2 . If the mean of the following distribution is 6 . Find the value of p ? x 2 4 6 10 P + 5 f 3 2 3 1 2 -Maths 9th

2 . If the mean of the following distribution is 6 . Find the value of p ? x 2 4 6 10 P + 5 f 3 2 3 1 2 -Maths 9th
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2 . If the mean of the following distribution is 6 . Find the value of p ? x 2 4 6 10 P + 5 f 3 2 3 1 2 -Maths 9th

Description : 2 . If the mean of the following distribution is 6 . Find the value of p ? x 2 4 6 10 P + 5 f 3 2 3 1 2 -Maths 9th

Last Answer : Here's ur answer...

If p(x) is a common multiple of degree 6 of the polynomials f(x) = x^3 + x^2 – x – 1 and g(x) = x^3 – x^2 + x – 1, then which -Maths 9th

Description : If p(x) is a common multiple of degree 6 of the polynomials f(x) = x^3 + x^2 – x – 1 and g(x) = x^3 – x^2 + x – 1, then which -Maths 9th

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In a frequency distribution, the mid value of a class is 10 and the width of the class is 6. The lower limit of the class is -Maths 9th

Description : In a frequency distribution, the mid value of a class is 10 and the width of the class is 6. The lower limit of the class is -Maths 9th

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In a frequency distribution, the mid value of a class is 10 and the width of the class is 6. The lower limit of the class is -Maths 9th

Description : In a frequency distribution, the mid value of a class is 10 and the width of the class is 6. The lower limit of the class is -Maths 9th

Last Answer : (b) Let x and y be the upper and lower class limit in a frequency distribution. Now, mid value of a class (x + y )/2=10 [given] ⇒ x + y = 20 (i) Also, given that, width of class x- y = 6 (ii) On ... putting x = 13 in Eq. (i), we get 13+y = 20 ⇒ y = 7 Hence, the lower limit of the class is 7.

If the polynomial x^6 + px^5 + qx^4 – x^2 – x – 3 is divisible by x^4 – 1, then the value of p^2 + q^2 is : -Maths 9th

Description : If the polynomial x^6 + px^5 + qx^4 – x^2 – x – 3 is divisible by x^4 – 1, then the value of p^2 + q^2 is : -Maths 9th

Last Answer : The divisor is x4−1=(x−1)(x+1)(x2+1) By factor theorem, f(1)=f(−1)=0 Thus, 1+p+q−1−1−3=0 and 1+q−1−3=p−1 i.e., p+q=4 and p−q=−2 Adding the two, 2p=2 i.e. p=1 and ∴ q=3. ∴ p2+q2=1+9=10

Find the value of the polynomial p(x) = x^3-3x^2-2x+6 at x = underroot 2 -Maths 9th

Description : Find the value of the polynomial p(x) = x^3-3x^2-2x+6 at x = underroot 2 -Maths 9th

Last Answer : In this chapter, we shall proceed with recalling some of the constructions already learnt in the earlier classes and deal with some more. Here in this section, we will construct some of these ... be done? 2. Always explain the construction. Write the sequence of steps that are actually taken.

Two students A and B solve an equation of the form x^2 + px + q = 0. A starts with a wrong value of p and obtains the roots as 2 and 6. -Maths 9th

Description : Two students A and B solve an equation of the form x^2 + px + q = 0. A starts with a wrong value of p and obtains the roots as 2 and 6. -Maths 9th

Last Answer : Let αα and ββ be the roots of the quadratic equation x2+px+q=0x2+px+q=0 Given that, A starts with a wrong value of p and obtains the roots as 2 and 6. But this time q is correct. i.e., a product of roots ... 1 Now, from Eqs. (ii) and (iii), we get α=−3 and β=−4α=−3 and β=−4 which are correct roots.

