(square) + 3x - 5? -Maths 9th

Description : What must be added to 2x(square) - 5x + 6 to get x(cube) - 3x(square) + 3x - 5? -Maths 9th

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If both (x -2) and (x - 1/2) are factors of px2 + 5x + r, show that p = r. -Maths 9th

Description : If both (x -2) and (x - 1/2) are factors of px2 + 5x + r, show that p = r. -Maths 9th

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If the expressions (px^3 + 3x^2 – 3) and (2x^3 – 5x + p) when divided by (x – 4) leave the same remainder, then what is the value of p ? -Maths 9th

Description : If the expressions (px^3 + 3x^2 – 3) and (2x^3 – 5x + p) when divided by (x – 4) leave the same remainder, then what is the value of p ? -Maths 9th

Last Answer : Given that the following polynomials leave the same remainder when divided by (x - 4) : We are to find the value of a. Remainder theorem: When (x - b) divides a polynomial p(x), then the remainder is p(b). So, from (i) and (ii), we get Thus, the required value of a is 1.

If x^3 + 5x^2 + 10k leaves remainder – 2x when divided by x^2 + 2, then what is the value of k ? -Maths 9th

Description : If x^3 + 5x^2 + 10k leaves remainder – 2x when divided by x^2 + 2, then what is the value of k ? -Maths 9th

Last Answer : x3+5x2+10k =(x2+2)(x+5)+10k−2x−10 ⇒10k−2x−10=−2x ⇒10k−10=0 or k=1.

If both (x – 2) and (x – 1/2) are factors of px^2 + 5x + r, then: -Maths 9th

Description : If both (x – 2) and (x – 1/2) are factors of px^2 + 5x + r, then: -Maths 9th

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What is (x^2-3x+2)/(x^2-5x +6)÷(x^2-5x+4)/(x^2-7x+12) equal to? -Maths 9th

Description : What is (x^2-3x+2)/(x^2-5x +6)÷(x^2-5x+4)/(x^2-7x+12) equal to? -Maths 9th

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Find the value of the polynomial 5x – 4x2 + 3 at x = 2 and x = –1 -Maths 9th

Description : Find the value of the polynomial 5x – 4x2 + 3 at x = 2 and x = –1 -Maths 9th

Last Answer : Let the polynomial be f(x) = 5x – 4x2 + 3 Now, for x = 2, f(2) = 5(2) – 4(2)2 + 3 => f(2) = 10 – 16 + 3 = –3 Or, the value of the polynomial 5x – 4x2 + 3 at x = 2 ... (–1) = –5 –4 + 3 = -6 The value of the polynomial 5x – 4x2 + 3 at x = -1 is -6.

Write a solution of the linear equation 5x + 0y +8 = 0 in two variables. -Maths 9th

Description : Write a solution of the linear equation 5x + 0y +8 = 0 in two variables. -Maths 9th

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For what value of k will the roots of the equation kx^2 – 5x + 6 = 0 be in the ratio 2 : 3 ? -Maths 9th

Description : For what value of k will the roots of the equation kx^2 – 5x + 6 = 0 be in the ratio 2 : 3 ? -Maths 9th

Last Answer : (b) 1 Let the roots of the equation kx2 - 5x + 6 = 0 be α and β. Then, α + β = \(\frac{5}{k}\) ...(i) αβ = \(\frac{6}{k}\) ...(ii) Given \(\ ... frac{9}{k}\) ⇒ 9k2 - 9k = 0 k(k - 1) = 0 ⇒ k = 0 or 1 But k = 0 does not satisfy the condition, so k = 1.

The point (2,3) lies on the graph of the linear equation 3x - (a -1)y =2a -1. If the same point also lies on the graph of the linear equation 5x + (1-2a)y = 3b, then find the value of b. -Maths 9th

Description : The point (2,3) lies on the graph of the linear equation 3x - (a -1)y =2a -1. If the same point also lies on the graph of the linear equation 5x + (1-2a)y = 3b, then find the value of b. -Maths 9th

Last Answer : Given, point (2,3) lies on the line. So, the point (2, 3) is the solution of 3x - (a -1) y = 2a - 1 On putting x = 2 and y = 3 in given solution. ∴ 3 2 - (a-1) 3 = 2a - 1 ⇒ 6 - 3a + 3 = 2a - 1 ⇒ - 3a ... 2 2) 3 = 3b ⇒ 10 - 9 = 3b ⇒ 1 = 3b ⇒ 1 / 3 = b Hence, the value of b is 1 / 3.

