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The mean weight of a class of 34 students is 46.5 kg. If the weight of the new boy is included, the mean is rises by 500 g. Find the weight of the new boy.

Posted by Anshika Choudhary 1 year, 8 months ago

CBSE > Class 09 > Mathematics

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If the mean of six observations y, y + 1, y + 4, y + 6, y + 8, y + 5 is 13, find the value of y.

Posted by Anshika Choudhary 1 year, 8 months ago

CBSE > Class 09 > Mathematics

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Harsh Singh Rajput Harsh 1 year, 8 months ago

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The daily minimum questions solved by a student during a week were as under: Monday Tuesday Wednesday Thursday Friday Saturday 35 30 27 32 23 28 Find the mean.

Posted by Anshika Choudhary 1 year, 8 months ago

CBSE > Class 09 > Mathematics

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The following is the list of number of coupons issued in a school canteen during a week: 105, 216, 322, 167, 273, 405 and 346. Find the average no. of coupons issued per day.

Posted by Anshika Choudhary 1 year, 8 months ago

CBSE > Class 09 > Mathematics

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The number of family members in 10 flats of society are 2, 4, 3, 3, 1,0,2,4,1,5. Find the mean number of family members per flat.

Posted by Anshika Choudhary 1 year, 8 months ago

CBSE > Class 09 > Mathematics

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The height of 20 students of class V are noted as follows 4, 4.5, 5, 5.5, 4, 4, 4.5, 5, 5.5, 4, 3.5, 3.5, 4.2, 4.6, 4.2, 4.7, 5.5, 5.3, 5, 5.5. Make a frequency distribution table for the above data. Which is the most common height and which is the rarest height among these students?

Posted by Anshika Choudhary 1 year, 8 months ago

CBSE > Class 09 > Mathematics

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Give three examples of data which you can get from your day-to-day life.

Posted by Anshika Choudhary 1 year, 8 months ago

CBSE > Class 09 > Mathematics

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A rectangular plot is given for constructing a house, having a measurement of 40 m long and 15 m in the front. According to the laws, a minimum of 3 m, wide space should be left in the front and back each and 2 m wide space on each of other sides. Find the largest area where a house can be constructed.

Posted by Anshika Choudhary 1 year, 8 months ago

CBSE > Class 09 > Mathematics

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How much paper of each shade is needed to make a kite given in the figure, in which ABCD is a square with diagonal 44 cm.

Posted by Anshika Choudhary 1 year, 8 months ago

CBSE > Class 09 > Mathematics

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The perimeter of an isosceles triangle is 32 cm. The ratio of the equal side to its base is 3: 2. Find the area of the triangle.

Posted by Anshika Choudhary 1 year, 8 months ago

CBSE > Class 09 > Mathematics

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Find the cost of laying grass in a triangular field of sides 50 m, 65 m and 65 m at the rate of Rs 7 per m2.

Posted by Anshika Choudhary 1 year, 8 months ago

CBSE > Class 09 > Mathematics

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A rhombus-shaped field has green grass for 18 cows to graze. If each side of the rhombus is 30 m and its longer diagonal is 48 m, how much area of grass field will each cow be getting?

Posted by Anshika Choudhary 1 year, 8 months ago

CBSE > Class 09 > Mathematics

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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.

Posted by Anshika Choudhary 1 year, 8 months ago

CBSE > Class 09 > Mathematics

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The sides of a triangle are in the ratio of 12: 17: 25 and its perimeter is 540cm. Find its area.

Posted by Anshika Choudhary 1 year, 8 months ago

CBSE > Class 09 > Mathematics

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Find the area of a triangle whose two sides are 18 cm and 10 cm and the perimeter is 42cm.

Posted by Anshika Choudhary 1 year, 8 months ago

CBSE > Class 09 > Mathematics

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Show that the quadrilateral, formed by joining the mid-points of the sides of a square is also a square.

Posted by Anshika Choudhary 1 year, 8 months ago

CBSE > Class 09 > Mathematics

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Find the area of an isosceles triangle each of whose equal sides measures 13 cm and whose base measures 20 cm. in class 9 in RS Agarwal book chap 14

Posted by Nitin Mishra 1 year, 8 months ago

CBSE > Class 09 > Mathematics

1 answers

Mansu . 1 year, 8 months ago

We have, Area of a triangle by heron's formula= √s(s-a)(s-b) (s-c) Now we know, s = perimeter of triangle/2 So, s=13+13+20/2=23 Let, put the value of s in formula =√23(23-13) (23-13) (23-20) =√23*10*10*3 =√23*2*5*2*5*3 Area = 10√69 cm square

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Find the class mark of the class 100 to 120

Posted by Sabnam Ara 1 year, 8 months ago

CBSE > Class 09 > Mathematics

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Anshika Choudhary 1 year, 8 months ago

We have given interval is 100 - 120. Here upper limit is 120 and lower limit is 100. Hence, answer is 110.

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a =√5+√2/√5-√2. and b =√5-√2/√5+√2 then find the value of a²+ab+b²/a²-ab+b²

Posted by Deshu Kaur 1 year, 8 months ago

CBSE > Class 09 > Mathematics

1 answers

Anshika Choudhary 1 year, 8 months ago

a = (√5 + 2)/(√5 – 2) b = (√5 – 2)/( √5 + 2) Concept: First rationalize the values of ‘a’ and ‘b’ and then proceed. Formula used: (a + b)2 = a2 + b2 + 2ab (a + b)(a – b) = a2 – b2 Calculation: ∵ a = (√5 + 2)/(√5 – 2) ⇒ a = [(√5 + 2)( (√5 + 2)]/[(√5 – 2)(√5 + 2)] ⇒ a = [(√5 + 2)2]/[(√5)2 – (2)2] ⇒ a = [5 + 4 + 4√5]/(5 – 4) ⇒ a = (9 + 4√5) Similarly; b = (√5 – 2)/( √5 + 2) ⇒ b = [(√5 – 2)(√5 - 2)]/[(√5 + 2)(√5 – 2)] ⇒ b = [(√5 – 2)2]/[(√5)2 – (2)2] ⇒ b = [5 + 4 – 4√5]/(5 – 4) ⇒ b = (9 – 4√5) Now, a2 + b2 + ab = (a + b)2 – ab ⇒ (9 + 4√5 + 9 – 4√5)2 – [(9 + 4√5)( 9 – 4√5)] ⇒ (18)2 – [(9)2 – (4√5)2] ⇒ 324 – (81 – 80) ⇒ 324 – 1 ⇒ 323 ∴ The value of the given expression is 323.

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Ifx+y=8and xy=15then x²+y²

Posted by Raghav Gurjar 1 year, 8 months ago

CBSE > Class 09 > Mathematics

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Anurag Singh 1 year, 8 months ago

x+y=8 xy=15, using : (x+y)^2 = x^2 + y^2 +2(x)(y) , (8)^2 = x^2 + y^2 + 2(15) , 64 = x^2 + y^2 + 30, 64 - 30 = x^2 + y^2 , 34 = x^2 + y^2

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Anshika Choudhary 1 year, 8 months ago

x+y=8 xy=15 using : (x+y)^2 = x^2 + y^2 +2(x)(y) (8)^2 = x^2 + y^2 + 2(15) 64 = x^2 + y^2 + 30 64 - 30 = x^2 + y^2 34 = x^2 + y^2

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Anurag Singh 1 year, 8 months ago

x+y=8 xy=15 using : (x+y)^2 = x^2 + y^2 +2(x)(y) (8)^2 = x^2 + y^2 + 2(15) 64 = x^2 + y^2 + 30 64 - 30 = x^2 + y^2 34 = x^2 + y^2

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O.9bar in p/q form

Posted by Aditya Sharma 1 year, 8 months ago

CBSE > Class 09 > Mathematics

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Abhijit Garai 1 year, 8 months ago

Let x = 0.999 10x = 9.999 10x - x =9.999 -0.999 9x = 9 x = 9/9 x = 1

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Simplify (√5+√2)`2

Posted by Akshat Sanghi 1 year, 8 months ago

CBSE > Class 09 > Mathematics

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Jasmeet Kaur 1 year, 8 months ago

( √ 5 + √ 2 ) 2 = ( √ 5 ) 2 + ( √ 2 ) 2 + 2 × √ 5 × √ 2 [ ∵ ( a + b ) 2 = a 2 + b 2 + 2 a b ] = 5 + 2 + 2 × √ 5 × √ 2 = 7 + 2 √ 10

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IF p(x)=x²-3x+2then findp(1)+p(-1)+p(2)

Posted by Sachin Yadav 1 year, 8 months ago

CBSE > Class 09 > Mathematics

3 answers

Binod Binod 1 year, 8 months ago

In jee 2024

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Binod Binod 1 year, 8 months ago

Have you cracked because I have received AIR -977

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Ashis Shome 1 year, 8 months ago

P(1)=0 P(-1)=6 P(2)=0 P1+p-1+p2=6

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1/√7+√3+√2

Posted by Shreya Singhal 1 year, 8 months ago

CBSE > Class 09 > Mathematics

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Abhijit Garai 1 year, 8 months ago

Don't post personal information, mobile numbers and other details.
Don't use this platform for chatting, social networking and making friends. This platform is meant only for asking subject specific and study related questions.
Be nice and polite and avoid rude and abusive language. Avoid inappropriate language and attention, vulgar terms and anything sexually suggestive. Avoid harassment and bullying.
Ask specific question which are clear and concise.

Remember the goal of this website is to share knowledge and learn from each other. Ask questions and help others by answering questions.

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The edges of a triangular board are 6cm 8cm 10cm.t he cost of painting it at the rate 9 paise per cm square is

Posted by Krishna Shrivastav 1 year, 8 months ago

CBSE > Class 09 > Mathematics

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Anshika Choudhary 1 year, 8 months ago

Hence, the cost of painting the triangular board at the rate of 9 paise per cm2 is Rs 2.16.

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Show that the quadrilateral formed by joining the mid point of the sides of a square is also a square

Posted by Preet Rajput 1 year, 8 months ago

CBSE > Class 09 > Mathematics

1 answers

Anshika Choudhary 1 year, 8 months ago

In a square A B C D , P , Q , R and S are the mid-points of A B , B C , C D and D A respectively. ⇒ A B = B C = C D = A D [ Sides of square are equal ] In △ A D C , S R ∥ A C and S R = 1 2 A C [ By mid-point theorem ] ---- ( 1 ) In △ A B C , P Q ∥ A C and P Q = 1 2 A C [ By mid-point theorem ] ---- ( 2 ) From equation ( 1 ) and ( 2 ), S R ∥ P Q and S R = P Q = 1 2 A C ---- ( 3 ) Similarly, S P ∥ B D and B D ∥ R Q ∴ S P ∥ R Q and S P = 1 2 B D and R Q = 1 2 B D ∴ S P = R Q = 1 2 B D Since, diagonals of a square bisect each other at right angle. ∴ A C = B D ⇒ S P = R Q = 1 2 A C ----- ( 4 ) From ( 3 ) and ( 4 ) S R = P Q = S P = R Q We know that the diagonals of a square bisect each other at right angles. ∠ E O F = 90 o . Now, R Q ∥ D B R E ∥ F O Also, S R ∥ A C ⇒ F R ∥ O E ∴ O E R F is a parallelogram. So, ∠ F R E = ∠ E O F = 90 o (Opposite angles are equal) Thus, P Q R S is a parallelogram with ∠ R = 90 o and S R = P Q = S P = R Q ∴ P Q R S is a square.

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ABCD is a rhombus such that <ABC=50 then find <ADB

Posted by Chetan Anand Pabboju 1 year, 8 months ago

CBSE > Class 09 > Mathematics