Find the value of f(x) = 2x(square) + 7x + 3 at x= -2. -Maths 9th

Description : Find the value of f(x) = 2x(square) + 7x + 3 at x= -2. -Maths 9th

Last Answer : Solution :-

A cubic polynomial f(x) is such that f(1) = 1, f(2) = 2, f(3) = 3 and f(4) = 5, then f(6) equals : -Maths 9th

Description : A cubic polynomial f(x) is such that f(1) = 1, f(2) = 2, f(3) = 3 and f(4) = 5, then f(6) equals : -Maths 9th

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When a polynomial f(x) is divided by (x – 3) and (x + 6), the respective remainders are 7 and 22. What is the remainder when f(x) is -Maths 9th

Description : When a polynomial f(x) is divided by (x – 3) and (x + 6), the respective remainders are 7 and 22. What is the remainder when f(x) is -Maths 9th

Last Answer : answer:

Find the value of x, if log2 (5.2^x + 1), log4(2^(1–x) + 1) and 1 are in A.P. -Maths 9th

Description : Find the value of x, if log2 (5.2^x + 1), log4(2^(1–x) + 1) and 1 are in A.P. -Maths 9th

Last Answer : (b) 1 - log25 Given, log2 (5.2x + 1), log4 (21- x + 1), 1 are in A.P. ⇒ log2 (5.2x + 1) + 1 = 2 log4 (21 - x + 1) ⇒ log2 (5.2x + 1) + log22 = 2 log22 (21-x + 1)⇒ log2 (5.2x + 1).2 = 2 x \(rac12\) ... a=-rac{1}{2}\big)\)⇒ log 2x = log \(rac{2}{5}\)⇒ x log2 2 = log2 2 - log2 5 ⇒ \(x\) = 1 - log2 5.

If log32, log3 (2^x – 5) and log^3 (2^x – 7/2) are in A.P., then what is the value of x ? -Maths 9th

Description : If log32, log3 (2^x – 5) and log^3 (2^x – 7/2) are in A.P., then what is the value of x ? -Maths 9th

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Determine the ratio in which the point P(m, 6) divides the join of A(– 4, 3) and B(2, 8). Also find the value of m. -Maths 9th

Description : Determine the ratio in which the point P(m, 6) divides the join of A(– 4, 3) and B(2, 8). Also find the value of m. -Maths 9th

Last Answer : (b) 1 : 2Any point on the x-axis is (a, 0).Let the point (a, 0) divide the join of A(2, -3) and B(5, 6) in the ratio k : 1. Then the co-ordinates of the point of division are \(\bigg(rac{5k+2}{k+1},rac{6k-3}{k+1}\ ... 6k - 3 = 0 ⇒ k = \(rac{1}{2}\)Required ratio is k : 1 ⇒ \(rac{1}{2}\) : 1 = 1 : 2.

The width of each of five continuous classes in a frequency distribution is 5 and the lower class limit of the lowest class is 10. -Maths 9th

Description : The width of each of five continuous classes in a frequency distribution is 5 and the lower class limit of the lowest class is 10. -Maths 9th

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The width of each of five continuous classes in a frequency distribution is 5 and the lower class limit of the lowest class is 10. -Maths 9th

Description : The width of each of five continuous classes in a frequency distribution is 5 and the lower class limit of the lowest class is 10. -Maths 9th

Last Answer : (c) Let x and y be the upper and lower class limit of frequency distribution. Given, width of the class = 5 ⇒ x-y= 5 (i) Also, given lower class (y) = 10 On putting y = 10 in Eq. (i), we get ... 20-25, 25-30 and 30-35. Thus, the highest class is 30-35, Hence, the upper limit of this class is 35.

If p = 100r – t, find the value of p when r = 0.25 and t = 10. -Maths 9th

Description : If p = 100r – t, find the value of p when r = 0.25 and t = 10. -Maths 9th

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For what value of k,(x+1) is a factor of p(x) - kx(square) - x - 4 ? -Maths 9th

Description : For what value of k,(x+1) is a factor of p(x) - kx(square) - x - 4 ? -Maths 9th

Last Answer : Solution :-

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

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

Last Answer : Solution :-

Let p and q be the roots of the quadratic equation x^2 – (a – 2)x – a – 1 = 0. What is the minimum possible value of p^2 + q^2 ? -Maths 9th

Description : Let p and q be the roots of the quadratic equation x^2 – (a – 2)x – a – 1 = 0. What is the minimum possible value of p^2 + q^2 ? -Maths 9th

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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:

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 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

If the mean of the following data is 20.2, find the value of p: -Maths 9th

Description : If the mean of the following data is 20.2, find the value of p: -Maths 9th

Last Answer : x̅ = ∑fx/∑f ∴ 20.2 = 610 + 20p/30 + p ⇒ 20.2(30 + p) = 610 + 20p ⇒ 606 + 20.2p = 610 + 20p ⇒ 20.2p - 20p = 610 - 606 ⇒ 0.2p = 4 ⇒ p = 4/0.2 = 40/2 ⇒ p = 20

the value f (249)2-(248)2 is -Maths 9th

Description : the value f (249)2-(248)2 is -Maths 9th

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the value f (249)2-(248)2 is -Maths 9th

