root( 5+2root6) -Maths 9th

Last Answer : Root 72+root 800-root18 Root 4×2×9+root 4×2×10root 9×2 2×3root2+2×10root2-3root2 6root2+20root2-3root2 26root2-3root2 23root2 hence the answer.

Identify its surd or not and justify 3 root 7 -Maths 9th

Description : Identify its surd or not and justify 3 root 7 -Maths 9th

Last Answer : Yes,it is surd because a surd needs to be of the form nth root of a (unable to type exactly)where n is a positive integer and a is positive rational number.3 root 7 can be written as square root of 63 .Hence it is surd...

Find the value of a and b if root 3+1/root 3-1=a+b root 3 -Maths 9th

Description : Find the value of a and b if root 3+1/root 3-1=a+b root 3 -Maths 9th

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

Simplify: x to the power 4 whole to the power 1/3 under root 12. -Maths 9th

Description : Simplify: x to the power 4 whole to the power 1/3 under root 12. -Maths 9th

Last Answer : Solution :-

Divide: 12 15 under root 4 by 8 cube root 3. -Maths 9th

Description : Divide: 12 15 under root 4 by 8 cube root 3. -Maths 9th

Last Answer : Solution :-

Calculate 4th root of 193 + underroot 2178 -Maths 9th

Description : Calculate 4th root of 193 + underroot 2178 -Maths 9th

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

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...

Find the condition that one root of ax^2 + bx + c = 0 may be four times the other. -Maths 9th

Description : Find the condition that one root of ax^2 + bx + c = 0 may be four times the other. -Maths 9th

Last Answer : answer:

Find the values of k for which the equations x^2 – kx – 21 = 0 and x^2 – 3kx + 35 = 0 will have a common root? -Maths 9th

Description : Find the values of k for which the equations x^2 – kx – 21 = 0 and x^2 – 3kx + 35 = 0 will have a common root? -Maths 9th

Last Answer : Let the common root be α ⟹α2−kα−21=0......(1) α2−3kα+35=0........(2) (1)−(2)⟹2kα=56⟹α=k28Substituting α=k28 in (1) (k28)2−28−21=0 ⟹k=±4

If two equations x^2 + a^2 = 1 – 2ax and x^2 + b^2 = 1 – 2bx have only one common root, then -Maths 9th

Description : If two equations x^2 + a^2 = 1 – 2ax and x^2 + b^2 = 1 – 2bx have only one common root, then -Maths 9th

Last Answer : x2+b2=1−2bx ...... (i) x2+b2+2bx=1 (x+b)2=1 x+b= 1 ∴x=1−b,−1−b x2+a2=1−2ax ...... (ii) x2+a2+2ax=1 (x+a)2=1 ∴x=1−a,−1−a It is given that the equations have only one root in ... or −1−a=−b−1 we get a=b but then both roots will be common, which is not possible. Hence, options A,B and C are correct.

If one root of the equation (x^2/a)+ (x/a)+(1/c) = 0is reciprocal of the other, then which of the following is correct ? -Maths 9th

Description : If one root of the equation (x^2/a)+ (x/a)+(1/c) = 0is reciprocal of the other, then which of the following is correct ? -Maths 9th

Last Answer : answer:

If one root of the equation ax^2 + x – 3 = 0 is –1, then what is the other root ? -Maths 9th

Description : If one root of the equation ax^2 + x – 3 = 0 is –1, then what is the other root ? -Maths 9th

Last Answer : answer:

One root of x^2 + kx – 8 = 0 is the square of the other, then the value of k is : -Maths 9th

Description : One root of x^2 + kx – 8 = 0 is the square of the other, then the value of k is : -Maths 9th

Last Answer : answer:

If x^2 + mx + n = 0 and x^2 + px + q = 0 have a common root, then the common root is -Maths 9th

Description : If x^2 + mx + n = 0 and x^2 + px + q = 0 have a common root, then the common root is -Maths 9th

Last Answer : Let α be the common root ∴α2+pα+q=0 ...........(1) and α2+qα+p=0 ........ (2) Solving (1) & (2), we get, p2−q2α2=q−pα=q−p1∴α=q−pp2−q2 and α=1 ⇒q−pp2−q2=1 ⇒p2−q2=q−p (or) (p2−q2)+(p−q)=0 ⇒(p−q)[p+q+1]=0 ⇒p−q=0 or p+q+1=0

