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Ask QuestionPosted by Prince Garg 6 years, 11 months ago
- 4 answers
Nitish Kumar 6 years, 11 months ago
Posted by Sadhana Gaur 6 years, 11 months ago
- 2 answers
Poonam Bais 6 years, 11 months ago
Posted by Aarti Kushwah 6 years, 11 months ago
- 1 answers
Ram Kushwah 6 years, 11 months ago
(1+cosA/sinA-1/sinA)(1+sinA/cosA+2/cosA)
=1/sinAcosA(sinA+cosA-1)(sinA+cosA+1)
=1/sinAcosA{(sinA+cosA)2-1)}
=1/sinAcosA(sin2A+cos2 A+2sinAcosA-1)
=2sinAcosA/sinAcosA=2 proved
Posted by Niku Jatt 6 years, 11 months ago
- 4 answers
Abhishek Yadav 6 years, 11 months ago
Posted by Raj Biswas 6 years, 11 months ago
- 2 answers
Posted by Aswany Suresh 6 years, 11 months ago
- 1 answers
Posted by Appu Appu 6 years, 11 months ago
- 2 answers
Ram Kushwah 6 years, 11 months ago
x2 +ax-b=0
a+b=-a so b=-2a
And ab=-b so a=-b/b=-1
and b=-2a=2
Posted by Mohit Macpherson 6 years, 11 months ago
- 3 answers
Ram Kushwah 6 years, 11 months ago
Sorry there was mistaken
Sum even numbers from 1to 500 are
S2=1/4×500×502=62750......(1)
The multiples of 5 not included are
5,15,25,.....495
If n are terms then 495=5+(n-1)×10
490=10n-10. n=50
So S5=50/2×(5+495)=25×500=12500
Total sum S2+S5=
=62750+12500=75250
Ram Kushwah 6 years, 11 months ago
Multiples of 2 are
2,4,6....500
S2 =250/2×(2+500)=125×251=31375
Multiples of not included are
5,15,25,35......495
If n are the numbers then
495=5+(n-2)×10
490+20=5n, n=510/2=255
S5 =255/2×(5+595)=600×255/5=76500
Total sum =32375+76500=108875
Posted by Tridev Kumar 6 years, 11 months ago
- 4 answers
Ram Kushwah 6 years, 11 months ago
Here r=p h=s
Triyak unchayee
{tex}l=\sqrt (p^2+s^2){/tex}
Raunak Pandey ?? 6 years, 11 months ago
Posted by Music_ Lover?? 6 years, 11 months ago
- 5 answers
Posted by Sanjay Harsha 6 years, 11 months ago
- 2 answers
Yogita Ingle 6 years, 11 months ago
Here, a = 7, d = 3.5 and last term = 84
Number of terms can be calculated as follows;
{tex}a_n = a+(n−1)d{/tex}
Or, {tex}84 =7+(n−1)3.5{/tex}
Or, {tex}(n−1)3.5 = 84−7{/tex}
Or, {tex}n−1 = 77 /3.5 = 22{/tex}
Or, {tex}n = 23{/tex}
Posted by Naresh Pandit 6 years, 11 months ago
- 0 answers
Posted by Shivani Mishra 6 years, 11 months ago
- 2 answers
Posted by Nidhi Kumari 6 years, 11 months ago
- 1 answers
Posted by Shalini Agrawal 6 years, 11 months ago
- 0 answers
Posted by Ankit Thakur 6 years, 11 months ago
- 0 answers
Posted by Abhimanyu Yadav 6 years, 11 months ago
- 3 answers
Prashant Chaudhary 6 years, 11 months ago
Posted by Manish Yadav 6 years, 11 months ago
- 3 answers
Ram Kushwah 6 years, 11 months ago
Let one no. Is xthe other is 16-x
And 1/x + 1/(16-x)=1/3
3(16-x)+3x=x(16-x)
48-3x+3x=16x-x2
x2 -16x+48=0
(x-12)(x-4)=0
X=12,4
So no. are 12 and 4
Dev Singh Gehlot 6 years, 11 months ago
Posted by Sabir Ali 6 years, 11 months ago
- 1 answers
Posted by ????? ❤️ 6 years, 11 months ago
- 3 answers
Susmita Mandal 6 years, 11 months ago
????? ❤️ 6 years, 11 months ago
Posted by Aj Raghav 6 years, 11 months ago
- 3 answers
Aj Raghav 6 years, 11 months ago
Posted by Annu Thakur 6 years, 11 months ago
- 1 answers
Shruti Singh 6 years, 11 months ago
Posted by Faizan Nabi 6 years, 11 months ago
- 2 answers
Gaurav Seth 6 years, 11 months ago
7th term = (a+6d)
11th term = (a+10d)
(a+6d)7=(A+10d)11 (given)
7a + 42d =11a +110d (on solving)
7a - 11a = 110d - 42d
-4a = -68d
a= -17d
we know 18th term = (a+17d)........................... (i)
now we substitute the value of a in (i)
which emplies,
-17d + 17d
= 0
hence proved
Posted by Shreyas Jha Sj 6 years, 11 months ago
- 1 answers
Yaar 6 years, 11 months ago
Posted by Dev Singh Gehlot 6 years, 11 months ago
- 0 answers
Posted by Misha Singh 6 years, 11 months ago
- 4 answers
Divya Garg 6 years, 11 months ago
Prisha Rathore 6 years, 11 months ago
Posted by Keerthana Keerthana 6 years, 11 months ago
- 7 answers
Divya Garg 6 years, 11 months ago
@ Aashu 6 years, 11 months ago
Yaar 6 years, 11 months ago
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Prince Garg 6 years, 11 months ago
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