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Gaurav Seth 6 years, 9 months ago
Question: 5 pencils and 7 pens together cost Rs. 50 , where as 7 pencils and 5 pens together cost Rs. 46. Find the cost of one pencil and one pen graphically.
Posted by Sheetal Kumari 6 years, 4 months ago
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Sia ? 6 years, 4 months ago
Let n = 4q + 1 (an odd integer)
{tex}\therefore \quad n ^ { 2 } - 1 = ( 4 q + 1 ) ^ { 2 } - 1{/tex}
{tex}= 16 q ^ { 2 } + 1 + 8 q - 1 \quad \text { Using Identity } ( a + b ) ^ { 2 } = a ^ { 2 } + 2 a b + b ^ { 2 }{/tex}
{tex}= 16{q^2} + 8q{/tex}
{tex}= 8 \left( 2 q ^ { 2 } + q \right){/tex}
= 8m, which is divisible by 8.
Posted by Arpit Patel 6 years, 9 months ago
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Raunak Pandey ?? 6 years, 9 months ago
Lionel Messi⚽️ 6 years, 9 months ago
Posted by Simran Kaur 6 years, 9 months ago
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Gaurav Seth 6 years, 9 months ago
Given n is a natural number, then 92n – 42n
If n = 1 then 81 - 16 = 65 is always divisible by both 5 and 13.
if n = 2 then 6561 - 256 = 6305 is always divisible by both 5 and 13.
Posted by Kanak Garg 6 years, 9 months ago
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Sia ? 6 years, 4 months ago
We have to prove that {tex} \frac { 1 + \sec \theta - \tan \theta } { 1 + \sec \theta + \tan \theta } = \frac { 1 - \sin \theta } { \cos \theta }{/tex}
Recall identity sec2 θ – tan2θ = 1
Here, LHS {tex}= \frac { 1 + \sec \theta - \tan \theta } { 1 + \sec \theta + \tan \theta }{/tex}
{tex}= \frac { \sec ^ { 2 } \theta - \tan ^ { 2 } \theta + \sec \theta - \tan \theta } { 1 + \sec \theta + \tan \theta }{/tex}
{tex}= \frac { ( \sec \theta - \tan \theta ) ( \sec \theta + \tan \theta ) + ( \sec \theta - \tan \theta ) } { 1 + \sec \theta + \tan \theta }{/tex} [ because, a2 – b2 = (a – b)(a + b)]
{tex}= \frac { ( \sec \theta - \tan \theta ) [ \sec \theta + \tan \theta + 1 ] } { ( \sec \theta + \tan \theta + 1 ) }{/tex}
= sec θ – tan θ
{tex}= \frac { 1 } { \cos \theta } - \frac { \sin \theta } { \cos \theta }{/tex}
{tex}= \frac { 1 - \sin \theta } { \cos \theta }{/tex} = RHS
Hence, proved.
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Sia ? 6 years, 4 months ago
You can check the formulae in the notes : https://mycbseguide.com/cbse-revision-notes.html
Posted by Sonu Bisht 6 years, 9 months ago
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Ram Kushwah 6 years, 9 months ago
We assume that 1/{tex}\sqrt2{/tex} is a rational no
So {tex}\frac ab=\frac{1}{\sqrt2}{/tex}
( Where and b are co-primes)
b=a{tex}\sqrt 2{/tex}
b2=2a2--------(1)
b2 is divisible by 2 ,so b is divisible by 2-----(2)
Let b=2c( c is any integer)
b2=4c2
2a2=4c2
a2=2c2
So a2 is divisible by 2,So a is divisible by 2-----(3)
From (2) and (3)
a and b are divisible by 2
But this contradict the assumption that a and b are co-prime ,So no common factor in and b except 1
Hence our assumption that 1/ {tex}\sqrt2{/tex} is wrong ,
Hence 1/{tex}\sqrt2{/tex} is a non prime number
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Sia ? 6 years, 4 months ago
For any two given positive integers a and b there exist unique whole numbers {tex}q\ and\ r {/tex} such that
{tex}a = bq + r,{/tex} where {tex}0 \leq r < b{/tex}
We call 'a' as dividend, b as divisor, q as quotient and r as remainder.
{tex}Dividend = (divisor \times quotient) + remainder{/tex}
Posted by Sudharani Agrawal 6 years, 4 months ago
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Sia ? 6 years, 4 months ago
Let n be an arbitrary positive integer.
On dividing n by 3, let q be the quotient and r be the remainder.
Then, by Euclid's division lemma, we have
n = 3q + r, where {tex}0 \leq r < 3{/tex}.
The possibilities of remainder = 0,1 or 2
n2 = (3q + r)2 [∵ (a + b) 2 = a2 + 2ab + b2]
{tex}\therefore{/tex} n2 = 9q2 + r2 + 6qr ... ....(i), where {tex}0 \leq r < 3{/tex}.
Case I When r = 0.
Putting r = 0 in (i), we get
n2 = 9q2
= 3(3q2)
n2 = 3m, where m = 3q2 is an integer.
Case II When r = 1.
Putting r = 1 in (i), we get
n2 = (9q2 + 1 + 6 q)
= 3(3q2 + 2q) + 1
n2= 3 m + 1, where m = (3q2 + 2q) is an integer.
Case lll When r = 2.
Putting r = 2 in (i), we get
n2 = (9q2 + 4 + 12q)
= 3(3q2 + 4q + 1) + 1
n2= 3m + 1, where m = (3q2 + 4q + 1) is an integer.
From all the above cases it is clear that the square of any positive integer is of the form 3m or (3m + 1) for some integer m.
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