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Ask QuestionPosted by Garima Chahar 6 years, 9 months ago
- 0 answers
Posted by Rohit Sharma 6 years, 9 months ago
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Posted by Udit Mehandiratta 6 years, 9 months ago
- 0 answers
Posted by Anshu Trivedi 5 years, 9 months ago
- 2 answers
Posted by Ash Sunil 6 years, 9 months ago
- 2 answers
Yogita Ingle 6 years, 9 months ago
For a pair of given positive integers ‘a’ and ‘b’, there exist unique integers ‘q’ and ‘r’ such that
a = bq+r , where 0 ≤r < b
Explanation:
Thus, for any pair of two positive integers a and b; the relation
a = bq + r , where 0≤r<b
will be true where q is some integer.
Posted by Aapka Channel. 6 years, 9 months ago
- 1 answers
Yogita Ingle 5 years, 9 months ago
tanx = 3cotx
But cotx = 1/tanx
⇒ tanx = 3/tanx
Multiply both side by tanx
=> tan 2x = 3
tanx = ±√3
X = 60 when tan x is +3
X = 120 when tan x is -3
Posted by Abhay Mathur 6 years, 9 months ago
- 1 answers
Abhay Mathur 6 years, 9 months ago
Posted by Kunal D 6 years, 9 months ago
- 2 answers
Posted by Mahak Mishra 6 years, 9 months ago
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Posted by Tm Priya 6 years, 9 months ago
- 1 answers
Gaurav Seth 6 years, 9 months ago
ΔABC ~ ΔPQR
AB / PQ = BC / QR = AC / PR = 1 / 3
and AB / PQ = BC / PQ = AC / PR = 1 / 3
ar ΔABC / ar ΔPQR = (1 / 3)2 = 1 / 9
ar ΔABC / ar ΔPQR = 1 / 9
Posted by Divyansh Srivastava 6 years, 9 months ago
- 2 answers
Gaurav Seth 6 years, 9 months ago
Let two digit numbers are 10, 11, 12 , -------------99
which are divisible by 3 are 12, 15 ...............,99.
Here a = 12 and common dofference = 3 , an = l = 99
nth term (tn) = a + ( n - 1) d
⇒ 99 = 12 + ( n - 1) 3
⇒ (n - 1)3 = 99 - 12 = 87
⇒ ( n -1) = 87 / 3
⇒ n - 1 = 29
∴ n = 30
∴ 30 two digit numbers which are divisible by 3.
Posted by Mandeep Singh 6 years, 9 months ago
- 2 answers
Gaurav Seth 6 years, 9 months ago
Let the zeroes of the polynomial be a and b.
Given Zeroes of the Quadratic polynomial are
Therefore the required quadratic polynomial is:
Posted by Saurabh Kumar 6 years, 9 months ago
- 2 answers
Gaurav Seth 6 years, 9 months ago
Solution: We know that any positive integer is of the form 3q or 3q + 1 or 3q + 2 for some integer q & one and only one of these possibilities can occur
Case I : When n = 3q
In this case, we have,
n=3q, which is divisible by 3
n=3q
= adding 2 on both sides
n + 2 = 3q + 2
n + 2 leaves a remainder 2 when divided by 3
Therefore, n + 2 is not divisible by 3
n = 3q
n + 4 = 3q + 4 = 3(q + 1) + 1
n + 4 leaves a remainder 1 when divided by 3
n + 4 is not divisible by 3
Thus, n is divisible by 3 but n + 2 and n + 4 are not divisible by 3
Case II : When n = 3q + 1
In this case, we have
n = 3q +1
n leaves a reaminder 1 when divided by 3
n is not divisible by 3
n = 3q + 1
n + 2 = (3q + 1) + 2 = 3(q + 1)
n + 2 is divisible by 3
n = 3q + 1
n + 4 = 3q + 1 + 4 = 3q + 5 = 3(q + 1) + 2
n + 4 leaves a remainder 2 when divided by 3
n + 4 is not divisible by 3
Thus, n + 2 is divisible by 3 but n and n + 4 are not divisible by 3
Case III : When n = 3q + 2
In this case, we have
n = 3q + 2
n leaves remainder 2 when divided by 3
n is not divisible by 3
n = 3q + 2
n + 2 = 3q + 2 + 2 = 3(q + 1) + 1
n + 2 leaves remainder 1 when divided by 3
n + 2 is not divsible by 3
n = 3q + 2
n + 4 = 3q + 2 + 4 = 3(q + 2)
n + 4 is divisible by 3
Thus, n + 4 is divisible by 3 but n and n + 2 are not divisible by 3 .
