Ask questions which are clear, concise and easy to understand.
Ask QuestionPosted by Seema Yadav 5 years ago
- 4 answers
Paru ? 5 years ago
Posted by Aditya Singh 5 years ago
- 1 answers
Yogita Ingle 5 years ago
Steps of construction:
- Draw circle with centre O and radius OA = 5 cm, .
- Mark another point B on the circle such that ∠AOB = 120°, supplementary to the angle between the tangents. Since the angle between the tangents to be constructed is 60°.
∴ ∠AOB = 180° – 60° = 120°.
3.Construct angles of 90° at A and B and extend the lines so as to intersect at point P.
4. Thus, AP and BP are the required tangents to the circle.
Posted by Mudit Sharma 5 years ago
- 4 answers
Posted by Vishnupriya Ravi 5 years ago
- 4 answers
Posted by Gautam Sinha 5 years ago
- 2 answers
Pardhan Ji 5 years ago
Posted by Sukhveer Brar 5 years ago
- 5 answers
Posted by Syed Irfan 5 years ago
- 1 answers
Gaurav Seth 5 years ago
Answer:
x = 2 and y = 3
Step-by-step explanation:
Given :
To Find : solve using substitution method
Solution :
-----1
------2
Substitute the value of x from 1 in 2
So, x = 2 and y = 3
Posted by Anant Singh 5 years ago
- 2 answers
Gaurav Seth 5 years ago
Let AB and CD be the multi-storeyed building and the building respectively.
Let the height of the multi-storeyed building= h m and
the distance between the two buildings = x m.
AE = CD = 8 m [Given]
BE = AB – AE = (h – 8) m
and
AC = DE = x m [Given]
Also,
∠FBD = ∠BDE = 30° (Corresponding angles)
∠FBC = ∠BCA = 45° (Corresponding angles)
Now,
In Δ ACB,
In Δ BDE,
From (i) and (ii), we get,
Distance between the two building
King Adithya H M... 5 years ago
Posted by Shivani A Patil 5 years ago
- 1 answers
Shivani A Patil 5 years ago
Posted by Priyanshu Raj 5 years ago
- 1 answers
Posted by Vansh Patidar 5 years ago
- 2 answers
Vansh Patidar 5 years ago
Posted by Chinmay Kumar Sahoo 5 years ago
- 4 answers
Posted by Shivam Chaurasiya 5 years ago
- 2 answers
Yogita Ingle 5 years ago
15m+3n=6 ..... (i)
6m+3n=1..... (ii)
Subtract (i) and (ii)
15m + 3n = 6
- 6m + 3n=1
-------------------
9m = 5
m = 5/9
Put m = 5/9 in (ii), we get
6m + 3n = 1
6(5/9) + 3n = 1
10/3 + 3n = 1
3n = 1 - 10/3
3n = (3 - 10)/3
3n = -7/3
n = -7/9
Posted by Revanthkumar Mentada 5 years ago
- 0 answers
Posted by Manas Oriya 4 years, 7 months ago
- 0 answers
Posted by Ankush Gupta 5 years ago
- 5 answers
Posted by Ramanpreet Dhaliwal 5 years ago
- 1 answers
Yogita Ingle 5 years ago
Let AB be the vertical pole and CA be the rope. Then,
∠ACB=30o and AC=20 m
In right △ ABC,
sin30o = AB/AC
1/2 =AB/20
AB=10 m
Therefore, the height of the pole is 10 m.
Posted by Yash Motwani 5 years ago
- 1 answers
Gaurav Seth 5 years ago
Coordinate geometry deals with graphing (or plotting) and analyzing points, lines, and areas on the coordinate plane (coordinate graph). Each point on a number line is assigned a number. In the same way, each point in a plane is assigned a pair of numbers.
Posted by Shreya Chatterjee 5 years ago
- 3 answers
Posted by Sahid Alam 5 years ago
- 1 answers
Gaurav Seth 5 years ago
Statement: If a line is drawn parallel to one side of a triangle intersecting the other two sides, then it divides the two sides in the same ratio.
Given: A triangle ABC in which DE || BC, and intersects AB in D and AC in E.
Posted by Pratik Kumbhakar 5 years ago
- 2 answers
Bro . 5 years ago
Posted by Shivani A Patil 5 years ago
- 1 answers
Gaurav Seth 5 years ago
We have given that
Sum of the first n terms = 4n – n2
⇒ Sn = 4n – n2
Put, n = 1, then
S1 = 4(1) – (1)2 = 4 – 1 = 3
⇒ d1 = 3
Hence, the first term is
Put n = 2, then
S2 = 4(2) – (2)2
= 8 – 4 = 4
Hence, the sum of two terms is 4.
Now, second term = S2 – S1 = 4 – 3 = 1
Put n = 3, then
S3 = 4(3) – (3)2
= 12 – 9 = 3
Third term = S3 – S2 = 3 – 4 = –1
Put n = 9, 10
S9= 4(9) – (9)2
= 36 – 81 = –45
S10= 4(10) – (10)2
= 40 –100 = –60
∴ Tenth term = S10 – S9
= – 60 – (–45)
= – 60 + 45 = –15
Now, Sn –1 = 4(n – 1) – (n – 1)2
= (4n – 4) – (n2 + 1 – 2n)
= 4n – 4 – n2 – 1 + 2n
= 6n – n2 – 5
∴ nth term = Sn – Sn – 1
= (4n – n2) – (6n – n2 – 5)
= 4n – n2 – 6n + n2 + 5
= 5 –2n
Posted by Shivani A Patil 5 years ago
- 0 answers
Posted by Sahil Khan 5 years ago
- 1 answers
Posted by Pranjali Tiwari 5 years ago
- 5 answers
Posted by Abhi Sarkar 5 years ago
- 3 answers
Sumesh ☺️☺️☺️ 5 years ago
Posted by Harman Kaur 5 years ago
- 1 answers
Posted by Deepanshu Sablaniya 5 years ago
- 1 answers
Posted by Sajil Rahman 5 years ago
- 0 answers
Posted by Aniket Jariwala 5 years ago
- 1 answers
Gaurav Seth 5 years ago
Let a be an arbitrary positive integer. Then, by Euclid’s division algorithm, corresponding to the positive integers ‘a’ and 6, there exist non-negative integers and r such that a = 6 q + r, where, 0 < r < 6
Hence, the cube of a positive integer of the form 6q + r,q is an integer and r = 0,1,2, 3, 4,5 is also of the forms 6m, 6m + 1, 6m + 2, 6m + 3,6m + 4 and 6m + 5 i.e.,6m + r.
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
Supriya Mishra 5 years ago
0Thank You