The length, breadth and height of a room are 5 m, 4 m and 3 m respectively. Find the cost of white washing the walls of the room and the ceiling at the rate of ₹7.50 m². -Maths 9th

Description : The length, breadth and height of a room are 5 m, 4 m and 3 m respectively. Find the cost of white washing the walls of the room and the ceiling at the rate of ₹7.50 m². -Maths 9th

Last Answer : Length of a room (l) = 5m Breadth (b) = 4 m and height (h) = 3 m ∴ Area of 4 walls = 2(l + b) x h = 2(5 + 4) x 3 = 6 x 9 = 54 m² and area of ceiling = l x b = 5 x ... ∴ Total area = 54 + 20 = 74 m2 Rate of white washing = 7.50 per m² ∴ Total cost = ₹74 x 7.50 = ₹555

A cylindrical pillar is 50 cm in diameter and 3.5 m in height. Find the cost of painting the curved surface of the pillar at the rate of ₹12.50 per m2 . -Maths 9th

Description : A cylindrical pillar is 50 cm in diameter and 3.5 m in height. Find the cost of painting the curved surface of the pillar at the rate of ₹12.50 per m2 . -Maths 9th

Last Answer : Diameter of the pillar = 50 cm ∴ Radius (r) = 502m = 25 m = 14m and height (h) = 3.5m Curved surface area of a pillar = 2πrh ∴ Curved surface area to be painted = 112m2 ∴ Cost of painting of 1 m2 pillar = Rs. 12.50 ∴ Cost of painting of 112 m2 pillar = Rs. ( 112 x 12.50 ) = Rs. 68.75.

Find the area of a quadrilateral ABCD in which AB = 3 cm, BC = 4 cm, CD = 4 cm, DA = 5 cm and AC = 5 cm. -Maths 9th

Description : Find the area of a quadrilateral ABCD in which AB = 3 cm, BC = 4 cm, CD = 4 cm, DA = 5 cm and AC = 5 cm. -Maths 9th

Last Answer : Given a quadrilateral ABCD with AB = 3 cm, BC = 4 cm, CD = 4 cm, DA = 5 cm and AC = 5 cm. For ∆ABC, a = AB = 3 cm, b = BC = 4 cm and c = AC = 5 cm Now, area of quadrilateral ABCD = area of ∆ABC + area of ∆ACD = 6 cm2 + 9.2 cm2 = 15.2 cm2 (approx.)

The capacity of a cuboidal tank is 50000 litres of water. Find the breadth of the tank, if its length and depth are respectively 2.5 m and 10 m -Maths 9th

Description : The capacity of a cuboidal tank is 50000 litres of water. Find the breadth of the tank, if its length and depth are respectively 2.5 m and 10 m -Maths 9th

Last Answer : Length (l) and depth (h) of tank is 2.5 m and 10 m respectively. To find: The value of breadth, say b. Formula to find the volume of a tank = l b h = (2.5 b 10) m3= 25b m3 Capacity ... of water (Given) Therefore, 25000 b = 50000 This implies, b = 2 Therefore, the breadth of the tank is 2 m.

Find the surface area of a sphere of radius: (i) 10.5cm (ii) 5.6cm (iii) 14cm -Maths 9th

Description : Find the surface area of a sphere of radius: (i) 10.5cm (ii) 5.6cm (iii) 14cm -Maths 9th

Last Answer : Formula: Surface area of sphere (SA) = 4πr2 (i) Radius of sphere, r = 10.5 cm SA = 4 (22/7) 10.52 = 1386 Surface area of sphere is 1386 cm2 (ii) Radius of sphere, r = 5.6cm Using formula, SA = 4 (22 ... 75 cm Surface area of sphere = 4πr2 = 4 (22/7) 1.752 = 38.5 Surface area of a sphere is 38.5 cm2

he frame has a base diameter of 20 cm and height of 30 cm. A margin of 2.5 cm is to be given for folding it over the top and bottom of the frame. -Maths 9th

Description : he frame has a base diameter of 20 cm and height of 30 cm. A margin of 2.5 cm is to be given for folding it over the top and bottom of the frame. -Maths 9th

Last Answer : Say h = height of the frame of lampshade, looks like cylindrical shape r = radius Total height is h = (2.5+30+2.5) cm = 35cm and r = (20/2) cm = 10cm Use curved surface area formula to find the ... 2πrh = (2 (22/7) 10 35) cm2 = 2200 cm2 Hence, 2200 cm2 cloth is required for covering the lampshade.

