Ask questions which are clear, concise and easy to understand.
Ask QuestionPosted by Shreya Chatterjee 4 years, 9 months ago
- 1 answers
Posted by Kamlesh Prajapati 4 years, 9 months ago
- 3 answers
Posted by Ayush Biswas 4 years, 9 months ago
- 0 answers
Posted by Nalshatra Singh Rajawat 4 years, 9 months ago
- 2 answers
Tanu Man 4 years, 9 months ago
Swayam Gupta 4 years, 9 months ago
Posted by Himanshu Sharma 4 years, 9 months ago
- 5 answers
Posted by Deepali Gupta 4 years, 9 months ago
- 2 answers
Posted by Aman Kumar 4 years, 9 months ago
- 4 answers
Posted by Ranveer Singh Sidhu 4 years, 9 months ago
- 5 answers
Swayam Gupta 4 years, 9 months ago
Posted by Divyanshu Kishor 4 years, 9 months ago
- 2 answers
Posted by Adarsh M 4 years, 9 months ago
- 2 answers
Posted by Gopika Satheesh 4 years, 9 months ago
- 2 answers
Yogita Ingle 4 years, 9 months ago
(i) Draw a circle of radius 3cm.
(ii) Draw a straight line of 8cm from the centre and give it a name say OA.
(iii) Now, draw a perpendicular bisector of the line OA.
(iv) Place the compass to the midpoint of a line, adjust its length to reach the endpoint and draw the arcs to the circle.
(v) Draw the lines which passes through the arc and endpoints of the bisector.
AR and AT are the required tangents.
Posted by Lisa Anto 4 years, 9 months ago
- 0 answers
Posted by Ansh Tyagi 4 years, 9 months ago
- 2 answers
Yogita Ingle 4 years, 9 months ago
Length of cube =a
a3=64
⇒a=4
Length of cuboid = 8=l
breadth of cuboid = 4=b
height of cuboid = 4=h
Surface area =2(lb+bh+lh)
=2((8×4)+(4×4)+(8×4))
=2(32+16+32)
=160 cm2
Gaurav Seth 4 years, 9 months ago
Given,
Volume of each cube = 64cm^3
length of the cuboid =4+4= 8cm
Breadth of the cuboid =4 cm
Height of the cuboid = 4 cm
T.S.A of cuboid =2(lb+bh+hl)
=2(8×4+4×4+4×8)
=2(32+16+32)
=2(80)
Posted by Kushala V 4 years, 9 months ago
- 2 answers
Gaurav Seth 4 years, 9 months ago
The two equations are
3x + 2y = 5 …… (1)
2x − 3y = 7 …… (2)
Here,
Therefore, they will intersect each other at a unique point and thus, there will be a unique solution for these equations. That is, there is only one solution.
Posted by Priyanshu Raj 4 years, 9 months ago
- 2 answers
Gaurav Seth 4 years, 9 months ago
Let the number of terms required to make the sum of 636 be n and common difference be d.
Given Arithmetic Progression : 9 , 17 , 25 ....
First term = a = 9
Second term = a + d = 17
Common difference = d = a + d - a = 17 - 9 = 8
From the indentities of arithmetic progressions, we know : -
, where
is the sum of first term to nth term of the AP, a is the first term, d is the common difference and n is the number of terms of AP.
In the given Question, sum of APs is 636.
Therefore,
= > 4n² + 5n - 636 = 0
= > 4n² + ( 53 - 48 )n - 636 = 0
= > 4n² + 53n - 48n - 636 = 0
= > 4n² - 48n + 53n - 636 = 0
= > 4n( n - 12 ) + 53( n - 12 ) = 0
= > ( n - 12 )( 4n + 53 ) = 0
By Zero Product Rule,
= > n - 12 = 0
= > n = 12
Hence,
Number of terms of the AP [ 9 , 17 , 25 ] which are required to make the sum of 636 is 12.
Posted by Yash Vani 4 years, 9 months ago
- 3 answers
Priyanshu Raj 4 years, 9 months ago
Gaurav Seth 4 years, 9 months ago
HRD Minister Ramesh Nishank announced a major CBSE syllabus reduction for the new academic year 2020-21 on July 7 which was soon followed by an official notification by CBSE on the same.
Considering the loss of classroom teaching time due to the Covid-19 pandemic and lockdown, CBSE reduced the syllabus of classes 9 to 12 with the help of suggestions from NCERT.
The CBSE syllabus has been rationalized keeping intact the learning outcomes so that the core concepts of students can be retained.
Deleted syllabus of CBSE Class 10 Mathematics
Posted by Gupta Binita 4 years, 9 months ago
- 2 answers
Yogita Ingle 4 years, 9 months ago
It is in AP with common difference d=52−1=24 and a=1,
Next three terms are
a+(5−1)d=97,
a+(6−1)d=121,
a+(7−1)d=145
Posted by Monalisha Pradhani 4 years, 9 months ago
- 1 answers
Gaurav Seth 4 years, 9 months ago
Let speed of the boat in still water = x km/hr, and
Speed of the current = y km/hr
Downstream speed = (x+y) km/hr
Upstream speed = (x - y) km/hr
... (1)
... (2)
Putting 
the equations become:
24u + 16v = 6
Or, 12u + 8v = 3... (3)
36u + 12v = 6
Or, 6u + 2v = 1... (4)
Multiplying (4) by 4, we get,
24u + 8v = 4… (5)
Subtracting (3) by (5), we get,
12u = 1
u = 
Putting the value of u in (4), we get, v = 
Thus, speed of the boat upstream = 4 km/hr
Speed of the boat downstream = 12 km/hr
Posted by Dharmender Kumar 4 years, 9 months ago
- 0 answers
Posted by Akshit Kumar 4 years, 9 months ago
- 0 answers
Posted by Sakshi Tiwari 4 years, 9 months ago
- 3 answers
Priyanshu Raj 4 years, 9 months ago
Gaurav Seth 4 years, 9 months ago
HRD Minister Ramesh Nishank announced a major CBSE syllabus reduction for the new academic year 2020-21 on July 7 which was soon followed by an official notification by CBSE on the same.
Considering the loss of classroom teaching time due to the Covid-19 pandemic and lockdown, CBSE reduced the syllabus of classes 9 to 12 with the help of suggestions from NCERT.
The CBSE syllabus has been rationalized keeping intact the learning outcomes so that the core concepts of students can be retained.
Deleted syllabus of CBSE Class 10 Mathematics
D
Posted by Gopika Satheesh 4 years, 9 months ago
- 1 answers
Gaurav Seth 4 years, 9 months ago
Height of first Airplane = AB = 4000 m
Height of another plane = BC
The angles of elevation of two planes from the same point on the ground are 60° and 45°. i.e.∠ADB = 60° and ∠CDB = 45°
Let AC be h
CB = AB - AC = 4000-h
In ΔABD
In ΔCBD
Hence the vertical distance between the aeroplane at that instant is 1690.599 m
Posted by Sindhu Nallapaneni 4 years, 9 months ago
- 2 answers
Posted by Riya Choudhary 4 years, 9 months ago
- 4 answers
ಹರ್ಷನಂದ ಹರ್ಷನಂದ 4 years, 9 months ago
Gaurav Seth 4 years, 9 months ago
Unit-wise wieghtage to be followed in the CBSE Class 10 Maths Exam 2021 is as follows:
|
Unit |
Marks |
|
I. NUMBER SYSTEMS |
06 |
|
II. ALGEBRA |
20 |
|
III. COORDINATE GEOMETRY |
06 |
|
IV. GEOMETRY |
15 |
|
V. TRIGONOMETRY |
12 |
|
VI. MENSURATION |
10 |
|
VII. STATISTICS & PROBABILITY |
11 |
|
Total |
80 |
Posted by Aarchi Shah 4 years, 9 months ago
- 1 answers
Gaurav Seth 4 years, 9 months ago
Find the area of the shaded region in Fig.12.48, where arc(APD, AQB, BRC and CSD) are semlcircles of diameter 14cm, 3.5cm,7cm and 3.5cm respectively. (Use π=22/7)
Solution is..........
Posted by Nandini Ghatkar 4 years, 9 months ago
- 2 answers
Posted by Rishika Goenka 4 years, 9 months ago
- 1 answers
Gaurav Seth 4 years, 9 months ago
Given:
The roots of the quadratic equation :
p(q-r)x²+q(r-p)x+r(p-q)=0 are equal.
To prove :
Solution:
Compare given Quadratic equation with ax²+bx+c=0, we get
a = p(q-r), b = q(r-p), c = r(p-q)
Discreminant (D) = 0
/* roots are equal given */
=> b²-4ac=0
=>[q(r-p)]²-4×p(q-r)×r(p-q)=0
=>(qr-pq)²-4pr(q-r)(p-q)=0
=> (qr)²+(pq)²-4(qr)(pq)-4pr(pq-q²-pr+qr)=0
=> (qr)²+(pq)²-4pq²r-4p²qr+4prq²+4p²r²-4pqr²=0
=> (qr)²+(pq)²+(-2pr)²+2pq²r-4p²qr-4pqr²=0
=> (qr)²+(pq)²+(-2pr)²+2(qr)(pq)+2(pq)(-2pr)+2(-2pr)(qr)=0
/* we know the algebraic identity*/
/*a²+b²+c²+2ab+2bc+2ca=(a+b+c)² */
=> (qr+pq-2pr)² = 0
=> qr+pq-2pr = 0
Divide each term by pqr , we get
Therefore,
Posted by Mohd Sufiyan 4 years, 9 months ago
- 1 answers
Yogita Ingle 4 years, 9 months ago
Given :∆ABC ~ ∆PQR
RTP:
Construction:
Draw AM perpendicular to BC and PN perpendicular to QR.
Proof:
∆ABM~∆PQN ( By AA similarity )
Also ∆ABC ~ ∆PQR (given)
\* From (1),(2) and (3)
Now, by using (3) , we get
myCBSEguide
Trusted by 1 Crore+ Students
Test Generator
Create papers online. It's FREE.
CUET Mock Tests
75,000+ questions to practice only on myCBSEguide app
Gaurav Seth 4 years, 9 months ago
The given equations are:
(k+1)x+3ky+15=0, and
5x+ky+5=0
Since, the given equations are coincident, therefore
{tex}\frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}}=\frac{c_{1}}{c_{2}} \frac{k+1}{5}=\frac{3k}{k}=\frac{15}{5}{/tex}
Taking the first two terms, we have
k=14
Or
The given equations are:
Since, the given equations are coincident, therefore
Taking the first two terms, we have
2Thank You