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Ask QuestionPosted by Nikita Arora 7 years, 4 months ago
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Dev Tera Yaar Jaat☢️ 7 years, 4 months ago
Posted by Dhruvika Goyal 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 Sujal Jha 7 years, 4 months ago
- 1 answers
Durgendra Singh 7 years, 4 months ago
Posted by Aditi Singh 6 years, 5 months ago
- 1 answers
Sia ? 6 years, 5 months ago
The given equations may be written as
{tex} \frac { x + 1 } { 2 } + \frac { y - 1 } { 3 } = 8 {/tex}
⇒ 3(x + 1) + 2 (y - 1) = 48
⇒ 3x + 3 + 2y - 2 = 48
⇒ 3x + 2y + 1= 48
⇒ 3x+2y = 47 ... (i)
{tex}\frac { x - 1 } { 3 } + \frac { y + 1 } { 2 } = 9{/tex}
⇒ 2(x -1 ) + 3(y +1) = 54
⇒ 2x - 2 + 3y + 3 = 54
⇒ 2x + 3y + 1 = 54
⇒ 2x + 3y = 53. ... (ii)
Multiplying (i) by 2 and (ii) by 3 and subtracting, we get
(4 - 9)y = 94-159
{tex} \Rightarrow - 5 y = - 65 {/tex}
{tex}\Rightarrow y = \frac { - 65 } { - 5 } {/tex}
{tex}\Rightarrow y = 13{/tex}
Putting y = 13 in (i), we get
3x + (2 {tex} \times{/tex} 13) = 47
{tex} \Rightarrow{/tex} 3x + 26 = 47
{tex} \Rightarrow{/tex} 3x = (47 - 26)
{tex} \Rightarrow{/tex}3x = 21
{tex} \Rightarrow \quad x = \frac { 21 } { 3 } = 7{/tex}
So, x = 7
Hence, x = 7 and y = 13
Posted by Aditi Singh 6 years, 5 months ago
- 1 answers
Sia ? 6 years, 5 months ago
{tex}\frac { x + 1 } { x - 1 } + \frac { x - 2 } { x + 2 } = 4 - \frac { 2 x + 3 } { x - 2 } {/tex}
{tex}\frac { x ^ { 2 } + 3 x + 2 + x ^ { 2 } - 3 x + 2 } { x ^ { 2 } + x - 2 }{/tex} = {tex}\frac { 4 x - 8 - 2 x - 3 } { x - 2 }{/tex}
(2x2 + 4) (x - 2) = (2x -11 ) (x2 + x - 2)
or, 5x2 + 19x- 30 = 0
or, (5x - 6) (4x - 5) = 0
or, x = {tex}\frac{5}{4}{/tex}, {tex}\frac 65{/tex}
Posted by Komal Jha 7 years, 4 months ago
- 2 answers
Posted by Palak Sharma 7 years, 4 months ago
- 1 answers
Posted by Deepesh Garg 7 years, 4 months ago
- 1 answers
Vivek Gupta 7 years, 4 months ago
Posted by Kehkasha Khan 7 years, 4 months ago
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Posted by Uttu Pandey 7 years, 4 months ago
- 1 answers
Uttu Pandey 7 years, 4 months ago
Posted by Nitin Gupta 7 years, 4 months ago
- 1 answers
Posted by Ujjwal Kansra 7 years, 4 months ago
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Posted by Disha Rathore 6 years, 4 months ago
- 1 answers
Sia ? 6 years, 4 months ago
Let the speed of the stream = x km/hr
Speed of the boat in upstream = (5 - x) km/hr
Speed of the boat in downstream = (5 + x) km/hr
Distance = 40 km
Time taken in upstream = {tex}{40 \over (5 - x)} hr{/tex}
Time taken in downstream = {tex}{40\over (5 + x)} hr{/tex}
According to the question,
{tex}\frac{40}{{5 - x}} = 3\left( {\frac{40}{{5 + x}}} \right){/tex}
⇒ (5 + x) = 3(5 - x)
⇒ 5 + x = 15 - 3x
⇒ 4x = 10
⇒ x = 2.5
Therefore, speed of the stream = 2.5 km/hr.