Diameter of the base of a cone is 10.5 cm and its slant height is 10 cm. Find its curved surface area -Maths 9th

Description : Diameter of the base of a cone is 10.5 cm and its slant height is 10 cm. Find its curved surface area -Maths 9th

Last Answer : Radius of the base of cone = diameter/ 2 = (10.5/2)cm = 5.25cm Slant height of cone, say l = 10 cm CSA of cone is = πrl = (22/7)×5.25×10 = 165

In a hot water heating system, there is cylindrical pipe of length 28 m and diameter 5 cm. Find the total radiating surface in the system. -Maths 9th

Description : In a hot water heating system, there is cylindrical pipe of length 28 m and diameter 5 cm. Find the total radiating surface in the system. -Maths 9th

Last Answer : Height of cylindrical pipe = Length of cylindrical pipe = 28m Radius of circular end of pipe = diameter/ 2 = 5/2 cm = 2.5cm = 0.025m Now, CSA of cylindrical pipe = 2πrh, where r = radius and h = height of ... = 2 (22/7) 0.025 28 m2 = 4.4m2 The area of the radiating surface of the system is 4.4m2.

The paint in a certain container is sufficient to paint an area equal to 9.375 m2. How many bricks of dimensions 22.5 cm×10 cm×7.5 cm can be painted out of this container? -Maths 9th

Description : The paint in a certain container is sufficient to paint an area equal to 9.375 m2. How many bricks of dimensions 22.5 cm×10 cm×7.5 cm can be painted out of this container? -Maths 9th

Last Answer : Total surface area of one brick = 2(lb +bh+lb) = [2(22.5 10+10 7.5+22.5 7.5)] cm2 = 2(225+75+168.75) cm2 = (2 468.75) cm2 = 937.5 cm2 Let n bricks can be painted out by the ... 93750 cm2 So, we have, 93750 = 937.5n n = 100 Therefore, 100 bricks can be painted out by the paint of the container.

The length, breadth and height of a room are 5 m, 4 m and 3 m respectively. Find the cost of white washing the walls of the room and ceiling at the rate of Rs 7.50 per m2. -Maths 9th

Description : The length, breadth and height of a room are 5 m, 4 m and 3 m respectively. Find the cost of white washing the walls of the room and ceiling at the rate of Rs 7.50 per m2. -Maths 9th

Last Answer : Length (l) of room = 5m Breadth (b) of room = 4m Height (h) of room = 3m It can be observed that four walls and the ceiling of the room are to be white washed. Total area to be white washed = Area of walls + ... m2 area = Rs.7.50 (Given) Cost of white washing 74 m2 area = Rs. (74 7.50) = Rs. 555

The sides of a triangle are in the ratio of 3 : 4 : 5 and its perimeter is 510 m. What is the measure of its greatest side? -Maths 9th

Description : The sides of a triangle are in the ratio of 3 : 4 : 5 and its perimeter is 510 m. What is the measure of its greatest side? -Maths 9th

Last Answer : Let the sides of triangle be 3x,4x,5x Perimeter =3x + 4x + 5x=144 cm 12x=144 ∴x=12 Then sides of triangle are 3x=3 12=36 cm, 4x=4 12=48 cm, 5x=5 12=60 cm. Now, Semi perimeter, s=2 Sum of sides of ... , Area of triangle =s (s−a)(s−b)(s−c) = 72(72−36)(72−48)(72−60) = 72 36 24 12 = 864 cm2

Find the area of a triangle whose sides are 4.5 cm and 10 cm and perimeter 10.5 cm. -Maths 9th

Description : Find the area of a triangle whose sides are 4.5 cm and 10 cm and perimeter 10.5 cm. -Maths 9th

Last Answer : Step-by-step explanation: ◾As we have given the two sides of triangle, let the three sides of triangle are (a) , (b), (c) . ◾And perimeter of given triangle is 10.5 cm ◾were, let us assume the sides are, ... . ◾So, the area of a triangle whose sides are 4.5 cm and 10 cm and perimeter 10.5 [Area ]=

5. In a parallelogram ABCD, E and F are the mid-points of sides AB and CD respectively (see Fig. 8.31). Show that the line segments AF and EC trisect the diagonal BD. -Maths 9th

Description : 5. In a parallelogram ABCD, E and F are the mid-points of sides AB and CD respectively (see Fig. 8.31). Show that the line segments AF and EC trisect the diagonal BD. -Maths 9th

Last Answer : . Solution: Given that, ABCD is a parallelogram. E and F are the mid-points of sides AB and CD respectively. To show, AF and EC trisect the diagonal BD. Proof, ABCD is a parallelogram , AB || CD also, ... (i), DP = PQ = BQ Hence, the line segments AF and EC trisect the diagonal BD. Hence Proved.

