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Ask QuestionPosted by Himangi Mishra 8 years, 7 months ago
- 1 answers
Posted by Mohit Mathur 8 years, 7 months ago
- 2 answers
Saumya Singh 8 years, 7 months ago
Sumit Ahirwar 8 years, 7 months ago
Posted by Lavish Kumar 8 years, 7 months ago
- 1 answers
Posted by Ishan Yadav 8 years, 7 months ago
- 5 answers
Hans Raj 8 years, 7 months ago
No , 1 is not a prime number
because a prime number is a number GREATER THAN ONE which can only be divided by 1 and itself, which are 2 numbers but in case of 1 , there is only 1 number that is 1 , hence 1 is not a prime number
Posted by Ishika Bhardwaj 8 years, 7 months ago
- 2 answers
Ishika Bhardwaj 8 years, 7 months ago
Payal Singh 8 years, 7 months ago
False – A number line may have negative or positive number. Since, no negative can be the sqauare root of a natural number, thus every point on the the number line cannot be in the form of {tex}\sqrt m{/tex} , where m is a natural number.
Posted by Master Aishwarya Jivtode 8 years, 7 months ago
- 2 answers
Hans Raj 8 years, 7 months ago
2{tex} \sqrt{3}{/tex} + {tex} \sqrt{3}{/tex}
Taking out {tex} \sqrt{3}{/tex} which is common in both
= {tex} \sqrt{3}{/tex} ( 2 + 1)
= 3 {tex} \sqrt{3}{/tex}
Posted by Abir Hazarika 8 years, 7 months ago
- 2 answers
Hans Raj 8 years, 7 months ago
since we want six rational numbers , we write 1 and 2 as rational numbers with denominator 6 + 1 = 7
so multiply in numerator and denominator by 7 we get
1/100 x 7/7 = 7/700 3/100 x 7/7 = 21/700
we know that 7 < 8 <9 <10 <11 <12 <13 <21
7/700 < 8/700 < 9/700 < 10/700 <11/700 < 12/700 < 13/700 < 21/700
hence stx rational numbers between .01 = 7/700 and .03 = 21/700 are
8/700 , 9/700, 10/700, 11/700, 12/700, 13/700
Sumit Ahirwar 8 years, 7 months ago
Posted by Reena Karim 8 years, 7 months ago
- 2 answers
Dharmendra Kumar 8 years, 7 months ago
Given x=3+{tex}{\sqrt 3}{/tex}
Then {tex}1\over x{/tex} = {tex}1\over 3+{\sqrt 3}{/tex} ={tex}{1\over 3+{\sqrt 3}}×{{3-{\sqrt 3}}\over 3-{\sqrt 3}}{/tex} [by rationalisation]
={tex}3-{\sqrt 3}\over 3^2-({\sqrt 3})^2{/tex} [by using identity (a+b)(a-b)=a2-b2]
= {tex}3-{\sqrt 3}\over 9-3{/tex}
= {tex}3-{\sqrt 3}\over 6{/tex}
So x+{tex}1\over x{/tex} ={tex}3+{\sqrt 3}{/tex} {tex}+{{3-{\sqrt 3}}\over 6}{/tex}
= {tex}6×(3+{\sqrt 3})+3-{\sqrt 3}\over 6{/tex}
= {tex}18+6{\sqrt 3}+3-{\sqrt 3}\over 6{/tex}
= {tex}21+5{\sqrt 3}\over 6{/tex}
Poulami Dasgupta 8 years, 7 months ago
x=3+{tex}\sqrt3{/tex}
So, {tex}x+1\over x{/tex}={tex}3+\sqrt3+1\over3+\sqrt3{/tex}={tex}4+\sqrt3\over3+\sqrt3{/tex}={tex}(4+\sqrt3)(3—\sqrt3)\over(3+\sqrt3)(3—\sqrt3){/tex}
={tex}12—4\sqrt3+3\sqrt3—3\over(3)\ ^2—(\sqrt3)\ ^2{/tex}
