If X^3-1/x^3=14 then find x-1/x=? -Maths 9th

If X^3-1/x^3=14 then find x-1/x=? -Maths 9th
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If X^3-1/x^3=14 then find x-1/x=? -Maths 9th

Description : If X^3-1/x^3=14 then find x-1/x=? -Maths 9th

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

In Fig. 6.14, if x+y = w+z,then then prove that AOB is a line. -Maths 9th

Description : In Fig. 6.14, if x+y = w+z,then then prove that AOB is a line. -Maths 9th

Last Answer : Solution :-

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

Ten observations 6, 14, 15, 17, x + 1, 2x – 13, 30, 32, 34, 43 are written in ascending order. -Maths 9th

Description : Ten observations 6, 14, 15, 17, x + 1, 2x – 13, 30, 32, 34, 43 are written in ascending order. -Maths 9th

Last Answer : Here, the arranged data is 6, 14, 15, 17, x + 1, 2x - 13, 30, 32, 34, 43 Total number of observations = 10 Here, 10 is an even number , therefore median will be the mean of (10 / 2)th and (10 / 2 + 1)th observation. ... ⇒ 3x + 12 / 2 = 24 ⇒ 3x - 12 = 48 ⇒ 3x = 60 ⇒ x = 20 ∴ The value of x = 20

Ten observations 6, 14, 15, 17, x + 1, 2x – 13, 30, 32, 34, 43 are written in ascending order. -Maths 9th

Description : Ten observations 6, 14, 15, 17, x + 1, 2x – 13, 30, 32, 34, 43 are written in ascending order. -Maths 9th

Last Answer : Here, the arranged data is 6, 14, 15, 17, x + 1, 2x - 13, 30, 32, 34, 43 Total number of observations = 10 Here, 10 is an even number , therefore median will be the mean of (10 / 2)th and (10 / 2 + 1)th observation. ... ⇒ 3x + 12 / 2 = 24 ⇒ 3x - 12 = 48 ⇒ 3x = 60 ⇒ x = 20 ∴ The value of x = 20

Ten observations 6, 14, 15, 17, x+1, 2x -13, -Maths 9th

Description : Ten observations 6, 14, 15, 17, x+1, 2x -13, -Maths 9th

Last Answer : 6, 14, 15, 17, x + 1, 2x -13, 30, 32, 34, 43, Here, n = 10 Since the number of observations is 10 (an even number), therefore, the median = (10/2)th observation + (10/2 + 1)th observation/2 = 5th observation + 6th ... = x + 1 + 2x - 13/2 ⇒ 48 = 3x - 12 ⇒ 3x = 48 + 12 = 60 ⇒ x = 20

If x^5 – 9x^2 + 12x – 14 is divisible by (x – 3), what is the remainder ? -Maths 9th

Description : If x^5 – 9x^2 + 12x – 14 is divisible by (x – 3), what is the remainder ? -Maths 9th

Last Answer : answer:

What is the equation of the straight line which passes through (3, 4) and the sum of whose x-intercept and y-intercept is 14 ? -Maths 9th

Description : What is the equation of the straight line which passes through (3, 4) and the sum of whose x-intercept and y-intercept is 14 ? -Maths 9th

Last Answer : (a) 4x + 3y = 24 Let the x-intercept = a. Then, y-intercept = 14 - a ∴ Eqn of the straight line is \(rac{x}{a}\) + \(rac{y}{14-a}\) = 1Since it passes through (3, 4), so\(rac{3}{a}\) + \(rac{4}{14-a}\) = 1⇒ 3(14 - ... = 1 ⇒ x + y = 7or \(rac{x}{6}\) + \(rac{y}{8}\) = 1 ⇒ 8x + 6y = 48 ⇒ 4x + 3y = 24.

A field is in the shape of a trapezium whose parallel sides are 25 m and 10 m. The non-parallel sides are 14 m and 13 m. Find the area of the field. -Maths 9th

Description : A field is in the shape of a trapezium whose parallel sides are 25 m and 10 m. The non-parallel sides are 14 m and 13 m. Find the area of the field. -Maths 9th

Last Answer : Let the given field is in the form of a trapezium ABCD such that parallel sides are AB = 10 m and DC = 25 m Non-parallel sides are AD = 13 m and BC = 14 m. We draw BE || AD, such that BE = 13 m. ... = 112 m2 So, area of the field = area of ∆BCE + area of parallelogram ABED = 84 m2 + 112 m2 = 196 m2

The slant height and base diameter of conical tomb are 25m and 14 m respectively. Find the cost of white-washing its curved surface at the rate of Rs. 210 per 100 m2 -Maths 9th

Description : The slant height and base diameter of conical tomb are 25m and 14 m respectively. Find the cost of white-washing its curved surface at the rate of Rs. 210 per 100 m2 -Maths 9th

Last Answer : Slant height of conical tomb, l = 25m Base radius, r = diameter/2 = 14/2 m = 7m CSA of conical tomb = πrl = (22/7)×7×25 = 550 CSA of conical tomb= 550m2 Cost of white-washing 550 m2 area, which is Rs (210×550)/100 = Rs. 1155 Therefore, cost will be Rs. 1155 while white-washing tomb.

Curved surface area of a cone is 308 cm2 and its slant height is 14 cm. Find (i) radius of the base -Maths 9th

Description : Curved surface area of a cone is 308 cm2 and its slant height is 14 cm. Find (i) radius of the base -Maths 9th

Last Answer : Slant height of cone, l = 14 cm Let the radius of the cone be r. (i) We know, CSA of cone = πrl Given: Curved surface area of a cone is 308 cm2 (308 ) = (22/7) r 14 308 = 44 r r = 308 ... Total surface area of cone = 308+(22/7) 72 = 308+154 Therefore, the total surface area of the cone is 462 cm2.

The curved surface area of a right circular cylinder of height 14 cm is 88 cm2. Find the diameter of the base of the cylinder. (Assume π =22/7 ) -Maths 9th

Description : The curved surface area of a right circular cylinder of height 14 cm is 88 cm2. Find the diameter of the base of the cylinder. (Assume π =22/7 ) -Maths 9th

Last Answer : Height of cylinder, h = 14cm Let the diameter of the cylinder be d Curved surface area of cylinder = 88 cm2 We know that, formula to find Curved surface area of cylinder is 2πrh. So 2πrh =88 cm2 (r is the ... 88 cm2 2r = 2 cm d =2 cm Therefore, the diameter of the base of the cylinder is 2 cm.

Calculate the surface area of a hemispherical dome of a temple with radius 14 m to be whitewashed from outside. -Maths 9th

Description : Calculate the surface area of a hemispherical dome of a temple with radius 14 m to be whitewashed from outside. -Maths 9th

Last Answer : Here, radius of hemispherical dome (r) = 14 m Surface area of dome = 2πr2 = 2 × 22 / 7 × 14 × 14 = 1232 m2 Hence , total surface area to be whitewashed from outside is 1232 m2

Find the mode of the following scores : 14, 25, 14, 28, 18, 17, 18, 14, 23, 22, 14, 18 -Maths 9th

Description : Find the mode of the following scores : 14, 25, 14, 28, 18, 17, 18, 14, 23, 22, 14, 18 -Maths 9th

Last Answer : 14 repeat maximum number of times (4 times) in the given data. Mode = 14

Given are the scores (out of 25) of 9 students in a Monday test : 14, 25, 17, 22, 20, 19, 10, 8 and 23 -Maths 9th

Description : Given are the scores (out of 25) of 9 students in a Monday test : 14, 25, 17, 22, 20, 19, 10, 8 and 23 -Maths 9th

Last Answer : Ascending orders of scores is : 8, 10, 14, 17, 19, 20, 22, 23, 25 Now, new score = 8 + 10 + 14 + 17 + 19 + 20 + 22 + 23 + 25 / 9 = 158 / 9 = 17.5 marks Median = (n + 1 / 2)th observation because n is odd = (9 + 1 / 2)th observation = 5th observation = 19 marks.

Calculate the surface area of a hemispherical dome of a temple with radius 14 m to be whitewashed from outside. -Maths 9th

Description : Calculate the surface area of a hemispherical dome of a temple with radius 14 m to be whitewashed from outside. -Maths 9th

Last Answer : Here, radius of hemispherical dome (r) = 14 m Surface area of dome = 2πr2 = 2 × 22 / 7 × 14 × 14 = 1232 m2 Hence , total surface area to be whitewashed from outside is 1232 m2

Find the mode of the following scores : 14, 25, 14, 28, 18, 17, 18, 14, 23, 22, 14, 18 -Maths 9th

Description : Find the mode of the following scores : 14, 25, 14, 28, 18, 17, 18, 14, 23, 22, 14, 18 -Maths 9th

Last Answer : 14 repeat maximum number of times (4 times) in the given data. Mode = 14

Given are the scores (out of 25) of 9 students in a Monday test : 14, 25, 17, 22, 20, 19, 10, 8 and 23 -Maths 9th

Description : Given are the scores (out of 25) of 9 students in a Monday test : 14, 25, 17, 22, 20, 19, 10, 8 and 23 -Maths 9th

Last Answer : Ascending orders of scores is : 8, 10, 14, 17, 19, 20, 22, 23, 25 Now, new score = 8 + 10 + 14 + 17 + 19 + 20 + 22 + 23 + 25 / 9 = 158 / 9 = 17.5 marks Median = (n + 1 / 2)th observation because n is odd = (9 + 1 / 2)th observation = 5th observation = 19 marks.

The sides of a quadrilateral ABCD are 6 cm, 8 cm, 12 cm and 14 cm (taken in order), respectively and the angle between the first two sides is a right angle. -Maths 9th

Description : The sides of a quadrilateral ABCD are 6 cm, 8 cm, 12 cm and 14 cm (taken in order), respectively and the angle between the first two sides is a right angle. -Maths 9th

Last Answer : Given ABCD is a quadrilateral having sides AB=6cm, BC=8cm, CD=12cm and DA=14 cm. Now. Join AC. We have, ABC is a right angled triangle at B. Now, AC2=AB2+BC2 [by Pythagoras theorem]Now, AC2=AB2+BC2 ... =24(1+6-√)cm2=24+246=24(1+6)cm2 Hence, the area of quadrilateral is 241+6-√−−−−−−√cm2241+6cm2 .

The mode of given data 15, 14, 19, 20, 14, 15, 16, 14, 15, 18, 14, 19, 15,17 and 15 is -Maths 9th

Description : The mode of given data 15, 14, 19, 20, 14, 15, 16, 14, 15, 18, 14, 19, 15,17 and 15 is -Maths 9th

Last Answer : We first arrange the given data in ascending order as follows 14, 14, 14, 14, 15, 15, 15, 15, 15, 16, 17, 18, 19, 19, 20 From above, we see that 15 occurs most frequently i.e., 5 times. Hence, the mode of the given data is 15.

A child says that the median of 3, 14, 18, 20 and 5 is 18. What does not the child understand about finding the median? -Maths 9th

Description : A child says that the median of 3, 14, 18, 20 and 5 is 18. What does not the child understand about finding the median? -Maths 9th

Last Answer : NEED ANSWER

The class marks of a continuous distribution are 1.04, 1.14, 1.24, 1.34, 1.44,1.54 and 1.64. -Maths 9th

Description : The class marks of a continuous distribution are 1.04, 1.14, 1.24, 1.34, 1.44,1.54 and 1.64. -Maths 9th

Last Answer : NEED ANSWER

The sides of a quadrilateral ABCD are 6 cm, 8 cm, 12 cm and 14 cm (taken in order), respectively and the angle between the first two sides is a right angle. -Maths 9th

Description : The sides of a quadrilateral ABCD are 6 cm, 8 cm, 12 cm and 14 cm (taken in order), respectively and the angle between the first two sides is a right angle. -Maths 9th

Last Answer : Given ABCD is a quadrilateral having sides AB = 6 cm, BC = 8 cm, CD = 12 cm and DA = 14 cm. Now, join AC.

The mode of given data 15, 14, 19, 20, 14, 15, 16, 14, 15, 18, 14, 19, 15,17 and 15 is -Maths 9th

Description : The mode of given data 15, 14, 19, 20, 14, 15, 16, 14, 15, 18, 14, 19, 15,17 and 15 is -Maths 9th

Last Answer : (b) We first arrange the given data in ascending order as follows 14, 14, 14, 14, 15, 15, 15, 15, 15, 16, 17, 18, 19, 19, 20 From above, we see that 15 occurs most frequently i.e., 5 times. Hence, the mode of the given data is 15.

A child says that the median of 3, 14, 18, 20 and 5 is 18. What does not the child understand about finding the median? -Maths 9th

Description : A child says that the median of 3, 14, 18, 20 and 5 is 18. What does not the child understand about finding the median? -Maths 9th

Last Answer : The child does not understand, that data has to be arranged in ascending or descending order before finding the median.

The class marks of a continuous distribution are 1.04, 1.14, 1.24, 1.34, 1.44,1.54 and 1.64. -Maths 9th

Description : The class marks of a continuous distribution are 1.04, 1.14, 1.24, 1.34, 1.44,1.54 and 1.64. -Maths 9th

Last Answer : It is not correct. Because the difference between two consecutive class marks should be equal to the class size. Here, difference between two consecutive marks is 0.1 and class size of 1.55-1.73 is 0.18, which are not equal.

If z2+ 1/z2 = 14, find the value of z3+ 1/z3. -Maths 9th

Description : If z2+ 1/z2 = 14, find the value of z3+ 1/z3. -Maths 9th

Last Answer : Solution :-

Nidhi has to find the area of a sphere whose diameter was 14 cm. -Maths 9th

Description : Nidhi has to find the area of a sphere whose diameter was 14 cm. -Maths 9th

Last Answer : Area is two-dimensional while 4 πr represents a length.

30 circular plates, each of radius 14 cm -Maths 9th

Description : 30 circular plates, each of radius 14 cm -Maths 9th

Last Answer : Height of the cylinder formed (h) = 30 x 3 = 90 cm Radius of the base of the cylinder formed (r) = 14 cm (i) Total surface area of the cylinder = 2 πr(r + h) = 2 x 22/7 x 14(14 + 90) = 2 x 22/7 x 14 x 104 = 9152 cm2 (ii) Volume of the cylinder formed = πr2h = 22//7 x 14 x 14 x 90 = 55440 cm3

A child says that the median of 3, 14, 18, 20, 5 is 18. -Maths 9th

Description : A child says that the median of 3, 14, 18, 20, 5 is 18. -Maths 9th

Last Answer : In order to find the median, data has to be arranged in ascending or descending order before finding the median.

MCQ Questions for Class 9 Maths Chapter 14 Statistics with answers -Maths 9th

Description : MCQ Questions for Class 9 Maths Chapter 14 Statistics with answers -Maths 9th

Last Answer : Below you will find MCQ Questions of Chapter 14 Statistics Class 9 Maths Free PDF Download that will help you in gaining good marks in the examinations and also cracking competitive exams. These ... . These Statistics MCQ Questions will help you in practising more and more questions in less time.

A semi-circle of diameter 14 cm has three chords of equal length connecting the two end points of the diameter so as -Maths 9th

Description : A semi-circle of diameter 14 cm has three chords of equal length connecting the two end points of the diameter so as -Maths 9th

Last Answer : (c)\(\bigg(rac{147 imes\sqrt3}{4}\bigg)\) cm2Take the trapezoid ABCD in the semi-circle with centre O such that AD = DC = CB. Now, complete the circle and draw an identical trapezoid in the other semicircle also. Then, ADCBEF ... {1}{2}\)x \(rac{3\sqrt3}{2}\) x 49 = \(rac{147 imes\sqrt3}{4}\) cm2.

In how many ways can a team of 11 players be selected from 14 players when two of them play as goalkeepers only? -Maths 9th

Description : In how many ways can a team of 11 players be selected from 14 players when two of them play as goalkeepers only? -Maths 9th

Last Answer : As each team of 11 players has one goalkeeper and 10 team members, and out of 14 players there are 2 goalkeepers and 12 team members. = 12×112×2 = 132.

Water is following at the rate of 5 km/hr through a pipe of diameter 14 cm into a rectangular tank which is 50 m long -Maths 9th

Description : Water is following at the rate of 5 km/hr through a pipe of diameter 14 cm into a rectangular tank which is 50 m long -Maths 9th

Last Answer : Convert all to metres: 5 km = 5000 m 14 cm = 0.14 m 7 cm = 0.07 m Find the radius: Radius = Diameter 2 Radius = 0.14 2 = 0.07 m Find the amount of water that flowed out in an hour: Volume ... hours needed: Number of hours = 154 77 = 2 hours It takes 2 hours to fill up the tank to rise by 7 cm

The base of a right prism is a trapezium. The lengths of the parallel sides are 8 cm and 14 cm and the distance between the parallel -Maths 9th

Description : The base of a right prism is a trapezium. The lengths of the parallel sides are 8 cm and 14 cm and the distance between the parallel -Maths 9th

Last Answer : Area of trapezium =12×h(AB+CD) =12×8×(8+14)=12×8×(8+14) =4×22=88cm2=4×22=88cm2 = Volume of prism = Height of prism ×× area of base ⇒height×88=1056 (given)⇒height×88=1056 (given) ⇒height×88=105688⇒height×88=105688 ⇒12cm =12×h(AB+CD)

A cylindrical container of height 14 m and base 12 m contains oil. The oil is to be transferred to one cylindrical can, -Maths 9th

Description : A cylindrical container of height 14 m and base 12 m contains oil. The oil is to be transferred to one cylindrical can, -Maths 9th

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The volume of the metal of a cylindrical pipe is 748 cm^3. The length of the pipe is 14 cm and its external radius is 9 cm -Maths 9th

Description : The volume of the metal of a cylindrical pipe is 748 cm^3. The length of the pipe is 14 cm and its external radius is 9 cm -Maths 9th

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Find the value of 27X3 + 8y3 if (i) 3x + 2y = 14 and xy = 8 (ii) 3x + 2y = 20 and xy = 149 -Maths 9th

Description : Find the value of 27X3 + 8y3 if (i) 3x + 2y = 14 and xy = 8 (ii) 3x + 2y = 20 and xy = 149 -Maths 9th

Last Answer : answer:

Cbqs (case base study ) of chapter 14 Statistics of maths class 9th -Maths 9th

Description : Cbqs (case base study ) of chapter 14 Statistics of maths class 9th -Maths 9th

Last Answer : answer:

A boat covers 14 kms in upstream and 20 kms downstream in 7 hours. Also it covers 22 kms upstream and 34 kms downstream in 10 hours. Find the speed of the boat in still water and of that the stream. -Maths 9th

Description : A boat covers 14 kms in upstream and 20 kms downstream in 7 hours. Also it covers 22 kms upstream and 34 kms downstream in 10 hours. Find the speed of the boat in still water and of that the stream. -Maths 9th

Last Answer : Given, The boat covers 14 km upstream and 20km downstream . at time 7 hours also cover 22km ups. and 34km dwn in10 hours total speed = total distance/total time :.total distance = 14+20+22+34=90km and total time=7+10= ... =90/17 => 5.294km/h => 5.294km/h the speed of boat in still water is 5.29 km/h

If x+y=10 and x=z then show that z+y=10 by using appropriate eculids axioms? -Maths 9th

Description : If x+y=10 and x=z then show that z+y=10 by using appropriate eculids axioms? -Maths 9th

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

If x+y =10 and x=z then show that z+y =10 -Maths 9th

Description : If x+y =10 and x=z then show that z+y =10 -Maths 9th

Last Answer : It is given that x+y =10 Also x= z Therefore, x+y =10 Z+y =10 ( x = z)

In the given figure, if chords AB and CD of the circle intersect each other at right angles, then find x + y. -Maths 9th

Description : In the given figure, if chords AB and CD of the circle intersect each other at right angles, then find x + y. -Maths 9th

Last Answer : ∴ ∠CAO = ∠ODB = x [angles in same segment ] ---- (i) Now, in right angled ΔDOB , ∠ODB + ∠DOB + ∠OBD = 180° ⇒ x + 90° + y =180° (using equation i) ⇒ x + y = 90°

If a wooden box of dimensions 8 m x 7 m x 6 m is to carry boxes of dimensions 8 cm x 7 cm x 6 cm, then find the maximum number of boxes that can be carried in the wooden box. -Maths 9th

Description : If a wooden box of dimensions 8 m x 7 m x 6 m is to carry boxes of dimensions 8 cm x 7 cm x 6 cm, then find the maximum number of boxes that can be carried in the wooden box. -Maths 9th

Last Answer : Volume of wooden box = 800 cm × 700 cm × 600 cm Volume of box = 8 cm × 7 cm × 6 cm ∴ Number of boxes = volume of wooden box / volume of each box = 800 cm × 700 cm × 600 cm / 8 cm × 7 cm × 6 cm = 1000000

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 median of data (arranged in ascending order) 31, 33, 35, x, x+10, 48, 48, 50 is 40, then find value of x. -Maths 9th

Description : If the median of data (arranged in ascending order) 31, 33, 35, x, x+10, 48, 48, 50 is 40, then find value of x. -Maths 9th

Last Answer : Given data is 31, 33, 35, x, x+10, 48, 48, 50 Number of observation = 8 (even) Median = Value of (8/2)th observation + Value of (8/2+1)th observation / 2 Value of 4th observation + Value of 5th observation / 2 = x + x + 10 / 2 = x + 5 ∴ x + 5 = 40 ⇒ x = 35

In a class, there are x girls and y boys, a student is selected at random, then find the probability of selecting a boy. -Maths 9th

Description : In a class, there are x girls and y boys, a student is selected at random, then find the probability of selecting a boy. -Maths 9th

Last Answer : Number of boys = y Total students = (x + y) Thus,P(a boy)= y/(x+y)

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 x + 1 is a factor of the polynomial 2x2 + kx, then the value of k is -Maths 9th

Description : If x + 1 is a factor of the polynomial 2x2 + kx, then the value of k is -Maths 9th

Last Answer : (c) Let p(x) = 2x2 + kx Since, (x + 1) is a factor of p(x), then p(-1)=0 2(-1)2 + k(-1) = 0 ⇒ 2-k = 0 ⇒ k= 2 Hence, the value of k is 2.

If x51 + 51 is divided by x + 1, then the remainder is -Maths 9th

Description : If x51 + 51 is divided by x + 1, then the remainder is -Maths 9th

Last Answer : (d) Let p(x) = x51 + 51 . …(i) When we divide p(x) by x+1, we get the remainder p(-1) On putting x= -1 in Eq. (i), we get p(-1) = (-1)51 + 51 = -1 + 51 = 50 Hence, the remainder is 50.