Degree of the polynomial 4x4 + Ox3 + Ox5 + 5x+ 7 is -Maths 9th

Description : Degree of the polynomial 4x4 + Ox3 + Ox5 + 5x+ 7 is -Maths 9th

Last Answer : (a) Degree of 4x4 + Ox3 + Ox5 + 5x + 7 is equal to the highest power of variable x. Here, the highest power of x is 4, Hence, the degree of a polynomial is 4.

One of the factors of (25x2 – 1) + (1 + 5x)2 is -Maths 9th

Description : One of the factors of (25x2 – 1) + (1 + 5x)2 is -Maths 9th

Last Answer : (d) Now, (25x2 -1) + (1 + 5x)2 = 25x2 -1 + 1 + 25x2 + 10x [using identity, (a + b)2 = a2 + b2 + 2ab] = 50x2 + 10x = 10x (5x+ 1) Hence, one of the factor of given polynomial is 10x.

The point (2,3) lies on the graph of the linear equation 3x - (a -1)y =2a -1. If the same point also lies on the graph of the linear equation 5x + (1-2a)y = 3b, then find the value of b. -Maths 9th

Description : The point (2,3) lies on the graph of the linear equation 3x - (a -1)y =2a -1. If the same point also lies on the graph of the linear equation 5x + (1-2a)y = 3b, then find the value of b. -Maths 9th

Last Answer : Given, point (2,3) lies on the line. So, the point (2, 3) is the solution of 3x - (a -1) y = 2a - 1 On putting x = 2 and y = 3 in given solution. ∴ 3 2 - (a-1) 3 = 2a - 1 ⇒ 6 - 3a + 3 = 2a - 1 ⇒ - 3a ... 2 2) 3 = 3b ⇒ 10 - 9 = 3b ⇒ 1 = 3b ⇒ 1 / 3 = b Hence, the value of b is 1 / 3.

Degree of the polynomial 4x4 + Ox3 + Ox5 + 5x+ 7 is -Maths 9th

Description : Degree of the polynomial 4x4 + Ox3 + Ox5 + 5x+ 7 is -Maths 9th

Last Answer : (a) Degree of 4x4 + Ox3 + Ox5 + 5x + 7 is equal to the highest power of variable x. Here, the highest power of x is 4, Hence, the degree of a polynomial is 4.

One of the factors of (25x2 – 1) + (1 + 5x)2 is -Maths 9th

Description : One of the factors of (25x2 – 1) + (1 + 5x)2 is -Maths 9th

Last Answer : (d) Now, (25x2 -1) + (1 + 5x)2 = 25x2 -1 + 1 + 25x2 + 10x [using identity, (a + b)2 = a2 + b2 + 2ab] = 50x2 + 10x = 10x (5x+ 1) Hence, one of the factor of given polynomial is 10x.

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

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What is the equation of the line having the y-intercept –1 and parallel to the line y = 5x – 7 ? -Maths 9th

Description : What is the equation of the line having the y-intercept –1 and parallel to the line y = 5x – 7 ? -Maths 9th

Last Answer : Slope of AB = \(rac{2-4}{1-0}\) = -2, Slope of BC = \(rac{3-2}{3-1}\) = \(rac{1}{2}\)Slope of AC = \(rac{3-4}{3-0}\) = \(-rac{1}{3}\)Slope of AB × Slope of BC = -2 x \(rac{1}{2}\) = -1∴ AB ⊥ BC, i.e, ∠B = 90º ⇒ ΔABC is a right angled.

Solve the following equations `(i) sgn({[x]})=0` `(ii) sgn(x^(2)-2x-8)=-1` `(iii)sgn((x^(2)-5x+4)/({x}))=-1`

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Last Answer : (d) \(a^{-rac{4}{3}}\)Since logaxa = \(rac{1}{log_aax}\) = \(rac{1}{log_aa+log_ax}\) = \(rac{1}{1+log_ax}\) andloga2x a = \(rac{1}{log_aa^2x}\) = \(rac{1}{log_aa^2+log_{ax}x}\) = \(rac{1}{2log_aa+log_{ax}x}\)= \( ... 2}}\)When t = \(-rac{4}{3}\), logax = \(-rac{4}{3}\) ⇒ \(x\) = \(a^{-rac{4}{3}}\)

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Description : Solve the following in equalities `(i) |x+7| gt 5` `(ii) |x+3| lt 10` `(iii) (x+2) lt |x^(2)+3x+5|``(iv) |(2x-1)/(x-1)| gt 2` `(v) |x-6| le x^(2)-5x+9

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