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

If x + (1/x) = p, then x^6+(1/x^6) equals to : -Maths 9th
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Express 0.6 in the form p/q, where p and q are integers and q is not equals to 0. -Maths 9th

Description : Express 0.6 in the form p/q, where p and q are integers and q is not equals to 0. -Maths 9th

Last Answer : Let x = 0.6 recurring Then, x = 0.666..... ....(i) implies 10x = 6.666........ .....(ii) Substracting (i) from (ii),we get 9x = 6 implies x = 6/9 implies x = 2/3

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

Last Answer : answer:

Express 0.00323232... in the form p/q, where p and q are integers and q is not equals 0. -Maths 9th

Description : Express 0.00323232... in the form p/q, where p and q are integers and q is not equals 0. -Maths 9th

Last Answer : Solution :-

Express 0.35777... in the form p/q,where p and q are integers and q is not equals to 0. -Maths 9th

Description : Express 0.35777... in the form p/q,where p and q are integers and q is not equals to 0. -Maths 9th

Last Answer : Solution :-

Express:2.0151515... in the p/q form, where p and q are integers and q is not equals to 0. -Maths 9th

Description : Express:2.0151515... in the p/q form, where p and q are integers and q is not equals to 0. -Maths 9th

Last Answer : Solution :-

If x = log2a a, y = log3a2a, z = log4a3a, then xyz – 2yz equals -Maths 9th

Description : If x = log2a a, y = log3a2a, z = log4a3a, then xyz – 2yz equals -Maths 9th

Last Answer : (d) -1\(x\) = log2a a = \(rac{ ext{log}\,a}{ ext{log}\,2a}\), y = log3a 2a = \(rac{ ext{log}\,2a}{ ext{log}\,3a}\)z = log4a 3a = \(rac{ ext{log}\,3a}{ ext{log}\,4a}\)∴ xyz - 2yz = \(rac{ ext{log}\ ... \(rac{ ext{log}\,(4a)^{-1}}{ ext{log}\,(4a)}\) = \(rac{-1. ext{log}\,4a}{ ext{log}\,4a}\) = -1.

If x + y + z = 0, then x (y – z)^3 + y (z – x)^3 + z (x – y)^3 equals -Maths 9th

Description : If x + y + z = 0, then x (y – z)^3 + y (z – x)^3 + z (x – y)^3 equals -Maths 9th

Last Answer : answer:

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

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

Last Answer : answer:

If a = log12m and b = log18m, then (a-2b)/(b-2a) equals -Maths 9th

Description : If a = log12m and b = log18m, then (a-2b)/(b-2a) equals -Maths 9th

Last Answer : (a) log32\(rac{a-2b}{b-2a}\) = \(rac{ ext{log}_{12}\,m-\,2 ext{log}_{18}\,m}{ ext{log}_{18}\,m-\,2 ext{log}_{12}\,m}\)

If S denotes the area of the curved surface of a right circular cone of height h end semi-vertical angle a, then S equals -Maths 9th

Description : If S denotes the area of the curved surface of a right circular cone of height h end semi-vertical angle a, then S equals -Maths 9th

Last Answer : answer:

Draw a circle of diameter 6.4 cm. Then draw two tangents to the circle from a point P at a distance 6.4 cm from the centre of the circle. -Maths 9th

Description : Draw a circle of diameter 6.4 cm. Then draw two tangents to the circle from a point P at a distance 6.4 cm from the centre of the circle. -Maths 9th

Last Answer : clear .

If p (x) = x + 3, then p(x)+ p (- x) is equal to -Maths 9th

Description : If p (x) = x + 3, then p(x)+ p (- x) is equal to -Maths 9th

Last Answer : (d) Given p(x) = x+3, put x = -x in the given equation, we get p(-x) = -x+3 Now, p(x)+ p(-x) = x+ 3+ (-x)+ 3=6

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

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

Last Answer : Show that p = r.

If the perpendicular distance of a point P from the X-axis is 5 units and the foot of the perpendicular lies on the negative direction of X-axis, then the point P has -Maths 9th

Description : If the perpendicular distance of a point P from the X-axis is 5 units and the foot of the perpendicular lies on the negative direction of X-axis, then the point P has -Maths 9th

Last Answer : (d) We know that, the perpendicular distance of a point from the X-axis gives y-coordinate of that point. Here, foot of perpendicular lies on the negative direction of X-axis, so perpendicular distance can be measure in II quadrant or III quadrant. Hence, the point P has y-coordinate = 5 or -5.

