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Ask QuestionPosted by Abhinav Rana 6 years, 11 months ago
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Posted by ?Garima Sharma? 6 years, 11 months ago
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S Aniruddh 6 years, 11 months ago
Posted by Manpreet Rohilla 6 years, 11 months ago
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Yogita Ingle 6 years, 11 months ago
Cone is a three-dimensional structure having a circular base where a set of line segments, connecting all of the points on the base to a common point called apex.
Curved surface area of a cone = <nobr>πrl</nobr>
Total surface area of a cone = <nobr>πr(l+r)</nobr>
Where, r is the base radius, h is the height and l is the slant height of the cone.
Posted by Supriya Sahoo 6 years, 11 months ago
- 1 answers
Ram Kushwah 6 years, 11 months ago
Given That AD=2 cm , DB=3 cm DE= 4 cm BC=x
{tex}\begin{array}{l}As\;DE\parallel\;BC\\So\\\frac{AD}{AB}=\frac{DE}{BC}\\\frac2{2+3}=\frac4x\\2x=4\times5=20\\x=10\\\end{array}{/tex}
Posted by Akshita Khandelwal 6 years, 11 months ago
- 2 answers
Gaurav Seth 6 years, 11 months ago
Let the speed of sailor in still water is x km/hr and the speed of stream is y km/hr.
Now, the speed of boat (upstream) = (x - y) km/hr
and the speed of boat (downstream) = (x + y) km/hr
Now, according to question,
8/(x + y) = 40/60 {Since time = distance/speed}
=> 8/(x + y) = 4/6
=> 4(x + y) = 8*6
=> 4(x + y) = 48
=> x + y = 48/4
=> x + y = 12 ................1
Again, 8/(x - y) = 1
x - y = 8 ...............2
Add equation 1 and 2, we get
2x = 20
=> x = 20/2
=> x = 10
From equation 1, we get
10 + y = 12
=> y = 12 - 10
=> y = 2
Hence, the speed of sailor in still water is 10 km/hr and the speed of stream is 2 km/hr
Posted by Julfiqar Ali 6 years, 11 months ago
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Posted by Vivek Saini 5 years, 8 months ago
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Gaurav Seth 6 years, 11 months ago
Let the two numbers be A and B
=>A+B=15 .............(1)
and
1/A +1/B=3/10
⇒(A+B)/AB = 3/10
from (1)
(A+B)/AB = 15/AB
⇒15/AB = 3/10
⇒AB = 50 ............(2)
solving (1) and (2)
A = 10 and B = 5
or
A = 5 and B = 10
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Maths Lover **** √Π÷= 6 years, 11 months ago
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Sia ? 6 years, 6 months ago
{tex}\frac { \tan \theta } { 1 - \cot \theta } + \frac { \cot \theta } { 1 - \tan \theta } = 1 + \tan \theta + \cot \theta{/tex}
{tex}\text { L.H.S. } = \frac { \tan \theta } { 1 - \cot \theta } + \frac { \cot \theta } { 1 - \tan \theta }{/tex}
{tex}= \frac { \frac { \sin \theta } { \cos \theta } } { 1 - \frac { \cos \theta } { \sin \theta } } + \frac { \frac { \cos \theta } { \sin \theta } } { 1 - \frac { \sin \theta } { \cos \theta } }{/tex}
{tex}= \frac { \sin ^ { 2 } \theta } { \cos \theta ( \sin \theta - \cos \theta ) } - \frac { \cos ^ { 2 } \theta } { \sin \theta ( \sin \theta - \cos \theta ) }{/tex}
{tex}= \frac { \sin ^ { 3 } \theta - \cos ^ { 3 } \theta } { \sin \theta \cos \theta ( \sin \theta - \cos \theta ) }{/tex}
{tex}= \frac { ( \sin \theta - \cos \theta ) \left( \sin ^ { 2 } \theta + \cos ^ { 2 } \theta + \sin \theta \cos \theta \right) } { \sin \theta \cos \theta ( \sin \theta - \cos \theta ) }{/tex}{tex}\left[ {\because {a^3} - {b^3} = (a - b)({a^2} + ab + {b^2}} \right){/tex}
{tex}= \frac { \sin ^ { 2 } \theta } { \sin \theta \cos \theta } + \frac { \cos ^ { 2 } \theta } { \sin \theta \cos \theta } + \frac { \sin \theta \cos \theta } { \sin \theta \cos \theta }{/tex}
{tex}= \tan \theta + \cot \theta + 1 = 1 + \tan \theta + \cot \theta = R . H S \text { proved }{/tex}
Since, {tex}\tan A = \frac{{\sin A}}{{\cos A}}{/tex}
{tex}\cot A = \frac{{\cos A}}{{\sin A}}{/tex}
Posted by Abhishek Jain 6 years, 6 months ago
- 1 answers
Sia ? 6 years, 6 months ago
Let {tex}p(x) =ax^3 + 3x^2 - bx - 6{/tex}
{tex}\because{/tex} - 1 is a zero
{tex}\therefore{/tex} {tex}p(-1) =0{/tex}
{tex}\Rightarrow{/tex} {tex}a(-1)^3 + 3(-1)^2 - b(-1) - 6 = 0{/tex}
{tex}\Rightarrow{/tex} {tex}-a + 3 + b - 6 = 0{/tex}
{tex}\Rightarrow{/tex} {tex}-a + b = 3.{/tex}.........(i)
Also, -2 is another zero
{tex}\therefore{/tex} {tex}p(-2) =0{/tex}
{tex}\Rightarrow{/tex} {tex}a(-2)^3 + 3(-2)^2 - b(-2) - 6 = 0{/tex}
{tex}\Rightarrow{/tex} {tex}- 8a + 12 + 2b - 6 = 0{/tex}
{tex}\Rightarrow{/tex} {tex}- 8a + 2b = -6{/tex}
{tex}\Rightarrow{/tex} {tex}4a - b = 3{/tex}.........(ii)
Solving equation (i) and (ii), we get
a = 2 and b = 5
{tex}\therefore{/tex} p(x) = 2x3 + 3x2 - 5x - 6
Let third zero = k
Sum of zeroes = {tex}\frac { - 3 } { 2 }{/tex}
{tex}\Rightarrow{/tex} -1 + (-2) + k = {tex}\frac { - 3 } { 2 } \Rightarrow k = \frac { - 3 } { 2 } + 3 = \frac { 3 } { 2 }{/tex}
Therefore, third zero = {tex}\frac { 3 } { 2 }{/tex}
Posted by Eshu Rajwal 6 years, 11 months ago
- 3 answers
Pavan Kumar 6 years, 11 months ago
Aditya Maurya 6 years, 11 months ago
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Puja Sahoo? 6 years, 11 months ago
4Thank You