5. Show that if the diagonals of a quadrilateral are equal and bisect each other at right angles, then it is a square. -Maths 9th

Description : 5. Show that if the diagonals of a quadrilateral are equal and bisect each other at right angles, then it is a square. -Maths 9th

Last Answer : Solution: Given that, Let ABCD be a quadrilateral and its diagonals AC and BD bisect each other at right angle at O. To prove that, The Quadrilateral ABCD is a square. Proof, In ΔAOB and ΔCOD, AO = ... right angle. Thus, from (i), (ii) and (iii) given quadrilateral ABCD is a square. Hence Proved.

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...

Dhondoop's field has adjesent angles in the ratio 4:5. Find all the angles of his field. -Maths 9th

Description : Dhondoop's field has adjesent angles in the ratio 4:5. Find all the angles of his field. -Maths 9th

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

A chord of a circle of radius 7.5 cm with centre 0 is of length 9 cm. Find its distance from the centre. -Maths 9th

Description : A chord of a circle of radius 7.5 cm with centre 0 is of length 9 cm. Find its distance from the centre. -Maths 9th

Last Answer : ∵ PM = MQ = 1/2 = PQ = 45 cm and OP = 7.5 cm In right angled ΔOMP, using phthagoras theorem OM2 = OP2 - PM2 ⇒OM2 = 7.52 - 4.52 ⇒OM2 = 56.25 - 20.25 ⇒OM2 = 36 ∴ OM = √36 = 6 cm

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 metres of cloth 5 m wide will be required to make a conical tent, the radius of whose base is 7 m and whose height is 24 m? -Maths 9th

Description : How many metres of cloth 5 m wide will be required to make a conical tent, the radius of whose base is 7 m and whose height is 24 m? -Maths 9th

Last Answer : Given, radius (r) = 7 m and height (h) = 24m ∴ Slant height (l) = √h2 + r2 = √242 + 72 = √625 = 25 m ∴ Length of canvas required

The diameter of a garden roller is 1.4 m and it is 2 m long. How much area will it level in 5 revolutions? -Maths 9th

Description : The diameter of a garden roller is 1.4 m and it is 2 m long. How much area will it level in 5 revolutions? -Maths 9th

Last Answer : Radius of a garden roller (r) = 1.4 / 2 = 0.7 m and length of the garden roller (h) = 2 m ∴ Curved surface area of garden roller = 2πrh . So, the area levelled of a garden roller in one revolution is 8.8 m2. Now, area levelled of a garden in 5 revolutions = 8.8 x 5= 44m2

Find two irrational numbers between 2 and 2.5 . -Maths 9th

Description : Find two irrational numbers between 2 and 2.5 . -Maths 9th

Last Answer : The two irrational numbers between 2 and 2.5 are 2.101001000100001----- and 2.201 001 0001 00001-----

Find three irrational numbers between 2 and 2.5 . -Maths 9th

Description : Find three irrational numbers between 2 and 2.5 . -Maths 9th

Last Answer : If a and b are any two distinct positive rational numbers such that ab is not a perfect square , then the irrational number √ab lies between a and b. ∴ Irrational number between 2 and 2.5 is √ 2 2.5 , i.e √5 Irrational number ... 2.5 are √5 , 2(1/2) 5(1/4) and (1/2) 5 3/4 21/2 .

Find the remainder when y3 + y2 - 2y + 5 is divided by y - 5. -Maths 9th

Description : Find the remainder when y3 + y2 - 2y + 5 is divided by y - 5. -Maths 9th

Last Answer : Remainder = 145 Again, we should evaluate p(5) Let p(y) = y3 + y2 - 2y + 5 ∴ p(5) = 53 + 52 - 2 x 5 + 5 = 125 + 25 - 10 + 5 = 145 Thus , we find that p(5) is the remainder when p(y) is divided by y - 5 .

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

root( 5+2root6) -Maths 9th

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