Ask questions, doubts, problems and we will help you..
Ask questions which are clear, concise and easy to understand.
Ask Question
CBSE, JEE, NEET, CUET
Question Bank, Mock Tests, Exam Papers
NCERT Solutions, Sample Papers, Notes, Videos
Posted by Vandana Bari 5 years ago (9769016)
- 2 answers
Disha S 5 years ago (4392449)
Posted by Tejas Pawar 5 years ago (10403486)
- 1 answers
Posted by Anushri Rathi 5 years ago (8983274)
- 2 answers
Vivek Agrawal Gaming 4 years, 11 months ago (10345644)
Posted by C Akhilesh 5 years ago (9172127)
- 1 answers
Yash Bansal 5 years ago (7321736)
Posted by C Akhilesh 5 years ago (9172127)
- 2 answers
Yash Bansal 5 years ago (7321736)
Posted by C Akhilesh 5 years ago (9172127)
- 1 answers
Posted by C Akhilesh 4 years, 7 months ago (9172127)
- 1 answers
Sia ? 4 years, 7 months ago (6945213)
Now, the point is on x-axis, so the value 'y' of the given point will be zero.
Let, the value of the 'x' value of the given point = x
[Assume,x as a variable to do the further mathematical calculations.]
So,the point is = (x,0)
As mentioned in the question,the two points are equidistant from (x,0).
So,
Distance between (x,0) and (2,-5)
= ✓(x-2)²+(0+5)²
Distance between (x,0) and (-2,9).
=✓(x+2)²+(0-9)²
Now,the points are equidistant.
So,
✓(x-2)²+(0+5)² = ✓(x+2)²+(0-9)²
(x-2)²+(0+5)² = (x+2)²+(0-9)²
x²-4x+4+25 = x²+4x+4+81
x²-4x-x²-4x = 4+81-4-25
-8x = 56
x = -7
Posted by C Akhilesh 4 years, 7 months ago (9172127)
- 1 answers
Sia ? 4 years, 7 months ago (6945213)
The traffic light at three different road crossing change after every 48 seconds,72 seconds and 108 seconds, respectively.
So let us take the LCM of the given time that is 48 seconds, 72 seconds, 108 seconds
⇒ 48 = 2 × 2 × 2 × 2 × 3
⇒ 72 = 2 × 2 × 2 × 3 × 3
⇒ 108 = 2 × 2 × 3 × 3 × 3
Hence, LCM of 48, 72 and 108 = (2 × 2 × 2 × 2 × 3 × 3 × 3)
LCM of 48, 72 and 108 = 432
So after 432 seconds, they will change simultaneously
We know that
60 seconds = 1 minute
so on dividing 432 / 60, we get 7 as quotient and 12 as a reminder
Hence, 432 seconds = 7 min 12 seconds
∴ The time = 7 a.m. + 7 minutes 12 seconds
Hence, the lights change simultaneously at 7:07:12 a.m
Posted by C Akhilesh 4 years, 7 months ago (9172127)
- 1 answers
Sia ? 4 years, 7 months ago (6945213)
Let O be the mid point,
As = Ap ( tangent to the given circle)
and since L SAp is 90°
As = OP = Ap = 10cm ( radius)
Now, Let t be the point where Ap meets Cq
and we know that Cq = PR = 27 (tangent)
and Ct = 38 ( given in the fig)
» Cq + qt = 38
» 27cm + qt = 38cm
» qt = 38cm-27cm = 11cm
and now tp = tq ( tangent)
tp = 11cm
So to find x , i.e Apt
we have Apt = Ap + pt
= 10cm + 11cm = 21 cm
Posted by C Akhilesh 5 years ago (9172127)
- 2 answers
Bhoomika Verma 5 years ago (9985395)
Yash Bansal 5 years ago (7321736)
Posted by C Akhilesh 4 years, 7 months ago (9172127)
- 1 answers
Sia ? 4 years, 7 months ago (6945213)
ABC is an equilateral triangle , where D point on side BC in such a way that BD = BC/3 . Let E is the point on side BC in such a way that AE⊥BC .