Posted by Pratikshya Sahoo 6 years, 9 months ago
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Posted by Vinod Thamke 6 years, 9 months ago
- 3 answers
Posted by Khushi Narwal 6 years, 9 months ago
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Pranshu Joshi 6 years, 9 months ago
Posted by Soham Bhatkar 6 years, 9 months ago
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Posted by Anand Singh 6 years, 9 months ago
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Posted by Prajwal Banu 6 years, 4 months ago
- 1 answers
Sia ? 6 years, 4 months ago
Let a be the positive integer and b = 6.
Then, by Euclid’s algorithm, a = 6q + r for some integer q ≥ 0 and r = 0, 1, 2, 3, 4, 5 because 0 ≤ r < 5.
So, a = 6q or 6q + 1 or 6q + 2 or 6q + 3 or 6q + 4 or 6q + 5.
(6q)2 = 36q2 = 6(6q2)
= 6m, where m is any integer.
(6q + 1)2 = 36q2 + 12q + 1
= 6(6q2 + 2q) + 1
= 6m + 1, where m is any integer.
(6q + 2)2 = 36q2 + 24q + 4
= 6(6q2 + 4q) + 4
= 6m + 4, where m is any integer.
(6q + 3)2 = 36q2 + 36q + 9
= 6(6q2 + 6q + 1) + 3
= 6m + 3, where m is any integer.
(6q + 4)2 = 36q2 + 48q + 16
= 6(6q2 + 7q + 2) + 4
= 6m + 4, where m is any integer.
(6q + 5)2 = 36q2 + 60q + 25
= 6(6q2 + 10q + 4) + 1
= 6m + 1, where m is any integer.
Hence, The square of any positive integer is of the form 6m, 6m + 1, 6m + 3, 6m + 4 and cannot be of the form 6m + 2 or 6m + 5 for any integer m.
Posted by Itika Singhal 6 years, 9 months ago
- 3 answers
Palak Sahoo?? 6 years, 9 months ago
Chetna ☺️ 6 years, 9 months ago
Posted by Mayank Lukhar 6 years, 9 months ago
- 1 answers
Yogita Ingle 6 years, 9 months ago
We applied Euclid Division algorithm on n and 3.
a = bq +r on putting a = n and b = 3
n = 3q +r , 0<r<3
i.e n = 3q -------- (1),n = 3q +1 --------- (2), n = 3q +2 -----------(3)
n = 3q is divisible by 3
or n +2 = 3q +1+2 = 3q +3 also divisible by 3
or n +4 = 3q + 2 +4 = 3q + 6 is also divisible by 3
Hence n, n+2 , n+4 are divisible by 3.
Posted by Parvej Malik 6 years, 4 months ago
- 1 answers
Sia ? 6 years, 4 months ago
If any number ends with the digit 0, it should be divisible by 10 or in other words, it will also be divisible by 2 and 5 as 10 = 2 × 5
Prime factorisation of 6n = (2 ×3)n
It can be observed that 5 is not in the prime factorisation of 6n.
Hence, for any value of n, 6n will not be divisible by 5.
Therefore, 6n cannot end with the digit 0 for any natural number n.
Posted by Da Ya 5 years, 9 months ago
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Posted by Rithiga Mohan Raj 6 years, 9 months ago
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Posted by Jumden Rongkup 6 years, 9 months ago
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Posted by Tejbeer Kaushal 6 years, 4 months ago
- 1 answers
Sia ? 6 years, 4 months ago
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