CBSE > Class 09 > Mathematics | CBSE Board | Homework Help

$\begin{aligned} & \Longrightarrow \frac{2 \sqrt{6}}{\sqrt{2}+\sqrt{3}} \times \frac{\sqrt{2}-\sqrt{3}}{\sqrt{2}-\sqrt{3}}+\frac{6 \sqrt{2}}{\sqrt{6}+\sqrt{3}} \times \frac{\sqrt{6}-\sqrt{3}}{\sqrt{6}-\sqrt{3}}-\frac{8 \sqrt{3}}{\sqrt{6}+\sqrt{2}} \times \frac{\sqrt{6}-\sqrt{2}}{\sqrt{6}-\sqrt{2}} \\ & \Longrightarrow \frac{4 \sqrt{3}-6 \sqrt{2}}{2-3}+\frac{12 \sqrt{3}-6 \sqrt{6}}{6-3}-\frac{24 \sqrt{2}-8 \sqrt{6}}{6-2} \\ & \Longrightarrow-4 \sqrt{3}+6 \sqrt{2}+\frac{12 \sqrt{3}-6 \sqrt{6}}{3}-\frac{24 \sqrt{2}-8 \sqrt{6}}{4} \\ & \Longrightarrow-4 \sqrt{3}+6 \sqrt{2}+4 \sqrt{3}-2 \sqrt{6}-6 \sqrt{2}+2 \sqrt{6} \\ & \Longrightarrow-4 \sqrt{3}+6 \sqrt{2}+4 \sqrt{3}-2 \sqrt{6}-6 \sqrt{2}+2 \sqrt{6} \\ & \Longrightarrow 0 \end{aligned}$

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Ajay lives in delhi.

Posted by Ujjwal Thakur 2 years, 3 months ago

CBSE > Class 09 > Mathematics

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In figure de is parallel to qr and ap and bpare bisectors of angle eab and rba respectively find angoe apb

Posted by Ayisha A 2 years, 3 months ago

CBSE > Class 09 > Mathematics

1 answers

Aryan Chauhan 2 years, 3 months ago

Wojciech is duo is duo unlike to ke stop Robert Unicoi d do controversial

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the cost of turfing a triangular feild at the rate of ₹5 per sq. meter is ₹6720. if the side of the field area in the ratio 4:7:9 find question of herons formula

Posted by Ayush Kumar 2 years, 3 months ago

CBSE > Class 09 > Mathematics

1 answers

Aryan Chauhan 2 years, 3 months ago

Let 4:7:9 be x Perimet

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2+2-2√2%2

Posted by Ashutosh Yadav 2 years, 3 months ago

CBSE > Class 09 > Mathematics

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Preeti Dabral 2 years, 3 months ago

3.17157287525

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SAS congruence

Posted by S. Kolvalai 2 years, 3 months ago

CBSE > Class 09 > Mathematics

2 answers

Preeti Dabral 2 years, 3 months ago

SAS congruence criterion: If two sides and included an angle of one triangle are equal to the corresponding two sides and included angle of another triangle then the two triangles are said to be congruent.

Consider $\triangle$ ABC, $\triangle$ PQR
here AB = PQ, AC = PR and $\angle$ A = $\angle$ P
Hence $\triangle$ ABC $\cong$ $\triangle$ PQR by SAS congruence criterion.

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Manju Rani 2 years, 3 months ago

But how we know

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If x=(√2+1)/(√2-1) and y=(√2-1)/(√2+1) then find the value of x2+y2+2xy

Posted by Yogesh Shekhadar 2 years, 3 months ago

CBSE > Class 09 > Mathematics

1 answers

Dikshita Chowdhury 2 years, 3 months ago

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8*1/3

Posted by Sumedh Wani 2 years, 3 months ago

CBSE > Class 09 > Mathematics

1 answers

Suju Apee 2 years, 3 months ago

8.33..

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An exterior angle of triangle is 120 degree one of the interior angle is 40 degree find two other of a triangle opposite

Posted by Himesh Patel 2 years, 3 months ago

CBSE > Class 09 > Mathematics

2 answers

Suju Apee 2 years, 3 months ago

80 and 60 degree

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Sumedh Wani 2 years, 3 months ago

Base angle 80° and tip angle 60°

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Classroom classroom 7 m long 6.5 m wide and 4 m high it has 123 m into 1.4 m and the three windows each measure into metre into 1 m the interior walls are to be colour washed the contractor charges rs 15 per square metre find the cost of colour and washing

Posted by Harini Murugan Harini 2 years, 3 months ago

CBSE > Class 09 > Mathematics

1 answers

Preeti Dabral 2 years, 3 months ago

For class room : l = 7m, b = 6.5 m, h = 4 m
∴ Area of walls of the room = 2(l + b)h = 2(7 + 6.5)4 = 108 m²
Area of door = 3 × 1.4 = 4.2 m²
Area of one window = 2 × 1 = 2 m²
∴ Area of 3 windows = 3 × 2 = 6 m²
∴ Area of the walls of the room to be colour washed = 108 – (4.2 + 6)
= 108 – 10.2 = 97.8 m²
∴ Cost of colour washing the classroom at Rs. 15 per square metre = Rs 97.8 × 15 = Rs. 1467.

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Construct a triangle abc in which ab =5.8cm bc+ca=8.4cm and angle b =45 degree

Posted by Sakshi Jaiswal 2 years, 3 months ago

CBSE > Class 09 > Mathematics

1 answers

Preeti Dabral 2 years, 3 months ago

Constructing triangle ABC in which AB = 5.8 cm , BC + CA = 8.4 cm and ∠B = 45°.

Step 1: Draw a line segment AB 5.8 cm.

Step 2 : Draw ∠B = 45°.

Step 3 : With Center B and radius 8.4 cm, make an arc which intersects BX at D.

Step 4 : Join D to A.

Step 5 : Draw a perpendicular bisector of segment DA it intersect the line segment BD at point C.

Step 6 : Join C to A.

ABC is the required triangle.

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2xxxx-2xxxxx

Posted by Archita Singh 2 years, 3 months ago

CBSE > Class 09 > Mathematics

2 answers

Raunak Yadav 2 years, 3 months ago

Hi archita

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Vijay Reddy 2 years, 3 months ago

Exercise 15.1 6,7,8,11,12,13

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Calculate the mass of 3.011×10^23 no. Of an atoms

Posted by Rishika Minj 2 years, 4 months ago

CBSE > Class 09 > Mathematics

1 answers

Ansh Upadhayay 2 years, 4 months ago

Ch 7

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405396

Posted by Amit Chaudhary 2 years, 4 months ago

CBSE > Class 09 > Mathematics

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If a chord subtend an angle of 98° at the centre of the circle , find out the measure of the angle subtended by the chord at the major arc of the circle

Posted by Aditi Gogoi 2 years, 4 months ago

CBSE > Class 09 > Mathematics

1 answers

Sheetal Gupta 2 years, 4 months ago

😅

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Find three rational and irrational number between -2 and -3

Posted by Aashika Bi 2 years, 3 months ago

CBSE > Class 09 > Mathematics

1 answers

Preeti Dabral 2 years, 3 months ago

Since we know that rational number between two rational numbers is obtained by $\frac{1}{2}$ [a+b]
A rational number lying between -2 and -3 is $\frac{1}{2}$ [(-2) + (-3)],
i.e., $-\frac{5}{2}$
Now, a rational number lying between -2 and $-\frac{5}{2}$ is
$\left.\frac{1}{2}[(-2)+\left(-\frac{5}{2}\right)\right ]$
= $\frac{1}{2} \times\left(-\frac{9}{2}\right)$
= $-\frac{9}{4}$
And, a rational number lying between $-\frac{5}{2}$ and -3 is
$\frac{1}{2}\left(\left(-\frac{5}{2}\right)+(-3)\right)$
$\frac{1}{2} \times\left(-\frac{11}{2}\right)$
$-\frac{11}{4}$
Thus, we have $-2>-\frac{9}{4}>-\frac{5}{2}>-\frac{11}{4}>3$
Hence, three rational numbers between -2 and -3 are $-\frac{9}{4},-\frac{5}{2}$ and $-\frac{11}{4}$

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The external diameter of an iron pipe is 25 cm and it's length is 20cm. If the thickness of the pipe is 1cm find the total surface area of the pipe.

Posted by Kashish Khan 2 years, 4 months ago

CBSE > Class 09 > Mathematics

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Two friends jahanvi and Leeza are making investment in a business. The business is related to cars. The investment made by jahanvi is given by polynomial 1: x^2-3.

Posted by Bhavnoor Kaur 1 year, 5 months ago

CBSE > Class 09 > Mathematics

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(-2-root3)(-2+root3)

Posted by Mukund Vishwakarma 2 years, 4 months ago

CBSE > Class 09 > Mathematics

1 answers

Kunal Barhate 2 years, 4 months ago

(-2-√3) (-2+√3) (-2) ^2 - (√3) ^2 4-3 1

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Draw different pairs of circle how many points does each pair have in common what is the maximum number of common points

Posted by Bonty Clock Gaming 2 years, 3 months ago

CBSE > Class 09 > Mathematics

1 answers

Preeti Dabral 2 years, 3 months ago

From the figure, we observe that when different pairs of circles are drawn, each pair have two points (say A and B) in common.
Maximum number of common points are two in number.

Suppose two circles C (O) and C (O’) intersect each other in three points, say A, B and C.
Then A, B and C are non-collinear points.
We know that:
There is one and only one circle passing through three non-collinear points.
Therefore, a unique circle passes through A, B and C.
$\Rightarrow$ O’ coincides with O and s = r where s and r are the radii of two circles C(O) and C(O')
A contradiction to the fact that C (O’,) $\ne$ C (O,)
$\therefore$ Our supposition is wrong.
Hence two different circles cannot intersect each other at more than two points.

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X-cordinate are called

Posted by Ashita Tamang 2 years, 4 months ago

CBSE > Class 09 > Mathematics

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Vansh Katiyar 2 years, 4 months ago

Abccica

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Rohit Meena 2 years, 4 months ago

Abscessena

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A right circular just encloses a sphere .find I) surface area of the sphere II) curved surface area of the cylinder III) ratio of the areas obtained in (I) and (II)

Posted by Yash Kaur 2 years, 3 months ago

CBSE > Class 09 > Mathematics

1 answers

Preeti Dabral 2 years, 3 months ago

Surface area of the sphere = $4\pi r^2$
For cylinder
Radius of the base = r
Height = 2r
$\therefore$ Curved surface area of the cylinder = $2\pi rh=2\pi (r)(2r)=4\pi r^2$
Ratio of the areas obtained in (i) and (ii)
$={{surface\ area\ of\ the\ sphere}\over{curved\ surface\ area\ of\ the\ cylinder}}$
$={{4\pi r^2}\over{4\pi r^2}}={1\over1}=1:1$