If P (5,1), Q (8, 0), R(0, 4), S(0, 5) and O(0, 0) are plotted on the graph paper, then the points on the X-axis is/are -Maths 9th

Description : If P (5,1), Q (8, 0), R(0, 4), S(0, 5) and O(0, 0) are plotted on the graph paper, then the points on the X-axis is/are -Maths 9th

Last Answer : (d) We know that, a point lies on X-axis, if its y-coordinate is zero. So, on plotting the given points on graph paper, we get Q and O lie on the X-axis.

If p (x) = x + 3, then p(x)+ p (- x) is equal to -Maths 9th

Description : If p (x) = x + 3, then p(x)+ p (- x) is equal to -Maths 9th

Last Answer : (d) Given p(x) = x+3, put x = -x in the given equation, we get p(-x) = -x+3 Now, p(x)+ p(-x) = x+ 3+ (-x)+ 3=6

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

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

Last Answer : Show that p = r.

If the perpendicular distance of a point P from the X-axis is 5 units and the foot of the perpendicular lies on the negative direction of X-axis, then the point P has -Maths 9th

Description : If the perpendicular distance of a point P from the X-axis is 5 units and the foot of the perpendicular lies on the negative direction of X-axis, then the point P has -Maths 9th

Last Answer : (d) We know that, the perpendicular distance of a point from the X-axis gives y-coordinate of that point. Here, foot of perpendicular lies on the negative direction of X-axis, so perpendicular distance can be measure in II quadrant or III quadrant. Hence, the point P has y-coordinate = 5 or -5.

If P (5,1), Q (8, 0), R(0, 4), S(0, 5) and O(0, 0) are plotted on the graph paper, then the points on the X-axis is/are -Maths 9th

Description : If P (5,1), Q (8, 0), R(0, 4), S(0, 5) and O(0, 0) are plotted on the graph paper, then the points on the X-axis is/are -Maths 9th

Last Answer : (d) We know that, a point lies on X-axis, if its y-coordinate is zero. So, on plotting the given points on graph paper, we get Q and O lie on the X-axis.

If 1, log9 (3^(1–x) + 2) and log3 (4.3^x –1) are in A.P., then x is equal to -Maths 9th

Description : If 1, log9 (3^(1–x) + 2) and log3 (4.3^x –1) are in A.P., then x is equal to -Maths 9th

Last Answer : (d) log3\(\big(rac{3}{4}\big)\) 1, log9 (31 - x + 2), log3 (4.3x -1) are in A.P. ⇒ log33 , log3 (31- x + 2)1/2, log3 (4.3x - 1) are in A.P. ⇒ 3, (31 - x + 2)1/2, (4.3x - 1) are in G.P.\ ... (rac{1}{3}\), \(rac{3}{4}\)∴ Rejecting the negative value, we have 3x = \(rac{3}{4}\)⇒ x = log3\(rac{3}{4}.\)

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 expression (px^3 + x^2 – 2x – q) is divisible by (x – 1) and (x + 1), then the values of p and q respectively are ? -Maths 9th

Description : If the expression (px^3 + x^2 – 2x – q) is divisible by (x – 1) and (x + 1), then the values of p and q respectively are ? -Maths 9th

Last Answer : Let f(x)=px3+x2−2x−q Since f(x) is divisible by (x−1) and (x+1) so x=1 and −1 must make f(x)=0. Therefore, p+1−2−q=0, i.e., p−q=1; and −p+1+2−q=0, i.e., p+q=3 Thus p=2 and q=1

If x^3 + px + q and x^3 + qx + p have a common factor, then which of the following is correct ? -Maths 9th

Description : If x^3 + px + q and x^3 + qx + p have a common factor, then which of the following is correct ? -Maths 9th

Last Answer : x3+px+q 3x2+p have common root Let common root =a a3+ap+q=0 a2=3−p​a=3−p​​4p3+27q2=? a3(a2+p)+q=0 ⇒3−p​​[(3−p​)+p]+q=0 3−p​​(3+2p​)=−q ⇒27−p(4p2)​=q2

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

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

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

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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...

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.

