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Kaarunya Sai 5 years, 2 months ago

y=mx, m is the gradiant

ANSWER

√56

Posted by Saranya B 5 years, 2 months ago

CBSE > Class 09 > Mathematics

1 answers

Arohi Raj 5 years, 2 months ago

Hiii

ANSWER

Find the five rational numbers between 3/5 and 4/5

Posted by Teju Kn 5 years, 2 months ago

CBSE > Class 09 > Mathematics

2 answers

Shubham Singh 5 years, 2 months ago

3/5 and 4/5 3/5×6/6=18/30 4/5×6/6=24/30 = 19/30,20/30,21/30,22/30,23/30,24/30 Thank you very much for these pretty question

Yogita Ingle 5 years, 2 months ago

Since we want five numbers, we write 3/5 and 4/5 So multiply in numerator and denominator by 5+1 =6 we get

ANSWER

Jcuvyxxyvk dkgydmviycjlvitxhvtxojv

Posted by Anuj Yadav 5 years, 2 months ago

CBSE > Class 09 > Mathematics

0 answers

ANSWER

Jcuvyxxyvk dkgydmviycjlvitxhvtxojv

Posted by Anuj Yadav 5 years, 2 months ago

CBSE > Class 09 > Mathematics

1 answers

Shridhar Kumbar 5 years, 1 month ago

अरे सज्जन तुम्हें क्या तकलीफ़ है भाय ।

ANSWER

If base of parallelogram of height 10cm is 3 cm. find its area

Posted by Avneesh Kumar 5 years, 2 months ago

CBSE > Class 09 > Mathematics

1 answers

Yogita Ingle 5 years, 2 months ago

base = 10 cm and height = 3 cm

Area of parallelogram = base × height sq. units

= 10 × 3

= 30 sq. cm

ANSWER

Prove that: 2n+2n+1/2n+1-2n=3/2

Posted by Vanshika Dik 5 years, 2 months ago

CBSE > Class 09 > Mathematics

1 answers

Indhu Indhu 5 years, 2 months ago

2n+2n−1 =2n.21−2n2n+2n.2−1 [a3ac=ab+c] ⇒2n[2−1]2n[1+1/2] =1+21 =23 Hence proved

ANSWER

Construct a triangle ABC in which BC=6cm angle B=60cm and of other two sides is 9cm

Posted by Ishita Budhwar 5 years, 2 months ago

CBSE > Class 09 > Mathematics

2 answers

Vishal Jat 5 years, 2 months ago

Wrong question

Yogita Ingle 5 years, 2 months ago

Step's of construction

(1) Draw BC = 6 cm.

(2) Construct ∠CBX = 60°.

(3) Along BX , set off BP = 9 cm.

(4) Join CP

(5) Draw the perpendicular bisector of CP tp intersect BP at A .

(6) Join AC. Then, ∆ABC is the required triangle.

ANSWER

Proof of vertically opposite angles

Posted by Asha Rani 5 years, 2 months ago

CBSE > Class 09 > Mathematics

1 answers

Gaurav Seth 5 years, 2 months ago

From the figure, we know that,

AB and CD intersect each other at point O.

Let the two pairs of vertically opposite angles be,

1^st pair – ∠AOC and ∠BOD

2^nd pair – ∠AOD and ∠BOC

To prove:

Vertically opposite angles are equal,

i.e., ∠AOC = ∠BOD, and ∠AOD = ∠BOC

From the figure,

The ray AO stands on the line CD.

We know that,

If a ray lies on a line then the sum of the adjacent angles is equal to 180°.

⇒ ∠AOC + ∠AOD = 180° (By linear pair axiom) … (i)

Similarly, the ray DO lies on the line AOB.

⇒ ∠AOD + ∠BOD = 180° (By linear pair axiom) … (ii)

From equations (i) and (ii),

We have,

∠AOC + ∠AOD = ∠AOD + ∠BOD

⇒ ∠AOC = ∠BOD – – – – (iii)

Similarly, the ray BO lies on the line COD.

⇒ ∠DOB + ∠COB = 180° (By linear pair axiom) – – – – (iv)

Also, the ray CO lies on the line AOB.

⇒ ∠COB + ∠AOC = 180° (By linear pair axiom) – – – – (v)

From equations (iv) and (v),

We have,

∠DOB + ∠COB = ∠COB + ∠AOC

⇒ ∠DOB = ∠AOC – – – – (vi)

Thus, from equation (iii) and equation (vi),

We have,

∠AOC = ∠BOD, and ∠DOB = ∠AOC

Therefore, we get, vertically opposite angles are equal.

Hence Proved.

ANSWER

A match box measures 10 cm x 6.5cm x 2.5cm. There are 5 dozen match boxes packed in a carton. Calculate the volume of the carton in m3.

Posted by Rakshit Kumar 5 years, 2 months ago

CBSE > Class 09 > Mathematics

2 answers

Kripa Daga 5 years, 2 months ago

Hope you get your answer

Kripa Daga 5 years, 2 months ago

0.00975m³

3Thank You

ANSWER

If each side of a triangle is increased by 5 times then it's area will increase by

Posted by Aarya Gupta 5 years, 2 months ago

CBSE > Class 09 > Mathematics

2 answers

Kripa Daga 5 years, 2 months ago

Sara Tonger 5 years, 2 months ago

5 times.

ANSWER

(2x + 1/3)^2 - (x - 1/2)^2

Posted by Diya Jogi 5 years, 2 months ago

CBSE > Class 09 > Mathematics

3 answers

Kripa Daga 5 years, 2 months ago

(3x-1/6)(x+5/6)

Sreevidhya Muralidharan 5 years, 2 months ago

(a^2-b^2)= (a+b)(a-b) =(2x+1/3+x-1/2)(2x+1/3-x+1/2) = (3x+1/6)(x+5/6)

Diya Jogi 5 years, 2 months ago

3x^2 + 8x + 4

4Thank You

ANSWER

Represent 9.3 on number line...

Posted by P R 5 years, 2 months ago

CBSE > Class 09 > Mathematics

2 answers

Sreevidhya Muralidharan 5 years, 2 months ago

ANSWER Steps: 1) Draw a line segment AB of length 9.3 units. 2) Extend the line by 1 unit more such that BC=1 unit . 3) Find the midpoint of AC. 4) Draw a line BD perpendicular to AB and let it intersect the semicircle at point D. 5) Draw an arc DE such that BE=BD. Therefore, BE= Root 9.3 units

Gaurav Seth 5 years, 2 months ago

Draw a line AB by measuring 9.3cm
From the point, B add 1cm and mark it as C
Mark the point of bisection by a compass and say it as ‘O’
Measure AO which is the radius and draws a semi-circle.
From B draw a perpendicular AB touching the semi-circle and mark as D
Draw an arc on the number line by taking compass pointer on B and pencil on D
The point which intersects the number line is the square root of 9.3

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ANSWER

Plz solve it

Posted by Jasnoor Kaur Sandha 5 years, 2 months ago

CBSE > Class 09 > Mathematics

2 answers

Suraj Yadav 5 years, 2 months ago

Solve kya krna hai

Sreevidhya Muralidharan 5 years, 2 months ago

What we need to solve?

ANSWER

Ch 10 ex 10.4 q2

Posted by Gargi Rajput 5 years, 2 months ago

CBSE > Class 09 > Mathematics

2 answers

Gaurav Seth 5 years, 2 months ago

If two equal chords of a circle intersect within the circle, prove that the segments of one chord are equal to corresponding segments of the other chord.

Solution:

Let AB and CD be two equal cords (i.e. AB = CD). In the above question, it is given that AB and CD intersect at a point, say, E.

It is now to be proven that the line segments AE = DE and CE = BE

Construction Steps:

Step 1: From the center of the circle, draw a perpendicular to AB i.e. OM ⊥ AB

Step 2: Similarly, draw ON ⊥ CD.

Step 3: Join OE.

Now, the diagram is as follows-

Proof:

From the diagram, it is seen that OM bisects AB and so, OM ⊥ AB

Similarly, ON bisects CD and so, ON ⊥ CD

It is known that AB = CD. So,

AM = ND — (i)

and MB = CN — (ii)

Now, triangles ΔOME and ΔONE are similar by RHS congruency since

OME = ONE (They are perpendiculars)

OE = OE (It is the common side)

OM = ON (AB and CD are equal and so, they are equidistant from the centre)

∴ ΔOME ΔONE

ME = EN (by CPCT) — (iii)

Now, from equations (i) and (ii) we get,

AM+ME = ND+EN

So, AE = ED

Now from equations (ii) and (iii) we get,

MB-ME = CN-EN

So, EB = CE (Hence proved).

Vikram Singh 5 years, 2 months ago

Chapter 6.2

ANSWER

Sardi ka answer se kya relation h sir ek to aapki Antoinette aur bhav kha rhe ho

Posted by Rajnish Tiwari 5 years, 2 months ago

CBSE > Class 09 > Mathematics

1 answers

Kripa Daga 5 years, 2 months ago

??????

ANSWER

Mean of ten observations is 56 each observations increased by 3what is the new mean

Posted by Puram Kowshika Gayathri 5 years, 2 months ago

CBSE > Class 09 > Mathematics

3 answers

Kripa Daga 5 years, 2 months ago

Joseph Debbarma 5 years, 2 months ago

What thing were

Rajnish Tiwari 5 years, 2 months ago

ANSWER

If the perimeter of an equilateral triangle is 180 find its area

Posted by Payal Kumari 5 years, 2 months ago

CBSE > Class 09 > Mathematics

5 answers

K. K. 5 years, 2 months ago

Given: Perimeter =180 cm. Equilateral triangle has all sides equal, therefore let the sides be x. Perimeter= 3x 180=3x 60=x By formula for finding area Root 3/4*(60)^2 =1558.8 cm^2

Veneet Rana 5 years, 2 months ago

Use this is approximately 1558.85 cm2

Veneet Rana 5 years, 2 months ago

Area of an equilateral traingle = √3/4 × ( side)2

Veneet Rana 5 years, 2 months ago

Sorry

Veneet Rana 5 years, 2 months ago

Abhe kucha aur puch le

ANSWER

Prove that a diagonal of a parallelogram divides it into congruent triangle theorm

Posted by Nitin Kumar Jain 5 years, 2 months ago

CBSE > Class 09 > Mathematics

1 answers

Gaurav Seth 5 years, 2 months ago

Given: A parallelogram ABCD and AC is its diagonal .

To prove : △ABC ≅ △CDA

Proof : In △ABC and △CDA, we have

∠DAC = ∠BCA [alt. int. angles, since AD | | BC]

AC = AC [common side]

and ∠BAC = ∠DAC [alt. int. angles, since AB | | DC]

∴ By ASA congruence axiom, we have

△ABC ≅ △CDA

ANSWER

In quadrilateral ABCD, AC=AD and AB bisects< A, show that angle ABD

Posted by Subhojit Mukherjee 5 years, 2 months ago

CBSE > Class 09 > Mathematics

1 answers

Sreevidhya Muralidharan 5 years, 2 months ago

In △ABC and △ABD, AC=AD (Given) ∠CAB=∠DAB (AB bisects ∠A) AB=AB (Common) ∴△ABC≅△ABD (By SAS congruence rule) ∴BC=BD (By CPCT) ∴, BC and BD are of equal lengths

ANSWER

Polynomial identities and formula

Posted by Suraj Yadav 5 years, 2 months ago

CBSE > Class 09 > Mathematics

4 answers

Suraj Yadav 5 years, 2 months ago

In this app, no one can solve my doubt so , I have to delete this app If anyone have solution of my doubt , so tell fast

Suraj Yadav 5 years, 2 months ago

many people see ?my question but anyone can not get answer

Suraj Yadav 5 years, 2 months ago

Is anyone here for help me

Suraj Yadav 5 years, 2 months ago

How to solve easily all questions and statements Is anyone tell

ANSWER

Simplify:- (3^1/3)^4

Posted by Aniket Kumar 5 years, 2 months ago

CBSE > Class 09 > Mathematics

4 answers

Aniket Kumar 5 years, 2 months ago

Chandni khan I can't understand this process of answer

Aniket Kumar 5 years, 2 months ago

Rajnish tiwari can you explain your answer u can't understand properly your answer

Rajnish Tiwari 5 years, 2 months ago

Answer is::1

Chandni Khan 5 years, 2 months ago

0.037037037 (3^1/3)^4 (3/3)^4 1^4 = 0.037037037

ANSWER

in figure angle OAB=30degree

Posted by Arul S 5 years, 2 months ago

CBSE > Class 09 > Mathematics

1 answers

Sreevidhya Muralidharan 5 years, 2 months ago

Question does not provide complete information

ANSWER

Visualise 3.765 on the number line , upto 4 decimal places.

Posted by P R 5 years, 2 months ago

CBSE > Class 09 > Mathematics

1 answers

Sreevidhya Muralidharan 5 years, 2 months ago

3.765 can be visualised as in the following steps. First, we draw a number line and mark points on it after that we will divide the number line between points 3 and 4. And then we will divide the points between 3.7 and 3.8 as the number is between both of them. And then divide the points points between 3.76 and 3.77.

ANSWER

Classify the triangles by the length of it's sides

Posted by Poorni Gowda 5 years, 2 months ago

CBSE > Class 09 > Mathematics

1 answers

Gaurav Seth 5 years, 2 months ago

Types of Triangles Based on Sides

According to the lengths of their sides, triangles can be classified into three types which are:

Scalene
Isosceles
Equilateral

If all three sides have the same length then it is an EQUILATERAL triangle, if only two sides have the same length then it is an ISOSCELES triangle and if there are no sides that have the same length then it is a SCALENE triangle.

ANSWER

how to find poot of 27 represented irrational natural number on a number line

Posted by Vijayalaxmi Malagitti 5 years, 2 months ago

CBSE > Class 09 > Mathematics

1 answers

Aakhya Verma 5 years, 2 months ago

Step-by-step explanation: 1) Draw a line of 27 cm. 2) Extend it to 1 cm. 3) Find the mid point of this line i.e, 28÷2=14cm. 4) Then draw an arc. through one end point to the other. 5) Now from the extended line draw a perpendicular. 6) Then drop the compaas to the number line. 7)Finally mark it as root 27 cm.

ANSWER

Chapter 7 ma ex 7.3 ma q1

Posted by Divyam Gupta 5 years, 2 months ago

CBSE > Class 09 > Mathematics

1 answers

Aakhya Verma 5 years, 2 months ago

In the above question, it is given that ΔABC and ΔDBC are two isosceles triangles. (i) ΔABD and ΔACD are similar by SSS congruency because: AD = AD (It is the common arm) AB = AC (Since ΔABC is isosceles) BD = CD (Since ΔDBC is isosceles) ∴ ΔABD ΔACD. (ii) ΔABP and ΔACP are similar as: AP = AP (It is the common side) PAB = PAC (by CPCT since ΔABD ΔACD) AB = AC (Since ΔABC is isosceles) So, ΔABP ΔACP by SAS congruency condition. (iii) PAB = PAC by CPCT as ΔABD ΔACD. AP bisects A. — (i) Also, ΔBPD and ΔCPD are similar by SSS congruency as PD = PD (It is the common side) BD = CD (Since ΔDBC is isosceles.) BP = CP (by CPCT as ΔABP ΔACP) So, ΔBPD ΔCPD. Thus, BDP = CDP by CPCT. — (ii) Now by comparing (i) and (ii) it can be said that AP bisects A as well as D. (iv) BPD = CPD (by CPCT as ΔBPD ΔCPD) and BP = CP — (i) also,BPD +CPD = 180° (Since BC is a straight line.) ⇒ 2BPD = 180° ⇒ BPD = 90° —(ii) Now, from equations (i) and (ii), it can be said that AP is the perpendicular bisector of BC.