Description : the value f (249)2-(248)2 is -Maths 9th

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

In Fig. 6.10, if m|n, then find the value of x. -Maths 9th

Description : In Fig. 6.10, if m|n, then find the value of x. -Maths 9th

Last Answer : Solution :-

When (x^3 – 2x^2 + px – q) is divided by (x^2 – 2x – 3), the remainder is (x – 6), What are the values of p and q respectively ? -Maths 9th

Description : When (x^3 – 2x^2 + px – q) is divided by (x^2 – 2x – 3), the remainder is (x – 6), What are the values of p and q respectively ? -Maths 9th

Last Answer : answer:

If x + (1/x) = p, then x^6+(1/x^6) equals to : -Maths 9th

Description : If x + (1/x) = p, then x^6+(1/x^6) equals to : -Maths 9th

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If the mean of 8,5,2,x,6,5 is 6 , then find the mean of the value of x -Maths 9th

Description : If the mean of 8,5,2,x,6,5 is 6 , then find the mean of the value of x -Maths 9th

Last Answer : (8+5+2+x+6+5)/6=6 (26+x)=6×6 26+x=36 X=36-26 X=10

Find the mean of the following distribution : -Maths 9th

Description : Find the mean of the following distribution : -Maths 9th

Last Answer : Now, mean =(x̅) = Σfx / Σf = 1900 / 100 = 19

Obtain the mean of the following distribution and also find the mode. -Maths 9th

Description : Obtain the mean of the following distribution and also find the mode. -Maths 9th

Last Answer : fixi/fi 270/55 Mean=4.9

Find the mean of the following distribution : -Maths 9th

Description : Find the mean of the following distribution : -Maths 9th

Last Answer : Now, mean =(x̅) = Σfx / Σf = 1900 / 100 = 19

Obtain the mean of the following distribution and also find the mode. -Maths 9th

Description : Obtain the mean of the following distribution and also find the mode. -Maths 9th

Last Answer : fixi/fi 270/55 Mean=4.9

Obtain the mean of the following distribution: -Maths 9th

Description : Obtain the mean of the following distribution: -Maths 9th

Last Answer : Solution :- ⇒ x̅ = ∑fx/∑f ⇒ x̅ = 332/40 ⇒ x̅ = 8.05

When f(x) = x4 - 2x3 + 3x2 - ax is divided by x + 1 and x - 1 , we get remainders as 19 and 5 respectively . -Maths 9th

Description : When f(x) = x4 - 2x3 + 3x2 - ax is divided by x + 1 and x - 1 , we get remainders as 19 and 5 respectively . -Maths 9th

Last Answer : When f(x) is divided by (x+1) and (x-1) , the remainders are 19 and 5 respectively . ∴ f(-1) = 19 and f(1) = 5 ⇒ (-1)4 - 2 (-1)3 + 3(-1)2 - a (-1) + b = 19 ⇒ 1 +2 + 3 + a + b = 19 ∴ a + b = 13 ------- ... + 3x2 - 5x + 8 ⇒ f(3) = 34 - 2 33 + 3 32 - 5 3 + 8 = 81 - 54 + 27 - 15 + 8 = 47

When f(x) = x4 - 2x3 + 3x2 - ax is divided by x + 1 and x - 1 , we get remainders as 19 and 5 respectively . -Maths 9th

Description : When f(x) = x4 - 2x3 + 3x2 - ax is divided by x + 1 and x - 1 , we get remainders as 19 and 5 respectively . -Maths 9th

Last Answer : When f(x) is divided by (x+1) and (x-1) , the remainders are 19 and 5 respectively . ∴ f(-1) = 19 and f(1) = 5 ⇒ (-1)4 - 2 (-1)3 + 3(-1)2 - a (-1) + b = 19 ⇒ 1 +2 + 3 + a + b = 19 ∴ a + b = 13 ------- ... + 3x2 - 5x + 8 ⇒ f(3) = 34 - 2 33 + 3 32 - 5 3 + 8 = 81 - 54 + 27 - 15 + 8 = 47

f(x) = x^4 – 2x^3 + 3x^2 – ax + b is a polynomial such that when it is divided by (x – 1) and (x + 1), the remainders are respectively 5 and 19. -Maths 9th