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:

Find the value of 7plus root five ÷by seven minus root five minus seven minus root five ÷by seven plus root five equals to a+7÷11root five b -Maths 9th

Description : Find the value of 7plus root five ÷by seven minus root five minus seven minus root five ÷by seven plus root five equals to a+7÷11root five b -Maths 9th

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

Three of the six vertices of a regular hexagon are chosen at random. The probability that the triangle with these vertices is equilateral equals : -Maths 9th

Description : Three of the six vertices of a regular hexagon are chosen at random. The probability that the triangle with these vertices is equilateral equals : -Maths 9th

Last Answer : (c) \(rac{1}{10}\)Let S be the sample space.Then n(S) = Number of triangles formed by selecting any three vertices of 6 vertices of a regular hexagon= 6C3 = \(rac{6 imes5 imes4}{3 imes2}\) = 20.Let A : Event that the ... Required probability = \(rac{n(A)}{n(S)}\) = \(rac{2}{20}\) = \(rac{1}{10}\).

(2a)/(a+b)+(2b)/(b+c) + (2c)/(c+a) + ((b-c)(c-a)(a-b))/((b+c)(c+a)(a+b))equals -Maths 9th

Description : (2a)/(a+b)+(2b)/(b+c) + (2c)/(c+a) + ((b-c)(c-a)(a-b))/((b+c)(c+a)(a+b))equals -Maths 9th

Last Answer : answer:

The length of the midline of a trapezoid equals 4 cm and the base angles are 40° and 50°. The length of the bases if the distance of their -Maths 9th

Description : The length of the midline of a trapezoid equals 4 cm and the base angles are 40° and 50°. The length of the bases if the distance of their -Maths 9th

Last Answer : answer:

If P,Q,R,S are respectively the mid - points of the sides of a parallelogram ABCD, if ar(||gm PQRS) = 32.5cm2 , then find ar(||gm ABCD). -Maths 9th

Description : If P,Q,R,S are respectively the mid - points of the sides of a parallelogram ABCD, if ar(||gm PQRS) = 32.5cm2 , then find ar(||gm ABCD). -Maths 9th

Last Answer : Join PR. ∵ △PSR and ||gm APRD are on the same base and between same parallel lines. ar(△PSR) = 1/2 ar(||gm APRD) Similarly, ar(△PQR) = 1/2 ar(||gm PBCR) ar(△PQRS) = ar(△PSR) + △(PQR) = 1/2 ar(||gm APRD) + 1 ... |gm PBCR) = 1/2 ar(||gm ABCD) ⇒ ar(||gm ABCD) = 2 ar(||gm PQRS) = 2 32.5 = 65cm2

The side AB of a parallelogram ABCD is produced to any point P. A line through A and parallel to CP meets CB produced at Q, then parallelogram PBQR is completed (see figure). -Maths 9th

Description : The side AB of a parallelogram ABCD is produced to any point P. A line through A and parallel to CP meets CB produced at Q, then parallelogram PBQR is completed (see figure). -Maths 9th

Last Answer : Join AC and QP, also it is given that AQ || CP ∴ △ACQ and △APQ are on the same base AQ and lie between the same parallels AQ || CP. ∴ ar(△ACQ) = ar(△APQ) or ar(△ABC) + ar(△ABQ) = ar(△BPQ) + ar(△ABQ) or ar(△ABC) = ar( △BPQ) or 1/2 ar(||gm ABCD) = 1/2 ar(||gm PBQR) or ar(||gm ABCD) = ar(||gm PBQR)

If P (event E) = 0.47, then find P(not E). -Maths 9th

Description : If P (event E) = 0.47, then find P(not E). -Maths 9th

Last Answer : P(not E) = 1 - P(E) ⇒ 1 - 0.47 = 0.53

If P(-l, 1), Q(3, -4), R(1, -1), S(-2, -3) and T(-4, 4) are plotted on the graph paper, then the point(s) in the fourth quadrant is/are -Maths 9th

Description : If P(-l, 1), Q(3, -4), R(1, -1), S(-2, -3) and T(-4, 4) are plotted on the graph paper, then the point(s) in the fourth quadrant is/are -Maths 9th

Last Answer : (b) In point P (-1, 1), x-coordinate is -1 unit and y-coordinate is 1 unit, so it lies in llnd quadrant. Similarly, we can plot all the points Q (3, -4), R (1, -1), S (-2, -3) and T (-4, 4), It is clear from the graph that points R and Q lie in fourth quadrant.

If the coordinates of the two points are P(-2, 3) and Q(-3, 5), then (Abscissa of P) – (Abscissa of Q) is -Maths 9th

Description : If the coordinates of the two points are P(-2, 3) and Q(-3, 5), then (Abscissa of P) – (Abscissa of Q) is -Maths 9th

Last Answer : (b) We have, points P(- 2, 3) and Q(- 3, 5) Here, abscissa of Pi.e., x-coordinate of Pis -2 and abscissa of Q i.e., x-coordinate of Q is -3. So, (Abscissa of P) – (Abscissa of Q) = - 2 - (-3) = -2 + 3 =1.

