Without actually calculating the cubes, find the value of: (-3/4)3 + (-5/8)3 + (11/8)3 -Maths 9th

Without actually calculating the cubes, find the value of: (-3/4)3 + (-5/8)3 + (11/8)3 -Maths 9th
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Without actually calculating the cubes, find the value of 36xy-36xy = 0 -Maths 9th

Description : Without actually calculating the cubes, find the value of 36xy-36xy = 0 -Maths 9th

Last Answer : Find the value of 36xy-36xy = 0.

Without actually calculating the cubes, find the value of 36xy-36xy = 0 -Maths 9th

Description : Without actually calculating the cubes, find the value of 36xy-36xy = 0 -Maths 9th

Last Answer : Find the value of 36xy-36xy = 0.

Without finding the cubes, factorise (x- 2y)3 + (2y – 3z)3 + (3z – x)3. -Maths 9th

Description : Without finding the cubes, factorise (x- 2y)3 + (2y – 3z)3 + (3z – x)3. -Maths 9th

Last Answer : We know that, a3 + b3 + c3 – 3 abc = (a + b + c)(a2 + b2 + c2 -ab-bc-ca) Also, if a + b + c = 0, then a3 + b3 + c3 = 3abc Here, we see that (x-2y) +(2y-3z)+ (3z-x) = 0 Therefore, (x-2y)3 + (2y-3z)3 + (3z-x)3 = 3(x-2y)(2y-3z)(3z-x).

Without finding the cubes, factorise (x- 2y)3 + (2y – 3z)3 + (3z – x)3. -Maths 9th

Description : Without finding the cubes, factorise (x- 2y)3 + (2y – 3z)3 + (3z – x)3. -Maths 9th

Last Answer : We know that, a3 + b3 + c3 – 3 abc = (a + b + c)(a2 + b2 + c2 -ab-bc-ca) Also, if a + b + c = 0, then a3 + b3 + c3 = 3abc Here, we see that (x-2y) +(2y-3z)+ (3z-x) = 0 Therefore, (x-2y)3 + (2y-3z)3 + (3z-x)3 = 3(x-2y)(2y-3z)(3z-x).

Without finding the cubes, factorise: (2r-3s)3 +(3s -5t)3+ (5t-2r)3. -Maths 9th

Description : Without finding the cubes, factorise: (2r-3s)3 +(3s -5t)3+ (5t-2r)3. -Maths 9th

Last Answer : Solution :-

If three cubes of copper, each with an edge 6 cm, 8 cm and 10 cm respectively are melted to form a single cube, -Maths 9th

Description : If three cubes of copper, each with an edge 6 cm, 8 cm and 10 cm respectively are melted to form a single cube, -Maths 9th

Last Answer : 20.8 cm Let the edge of the single cube be ‘a’ cm. Then, total volume melted = Volume of cube formed ⇒ (6)3 + (8)3 + (10)3 = a3 ⇒ a3 = 216 + 512 + 1000 = 1728 ... Diagonal of the new cube = 3–√a=(3–√×12)3a=(3×12) cm = 20.8 cm (approx.)

Three copper cubes whose edges measure 5 cm, -Maths 9th

Description : Three copper cubes whose edges measure 5 cm, -Maths 9th

Last Answer : Let a cm be the edge of new cube. Then volume of the new cube = Sum of the volumes of three cubes. ⇒ a3 = 53 + 43 + 33 = 125 + 64 + 27 ⇒ a3 = 216 ⇒ a3 = 63 ⇒ a = 6 cm ∴ Surface area of the new cube = 6a2 = 6 x 62 = 216 cm2

A 4 cm cube is cut into 1 cm cubes. Calculate the total surface area of all the small cubes. -Maths 9th

Description : A 4 cm cube is cut into 1 cm cubes. Calculate the total surface area of all the small cubes. -Maths 9th

Last Answer : Side of cube = 4 cm But cutting into 1 cm cubes, we get = 4 x 4 x 4 = 64 Now surface area of one cube = 6 x (1)² = 6 x 1=6 cm² and surface area of 64 cubes = 6 x 64 cm² = 384 cm²

Three equal cubes are placed adjacently in a row. Find the ratio of total surface area of the new cuboid to that of the sum of the surface areas of the three cubes. -Maths 9th

Description : Three equal cubes are placed adjacently in a row. Find the ratio of total surface area of the new cuboid to that of the sum of the surface areas of the three cubes. -Maths 9th

Last Answer : Let each side of a cube = a cm Then surface area = 6a² cm² and surface area of 3 such cubes = 3 x 6a² = 18a² cm² By placing three cubes side by side we get a cuboid whose ... + 3a²] = 14 a² ∴ Ratio between their surface areas = 14a² : 18a² = 7 : 9

Two cubes of side 2 cm each are joined end to end. Find the volume of the cuboid so formed. -Maths 9th

Description : Two cubes of side 2 cm each are joined end to end. Find the volume of the cuboid so formed. -Maths 9th

Last Answer : When two cubes of side 2 cm each are joined end to end then, Length (l) = (2 + 2) = 4cm Breadth (b) = 2 cm; Height (h) = 2 cm ∴ Volume of cuboid = lbh = 4 x 2 x 2 = 16 cm3

Two cubes of edge 6 cm are joined to form a cuboid. Find the total surface area of the cuboid. -Maths 9th

Description : Two cubes of edge 6 cm are joined to form a cuboid. Find the total surface area of the cuboid. -Maths 9th

Last Answer : When two cubes are joined end to end, then Length of the cuboid = 6 + 6 = 12 cm Breadth of the cuboid = 6 cm Height of the cuboid = 6 cm Total surface area of the cuboid = 2 (lb + bh + hi) = 2(12 x6 + 6×6 + 6×12) = 2(72 + 36 + 72) = 2(180) = 360 cm2

Two cubes of side 2 cm each are joined end to end. Find the volume of the cuboid so formed. -Maths 9th

Description : Two cubes of side 2 cm each are joined end to end. Find the volume of the cuboid so formed. -Maths 9th

Last Answer : When two cubes of side 2 cm each are joined end to end then, Length (l) = (2 + 2) = 4cm Breadth (b) = 2 cm; Height (h) = 2 cm ∴ Volume of cuboid = lbh = 4 x 2 x 2 = 16 cm3

Two cubes of edge 6 cm are joined to form a cuboid. Find the total surface area of the cuboid. -Maths 9th

Description : Two cubes of edge 6 cm are joined to form a cuboid. Find the total surface area of the cuboid. -Maths 9th

Last Answer : When two cubes are joined end to end, then Length of the cuboid = 6 + 6 = 12 cm Breadth of the cuboid = 6 cm Height of the cuboid = 6 cm Total surface area of the cuboid = 2 (lb + bh + hi) = 2(12 x6 + 6×6 + 6×12) = 2(72 + 36 + 72) = 2(180) = 360 cm2

How many small cubes each of 96 cm^2 surface area can be formed from the material obtained by melting a larger cube of 384 cm^2 surface area ? -Maths 9th

Description : How many small cubes each of 96 cm^2 surface area can be formed from the material obtained by melting a larger cube of 384 cm^2 surface area ? -Maths 9th

Last Answer : answer:

There are two identical cubes. Out of one cube, a sphere of maximum volume (VS) is cut off. Out of the second cube, a cone of maximum volume -Maths 9th

Description : There are two identical cubes. Out of one cube, a sphere of maximum volume (VS) is cut off. Out of the second cube, a cone of maximum volume -Maths 9th

Last Answer : answer:

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

How many bricks will be required to construct a wall 8 m long, 6 m high and 22.5 cm thick, if each brick measures 25 cm x 11.25 cm x 6 cm? -Maths 9th

Description : How many bricks will be required to construct a wall 8 m long, 6 m high and 22.5 cm thick, if each brick measures 25 cm x 11.25 cm x 6 cm? -Maths 9th

Last Answer : ∵ Volume of one brick = (25 × 11.25 × 6) cm3 and volume of the wall = (800 × 600 × 22.5) cm3 ∴ Number of bricks = Volume of the walls / Volume of one brick = 800 × 600 × 22.5 / 25 × 11.25 × 6 = 6400

How many bricks will be required to construct a wall 8 m long, 6 m high and 22.5 cm thick, if each brick measures 25 cm x 11.25 cm x 6 cm? -Maths 9th

Description : How many bricks will be required to construct a wall 8 m long, 6 m high and 22.5 cm thick, if each brick measures 25 cm x 11.25 cm x 6 cm? -Maths 9th

Last Answer : ∵ Volume of one brick = (25 × 11.25 × 6) cm3 and volume of the wall = (800 × 600 × 22.5) cm3 ∴ Number of bricks = Volume of the walls / Volume of one brick = 800 × 600 × 22.5 / 25 × 11.25 × 6 = 6400

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 :-

In Fig. 10.11, find the value of x and y. -Maths 9th

Description : In Fig. 10.11, find the value of x and y. -Maths 9th

Last Answer : Solution :- y = 2∠ ACB ⇒ y = 2 65° ⇒ y = 130° OA = ... Or 2x = 50° Or x = 25°

If 3x – 2y= 11 and xy = 12, find the value of 27x3 – 8y3. -Maths 9th

Description : If 3x – 2y= 11 and xy = 12, find the value of 27x3 – 8y3. -Maths 9th

Last Answer : Given, (a−b)3=a3−3ab(a−b)−b3 (3x−2y)3=27x3−8y3−3(3x)(2y)(3x−2y) 113=27x3−8y3−18(12)(11) 27x3−8y3=3707

Find the range of the given data : 25, 18, 20, 22, 16, 6, 17, 15, 12, 30, 32, 10, 19, 8, 11, 20 -Maths 9th

Description : Find the range of the given data : 25, 18, 20, 22, 16, 6, 17, 15, 12, 30, 32, 10, 19, 8, 11, 20 -Maths 9th

Last Answer : Here, the minimum and maximum values of given data are 6 and 32 respectively. Range = 32 – 6 = 26

Find the range of the given data : 25, 18, 20, 22, 16, 6, 17, 15, 12, 30, 32, 10, 19, 8, 11, 20 -Maths 9th

Description : Find the range of the given data : 25, 18, 20, 22, 16, 6, 17, 15, 12, 30, 32, 10, 19, 8, 11, 20 -Maths 9th

Last Answer : Here, the minimum and maximum values of given data are 6 and 32 respectively. Range = 32 – 6 = 26

The range of the data 25, 18, 20, 22, 16, 6, 17, 15, 12, 30, 32, 10, 19, 8, 11 and 20 is -Maths 9th

Description : The range of the data 25, 18, 20, 22, 16, 6, 17, 15, 12, 30, 32, 10, 19, 8, 11 and 20 is -Maths 9th

Last Answer : In conclusion, the range of data 25, 18, 20, 22, 16, 6, 17, 15, 12, 30, 32, 10, 19, 8, 11, and 20 is 26.

The range of the data 25, 18, 20, 22, 16, 6, 17, 15, 12, 30, 32, 10, 19, 8, 11 and 20 is -Maths 9th

Description : The range of the data 25, 18, 20, 22, 16, 6, 17, 15, 12, 30, 32, 10, 19, 8, 11 and 20 is -Maths 9th

Last Answer : (d) In a given data, maximum value = 32 and minimum value = 6 We know, range of the data = maximum value – minimum value = 32 – 6 = 26 Hence, the range of the given data is 26.

In the Fig.8.11. ABCD is a parallelogram.what is the sum of angles x,y and z? -Maths 9th

Description : In the Fig.8.11. ABCD is a parallelogram.what is the sum of angles x,y and z? -Maths 9th

Last Answer : Solution :-

A rectangular paper 11 cm by 8 cm can be exactly wrapped to cover the curved surface of a cylinder of height 8 cm . -Maths 9th

Description : A rectangular paper 11 cm by 8 cm can be exactly wrapped to cover the curved surface of a cylinder of height 8 cm . -Maths 9th

Last Answer : answer:

Two chords AB and CD of lengths 5 cm and 11 cm respectively of a circle are parallel to each other and are on opposite sides of its centre. If the A distance between AB and CD is 6 cm, find the radius of the circle. -Maths 9th

Description : Two chords AB and CD of lengths 5 cm and 11 cm respectively of a circle are parallel to each other and are on opposite sides of its centre. If the A distance between AB and CD is 6 cm, find the radius of the circle. -Maths 9th

Last Answer : Join OA and OC. Let the radius of the circle be r cm and O be the centre Draw OP⊥AB and OQ⊥CD. We know, OQ⊥CD, OP⊥AB and AB∥CD. Therefore, points P,O and Q are collinear. So, PQ=6 cm. Let OP=x. Then, ... r2=52+(2.5)2=25+6.25=31.25 ⇒r2=31.25⇒r=5.6 Hence, the radius of the circle is 5.6 cm

Let A = {(2, 5, 11)}, B = {3, 6, 10} and R be a relation from A to B defined by R = {(a, b) : a and b are co-prime}. Then R is -Maths 9th

Description : Let A = {(2, 5, 11)}, B = {3, 6, 10} and R be a relation from A to B defined by R = {(a, b) : a and b are co-prime}. Then R is -Maths 9th

Last Answer : (c) {(2, 3), (5, 3), (5, 6), (11, 3), (11, 6), (11, 10)}

A bag contains 5 green and 11 blue balls and the second one contains 3 green and 7 blue balls. Two balls are drawn from one of the bags. -Maths 9th

Description : A bag contains 5 green and 11 blue balls and the second one contains 3 green and 7 blue balls. Two balls are drawn from one of the bags. -Maths 9th

Last Answer : (c) \(rac{111}{240}\)P(Drawing of two balls of different colours from one of the bags)= P(choosing the 1st bag) P(Drawing 1 green out 5 green and 1 out of 11 blue balls) + P(choosing the 2nd bag) P(Drawing 1 green out ... (rac{11}{48}\) + \(rac{7}{30}\) = \(rac{55+56}{240}\) = \(rac{111}{240}\).

If the polynomial x^19 + x^17 + x^13 + x^11 + x^7 + x^5 + x^3 is divided by (x^2 + 1), then the remainder is : -Maths 9th

Description : If the polynomial x^19 + x^17 + x^13 + x^11 + x^7 + x^5 + x^3 is divided by (x^2 + 1), then the remainder is : -Maths 9th

Last Answer : answer:

If the points A(a, –11), B(5, b), C(2, 15) and D(1, 1) are the vertices of a parallelogram ABCD, find the values of a and b. -Maths 9th

Description : If the points A(a, –11), B(5, b), C(2, 15) and D(1, 1) are the vertices of a parallelogram ABCD, find the values of a and b. -Maths 9th

Last Answer : Let the x-axis divide the line joining the points (-2, 5) and (1, -9) in the ratio k : 1. Let the point of division on x-axis is P Then,\(x\) = \(rac{k-2}{k+1}\), y = \(rac{-9k+ ... (rac{5}{9}\)k being positive, the division is internal. ∴ x-axis divides the given line internally in the ratio 5 : 9.

Find the five rational no. Between5/7 and9/11 -Maths 9th

Description : Find the five rational no. Between5/7 and9/11 -Maths 9th

Last Answer : 5/7 and 9/11 find the lcm of 7 and 11 lcm =77. 5/7 *11/11=55/77 , 9/11*7/7=63/77 therefore five rational no are 56/77 57/77 58/77 59/77 60/77

If the lengths of the sides of a triangle are in the ratio 6:11:15 and it's perimeter is 96cm , then the height corresponding to the longest side is -Maths 9th

Description : If the lengths of the sides of a triangle are in the ratio 6:11:15 and it's perimeter is 96cm , then the height corresponding to the longest side is -Maths 9th

Last Answer : LET EACH SIDE BE X 6X+11X+15X=96 32X=96 X=3 SIDES=6 3=18 11 3=33 15 3=45 AREA OF TRIANGLE BY HERONS FORMULA=S=96/2=48 WHOLE UNDERROOT 48 48-18 48-33 48-45 UNDERROOT=12 4 30 15 3 4 3 15ROOT2 180 ... bh/2 180root2=18 h/2 360root2=18h h=20 root2 But root 2=1.4(approx) h=20 1.4(approx) h=28cm(approx).

Find the five rational no. Between5/7 and9/11 -Maths 9th

Description : Find the five rational no. Between5/7 and9/11 -Maths 9th

Last Answer : 5/7 and 9/11 find the lcm of 7 and 11 lcm =77. 5/7 *11/11=55/77 , 9/11*7/7=63/77 therefore five rational no are 56/77 57/77 58/77 59/77 60/77

If the lengths of the sides of a triangle are in the ratio 6:11:15 and it's perimeter is 96cm , then the height corresponding to the longest side is -Maths 9th

Description : If the lengths of the sides of a triangle are in the ratio 6:11:15 and it's perimeter is 96cm , then the height corresponding to the longest side is -Maths 9th

Last Answer : LET EACH SIDE BE X 6X+11X+15X=96 32X=96 X=3 SIDES=6 3=18 11 3=33 15 3=45 AREA OF TRIANGLE BY HERONS FORMULA=S=96/2=48 WHOLE UNDERROOT 48 48-18 48-33 48-45 UNDERROOT=12 4 30 15 3 4 3 15ROOT2 180 ... bh/2 180root2=18 h/2 360root2=18h h=20 root2 But root 2=1.4(approx) h=20 1.4(approx) h=28cm(approx).

In a diagnostic test in mathematics given to students, the following marks (out of 100) are recorded 46, 52, 48, 11, 41, 62, 54, 53, 96, 40, 98 and 44. -Maths 9th

Description : In a diagnostic test in mathematics given to students, the following marks (out of 100) are recorded 46, 52, 48, 11, 41, 62, 54, 53, 96, 40, 98 and 44. -Maths 9th

Last Answer : NEED ANSWER

In a diagnostic test in mathematics given to students, the following marks (out of 100) are recorded 46, 52, 48, 11, 41, 62, 54, 53, 96, 40, 98 and 44. -Maths 9th

Description : In a diagnostic test in mathematics given to students, the following marks (out of 100) are recorded 46, 52, 48, 11, 41, 62, 54, 53, 96, 40, 98 and 44. -Maths 9th

Last Answer : Median will be a good representative of the data, because 1.each value occurs once. 2.the data is influenced by extreme values.

If x to the power 11 + 101 is divided by x+1,what is the remainder? -Maths 9th

Description : If x to the power 11 + 101 is divided by x+1,what is the remainder? -Maths 9th

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Find the coordinate where the linear equation 3x -4y = 11 meets at x-axis. -Maths 9th

Description : Find the coordinate where the linear equation 3x -4y = 11 meets at x-axis. -Maths 9th

Last Answer : hope it helps

From fig. 4.11 answer the following: -Maths 9th

Description : From fig. 4.11 answer the following: -Maths 9th

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In a survey, it was found that 9 out of every 11 households -Maths 9th

Description : In a survey, it was found that 9 out of every 11 households -Maths 9th

Last Answer : 2/11 = 0.18(recurring), Non terminating repeating. People are becoming more social, helpful, cooperative and caring.

NCERT Solutions for class 9 Maths Chapter 11 Constructions Exercise 11.1 -Maths 9th

Description : NCERT Solutions for class 9 Maths Chapter 11 Constructions Exercise 11.1 -Maths 9th

Last Answer : 1: Construct an angle of 90º at the initial point of a given ray and justify the construction. 2: Construct an angle of 45º at the initial point of a given ray and justify the construction ... the construction. Let us construct an equilateral triangle, each of whose side = PQ Steps of construction:

NCERT Solutions for class 9 Maths Chapter 11 Constructions Exercise 11.2 -Maths 9th

Description : NCERT Solutions for class 9 Maths Chapter 11 Constructions Exercise 11.2 -Maths 9th

Last Answer : 1. Construct a triangle ABC in which BC = 7 cm, ∠B = 75º and AB + AC = 13 cm. Steps of construction: 2. Construct a triangle ABC in which BC = 8 cm, ∠ B = 45º and AB ∠ AC = 3 ... Construct a right triangle whose base is 12 cm and sum of its hypotenuse and other side is 18 cm. Steps of construction:

MCQ Questions for Class 9 Maths Chapter 11 Constructions with answers -Maths 9th

Description : MCQ Questions for Class 9 Maths Chapter 11 Constructions with answers -Maths 9th

Last Answer : Below you will find MCQ Questions of Chapter 11 Constructions Class 9 Maths Free PDF Download that will help you in gaining good marks in the examinations and also cracking competitive exams. ... These Constructions MCQ Questions will help you in practising more and more questions in less time.

Write any four solutions of the linear equation y = 4 x – 11. -Maths 9th

Description : Write any four solutions of the linear equation y = 4 x – 11. -Maths 9th

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A committee of 11 members sit at a round table. In how many ways can they be seated if the “President” and the “Secretary” choose to sit together. -Maths 9th

Description : A committee of 11 members sit at a round table. In how many ways can they be seated if the “President” and the “Secretary” choose to sit together. -Maths 9th

Last Answer : ∴ Required number of ways of seating = 9!

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.

7 persons enter an elevator on the ground floor of a 11 storey hotel. Any one of them can leave the elevator at any of the 10 floors. -Maths 9th

Description : 7 persons enter an elevator on the ground floor of a 11 storey hotel. Any one of them can leave the elevator at any of the 10 floors. -Maths 9th

Last Answer : answer: