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Posted by Kishan Krishna 7 years, 4 months ago
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Posted by Uday Kiran 6 years, 5 months ago
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Sia ? 6 years, 5 months ago
Let the length and breadth of the rectangle be x units and y units respectively.
Then, area of the rectangle = xy sq units.
Case I When the length is reduced by 5 units and the breadth is increased by 2 units.
Then, new length = {tex}(x - 5){/tex} units and
new breadth = {tex}(y+ 2){/tex} units.
{tex}\therefore{/tex} new area = {tex}(x - 5)(y + 2){/tex} sq units
{tex}\therefore{/tex} {tex}xy - ( x - 5)(y+ 2) = 80{/tex} {tex}\Rightarrow{/tex} {tex}5y - 2x = 70{/tex} ..... (i)
Case II When the length is increased by 10 units and the breadth is decreased by 5 units.
Then, new length = {tex}(x + 10){/tex} units
and new breadth = {tex}(y - 5){/tex} units.
{tex}\therefore{/tex} new area = {tex}(x + 10)(y - 5){/tex}sq units.
{tex}\therefore{/tex} {tex}(x + 10)(y - 5 ) - xy = 50{/tex}
{tex}\Rightarrow{/tex} {tex}10y - 5x = 100{/tex} {tex}\Rightarrow{/tex} {tex}2y - x = 20{/tex}. ...(ii)
On multiplying (ii) by 2 and subtracting the result from (i), we get y = 30.
Putting y = 30 in (ii), we get
{tex}(2\times 30) - x = 20{/tex} {tex}\Rightarrow{/tex} {tex}60 - x = 20{/tex}
{tex}\Rightarrow{/tex} {tex}x = (60 - 20) = 40.{/tex}
{tex}\therefore{/tex} {tex}x = 40\ and\ y = 30.{/tex}
Hence, length = 40 units and breadth = 30 units.
Posted by Abhijeet Singh 7 years, 4 months ago
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Dibya Ranjan Das Das 7 years, 4 months ago
Posted by Md Danish 6 years, 5 months ago
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Sia ? 6 years, 5 months ago
A, B, C, are interior angles of a {tex}\Delta {/tex}
{tex}\because A + B + C = 180 ^ { 0 }{/tex}
{tex}\Rightarrow B + C = 180 ^ { 0 } - A \Rightarrow \frac { B + C } { 2 } = 90 ^ { 0 } - \frac { A } { 2 }{/tex}
{tex}\Rightarrow \sin \frac { \mathrm { B } + \mathrm { C } } { 2 } = \sin \left( 90 ^ { \circ } - \frac { \mathrm { A } } { 2 } \right) \left[ \because \sin \left( 90 ^ { 0 } - \theta \right) = \cos \theta \right]{/tex}
{tex}\Rightarrow \sin \frac { \mathrm { B } + \mathrm { C } } { 2 } = \cos \frac { \mathrm { A } } { 2 } \text { proved }{/tex}
LHS = RHS
Posted by Ramendra Nath Tripathi 5 years, 8 months ago
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Sia ? 6 years, 4 months ago
Check NCERT solutions here : https://mycbseguide.com/ncert-solutions.html
Posted by K.N Girish Girish 7 years, 4 months ago
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Posted by Tushar Bansal 5 years, 8 months ago
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Anshika Mittal 7 years, 4 months ago
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Aman ??? 7 years, 4 months ago
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Sia ? 6 years, 6 months ago
The given equation is
{tex}\Rightarrow{/tex} 3x2 +11x + 10 = 0
Now, we have to find two numbers such that their sum is 11 and product is 3{tex}\times{/tex}10 = 30. Clearly, 5 +6 =11 and 5{tex}\times{/tex}6 = 30
{tex}\Rightarrow{/tex} 3x2 + 6x + 5x + 10 = 0
{tex}\Rightarrow{/tex} 3x(x + 2) +5(x + 2) = 0
{tex}\Rightarrow{/tex} (3x+5)(x+2) = 0
Either, 3x+5 =0 {tex}\Rightarrow{/tex} {tex}{/tex} {tex}x= - \frac{5}{3}{/tex}
Or, x+2 =0 {tex}\Rightarrow{/tex} x = -2
Hence, the required roots of given equation are -2 and {tex}-\frac{{5}}{3}{/tex}
Posted by Niyati Gupta 6 years, 6 months ago
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Sia ? 6 years, 6 months ago
Given,
{tex}\frac { 1 } { ( a + b + x ) } = \frac { 1 } { a } + \frac { 1 } { b } + \frac { 1 } { x }{/tex}
{tex}\Rightarrow \quad \frac { 1 } { ( a + b + x ) } - \frac { 1 } { x } = \frac { 1 } { a } + \frac { 1 } { b } \Rightarrow \frac { x - ( a + b + x ) } { x ( a + b + x ) } = \frac { b + a } { a b }{/tex}
{tex}\Rightarrow \quad \frac { - ( a + b ) } { x ( a + b + x ) } = \frac { ( a + b ) } { a b }{/tex}
On dividing both sides by (a+b)
{tex}\Rightarrow \quad \frac { - 1 } { x ( a + b + x ) } = \frac { 1 } { a b }{/tex}
Now cross multiply
{tex}\Rightarrow{/tex} x(a + b + x) = -ab
{tex}\Rightarrow{/tex} x2 + ax + bx + ab = 0
{tex}\Rightarrow{/tex} x(x +a) + b(x +a) = 0
{tex}\Rightarrow{/tex} (x + a) (x + b) = 0
{tex}\Rightarrow{/tex} x + a = 0 or x + b = 0
{tex}\Rightarrow{/tex} x = -a or x = -b.