If P, Q and R are the mid-points of the sides, BC, CA and AB of a triangle and AD is the perpendicular from A on BC, then prove that P, Q, R and D are concyclic. -Maths 9th

Description : If P, Q and R are the mid-points of the sides, BC, CA and AB of a triangle and AD is the perpendicular from A on BC, then prove that P, Q, R and D are concyclic. -Maths 9th

Last Answer : According to question prove that P, Q, R and D are concyclic.

If P,Q,R,S are respectively the mid - points of the sides of a parallelogram ABCD, if ar(||gm PQRS) = 32.5cm2 , then find ar(||gm ABCD). -Maths 9th

Description : If P,Q,R,S are respectively the mid - points of the sides of a parallelogram ABCD, if ar(||gm PQRS) = 32.5cm2 , then find ar(||gm ABCD). -Maths 9th

Last Answer : Join PR. ∵ △PSR and ||gm APRD are on the same base and between same parallel lines. ar(△PSR) = 1/2 ar(||gm APRD) Similarly, ar(△PQR) = 1/2 ar(||gm PBCR) ar(△PQRS) = ar(△PSR) + △(PQR) = 1/2 ar(||gm APRD) + 1 ... |gm PBCR) = 1/2 ar(||gm ABCD) ⇒ ar(||gm ABCD) = 2 ar(||gm PQRS) = 2 32.5 = 65cm2

The side AB of a parallelogram ABCD is produced to any point P. A line through A and parallel to CP meets CB produced at Q, then parallelogram PBQR is completed (see figure). -Maths 9th

Description : The side AB of a parallelogram ABCD is produced to any point P. A line through A and parallel to CP meets CB produced at Q, then parallelogram PBQR is completed (see figure). -Maths 9th

Last Answer : Join AC and QP, also it is given that AQ || CP ∴ △ACQ and △APQ are on the same base AQ and lie between the same parallels AQ || CP. ∴ ar(△ACQ) = ar(△APQ) or ar(△ABC) + ar(△ABQ) = ar(△BPQ) + ar(△ABQ) or ar(△ABC) = ar( △BPQ) or 1/2 ar(||gm ABCD) = 1/2 ar(||gm PBQR) or ar(||gm ABCD) = ar(||gm PBQR)

If P (event E) = 0.47, then find P(not E). -Maths 9th

Description : If P (event E) = 0.47, then find P(not E). -Maths 9th

Last Answer : P(not E) = 1 - P(E) ⇒ 1 - 0.47 = 0.53

If P(-l, 1), Q(3, -4), R(1, -1), S(-2, -3) and T(-4, 4) are plotted on the graph paper, then the point(s) in the fourth quadrant is/are -Maths 9th

Description : If P(-l, 1), Q(3, -4), R(1, -1), S(-2, -3) and T(-4, 4) are plotted on the graph paper, then the point(s) in the fourth quadrant is/are -Maths 9th

Last Answer : (b) In point P (-1, 1), x-coordinate is -1 unit and y-coordinate is 1 unit, so it lies in llnd quadrant. Similarly, we can plot all the points Q (3, -4), R (1, -1), S (-2, -3) and T (-4, 4), It is clear from the graph that points R and Q lie in fourth quadrant.

If the coordinates of the two points are P(-2, 3) and Q(-3, 5), then (Abscissa of P) – (Abscissa of Q) is -Maths 9th

Description : If the coordinates of the two points are P(-2, 3) and Q(-3, 5), then (Abscissa of P) – (Abscissa of Q) is -Maths 9th

Last Answer : (b) We have, points P(- 2, 3) and Q(- 3, 5) Here, abscissa of Pi.e., x-coordinate of Pis -2 and abscissa of Q i.e., x-coordinate of Q is -3. So, (Abscissa of P) – (Abscissa of Q) = - 2 - (-3) = -2 + 3 =1.

If P, Q and R are the mid-points of the sides, BC, CA and AB of a triangle and AD is the perpendicular from A on BC, then prove that P, Q, R and D are concyclic. -Maths 9th

Description : If P, Q and R are the mid-points of the sides, BC, CA and AB of a triangle and AD is the perpendicular from A on BC, then prove that P, Q, R and D are concyclic. -Maths 9th

Last Answer : According to question prove that P, Q, R and D are concyclic.

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