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Ask QuestionPosted by Yash Kumar 8 years, 4 months ago
- 1 answers
Posted by Nimrat Singla 8 years, 4 months ago
- 1 answers
Soumya Ghoshal 8 years, 4 months ago
{tex}x^3-1/x^3=(x-1/x)^3+3×x×1/x(x-1/x){/tex}
or,{tex}14= a^3+3a where x-1/x=a{/tex}
or,{tex}a^3+3a-14=0{/tex}
or,{tex}a³-2a²+2a²-4a+7a-14=0{/tex}
or,{tex}a²(a-2)+2a(a-2)+7(a-2)=0{/tex}
or,{tex}(a-2)(a²+2a+7)=0{/tex}
Then,{tex}a=2{/tex}
So,{tex} x-1/x=2{/tex}
Posted by Kavya Ps 8 years, 4 months ago
- 1 answers
Dharmendra Kumar 8 years, 4 months ago
No such triangle is possible because the sum of angles of a triangle is 180°.
Here anlge A+angle B+angleC=50+130+40=220>180.
Posted by Vaishnavi Yadav 8 years, 4 months ago
- 0 answers
Posted by Rahul Kunwar 8 years, 4 months ago
- 0 answers
Posted by Abhilasha Gahlot 8 years, 4 months ago
- 1 answers
Posted by Vikas Singh 8 years, 4 months ago
- 1 answers
Hans Raj 8 years, 4 months ago
Technically invented by many cultures around the world , Indians , Greeks, chinese , Ancient Roman used first math , i think Aristotle invented math but before him many used counting things e,g Albert Einstein
whereas Al-Khwarizmi is considered as father of Algebra
Posted by Brundha Brundha 8 years, 4 months ago
- 2 answers
Hans Raj 8 years, 4 months ago
({tex}{60 \over 1.70}{/tex}) x{tex}1.70{/tex}
= {tex}{60 \over 1.70}{/tex} x {tex}1.70{/tex}
= 60
Posted by Raja Jj 8 years, 4 months ago
- 1 answers
Poulami Dasgupta 8 years, 4 months ago
It should be angles
the space (usually measured in degrees) between two intersecting lines or surfaces at or close to the point where they meet.
Posted by Divya Gupta 8 years, 4 months ago
- 3 answers
Hans Raj 8 years, 4 months ago
The abscissa is the x or horizontal coordinae and the ordinate is the y or vertical co-ordinate
the horizontal x and vertical y axes on the Cartisian plane are perpendiculr to each other and they intersect at the point of origin whose co-ordinate is always at (0 , 0)
Abscissa is the x value and ordinate is the y value
the vertical axis of a graph is known as ordinate while horizontal axis of a graph is known as Abscissa
therefore while talking about a point (x , y ) on the graph , we refer to x value as the Abscissa and y value as the ordinate
Rahul Rajput 8 years, 4 months ago
Poulami Dasgupta 8 years, 4 months ago
Co-ordinate means the x axis value and y axis value written together like(x,y)
Abscissa is the value of x-coordinate
Posted by Rudresh Shukla 8 years, 4 months ago
- 4 answers
Hans Raj 8 years, 4 months ago
k = 8 - 2 /8
= 8 - 1/4 ( by using BODMAS rule i.e first solve D and then S )
= (32 - 1) / 4
k = 31/4
k = 7.75
Posted by Naman Shinde 8 years, 4 months ago
- 1 answers
Pratibha Behl 8 years, 4 months ago
Posted by Shuchita Bardhan 8 years, 4 months ago
- 0 answers
Posted by Sunil Kumar Pasupuleti 8 years, 4 months ago
- 0 answers
Posted by Harshita Mittal 8 years, 4 months ago
- 0 answers
Posted by Prince Raj 8 years, 4 months ago
- 1 answers
Hans Raj 8 years, 4 months ago
assuming that 30a=6b = 5c is given
then divide by 30 throughout
a = b/5=c/6
a;b;c = 1;5;6
1x + 5x + 6x = 180
12 x = 180
x = 15
a= 15
b = 75
c = 90
Posted by Ravi Kant Dwivedi 8 years, 4 months ago
- 0 answers
Posted by Ritika Agrawal 8 years, 4 months ago
- 1 answers
Hans Raj 5 years, 6 months ago
semi-perimeter = s =35 + 54 + 61/2=150/2= 75
Area= {tex} \sqrt{s(s - a) (s - b) (s - c)}{/tex} Heron's formula
= {tex}\sqrt75({75 - 35)(75 - 54)(75 - 61)}{/tex}
= {tex} \sqrt{75(40)(21)(14)}{/tex}
= {tex} \sqrt{}{/tex} 25x3x5x8x3x7x7x2
= {tex} \sqrt{}{/tex} 52x32x72x42x5
= 5x3x7x4x {tex} \sqrt{}{/tex}5
= 420 {tex} \sqrt{5}{/tex} cm2
Posted by Riya Sachan 8 years, 4 months ago
- 0 answers
Posted by Anna Maryjoseph 8 years, 4 months ago
- 0 answers
Posted by Sandra Fernandes 8 years, 4 months ago
- 1 answers
Rashmi Bajpayee 8 years, 4 months ago
{tex}{\left( {a + b} \right)^2} = 2{a^2} + 2{b^2}{/tex}
=> {tex}{a^2} + {b^2} + 2ab = 2{a^2} + 2{b^2}{/tex}
=> {tex}2ab = {a^2} + {b^2}{/tex}
=> {tex}{a^2} + {b^2} - 2ab = 0{/tex}
=> {tex}{\left( {a - b} \right)^2} = 0{/tex}
=> {tex}a-b=0{/tex}
=> {tex}a=b{/tex}
Hence proved
Posted by Navneet Kaur Ahuja 8 years, 4 months ago
- 0 answers
Posted by Raj Gaur 8 years, 4 months ago
- 0 answers
Posted by Ankush Sehrawat 8 years, 4 months ago
- 0 answers
Posted by Rohit Kumar 8 years, 4 months ago
- 0 answers
Posted by Adnan Guard 8 years, 4 months ago
- 0 answers
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Rashmi Bajpayee 8 years, 4 months ago
{tex}{{5 + 2\sqrt 3 } \over {7 + 4\sqrt 3 }} = a - b\sqrt 3 {/tex}
=> {tex}{{5 + 2\sqrt 3 } \over {7 + 4\sqrt 3 }} \times {{7 - 4\sqrt 3 } \over {7 - 4\sqrt 3 }} = a - b\sqrt 3 {/tex}
=> {tex}{{35 - 20\sqrt 3 + 14\sqrt 3 - 24} \over {49 - 48}} = a - b\sqrt 3 {/tex}
=> {tex}11 - 6\sqrt 3 = a - b\sqrt 3 {/tex}
On comparing, we get
{tex}a = 11,b = - 6{/tex}
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