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Ask QuestionPosted by D. Sanjai Nikkam 6 years, 5 months ago
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Posted by Rajashekarayya S Kulkarni Kulkarni 6 years, 5 months ago
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Sia ? 6 years, 5 months ago
= tan260° + tan245°
= ({tex}\sqrt 3{/tex})2 + (1)2
= 3 + 1
= 4
Posted by Vipin Pal 6 years, 5 months ago
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Parthiv Patel 6 years, 5 months ago
Posted by Sahil Slodhiya 6 years, 5 months ago
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Posted by Tripti Goyal 6 years, 5 months ago
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Kunja Bihari 6 years, 5 months ago
Posted by Nirmal Pradhan 6 years, 5 months ago
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Sia ? 6 years, 5 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 Nirmal Pradhan 6 years, 5 months ago
- 1 answers
Sia ? 6 years, 5 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 Piyush Utkarsh 6 years, 5 months ago
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Deepak Latesara 6 years, 5 months ago
Posted by Shantha D T 6 years, 5 months ago
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Posted by Om Chauhan 6 years, 5 months ago
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Posted by Kartik Singh Rajpoot 6 years, 5 months ago
- 1 answers
Isha Trivedi 6 years, 5 months ago
Posted by Naveen Kumar 6 years, 5 months ago
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Sia ? 6 years, 5 months ago
Check revision notes to check formulae : https://mycbseguide.com/cbse-revision-notes.html
Posted by Mahi K 6 years, 5 months ago
- 1 answers
Sia ? 6 years, 5 months ago
We have the following equation,
{tex}3x^2-5x+2k=0{/tex}
Putting x = -2
3(-2)2 - 5(-2) + 2k = 0
{tex}\Rightarrow{/tex} 12 + 10 + 2k = 0
{tex}\Rightarrow{/tex} 22 + 2k = 0
{tex}\Rightarrow{/tex} k = {tex}\frac{-22}{2}{/tex}
{tex}\Rightarrow{/tex} k = -11
Posted by Md Tafseel Ansari 6 years, 5 months ago
- 1 answers
Sia ? 6 years, 5 months ago
Check syllabus here : https://mycbseguide.com/cbse-syllabus.html
Posted by Padmanabhan K 6 years, 5 months ago
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Sia ? 6 years, 5 months ago
7 × 1 = 7
7 × 2 = 14
7 × 3 = 21
7 × 4 = 28
7 × 5 = 35
7 × 6 = 42
7 × 7 = 49
7 × 8 = 56
7 × 9 = 63
7 × 10 = 70
Posted by Jashanjot Dhillon 6 years, 5 months ago
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Posted by Vanashree Suresh Ingaladal 6 years, 5 months ago
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Posted by Shruti Jha 6 years, 5 months ago
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Ãkäsh Målhõträ 6 years, 5 months ago
Posted by Hajiya Zuha 6 years, 5 months ago
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Akhilesh Pandey 6 years, 5 months ago
Posted by Mohammad Dawood 6 years, 5 months ago
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Shruti Jha 6 years, 5 months ago
Posted by Niraj Kumar 6 years, 5 months ago
- 1 answers
Sia ? 6 years, 5 months ago
Check syllabus here : https://mycbseguide.com/cbse-syllabus.html
Posted by R.C. Khariwal 6 years, 5 months ago
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Posted by .. .? 6 years, 5 months ago
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Posted by Nilesh Bhardwaj 6 years, 5 months ago
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Posted by Nikhil Patil 6 years, 5 months ago
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Gaurav Seth 6 years, 5 months ago
56=40×1+16
40=16×2+8
16=8×2+0
HCF=8
404=8×50+4
8=4×2+0
HCF(56,96,404)=4
Posted by Kusum Malik 6 years, 5 months ago
- 1 answers
Gaurav Seth 6 years, 5 months ago
the triangle
( x1 , y1 ) = A ( 3 , - 7 )
( x2 , y2 ) = B ( - 8 , 6 )
( x3 , y3 ) = C ( 5 , 10 )
Centroid ( G )
= ( x1+x2+x3/3 , y1+y2+y3/3 )
= ( 3-8+5/3 , -7+6+10/3 )
= ( 0 , 9/3 )
= ( 0 , 3 )
G = ( 0 , 3 )
Posted by Kusum Malik 6 years, 3 months ago
- 1 answers
Sia ? 6 years, 3 months ago
According to question
PR = RQ or R is mid-point of PQ
Using mid-point formula x {tex}= \frac { x _ { 1 } + x _ { 2 } } { 2 }{/tex} and y {tex}= \frac { y _ { 1 } + y _ { 2 } } { 2 }{/tex}
{tex}\frac { b - 2 + 4 } { 2 }{/tex} = - 3
{tex}\Rightarrow{/tex} b + 2 = - 6
{tex}\Rightarrow{/tex} b = - 8
Posted by Tanish Singla 6 years, 5 months ago
- 1 answers
Sia ? 6 years, 5 months ago
D = b2 - 4ac
{tex}= ( - 9 ( a + b ) ) ^ { 2 } - 4 \times 9 \times \left( 2 a ^ { 2 } + 5 a b + 2 a b ^ { 2 } \right){/tex}
= 81(a + b)2 - 36(2a2 + 5ab + 2b2)
= 9[9(a2 + b2 + 2ab - 8a2 - 20ab - 8b2)]
= 9[a2 + b2 - 2ab]
= 9(a - b)2
{tex}x = \frac { - b \pm \sqrt { D } } { 2 a } = \frac { 9 ( a + b ) \pm \sqrt { 9 ( a - b ) ^ { 2 } } } { 2 \times 9 }{/tex}
{tex}{ \Rightarrow x = 3 \frac { [ 3 ( a + b ) \pm ( a - b ) ] } { 2 \times 9 } }{/tex}
{tex}{ \Rightarrow x = \frac { ( 3 a + 3 b ) \pm ( a - b ) } { 6 } }{/tex}
{tex}\Rightarrow x = \frac { 3 a + 3 b + a - b } { 6 } \text { or } x = \frac { 3 a + 3 b - a + b } { 6 }{/tex}
{tex}\Rightarrow x = \frac { 4 a + 2 b } { 6 } \text { or } x = \frac { 2 a + 4 b } { 6 }{/tex}
{tex}\Rightarrow x = \frac { 2 a + b } { 3 } \text { or } x = \frac { a + 2 b } { 3 }{/tex}
Posted by Anil Choudary 6 years, 5 months ago
- 1 answers
Sia ? 6 years, 5 months ago
The systems of equations are:
5ax + 6by = 28 .......................(i)
3ax + 4by = 18 .......................(ii)
From (i) and (ii), we get
a1 = 5a, b1 = 6b, c1 = -(28)
a2 = 3a, b2 = 4b, c2 = -(18)
By cross multiplication method, we get
{tex}\frac{x}{{-108b +112b}} = \frac{{ - y}}{{ - 90a + 84a }} = \frac{1}{{20ab - 18ab}}{/tex}
{tex}\frac{x}{{4b}} = \frac{{ - y}}{{ - 6a}} = \frac{1}{{2ab}}{/tex}
Now, {tex}\frac{x}{{4b}} = \frac{1}{{2ab}} {/tex}
{tex}⇒ x = \frac{2}{a}{/tex}
And, {tex}\frac{{ - y}}{{ - 6a}} = \frac{1}{{2ab}} {/tex}
{tex}⇒ y = \frac{3}{b}{/tex}
Therefore the solution of the given system of equation is {tex}\frac{2}{a}{/tex} and {tex}\frac{3}{b}{/tex}.
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Sia ? 6 years, 5 months ago
= cos 37° cosec 53°
= cos(90° - 53°) cosec 53°
= sin 53° cosec 53°
= sin 53° . {tex}\frac{1}{sin 53^ \circ}{/tex}
= 1
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