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Ask QuestionPosted by Puja Sahoo 7 years ago
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Rashi Kashyap 7 years ago
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Neha Jangir 7 years ago
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Puja Sahoo 7 years ago
Chētnà Pandey✌️ 7 years ago
Posted by Rishi Arora 7 years ago
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Ram Kushwah 7 years ago
{tex}\begin{array}{l}\frac{\sin\;^3A\;+\cos^3A}{\sin A+\cos A}+\;\sin AcosA\\={\textstyle\frac{\textstyle(\sin A\;+\;\cos A)(\sin^2A+\cos^2A-\sin A\cos A)}{\sin{\textstyle A}{\textstyle\;}{\textstyle+}{\textstyle\cos}{\textstyle A}}}\;+\;\sin AcosA\\=\sin^2A+\cos^2A-\sin A\cos A+\;\sin AcosA\\=1\\\end{array}{/tex}
Posted by Shiwansh Singh 7 years ago
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Posted by Shyamani Patel 7 years ago
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Posted by S P 7 years ago
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Posted by Rahul Dhot 6 years, 4 months ago
- 1 answers
Sia ? 6 years, 4 months ago
For sin A,
By using identity {tex}cosec ^ { 2 } A - \cot ^ { 2 } A = 1 \Rightarrow \cos e c ^ { 2 } A = 1 + \cot ^ { 2 } A{/tex}
{tex}\Rightarrow \frac { 1 } { \sin ^ { 2 } A } = 1 + \cot ^ { 2 } A{/tex}
{tex}\Rightarrow \sin A = \frac { 1 } { \sqrt { 1 + \cot ^ { 2 } A } }{/tex}
For secA,
By using identity {tex}\sec ^ { 2 } A - \tan ^ { 2 } A = 1 \Rightarrow \sec ^ { 2 } A = 1 + \tan ^ { 2 } A{/tex}
{tex}\Rightarrow \sec ^ { 2 } A = 1 + \frac { 1 } { \cot ^ { 2 } A } = \frac { \cot ^ { 2 } A + 1 } { \cot ^ { 2 } A } \Rightarrow \sec ^ { 2 } A = \frac { 1 + \cot ^ { 2 } A } { \cot ^ { 2 } A }{/tex}
{tex}\Rightarrow \sec A = \frac { \sqrt { 1 + \cot ^ { 2 } A } } { \cot A }{/tex}
For tanA,
{tex}\tan A = \frac { 1 } { \cot A }{/tex}
Posted by Puja Sahoo 7 years ago
- 3 answers
Sunita Awasthi 7 years ago
Abc . 7 years ago
Posted by Devashish Shah 7 years ago
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Posted by Ñítísh Ãñsh 7 years ago
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Posted by Jhankar Sharma 7 years ago
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Posted by Siddharth Chavan 6 years, 6 months ago
- 1 answers
Sia ? 6 years, 6 months ago
Let the usual speed of the plane be x km/hr. Then,
New speed = (x + 400) km/hr.
Time taken to cover 1600 km when the speed is x km/hr {tex} = \frac{{1600}}{x}{/tex} hrs.
Time taken to cover 1600 km when the speed is (x + 400) km/hr {tex} = \frac{{1600}}{{x + 400}}{/tex} hrs.
{tex}\therefore \frac{{1600}}{x} - \frac{{1600}}{{x + 400}} = \frac{{40}}{{60}}{/tex} {tex}\left[ {\because 40{\text{ minutes}} = \frac{{10}}{{60}}hr} \right]{/tex}
{tex} \Rightarrow \frac{{1600(x + 400) - 1600x}}{{x(x + 400)}} = \frac{2}{3}{/tex}
{tex} \Rightarrow \frac{{1600x + 640000 - 1600x}}{{{x^2} + 400x}} = \frac{2}{3}{/tex}
{tex} \Rightarrow \frac{{640000}}{{{x^2} + 400x}} = \frac{2}{3}{/tex}
{tex}\Rightarrow{/tex} 1920000 = 2(x2 + 400x)
{tex} \Rightarrow \frac{{1920000}}{2} = {x^2} + 400x{/tex}
{tex}\Rightarrow{/tex} 960000 = x2 + 400x
{tex}\Rightarrow{/tex} x2 + 400x - 960000 = 0
{tex}\Rightarrow{/tex} x2 + 1200x - 800x - 960000 = 0
{tex}\Rightarrow{/tex} x(x + 1200) - 800(x + 1200) = 0
{tex}\Rightarrow{/tex} (x + 1200)(x - 800) = 0
{tex}\Rightarrow{/tex} x - 800 = 0 [{tex}\because{/tex} Speed can not be negative {tex}\therefore{/tex} x + 1200 {tex}\ne{/tex} 0]
{tex}\Rightarrow{/tex} x = 800
Hence, the usual speed of the plane is 800 km/hr.
Posted by Amit Gupta 7 years ago
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Posted by Sanatombi Urikhimbam 7 years ago
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Asif Ali 7 years ago
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Posted by Priya Dhand 7 years ago
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Ram Kushwah 7 years ago
from 1 to 88 the numbers divid\sible by 8 are
8,16,24..............88
total 11 numbers
not divisible by 8 are =88-11=77
hence probability is {tex}\frac{77}{88}{/tex}
Posted by Keithellakpam Dilip 7 years ago
- 4 answers
Ram Kushwah 7 years ago
{tex}\begin{array}{l}CSA\;OF\;FUSTAM\;IS:\\\mathrm\pi(\sqrt{(\mathrm R-\mathrm r)2+\mathrm h^2)})(\mathrm R+\mathrm r)\\=3.14\times(\sqrt{(15-5)^2+}24^2)(15+5)\\=3.14\times(26\times20)=1632.8\;\mathrm{cm}^2\\\\\end{array}{/tex}
Posted by Juhi Sharma 7 years ago
- 1 answers
Yogita Ingle 7 years ago
The angle of elevation will 45 degrees as in the triangle formed by the head of man, end point of shadow and the point where leg touches ground. And the man is perpendicular to the groung so 90 degrees. As per the question heights of man and shadow are equal thus triangle will be isosceles thus answer will be 45 degree.
Posted by Divyanshu Jangid 7 years ago
- 3 answers
Poonam Bais 7 years ago
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