Posted by Arya Keshri 7 years, 4 months ago
- 1 answers
Hansika Verma 7 years, 4 months ago
Posted by Subham Gupta 6 years, 5 months ago
- 1 answers
Sia ? 6 years, 5 months ago
Let A(-2, -3) and B(5, 6) be the given points.
Suppose x-axis divides AB in the ratio k:1 at point P
Then, the coordinates of the point of division are
{tex}P \left[ \frac { 5 k - 2 } { k + 1 } , \frac { 6 k - 3 } { k + 1 } \right]{/tex}
Since P lies on x-axis, and y-coordinates of every point on x-axis is zero.
{tex}\therefore \quad \frac { 6 k - 3 } { k + 1 } = 0{/tex}
{tex}\Rightarrow{/tex} 6k - 3=0
{tex}\Rightarrow{/tex} 6k = 3
{tex}\Rightarrow{/tex} {tex}k = \frac { 3 } { 6 } \Rightarrow k = \frac { 1 } { 2 }{/tex}
Hence, the required ratio is 1:2.
Putting {tex}K = \frac { 1 } { 2 }{/tex}in the coordinates of P.
We find that its coordinates are {tex}\left( \frac { 1 } { 3 } , 0 \right).{/tex}
Posted by Sanjeet Parija 7 years, 4 months ago
- 1 answers
Prerika Lamba 7 years, 4 months ago
Posted by Jatin Lalwani 7 years, 4 months ago
- 0 answers
Posted by Shraddha Soni 6 years, 5 months ago
- 1 answers
Sia ? 6 years, 5 months ago
Let L1, L2, L3, L4, ...,L13 be the lengths of semicircles of radii 0.5 cm, 1 cm, 1.5 cm, 2 cm,... and {tex}\frac{{13}}{2}{/tex}cm respectively.
Then, we have
L1 = ({tex}\pi {/tex} {tex}\times{/tex} 0.5) cm = {tex}\frac{\pi }{2}{/tex}cm,
L2 = ({tex}\pi {/tex} {tex}\times{/tex} 1) cm = 2 ({tex}\frac{\pi }{2}{/tex})cm,
L3 = ({tex}\pi{/tex} {tex}\times{/tex} 1.5)cm = 3({tex}\frac{\pi }{2}{/tex})cm,
L4 = ({tex}\pi{/tex} {tex}\times{/tex} 2)cm = 4({tex}\frac{\pi }{2}{/tex})cm,....
L13 =({tex}\pi{/tex} {tex}\times{/tex}{tex}\frac{{13}}{2}{/tex})cm = 3({tex}\frac{\pi }{2}{/tex})cm.
{tex}\therefore{/tex} total length of the spiral
= L1 + L2 + L3 + L4 +... +13
={{tex}\frac{\pi }{2}{/tex}+2({tex}\frac{\pi }{2}{/tex})+3({tex}\frac{\pi }{2}{/tex})+4({tex}\frac{\pi }{2}{/tex})+....+13({tex}\frac{\pi }{2}{/tex})}cm
={tex}\frac{\pi }{2}{/tex}(1+2+3+4+...+13)cm
= {tex}\frac{\pi }{2}{/tex}{tex}\times{/tex}{tex}\frac{{13}}{2}{/tex}{tex}\times{/tex}(1+13)cm
=({tex}\frac{1}{2}{/tex}{tex}\times{/tex}{tex}\frac{{22}}{2}{/tex}{tex}\times{/tex}{tex}\frac{{13}}{2}{/tex}{tex}\times{/tex}14)cm=143cm.
Hence, the required length of the spiral is 143 cm.
Posted by Manikant Kumar 7 years, 4 months ago
- 5 answers
Bikram Murmu 7 years, 4 months ago
Posted by Heramb Tiwari 7 years, 4 months ago
- 4 answers
Posted by Manikant Kumar 7 years, 4 months ago
- 4 answers
Somya Choudhary 7 years, 4 months ago
Manikant Kumar 7 years, 4 months ago
Posted by Kunal Goel 7 years, 4 months ago
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Posted by Rishabh Pal 7 years, 4 months ago
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Posted by Alok Gupta 7 years, 4 months ago
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Posted by Nishita Gupta 7 years, 4 months ago
- 1 answers
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