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

The solution of the equation x - 2y = 4 is (4, 0) whereas (0, 2), (2, 0), (√2, 4√2) and (1, 1) are not solutions of this equation since for these values of x and y LHS ≠ RHS.

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Posted by Shyam Patel 1 year ago

CBSE > Class 09 > Mathematics

5 answers

Albhin Janish 1 year, 2 months ago

fast reply

5Thank You

Albhin Janish 1 year, 2 months ago

where is the answer

4Thank You

Zeus Yt 1 year, 7 months ago

227279

30Thank You

Krishna Modi 1 year, 9 months ago

where is the answer

70Thank You

Subham Mallick 1 year, 11 months ago

say

83Thank You

ANSWER

What is the correct meaning of corresponding angles?

Posted by Thrupthi Lalitha 2 years, 11 months ago

CBSE > Class 09 > Mathematics

1 answers

Raj Dhwaj Raj 2 years, 11 months ago

The correct ing mean of corresponding angle are if a transversal cuts two parralel line.

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ANSWER

Factorise 24x² - 41x + 12

Posted by Shaurya Garg 2 years, 11 months ago

CBSE > Class 09 > Mathematics

1 answers

Sia ? 2 years, 11 months ago

0Thank You

ANSWER

The area of the triangle formed by the points P(0,1),Q(0,5) and R(3,4) is

Posted by Saanjh Ambatkar 2 years, 11 months ago

CBSE > Class 09 > Mathematics

1 answers

Nivetha S 2 years, 11 months ago

Step-by-step explanation: Given 3 vertices of triangle that are P(0,1), Q(0,5)and R(3,4). We have to find out the area formed by 3 vertices of triangle. Here the triangle which is formed has base 4 units and height 3 units as shown in attachment. Area of triangle ABC is \frac{1}{2}\times base \times height21×base×height = \frac{1}{2}\times 4 \times 321×4×3 = 6 sq units

1Thank You

ANSWER

In a figure if angle x = angle y and ab = cb prove that ae = cd

Posted by Kamiya Shah 2 years, 11 months ago

CBSE > Class 09 > Mathematics

1 answers

Sia ? 2 years, 11 months ago

Consider the triangles AEB and CDB
{tex}\angle{/tex}EBA = {tex}\angle{/tex}DBC ... (Common angle) ... (i)
Further, we have:
{tex}\angle{/tex}BEA = 180 - y
{tex}\angle{/tex}BDC = 180 - x
Since x = y, we have:
180 - x = 180 - y
{tex}\angle{/tex}BEA = {tex}\angle{/tex}BDC ... (ii)
AB = CB ... (Given) ... (iii)
From (i), (ii) and (iii), we have:
{tex}\triangle{/tex}BDC{tex}\cong{/tex}{tex}\triangle{/tex}BEA ... (AAS criterion)
{tex}\therefore{/tex} AE = CD (CPCT)
Hence, proved.

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ANSWER

Square of 2

Posted by Akshat Singh 2 years, 11 months ago

CBSE > Class 09 > Mathematics

5 answers

Rishi Gour 2 years, 11 months ago

2Thank You

Lomashkumar Sahu 2 years, 11 months ago

2Thank You

Prajwal Sankanal 2 years, 11 months ago

1Thank You

Anshu Kumar 2 years, 11 months ago

Square of 2 is 4

3Thank You

Shaurya Kushwaha 2 years, 11 months ago

Square of 2 is nothing.

2Thank You

ANSWER

number system project file

Posted by Chhote Lal 2 years, 11 months ago

CBSE > Class 09 > Mathematics

0 answers

ANSWER

If ?A= 50° and ?B=60°, then the value of ?Y is

Posted by Akriti Kaundal 9A 2 years, 11 months ago

CBSE > Class 09 > Mathematics

5 answers

Mànyà Jain 2 years, 11 months ago

0Thank You

. Anisha 2 years, 11 months ago

70° 60°+50°= 180 +(angle sum property) 110°=180° =180°110° =70° Ans.

0Thank You

Manjeet Rajput 2 years, 11 months ago

70°

0Thank You

Kavita Singh 2 years, 11 months ago

70°

1Thank You

Aakash Rathor 2 years, 11 months ago

Samajik gyan

3Thank You

ANSWER

P (1,0) Q(4,0), S(1,3) find R in plotting

Posted by Vaibhav Dewangam 2 years, 11 months ago

CBSE > Class 09 > Mathematics

1 answers

Sia ? 2 years, 11 months ago

Coordinate of Point R is ( 4 , 3 ).

Step-by-step explanation:

Given Coordinates of the Points are

P ( 1 , 0 ) , Q( 4 , 0 ) and S( 1 , 3 )

We need to find: Coordinate of the point R such that PQRS is a Square.

First we mark all the given points on the Graph.

Then see where the Point R lies satisfying that PQRS is a Square.

1st Graph Attached has all 3 given point marked.

Clearly from graph 1. Coordinate of Point are must be on line x= 4 and y = 3

So, Coordinate of point R is intersection of lines x = 4 and y = 3

that is (4 , 3).

Therefore, Coordinate of Point R is ( 4 , 3 ).

0Thank You

ANSWER

5^4+5^9(7√21) -9^2/2√87

Posted by Shraddha Choudhary 2 years, 11 months ago

CBSE > Class 09 > Mathematics

1 answers

Shraddha Choudhary 2 years, 11 months ago

1972796.01

1Thank You

ANSWER

4√12× 7√6

Posted by Anak Yadav 2 years, 11 months ago

CBSE > Class 09 > Mathematics

5 answers

Anshika Singla 2 years, 11 months ago

4√12X 7√6 28√72 28√2X2 x3X3X2 28X2X3√2= 168√2

0Thank You

Siddharth Mohta 2 years, 11 months ago

32 underoot 70

0Thank You

Payal Rajput 2 years, 11 months ago

12√14

0Thank You

Shraddha Choudhary 2 years, 11 months ago

28√72 28*6√2 168√2

0Thank You

Anshu Kumar 2 years, 11 months ago

168√ 2

2Thank You

ANSWER

The Curved surface area of a right circular cylinder of height 14cm is 88cm². Find the diameter of the base of the cylinder

Posted by Adrash Yadav 2 years, 11 months ago

CBSE > Class 09 > Mathematics

4 answers

Sanket ... 2 years, 11 months ago

2cm

0Thank You

Aarti Sharma 2 years, 11 months ago

2cm

0Thank You

Shraddha Choudhary 2 years, 11 months ago

CSA= 2πrh 88=2×22/7×r×14 R=88/2×22×2 R=88/88 R=1cm Diameter=2×r =2×1 =2 cm

0Thank You

Adrash Yadav 2 years, 11 months ago

2cm

2Thank You

ANSWER

Root 2 into root 15

Posted by Kavin Kavin 2 years, 11 months ago

CBSE > Class 09 > Mathematics

5 answers

Rutika Tilekar 2 years, 11 months ago

Root 30

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Arpit Gautam 2 years, 11 months ago

√30

0Thank You

Shraddha Choudhary 2 years, 11 months ago

√30

0Thank You

Liku Moni 2 years, 11 months ago

√2*√15 =√30

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Pragya J 2 years, 11 months ago

Root 30

0Thank You

ANSWER

3/√8 +1/√2

Posted by 𝒩𝒾𝓇𝒶𝓃𝒿𝒶𝓃𝒶 𝓱𝓪𝓻𝓲 2 years, 11 months ago

CBSE > Class 09 > Mathematics

2 answers

Anshu Kumar 2 years, 11 months ago

2√2

1Thank You

Anshu Kumar 2 years, 11 months ago

2√3

0Thank You

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AB is a line segment . AX and BY are two equal line segment drawn on the opposite sides of AB such that AX|| BY. If AB and YX intersect each other at P, prove that (1) ∆APX =~∆BPY (2) AB and XY bisect each other .

Posted by M.R. Logistics 2 years, 11 months ago

CBSE > Class 09 > Mathematics

0 answers

ANSWER

4400

Posted by Shreya Sharma 2 years, 11 months ago

CBSE > Class 09 > Mathematics

0 answers

ANSWER

comparing volume of cylinder and cone worksheet answers class 9

Posted by Taniya T 2 years, 11 months ago

CBSE > Class 09 > Mathematics

1 answers

Sia ? 2 years, 11 months ago

The volume of a cylinder is: π × r² × h
The volume of a cone is: 1 3 π × r² × h

0Thank You

ANSWER

A+b whole square

Posted by Anuj Vishwakarma 2 years, 11 months ago

CBSE > Class 09 > Mathematics

5 answers

Shraddha Choudhary 2 years, 11 months ago

a^2+b^2+2ab

0Thank You

Anshu Kumar 2 years, 11 months ago

a^2+b^2+2ab

0Thank You

Kanishq Baheti 2 years, 11 months ago

a² +b² +2ab

0Thank You

Ajij Khan 2 years, 11 months ago

a2 +b2+2ab

2Thank You

Radhika Tayal 2 years, 11 months ago

a^2+2ab+b^2

7Thank You

ANSWER

In a parallelogram PQRS if angle p=3x-5angleQ=2X+50 find x

Posted by Aditya Sharma 2 years, 11 months ago

CBSE > Class 09 > Mathematics

2 answers

Sia ? 2 years, 11 months ago

Similar as (Q=2x+15):

{tex}\angle P+\angle Q=180^{\circ}{/tex} (Angles on the same side of a transversal are supplementary)
{tex}\Rightarrow \ 3 x-5+2 x+15=180^{\circ}{/tex}
5x + 10 = 180^o
{tex}\Rightarrow{/tex} 5x + 170^o {tex}\Rightarrow{/tex} x = 34^o

4Thank You

Priyanka Kumari 2 years, 11 months ago

3x-5+2x+50=180 (angle on the same side of a transversal are supplementary) 5x+45=180 5x=135 X= 27

1Thank You

ANSWER

How many angle

Posted by Nikhil Gurjar 2 years, 11 months ago

CBSE > Class 09 > Mathematics