={tex}9—\sqrt 3\over 9—3{/tex}
={tex}9—\sqrt3\over6{/tex}
Posted by Sumit Ahirwar 8 years, 7 months ago
- 1 answers
Poulami Dasgupta 8 years, 7 months ago
Since x+2 and x—1 are the factors of {tex}x\ ^3 + 10 x\ ^2+mx+n{/tex}
So, p(-2)=0
{tex}(-2)\ ^3+10(-2)\ ^2+m(-2)+n{/tex}=0
—8+40—2m+n=0
32—2m= —n------------(i)
and,p(1)=0
{tex}(1)\ ^3+10(1)\ ^2+m(1)+n=0{/tex}
1+10+m+n=0
11+m=—n--------------(ii)
equating (i) and (ii) we get,
32—2m=11+m
32—11=m+2m
21=3m
{tex}21\over3{/tex}=m
m=7
Putting m=7 in (i) we get,
32—2m=—n
32—2(7)=—n
32—14=—n
18=—n
n=—18
Posted by Ravi Kumar 8 years, 7 months ago
- 2 answers
Posted by Suraj Pandat 8 years, 7 months ago
- 2 answers
Hans Raj 8 years, 7 months ago
let us take a right angled triangle ABC right Ld at B
AB = one side
BC = another side
AC = 3rd side of triangle opposite to right angle is the hypotenuse
Bhavesh Verma 8 years, 7 months ago
Posted by Mayank Agarwal 8 years, 7 months ago
- 0 answers
Posted by Navneet Singh Nippy 8 years, 7 months ago
- 0 answers
Posted by Rohit Dev 8 years, 7 months ago
- 1 answers
Poulami Dasgupta 8 years, 7 months ago
Let the angle be x
According to question
x= 14+90—x
x+x=104
2x=104
x={tex}104\over 2{/tex}=52
The angle is 52°
Posted by Keshav Baghel 8 years, 7 months ago
- 0 answers
Posted by Shivam Malik 8 years, 7 months ago
- 1 answers
Hans Raj 8 years, 7 months ago
using Heron's formula
let sides of triangle A = a, b, c and of tringle B = 2a , 2b , 2c
s = a+b+c/2 , 2s = a+b+c s' = 2a + 2b + 2c/2 = a+b+c
s' = 2s
area of A = {tex} \sqrt{}{/tex}s(s-a)(s-b)(s-c)
area of B ={tex} \sqrt{}{/tex} 2s(2s-2a)(2s-2b)(2s-2c) = 4 {tex} \sqrt{}{/tex}s(s-a)(s-b)(s-c)
increase in area = 3 {tex} \sqrt{}{/tex} s(s-a)(s-b)(s-c)
percentage increase = 3 x 100 = 300%
Posted by Riya Agrawal 8 years, 7 months ago
- 0 answers
Posted by Aayush Patidar 8 years, 7 months ago
- 0 answers
Posted by Utkarsh Singh 8 years, 7 months ago
- 0 answers
Posted by Divyansh Khandelwal 8 years, 7 months ago
- 0 answers
Posted by Suraj Kumar 8 years, 7 months ago
- 1 answers
Poulami Dasgupta 8 years, 7 months ago
{tex}{x+1\over x}=a+b{/tex}--------(i)
{tex}{x—1\over x}=a—b{/tex}--------(ii)
adding (i) and (ii)
{tex}{x+1\over x}+{x—1\over x}=2a{/tex}
{tex}{x+1+x—1\over x}=2a{/tex}
{tex}2x\over x{/tex}=2a
a=1
Subtracting (ii) from (i) we get,
{tex}{x+1\over x}—{x—1\over x}=2b{/tex}
{tex}{x+1—x+1\over x}=2b{/tex}
{tex}2\over x{/tex}=2b
{tex}b={1\over x}{/tex}
{tex}x={1\over b}{/tex}
Posted by Sahil Pandit 8 years, 7 months ago
- 1 answers
Posted by Pig Bhuvanesh 8 years, 7 months ago
- 6 answers
Posted by Bandari Sunny 8 years, 7 months ago
- 0 answers
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Aman Kumar 8 years, 7 months ago
1Thank You