Therefore, -a and -b are the roots of the equation.
Posted by Niyati Gupta 7 years, 4 months ago
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Raushan Kumar 7 years, 4 months ago
Posted by #Aditi~ Angel???? 7 years, 4 months ago
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Gourav Shrivastava 7 years, 4 months ago
Posted by Ayush Kumar Gupta 7 years, 4 months ago
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Srijani Mitra 7 years, 4 months ago
Posted by Ayush Kumar Gupta 6 years, 6 months ago
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Sia ? 6 years, 6 months ago
We have to prove that {tex} \sqrt6{/tex} is an irrational number.
Let {tex}\sqrt6{/tex} be a rational number.
{tex}\therefore \quad \sqrt { 6 } = \frac { p } { q }{/tex}
where p and q are co-prime integers and {tex}q \neq 0{/tex}
On squaring both the sides, we get,
or, {tex}6 = \frac { p ^ { 2 } } { q ^ { 2 } }{/tex}
or, p2 = 6q2
{tex}\therefore{/tex} p2 is divisible by 6.
p is divisible by 6........(i)
Let p = 6r for some integer r
or, p2 = 36r2
6q2= 362 [∵ p2 = 6q2]
or, q2 = 6r2
or, q2 is divisible by 6.
{tex}\therefore{/tex}q is divisible by 6..........(ii)
From (i) and (ii)
p and q are divisible by 6, which contradicts the fact that p and q are co-primes.
Hence, our assumption is wrong.
{tex}\therefore{/tex} {tex}\sqrt6{/tex} is irrational number.
Posted by Divyanshu Singh 7 years, 4 months ago
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Kannu Kranti Yadav 7 years, 4 months ago
Posted by Knv . 7 years, 4 months ago
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Raushan Kumar 7 years, 4 months ago
Posted by Anu Bagde 7 years, 4 months ago
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Posted by Rohan Vaishnav 6 years, 6 months ago
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Sia ? 6 years, 6 months ago
Let a be any odd positive integer, then on dividing a by b, we have
a = bq+r,{tex}0 \leqslant r < b{/tex}.....(i) [by Euclid's division lemma]
On putting b = 2 in Eq.(i), we get
a = 2q+r, {tex}0\leqslant r < 2{/tex}
⇒ r = 0 or 1
If r = 0, then a = 2q, which is divisible by 2. So, 2q is even.
If {tex}r = 1{/tex}, then {tex}a = 2q + 1{/tex}, which is not divisible by 2. So {tex}(2q+1){/tex} is odd.
Now, as a is odd, so it cannot be of the form 2q.
Thus, any odd positive integer a is of the form (2q+1).
Now, consider a2 = (2q+1)2 = 4q2 + 1 + 4q {tex}\lbrack\therefore(x+y)^2=x^2+y^2+2xy\rbrack{/tex}
= 4( q2 + q ) + 1 = 4m + 1, where m = q2 + q
Hence, for some integer m, the square of any odd integer is of the form 4m+1.
Hence Proved.
Posted by Priya Khanagwal 7 years, 4 months ago
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Sejal Ameta 7 years, 4 months ago
Posted by Šhiri Biswas 7 years, 4 months ago
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Sia ? 6 years, 4 months ago
Given equation is {tex}x^2 - 2kx + (7k - 12) = 0{/tex}.
Here {tex}a = 1,\ b = - 2k,\ c = 7k -12{/tex}
{tex}D = b^2 - 4ac{/tex}
{tex}\Rightarrow{/tex}D = (-2k)2 - 4 {tex}\times{/tex} 1 {tex}\times{/tex} (7k-12)
= 4k2 - 28k + 48
The equation has equal roots, then D = 0
{tex}\Rightarrow{/tex}{tex}4k^2 - 28k + 48 = 0{/tex}
{tex}\Rightarrow{/tex} k2 - 7k + 12 = 0
{tex}\Rightarrow {/tex}(k - 3)(k - 4) = 0
{tex}\Rightarrow{/tex} k = 3,4.
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