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Ask QuestionPosted by Gungun._ Lalwani? 6 years, 10 months ago
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Abhinav Dutta 6 years, 10 months ago
Irfan Alam 6 years, 10 months ago
Posted by Gungun._ Lalwani? 6 years, 10 months ago
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Poonam Jyoti? 6 years, 10 months ago
Honey ??? 6 years, 10 months ago
Gungun._ Lalwani? 6 years, 10 months ago
Honey ??? 6 years, 10 months ago
Posted by Utsa Purkayastha 6 years, 10 months ago
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Prashant Parihar 6 years, 10 months ago
Posted by Anjali Kumari? 6 years, 10 months ago
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Bhuvnesh Rajpurohit 6 years, 10 months ago
Affu 😊 6 years, 10 months ago
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Anshika / U Can Call Me Alien ???? 6 years, 10 months ago
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Yogita Ingle 6 years, 10 months ago
Covalent bond: The bond formed by mutual sharing of valency electrons in order to get stable electronic configuration of nearest inert gas is called covalent bond. Generally this bond is formed between non metals.
Based on type of atoms participated in the bonding, covalent bond is classified into two types. They are (i) polar covalent bonding (ii) Non-polar covalent bonding
Khushi Chaudhary 6 years, 10 months ago
Posted by Priyam Singh 6 years, 10 months ago
- 2 answers
Gaurav Seth 6 years, 10 months ago
Let x be any positive integer
Then x = 5q or x = 5q+1 or x = 5q+4 for integer x.
If x = 5q, x2 = (5q)2 = 25q2 = 5(5q2) = 5n (where n = 5q2 )
If x = 5q+1, x2 = (5q+1)2 = 25q2+10q+1 = 5(5q2+2q)+1 = 5n+1 (where n = 5q2+2q )
If x = 5q+4, x2 = (5q+4)2 = 25q2+40q+16 = 5(5q2 + 8q + 3)+ 1 = 5n+1 (where n = 5q2+8q+3 )
<font face="Cambria Math, serif"><font style="box-sizing: inherit; outline: none; max-width: 100%; overflow: hidden; color: rgba(0, 0, 0, 0.87); font-size: 16px; font-style: normal; font-variant-ligatures: normal; font-variant-caps: normal; font-weight: 400; letter-spacing: normal; orphans: 2; text-align: start; text-indent: 0px; text-transform: none; white-space: normal; widows: 2; word-spacing: 0px; -webkit-text-stroke-width: 0px; text-decoration-style: initial; text-decoration-color: initial;">∴</font></font>in each of three cases x2 is either of the form 5q or 5q+1 or 5q+4 and for integer q.
Posted by Priyam Singh 6 years, 10 months ago
- 2 answers
Posted by Anubav Shaw 6 years, 10 months ago
- 1 answers
Gaurav Seth 6 years, 10 months ago
Depreciation after 1st year =600,000 × (15/100) = Rs 90,000
Depreciation after 2nd year =600,000 × (13.5/100) = Rs 81,000
Depreciation after 10th year =600,000 × 1.5/100 = Rs 9,000
Total depreciation = 90000 + 81000 + 72000+ …+ 9000
Here a = 90000, d = 81000 – 90000 = – 9000, t10 = 9000
Sum of n terms of AP is
= 5 × 99000 = 495000
Value at the end of 10 years = Rs 600000 – Rs 495000 = Rs 1,05,000
Posted by Abhinav Sharma 5 years, 8 months ago
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Bhuvnesh Rajpurohit 6 years, 10 months ago
Posted by Durgesh Kumar Singh 6 years, 4 months ago
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Sia ? 6 years, 4 months ago
LHS = (1 + tan{tex}\theta{/tex} + cot{tex}\theta{/tex})(sin{tex}\theta{/tex} - cos{tex}\theta{/tex})
{tex}= \left( 1 + \frac { \sin \theta } { \cos \theta } + \frac { \cos \theta } { \sin \theta } \right) ( \sin \theta - \cos \theta ){/tex}
{tex}= \left( \frac { \cos \theta \sin \theta + \sin ^ { 2 } \theta + \cos ^ { 2 } \theta } { \cos \theta \sin \theta } \right) ( \sin \theta - \cos \theta ){/tex}
{tex}= \frac { ( \cos \theta \sin \theta + 1 ) } { \cos \theta \sin \theta } ( \sin \theta - \cos \theta ){/tex}
RHS = {tex}\left( \frac { \sec \theta } { \ cosec ^ { 2 } \theta } - \frac { \cos e c \theta } { \sec ^ { 2 } \theta } \right) = \left( \frac { \frac { 1 } { \cos \theta } } { \frac { 1 } { \sin ^ { 2 } \theta } } - \frac { \frac { 1 } { \sin \theta } } { \frac { 1 } { \cos ^ { 2 } \theta } } \right){/tex}
{tex}= \left( \frac { \sin ^ { 2 } \theta } { \cos \theta } - \frac { \cos ^ { 2 } \theta } { \sin \theta } \right) = \frac { \sin ^ { 3 } \theta - \cos ^ { 3 } \theta } { \cos \theta \sin \theta }{/tex}
{tex}= \frac { ( \sin \theta - \cos \theta ) \left( \sin ^ { 2 } \theta + \cos ^ { 2 } \theta + \cos \theta \sin \theta \right) } { \cos \theta \sin \theta }{/tex} [{tex}\because{/tex} a3 - b3 = (a - b) (a2 + ab + b2) ]
{tex}= \frac { ( \sin \theta - \cos \theta ) ( 1 + \cos \theta \sin \theta ) } { \cos \theta \sin \theta }{/tex}
{tex}\therefore{/tex} LHS = RHS
Posted by Divyanshi Varshney 6 years, 10 months ago
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Durgesh Kumar Singh 6 years, 10 months ago
Posted by Aditi Sharma 6 years, 10 months ago
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Posted by Akash Jinaralakar 6 years, 4 months ago
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Sia ? 6 years, 4 months ago
Let y-axis divides the line segment joining the points
(5, - 6) and (- 1, - 4) in the ratio k: 1
and the coordinates of the required point be (0, y).
Then, 0 = {tex}\frac { 5 \times ( 1 ) + k \times ( - 1 ) } { k + 1 }{/tex}
{tex}\Rightarrow{/tex} 5 - k = 0
{tex}\Rightarrow{/tex} k = 5
Thus, y-axis divides the line segment joining the points (- 5 , 6) and (- 1, -4) in the ratio 5 : 1.
Posted by Roushan & Suraj 6 years, 4 months ago
- 1 answers
Sia ? 6 years, 4 months ago
Given,
Volume of gold = 1 cm3 = 1000 mm3 ......(1)
Diameter of wire = 0.1 mm
Radius of wire = {tex}\frac { 0.1 } { 2 }{/tex} = 0.05 mm
Let, the length of the wire = h mm
According to the question,
Volume of wire = Volume of gold
{tex}\Rightarrow {/tex} πr2h = 1000 .( Since, wire is cylindrical & from (1) )
{tex}\Rightarrow π× ( 0.05 ) ^ { 2 } \times h{/tex} = 1000
{tex}\Rightarrow \frac { 22 } { 7 } \times 0.0025 \times h{/tex} = 1000
{tex}\Rightarrow h = \frac { 1000 \times 7 } { 22 \times 0.0025 }{/tex}mm
{tex}\Rightarrow{/tex} h = 127272.72 mm
{tex}\Rightarrow{/tex} h = 127.27 m(approx.)
Posted by Siva Siva 6 years, 10 months ago
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Yogita Ingle 6 years, 10 months ago
Let the cost of pen be x and pencil be y,
5x+6y=9 ...(i)
and 3x+2y-=5 ...(ii)
solving (i) and (ii) by eliminatory method we get;
5x+6y=9
3x+2y=5( Multiply by 3)
-> 9x+6y=15 ...(iii)
From (i) and (iii)
5x + 6y= 9
9x + 6y= 15
(-) (-) (-)
------------------------
4x = 6
------------------------
or x= 1.5
Substituting the value of x in (ii) we get
3(1.5)+ 2y=5
⇒ 4.5+2y=5
⇒ 2y= 5-4.5
⇒ y=0.5/2
⇒y =0.25
Therefore cost of pen is Rs1.5 and that of pencil is Rs 0.25.
1Thank You