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Ask QuestionPosted by Videos With Iqra 6 years, 2 months ago
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Posted by Din Mohd 6 years, 2 months ago
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Lucky Kanger 6 years, 2 months ago
Abhishek Kumar 6 years, 2 months ago
Posted by Anil Kumar 6 years, 2 months ago
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Posted by Anany Mittal 6 years, 2 months ago
- 10 answers
Yogita Ingle 6 years, 2 months ago
Given, sin43°cos47° + cos43°sin47°
⇒ sin43°cos(90 - 43)° + cos43°sin(90 - 43)°
⇒ sin43°sin43° + cos43°cos43°
⇒ sin243° + cos243° [Since, sin2θ + cos2θ = 1]
= 1
Posted by Abhishek Singh 6 years, 2 months ago
- 2 answers
Yogita Ingle 6 years, 2 months ago
Given A.P:
8,14,20,26, ...
First term (a) = 8
common difference (d) =a_{2}-a_{1}
= 14-8
= 6
d = 6
Let n th term of A.P will be 72 more than its 41 th term.
We know that,
{tex}\boxed {n^{th} term = a_{n}=a+(n-1)d}{/tex}
According to the problem given,
{tex}a_{n}-a_{41}=72{/tex}
{tex}\implies a+(n-1)d-[a+(41-1)d]=72{/tex}
{tex}\implies a+nd-d-(a+40d)=72{/tex}
{tex}\implies a+nd-d-a-40d=72{/tex}
{tex}\implies nd-41d = 72{/tex}
{tex}\implies d(n-41)=72{/tex}
{tex}\implies n-41 = \frac{72}{6}{/tex}
{tex}\implies n = 41 +12{/tex}
{tex}\implies n = 53{/tex}
Therefore,
n = 53
53rd term in given A.P will be 72 more than its 41th term.
Posted by Keshav Kumar 6 years, 2 months ago
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Posted by Swayam Acoustic 6 years, 2 months ago
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Posted by Ayush Rathi 6 years, 2 months ago
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Sangeeta Devi 6 years, 2 months ago
Posted by Yash Kumar 6 years, 2 months ago
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Posted by Najiya Shirin 6 years, 2 months ago
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Ravi Naulakha 6 years, 2 months ago
Posted by Harish Verma 6 years, 2 months ago
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Yogita Ingle 6 years, 2 months ago
In a leap year, total number of days = 366 days.
In 366 days, there are 52 weeks and 2 days.
Now two days may be
(i) Sunday and Monday
(ii) Monday and Tuesday
(iii) Tuesday and Wednesday
(iv) Wednesday and Thursday
(v) Thursday and Friday
(vi) Friday and Saturday
(vii) Saturday and Sunday
Now in total 7 possibilities, Sunday and Monday both come together is 1 time.
So probabilities of 53 Sunday and Monday in a leap year = 1/7
Posted by K Hema Sundar 6 years, 2 months ago
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Posted by Suryansh Rout 6 years, 3 months ago
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Posted by अनसुलझे प्रश्न 6 years, 3 months ago
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Posted by Jiaur Choudhury 6 years, 3 months ago
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Yogita Ingle 6 years, 2 months ago
The given quadratic equation is {tex}kx^{2} -6x-1=0{/tex}
Here, a = k, b = - 6 and c = - d
To find, the value of k = ?
{tex}∴ D=b^{2} -4ac{/tex}
{tex}=(-6)^{2} -4(k)(-1){/tex}
=36+4k, does not have real roots.
=36 + 4k < 0
⇒ 4k < -36
⇒ k < {tex}\dfrac{-36}{4}{/tex}
⇒ k < - 9
∴ k < - 9
Hence, the value of k is < - 9.
Posted by Ashvin Namdev 6 years, 3 months ago
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Vimal Singh 6 years, 2 months ago
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Pk . 6 years, 3 months ago
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Rani Mishra ??? 6 years, 2 months ago
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Heenal Kalal 6 years, 2 months ago
Akanksha Kumari? 6 years, 3 months ago
Akanksha Kumari? 6 years, 3 months ago
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