Description : f(x) = x^4 – 2x^3 + 3x^2 – ax + b is a polynomial such that when it is divided by (x – 1) and (x + 1), the remainders are respectively 5 and 19. -Maths 9th

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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

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

Find the remainder when f(x)=4x(cube) - 12x(square) +14x - 3 is divided by g(x) = (2x-1). -Maths 9th

Description : Find the remainder when f(x)=4x(cube) - 12x(square) +14x - 3 is divided by g(x) = (2x-1). -Maths 9th

Last Answer : ____2x2-5x+4________________ 2x-1 ) 4x3-12x2+14x-3( 4x3-2x2 - + ____________ 0 -10x2+14x-3 ... + ___________ X+1

Find the remainder when f(x)=9x(cube) -x 3x(square) + 14x - 3 is divided by g(x)=(3x-1). -Maths 9th

Description : Find the remainder when f(x)=9x(cube) -x 3x(square) + 14x - 3 is divided by g(x)=(3x-1). -Maths 9th

Last Answer : Solution :-

The values of a, b and c respectively for the expression f(x) = x^3 + ax^2 + bx + c, if f(1) = f(2) = 0 and f(4) = f(0) are : -Maths 9th

Description : The values of a, b and c respectively for the expression f(x) = x^3 + ax^2 + bx + c, if f(1) = f(2) = 0 and f(4) = f(0) are : -Maths 9th

Last Answer : f′(x)=3x2+2ax+6 f(x)⇒f′(x)≥0 3x2+2ax+6≥0 ⇒D≤0 4[a2−3b]≤0 a2≤3b ∴P=6×6×6(16)×6​=3616​=94​ Answer.

If f(x) = log ((1+x)/(1-x)), show that f (2x/(1+x^2)) = 2 f(x). -Maths 9th

Description : If f(x) = log ((1+x)/(1-x)), show that f (2x/(1+x^2)) = 2 f(x). -Maths 9th

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The value of 1.999… in the form of p/q, where p and q are integers and -Maths 9th

Description : The value of 1.999… in the form of p/q, where p and q are integers and -Maths 9th

Last Answer : Value of 1.999 is given below .

The value of 1.999… in the form of p/q, where p and q are integers and -Maths 9th

Description : The value of 1.999… in the form of p/q, where p and q are integers and -Maths 9th

Last Answer : Value of 1.999 is given below .

if (1.-2) is a solution of the equation 2x-y=p,then find the value of p. -Maths 9th

Description : if (1.-2) is a solution of the equation 2x-y=p,then find the value of p. -Maths 9th

Last Answer : x = 1 y = -2 2x-y = p Therefore, p = 2(1)-(-2) = 2 + 2 = 4

if (1.-2) is a solution of the equation 2x-y=p,then find the value of p. -Maths 9th

Description : if (1.-2) is a solution of the equation 2x-y=p,then find the value of p. -Maths 9th

Last Answer : 2x-y=p put x-1,y=-2 =2(1)-(-2)=p p=4

For what value of p the point (p, 2) lies on the line 3x + y = 11? -Maths 9th

Description : For what value of p the point (p, 2) lies on the line 3x + y = 11? -Maths 9th

Last Answer : Solution :-

If a, b, c be the p^th, q^th, r^th terms of a GP, then the value of (q – r) log a + (r – p) -Maths 9th

Description : If a, b, c be the p^th, q^th, r^th terms of a GP, then the value of (q – r) log a + (r – p) -Maths 9th

Last Answer : (a) 0Let h be the first term and k be the common ratio of a GP, then a = hkp - 1, b = hkq - 1, c = hkr - 1∴ (q - r) log a + (r - p) log b + (p - q) log c = log [hkp -1]q - r + log [hkq -1]r - p + log[hkr -1]p - ... r + r - p + p - q) (kp - 1)q - r (kq -1)r - p (kr -1)p - q = log(ho ko) = log 1 = 0.

If pqr = 1, the value of (1)/([1+p+q^-1]) + (1)/([1+q+r^-1])+ (1)/([1+r+p^-1]) will be equal to : -Maths 9th

Description : If pqr = 1, the value of (1)/([1+p+q^-1]) + (1)/([1+q+r^-1])+ (1)/([1+r+p^-1]) will be equal to : -Maths 9th

Last Answer : answer: