What is/are the factors of (x^29 – x^24 + x^13 – 1) ? -Maths 9th

What is/are the factors of (x^29 – x^24 + x^13 – 1) ? -Maths 9th
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The radius of a circle is 13 cm and the length of one of its chords is 24 cm. Find the distance of the chord from the centre. -Maths 9th

Description : The radius of a circle is 13 cm and the length of one of its chords is 24 cm. Find the distance of the chord from the centre. -Maths 9th

Last Answer : Let PQ be a chord of a circle with centre O and radius 13cm such that PQ = 24cm. From O, draw OM perpendicular PQ and join OP. As, the perpendicular from the centre of a circle to a chord bisects the chord. ∴ PM ... Hence, the distance of the chord from the centre is 5cm.

In Fig. 8.29, ABCD is a parallelogram with perimeter 40 cm. Find x and y. -Maths 9th

Description : In Fig. 8.29, ABCD is a parallelogram with perimeter 40 cm. Find x and y. -Maths 9th

Last Answer : Solution :-

1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32, 33, 34, 35, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 47, 48, 49, 50'. What number is missing? -Riddles

Description : 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32, 33, 34, 35, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 47, 48, 49, 50'. What number is missing? -Riddles

Last Answer : 22

RBI has imposed how much fine on 13 banks for violation of FEMA and lapses in know your client (KYC) rules? A) Rs 24 crore B) Rs 25 crore C) Rs 27 crore D) Rs 28 crore E) Rs 29 crore

Description : RBI has imposed how much fine on 13 banks for violation of FEMA and lapses in know your client (KYC) rules? A) Rs 24 crore B) Rs 25 crore C) Rs 27 crore D) Rs 28 crore E) Rs 29 crore

Last Answer : C) Rs 27 crore Explanation: The Reserve Bank of India has come down hard on 13 banks and fined them for Rs 27 crore for violation of FEMA and lapses in know your client (KYC) rules.

In 165 litres of mixtures of milk and water, water is only 28%. The milkman sold 40 litres of this mixture and then he added 30 litres of pure milk and 13 litres of pure water in the remaining mixture. What is the percentage of water in the final mixture? 1) 29.35% 2) 28. 57% 3) 24. 57% 4) 27. 75% 5) 26. 57%

Description : In 165 litres of mixtures of milk and water, water is only 28%. The milkman sold 40 litres of this mixture and then he added 30 litres of pure milk and 13 litres of pure water in the remaining mixture. What is the percentage ... mixture? 1) 29.35% 2) 28. 57% 3) 24. 57% 4) 27. 75% 5) 26. 57%

Last Answer : 2) 28. 57%

The ratio of the ages of Bala and Babu is 13: 11. The total of their ages is 3.6decades. The proportion of their ages after 0.95decades will be [1 Decade = 10 years] a) 24:23 b) 12:11 c) 27:24 d) 29:26

Description : The ratio of the ages of Bala and Babu is 13: 11. The total of their ages is 3.6decades. The proportion of their ages after 0.95decades will be [1 Decade = 10 years] a) 24:23 b) 12:11 c) 27:24 d) 29:26

Last Answer : D Let, Bala’s age = 13A and babu’s age = 11A Then 13A + 11A = 36 A = 1.5 Bala’s age = 19.5years and Babu’s age = 16.5 years Proportion of their ages after 9.5 is = (19.5+9.5) : (16.5 + 9.5) = 29 : 26

ABCD is a quadrilateral in which P, Q, R and S are mid-points of the sides AB, BC, CD and DA (see Fig 8.29). AC is a diagonal. Show that: (i) SR || AC and SR = 1/2 AC (ii) PQ = SR (iii) PQRS is a parallelogram. -Maths 9th

Description : ABCD is a quadrilateral in which P, Q, R and S are mid-points of the sides AB, BC, CD and DA (see Fig 8.29). AC is a diagonal. Show that: (i) SR || AC and SR = 1/2 AC (ii) PQ = SR (iii) PQRS is a parallelogram. -Maths 9th

Last Answer : . Solution: (i) In ΔDAC, R is the mid point of DC and S is the mid point of DA. Thus by mid point theorem, SR || AC and SR = ½ AC (ii) In ΔBAC, P is the mid point of AB and Q is the mid point of BC. ... ----- from question (ii) ⇒ SR || PQ - from (i) and (ii) also, PQ = SR , PQRS is a parallelogram.

The median of the data 26,56,32,33,60,17,34,29,45 is 33. If 26 is replaced by 62, then find the new median. -Maths 9th

Description : The median of the data 26,56,32,33,60,17,34,29,45 is 33. If 26 is replaced by 62, then find the new median. -Maths 9th

Last Answer : Here, the given data in ascending order is 17, 29, 32, 33, 34, 45, 56, 60, 62 Now median is (9 + 1 / 2)th term i.e. , 5th term Hence, new median is 34.

The median of the data 26,56,32,33,60,17,34,29,45 is 33. If 26 is replaced by 62, then find the new median. -Maths 9th

Description : The median of the data 26,56,32,33,60,17,34,29,45 is 33. If 26 is replaced by 62, then find the new median. -Maths 9th

Last Answer : Here, the given data in ascending order is 17, 29, 32, 33, 34, 45, 56, 60, 62 Now median is (9 + 1 / 2)th term i.e. , 5th term Hence, new median is 34.

In Fig.6.29, DE||QR and AP and BP are bisectors of ZEAB and ZRBA respectively. Find ZAPB. -Maths 9th

Description : In Fig.6.29, DE||QR and AP and BP are bisectors of ZEAB and ZRBA respectively. Find ZAPB. -Maths 9th

Last Answer : Solution :-

Write the following decimals in the place value table. (a) 0.29 (b) 2.08 (c) 19.60 (d) 148.32 (e) 200.812 -Maths 9th

Description : Write the following decimals in the place value table. (a) 0.29 (b) 2.08 (c) 19.60 (d) 148.32 (e) 200.812 -Maths 9th

Last Answer : (a) 0.29 = 0.2 + 0.09 = 2 / 10 + 9 / 100 (b) 2.08 = 2 + 0.08 = 2 + 8 / 100 (c) 19.60 = 19 + 0.60 = 10 + 9 + 6 / 10 (d) 148.32 = 148 + 0.3 + 0. ... / 100 (e) 200.812 = 200 + 0.8 + 0.01 + 0.002 =200 + 8 / 10 + 1 / 100 + 2 / 1000HundredsTensOnesTenthsHundredthsThousandths000290002080019600148320200812

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

In the linear equation y = 4x + 13, if x is the number of hours a labourer is on work and y are his wages in rupees then draw the graph. Also find the wages when work is done for 6 hours. -Maths 9th

Description : In the linear equation y = 4x + 13, if x is the number of hours a labourer is on work and y are his wages in rupees then draw the graph. Also find the wages when work is done for 6 hours. -Maths 9th

Last Answer : when the work is done for 6 hours x=6 y=4(6)+13 y=24+13 y=37 the labourer gets Rs.37 if he works for 6hrs

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

When x^13 + 1 is divided by x –1, the remainder is : -Maths 9th

Description : When x^13 + 1 is divided by x –1, the remainder is : -Maths 9th

Last Answer : answer:

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:

The line L is given by x/5 + y/b = 1 passes through the point (13, 32). The line K is parallel to L and has the equation -Maths 9th

Description : The line L is given by x/5 + y/b = 1 passes through the point (13, 32). The line K is parallel to L and has the equation -Maths 9th

Last Answer : (a) 45º The equations of the given lines are: A\(x\) + By = A + B ⇒ By = -A\(x\) + (A + B) ⇒ y = \(-rac{A}{B}x\) + \(rac{(A+B)}{B}\) ....(i)and A(\(x\) - y) + B(\(x\) ... (ii) = m2 = \(rac{(A+B)}{B-A}\)Let θ be the angle between both the lines, then∴ tan θ = 1 ⇒ θ = tan-1 (1) = 45°.

If both (x+1) and (x -1) are factors of ax3 + x2 - 2x + b , find a and b. -Maths 9th

Description : If both (x+1) and (x -1) are factors of ax3 + x2 - 2x + b , find a and b. -Maths 9th

Last Answer : Let p(x) = ax3 + x2 - 2x + b Since (x+1) and (x-1) are the factors of p(x), ∴ p(-1) = 0 and p(1) = 0 ∴ p(-1) = a(-1)3 + (-1)2 - 2 (-1) + b = 0 ⇒ - a + 1 + 2 + b = 0 ⇒ a - b = 3 ---- (i) ... 0 ⇒ a + 1 - 2 + b = 0 ⇒ a + b = 1 ----- (ii) solving equations (i) and (ii) we get a = 2 and b = -1

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 both (x+1) and (x -1) are factors of ax3 + x2 - 2x + b , find a and b. -Maths 9th

Description : If both (x+1) and (x -1) are factors of ax3 + x2 - 2x + b , find a and b. -Maths 9th

Last Answer : Let p(x) = ax3 + x2 - 2x + b Since (x+1) and (x-1) are the factors of p(x), ∴ p(-1) = 0 and p(1) = 0 ∴ p(-1) = a(-1)3 + (-1)2 - 2 (-1) + b = 0 ⇒ - a + 1 + 2 + b = 0 ⇒ a - b = 3 ---- (i) ... 0 ⇒ a + 1 - 2 + b = 0 ⇒ a + b = 1 ----- (ii) solving equations (i) and (ii) we get a = 2 and b = -1

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.

Find the factors of {1-x(cube)}. -Maths 9th

Description : Find the factors of {1-x(cube)}. -Maths 9th

Last Answer : Solution :-

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

Last Answer : Solution :-

Which one of the following is one of the factors of x^2 (y – z) + y^2 (z – x) – z (xy – yz – zx) ? -Maths 9th

Description : Which one of the following is one of the factors of x^2 (y – z) + y^2 (z – x) – z (xy – yz – zx) ? -Maths 9th

Last Answer : answer:

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

Last Answer : answer:

The factors of x^8 + x^4 + 1 are : -Maths 9th

Description : The factors of x^8 + x^4 + 1 are : -Maths 9th

Last Answer : answer:

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

A right circular cylinder just encloses a sphere of radius r (see fig. 13.22). Find (i) surface area of the sphere, (ii) curved surface area of the cylinder -Maths 9th

Description : A right circular cylinder just encloses a sphere of radius r (see fig. 13.22). Find (i) surface area of the sphere, (ii) curved surface area of the cylinder -Maths 9th

Last Answer : Surface area of sphere = 4πr2, where r is the radius of sphere (ii) Height of cylinder, h = r+r =2r Radius of cylinder = r CSA of cylinder formula = 2πrh = 2πr(2r) (using value of h) = 4πr2 (iii) Ratio ... sphere)/CSA of Cylinder) = 4r2/4r2 = 1/1 Ratio of the areas obtained in (i) and (ii) is 1:1.

The perimeter of a right triangle is 30 cm. If its hypotenuse is 13 cm, then what are two sides? -Maths 9th

Description : The perimeter of a right triangle is 30 cm. If its hypotenuse is 13 cm, then what are two sides? -Maths 9th

Last Answer : The other two sides of the triangle are 12 cm and 5 cm Explanation: Let the other two sides of triangle be x and y It's hypotenuse is 13 cm Perimeter of triangle = Sum of all sides ... When y = 12 x=17-y = 17-12 =5 So, the other two sides of the triangle are 12 cm and 5 cm

The mean of 25 observations is 36. Out of these observations, if the mean of first 13 observations is 32 and that of the last 13 observations is 40, the 13th observation is -Maths 9th

Description : The mean of 25 observations is 36. Out of these observations, if the mean of first 13 observations is 32 and that of the last 13 observations is 40, the 13th observation is -Maths 9th

Last Answer : NEED ANSWER

The mean of 25 observations is 36. Out of these observations, if the mean of first 13 observations is 32 and that of the last 13 observations is 40, the 13th observation is -Maths 9th

Description : The mean of 25 observations is 36. Out of these observations, if the mean of first 13 observations is 32 and that of the last 13 observations is 40, the 13th observation is -Maths 9th

Last Answer : (b) Given, mean of 25 observations = 36 ∴ Sum of 25 observations = 36 x 25 = 900 Now, the mean of first 13 observations = 32 ∴ Sum of first 13 observations = 13 x 32 = 416 and the mean of last 13 ... - (Sum of 25 observations) = (520 + 416)-900 = 936 - 900 = 36 Hence, the 13th observation is 36.

Write 3/13 in decimal form and find what kind of decimal expansion it has. -Maths 9th

Description : Write 3/13 in decimal form and find what kind of decimal expansion it has. -Maths 9th

Last Answer : Solution :-

Find the length of a chord which is at a distance of 12 cm from the centre of a circle of radius 13 cm. -Maths 9th

Description : Find the length of a chord which is at a distance of 12 cm from the centre of a circle of radius 13 cm. -Maths 9th

Last Answer : Let AB be a chord of circle with centre O and radius 13cm. Draw OM perpendicular AB and join OA. In the right triangle OMA, we have OA2 = OM2 + AM2 ⇒ 132 = 122 + AM2 ⇒ AM2 = 169 - 144 ... . As the perpendicular from the centre of a chord bisects the chord.Therefore, AB = 2AM = 2 x 5 = 10cm.

Find the area of a trapezium whose parallel sides are 25 cm and 13 cm long and the distance between them is 8 cm. -Maths 9th

Description : Find the area of a trapezium whose parallel sides are 25 cm and 13 cm long and the distance between them is 8 cm. -Maths 9th

Last Answer : Area of a trapezium = 1/2(sum of parallel sides) x (perpendicular diatance between them) = 1/2(25 + 13) x 8 = 152cm2

The lengths of the sides of a triangle are 7 cm, 13 cm and 12 cm. -Maths 9th

Description : The lengths of the sides of a triangle are 7 cm, 13 cm and 12 cm. -Maths 9th

Last Answer : Let, a = 7 cm, b = 13 cm, c = 12 cm ∴ s = (a + b + c)/2 = (7 +13 +12)/2 = 32/2 = 16 cm Area of △ ABC = under root( √s(s -a) (s - b)(s -c)) = under root( √16(16 - 7)(16 - 13)(16 - 12) = ... 24 √3 cm2 Also, Area of △ ABC = 1/2AC.BD 24 √3 = 1/2 x 12 x BD ⇒ BD = (24 √3 x 2)/12 = 4 √3 cm

NCERT Solutions for class 9 Maths Chapter 13 Surface Areas and Volumes Exercise 13.1 -Maths 9th

Description : NCERT Solutions for class 9 Maths Chapter 13 Surface Areas and Volumes Exercise 13.1 -Maths 9th

Last Answer : 1. A plastic box 1.5 m long, 1.25 m wide and 65 cm deep is to be made. It is opened at the top. Ignoring the thickness of the plastic sheet, determine: (i) The area of the sheet required for making the box ... of a cylinder = 2πrh + 2πr2 = 2πr(h + r) Note: Unless it is mentioned assume π = (22/7)

NCERT Solutions for class 9 Maths Chapter 13 Surface Areas and Volumes Exercise 13.2 -Maths 9th

Description : NCERT Solutions for class 9 Maths Chapter 13 Surface Areas and Volumes Exercise 13.2 -Maths 9th

Last Answer : 1. 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. 2. It is required to make a closed cylindrical tank of height 1 m ... open from the top. ∴ Surface area of a penholder (cylinder) = [Lateral surface area] + [Base area]

NCERT Solutions for class 9 Maths Chapter 13 Surface Areas and Volumes Exercise 13.3 -Maths 9th

Description : NCERT Solutions for class 9 Maths Chapter 13 Surface Areas and Volumes Exercise 13.3 -Maths 9th

Last Answer : 1. Diameter of the base of a cone is 10.5 cm and its slant height is 10 cm. Find its curved surface area. 2.Find the total surface area of a cone, if its slant height is 21 m and diameter of its base is 24 ... m2, what will be the cost of painting all these cones? (Use π = 3.14 and take √1.04= 1.02)

NCERT Solutions for class 9 Maths Chapter 13 Surface Areas and Volumes Exercise 13.4 -Maths 9th

Description : NCERT Solutions for class 9 Maths Chapter 13 Surface Areas and Volumes Exercise 13.4 -Maths 9th

Last Answer : 1. Find the surface area of a sphere of radius: (i) 10.5 cm (ii) 5.6 cm (iii) 14 cm 2. Find the surface area of a sphere of diameter: (i) 14 cm (ii) 21 cm (iii) 3.5 m 3. Find the ... area of the sphere, (ii) curved surface area of the cylinder, (iii) ratio of the areas obtained in (i) and (ii).

NCERT Solutions for class 9 Maths Chapter 13 Surface Areas and Volumes Exercise 13.5 -Maths 9th

Description : NCERT Solutions for class 9 Maths Chapter 13 Surface Areas and Volumes Exercise 13.5 -Maths 9th

Last Answer : 1. A matchbox measures 4 cm x 2.5 cm. x 1.5 cm. What will be the volume of a packet containing 12 such boxes ? 2. A cuboidal water tank is 6 m long, 5 m wide and 4.5 m deep. How many litres of ... 40 m wide is flowing at the rate of 2 km per hour. How much water will fall into the sea in a minute ?

NCERT Solutions for class 9 Maths Chapter 13 Surface Areas and Volumes Exercise 13.6 -Maths 9th

Description : NCERT Solutions for class 9 Maths Chapter 13 Surface Areas and Volumes Exercise 13.6 -Maths 9th

Last Answer : 1. The circumference of the base of a cylindrical vessel is 132 cm and its height is 25 cm. How many litres of water can it hold ? (1000 cm3 = 1 l) . 2. The inner diameter of a cylindrical ... with soup to a height of 4 cm, how much soup the hospital has to prepare daily to serve 250 patients ?

MCQ Questions for Class 9 Maths Chapter 13 Surface Area and Volumes with answers -Maths 9th

Description : MCQ Questions for Class 9 Maths Chapter 13 Surface Area and Volumes with answers -Maths 9th

Last Answer : Below you will find MCQ Questions of Chapter 13 Surface Areas and Volumes Class 9 Maths Free PDF Download that will help you in gaining good marks in the examinations and also cracking competitive ... Areas and Volumes MCQ Questions will help you in practising more and more questions in less time.

A right DABC with sides 5 cm, 12 cm and 13 cm is revolved about the side 12 cm. What is the volume of the solid so obtained ? -Maths 9th

Description : A right DABC with sides 5 cm, 12 cm and 13 cm is revolved about the side 12 cm. What is the volume of the solid so obtained ? -Maths 9th

Last Answer : From the figure it is clear that a cone is formed. Here, h = 12 cm, r = 5 cm Volume of cone = = 314 cm3

A right triangular prism of height 18 cm and of base sides 5 cm, 12 cm and 13 cm is transformed into another right triangular prism on a base -Maths 9th

Description : A right triangular prism of height 18 cm and of base sides 5 cm, 12 cm and 13 cm is transformed into another right triangular prism on a base -Maths 9th

Last Answer : Vol. of △ ular prism = Area of △ ular base × height. ∴ Area of triangular base = area of triangle PQR By heron's formula. S=S(s−a)(s−b)(s−c)​where S=2a+b+c​∴Areaof△PQR= S=23+4+5​=6 S=6(6−3)(6−4)(6−5)​=3×2×3×2×1​=6cm2 ∴ vol. of Prism =6×10 =60cm3Answer.

Point A(5, 1) is the centre of the circle with radius 13 units. AB ⊥ chord PQ. B is (2, –3). The length of chord PQ is -Maths 9th

Description : Point A(5, 1) is the centre of the circle with radius 13 units. AB ⊥ chord PQ. B is (2, –3). The length of chord PQ is -Maths 9th

Last Answer : (c) ParallelogramAB = \(\sqrt{(4-7)^2+(5-6)^2}\) = \(\sqrt{9+1}\) = \(\sqrt{10}\)BC = \(\sqrt{(7-4)^2+(6-3)^2}\) = \(\sqrt{9+9}\) = \(3\sqrt2\)CD =\(\sqrt{(4-1)^2+(3-2) ... = \(2\sqrt{13}\)AB = CD, BC = AD and AC ≠ BD ⇒ opposite sides are equal and diagonals are not equal. ⇒ ABCD is a parallelogram.

A straight line passes through the points (a, 0) and (0, b). The length of the line segment contained between the axes is 13 and the product of -Maths 9th

Description : A straight line passes through the points (a, 0) and (0, b). The length of the line segment contained between the axes is 13 and the product of -Maths 9th

Last Answer : (d) \(rac{23}{\sqrt{17}}\)The given lines are:L : \(rac{x}{5}+rac{y}{b}=1\) ....(i)K : \(rac{x}{c}+rac{y}{3}=1.\) ...(ii)Since line L passes through (13, 32),\(rac{13}{ ... between parallel lines ax + by + c1 = 0 and ax + by c2 = 0 is d = \(rac{|c_2-c_1|}{\sqrt{a^2+b^2}}\bigg)\)

If 2x + 3y = 13 and xy = 6, find the value of 8x3 + 21y3. -Maths 9th

Description : If 2x + 3y = 13 and xy = 6, find the value of 8x3 + 21y3. -Maths 9th

Last Answer : 2x +3y = 13-----(1) xy =6-----(2) 8x³ +27y³ = (2x)³ +(3y)³ = (2x+3y)³ - 3*2x*3y(2x+3y) [using (a+b)³ = a³+b³+3ab(a+b)] = (13)³- 18 * 6 *13 [ using (1) and (2)] = 2197 - 1404 =793

Cbqs (case base study ) of chapter 13 surface areas and volume of maths class 9th -Maths 9th

Description : Cbqs (case base study ) of chapter 13 surface areas and volume of maths class 9th -Maths 9th

Last Answer : CBQs Ch- 13 Surface area and volume - Maths Class 9th 1 . Dev was doing an experiment to find the radius r of a sphere.For this he took a cylindrical container with radius R = 7 cm and height 10 cm. He filled the container ... .14)*6²*8 = 301.44 m³ area of the floor = πR² = 3.14 (6)² = 113.04 m²

A conical tent is 10 m high and the radius of its base is 24 m. Find (i) slant height of the tent. -Maths 9th

Description : A conical tent is 10 m high and the radius of its base is 24 m. Find (i) slant height of the tent. -Maths 9th

Last Answer : : Ncert solutions class 9 chapter 13-5 Let ABC be a conical tent Height of conical tent, h = 10 m Radius of conical tent, r = 24m Let the slant height of the tent be l. (i) In right triangle ... (13728/7) 70 = Rs 137280 Therefore, the cost of the canvas required to make such a tent is Rs 137280.