Now, ∆ABE and ∆AEC
∠AEB = ∠ACE = 90°
AE is common side of both triangles ,
AB = AC [ all sides of equilateral triangle are equal ]
From R - H - S congruence rule ,
∆ABE ≡ ∆ACE
∴ BE = EC = BC/2
Now, from Pythagoras theorem,
∆ADE is right angle triangle ∴ AD² = AE² + DE² ------(1)
∆ABE is also a right angle triangle ∴ AB² = BE² + AE² ------(2)
From equation (1) and (2)
AB² - AD² = BE² - DE²
= (BC/2)² - (BE - BD)²
= BC²/4 - {(BC/2) - (BC/3)}²
= BC²/4 - (BC/6)²
= BC²/4 - BC²/36 = 8BC²/36 = 2BC²/9
∵AB = BC = CA
So, AB² = AD² + 2AB²/9
9AB² - 2AB² = 9AD²
Hence, 9AD² = 7AB²
Posted by C Akhilesh 4 years, 7 months ago (9172127)
- 1 answers
Sia ? 4 years, 7 months ago (6945213)
Given cosA/(1+sinA) +(1+sinA)/cosA
On taking the LCM we get,
={cos²A +(1+sinA)²}/cosA.(1+sinA)
Combining the like terms and adding and also we know that,use sin²A+cos²A =1
=(1+1+2sinA)/cosA(1+sinA)
=2(1+sinA)/cosA(1+sinA)
=2/cosA
=2secA
Posted by C Akhilesh 5 years ago (9172127)
- 2 answers
Nargish Akhtar 5 years ago (10097107)
Himansh Makkar 5 years ago (9402550)
Posted by C Akhilesh 5 years ago (9172127)
- 1 answers
Saniya Khanam 5 years ago (9914292)
Posted by Arin Ror Arin Ror 5 years ago (9416270)
- 3 answers
Saksham Singh 5 years ago (10423735)
Posted by C Akhilesh 4 years, 7 months ago (9172127)
- 1 answers
Posted by Parth Lambate 5 years ago (9451239)
- 1 answers
𝓼𝓷𝓮𝓱𝓪 𝓢𝓾𝓶𝓪𝓷 5 years ago (4547992)
Posted by Dhruv Kumar 5 years ago (10199979)
- 1 answers
Sia ? 4 years, 9 months ago (6945213)
Hidden goodwill. When the value of goodwill is not given in the question, it has to be calculated on the basis of total capital/net worth of the firm and profit sharing ratio. X and Y are partners with capitals of ₹ 10,000 each. They admit Z as a partner for 1/4th share in the profits of the firm.
Posted by Nainsee Garg 5 years ago (10409610)
- 2 answers
Posted by Disha Wagh 5 years ago (10152318)
- 0 answers
Posted by Shobhna Mavani 5 years ago (8911320)
- 0 answers
Posted by Naniya Hanam 5 years ago (10409706)
- 1 answers
Posted by Account Deleted 4 years, 7 months ago (9425752)
- 1 answers
Sia ? 4 years, 7 months ago (6945213)
A 91,
DEEPMALA SOCIETY,
JAIPUR, RAHASTHAN.
X-MONTH-20XX
TO
THE EDITOR
JAIPUR.
SUBJECT = APPRECIATING INDIAN GOVERNMENT FOR BANNING OF FIRE CRACKERS.
SIR, MAM,
THROUGH THE ESTEEM COLUMN OF YOUR NEWS PAPER I HAVE READ AN ARTICLE ABOUT BANNING OF FIRE CRACKERS. I FELT HAPPY AT THAT TIME AS I WAS THINKING THAT OUS SOCIETY IS GOING TO EXEMPTED FROM THIS POPULATED ENVIRONMENT AND HAZARDUOUS NOICE OF THE CHRACKERS.
I ALSO AGREE WITH THAT BUT THE POLLUTION IS NOT CONTROLLED YET.
JUST LIKE DELHI AND RAJASTHAN WE HAVE TO CHECK ALL PART OF THE COUNTRY AS WELL.NOT ONLY THE FIRE CRACKERS POLLUTE THE ENVIRONMENT BUT THE CARS AND AUTOMOBILE VEHICLES, FACTORIES, ETC ALSO. SO, WE SHOLD HAVE TO CONTROLL THIS BY USING PUBLIC TRANSPORT, CONSERVATION OF NATURAL RECOURCES ETC.
" LIFE IS ONE AND WE SHOUDNT LET THE POLLUTION (WHICH IS OBVIOUSLY SPREADED BY US) RUIN THIS "
I WILL WAIT FOR YOUR PROMPT REPLY
THANKYOU
YOUR SINCERELY,
ROHIT
Posted by Sakshi Kumar 5 years ago (10409700)
- 4 answers
Posted by Pihu ✨ 4 years, 7 months ago (10059176)
- 1 answers
Sia ? 4 years, 7 months ago (6945213)
Let ABCD be a quadrilateral circumscribing a circle with centre O.
Now join AO, BO, CO, DO.
From the figure, ∠DAO=∠BAO [Since, AB and AD are tangents]
Let ∠DAO=∠BAO=1
Also ∠ABO=∠CBO [Since, BA and BC are tangents]
Let ∠ABO=∠CBO=2
Similarly we take the same way for vertices C and D
Sum of the angles at the centre is 360o
Recall that sum of the angles in quadrilateral, ABCD = 360o
= 2(1+2+3+4)=360o
=1+2+3+4=180o
In ΔAOB, ∠BOA=180−(1+2)
In ΔCOD, ∠COD=180−(3+4)
∠BOA+∠COD=360−(1+2+3+4)
=360o–180o
=180o
Since AB and CD subtend supplementary angles at O.
Thus, opposite sides of a quadrilateral circumscribing a circle subtend supplementary angles at the centre of the circle.
Posted by Aditya Kumar 5 years ago (10384108)
- 1 answers
𝓼𝓷𝓮𝓱𝓪 𝓢𝓾𝓶𝓪𝓷 5 years ago (4547992)
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
Gargi Sarkar 5 years ago (8475856)
0 Thank You