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Kishore 04/06/2016 in  Chemistry (Class XI-XII Tuition (PUC)), Class XI-XII Tuition (PUC), Mathematics(Class XI-XII Tuition (PUC))

Need to score more marks in supplementary.. need to write exam in the month July

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Kamal replied | 02/08/2016

contact me

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Divya 28/05/2016 in  Mathematics(Class XI-XII Tuition (PUC))

can i get mr.kannana as my professor?

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Praveen Singh replied | 27/09/2016

contect me

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Sagar 26/05/2016 in  Class XI-XII Tuition (PUC), Mathematics(Class XI-XII Tuition (PUC))

can someone help me to solve this

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Abhishek C replied | 15/06/2016

M is 5 and N is -4

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Shalini 26/05/2016 in  Class XI-XII Tuition (PUC), Mathematics(Class XI-XII Tuition (PUC))

I received 5 recommendations for my search. How can i get more options, suggestions or choices?

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U.k. 01/05/2016 in  Class XI-XII Tuition (PUC), Mathematics(Class XI-XII Tuition (PUC))

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Chandan replied | 12/06/2016

5

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Chakraborty T replied | 19/06/2016

Answer is 4

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Rajesh replied | 28/05/2016

520

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Bimal replied | 01/06/2016

a+6d= 40 ....(1)
S13 = 13/2 [2a+(13-1)d]
= 13/2*2 [ a+6d] ........(2)
= 13 *40. From eq (1)
' = 520

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A

Ashutosh replied | 12/05/2016

0.20788
exp(i*x) = cos(x) + i*sin(x)
exp(i*Pi/2) = cos(Pi/2) + i*sin(Pi/2) = i
square both the sides now.
i^i = exp(i*i*Pi/2) = exp(-Pi/2) = 0.20788
Interesting fact is that i raised to the i-th power is actually a real number!

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Md Aslam Ali replied | 24/05/2016

we know that e^(ix)= cos(x)+isin(x)
where i=sqrt(-1)
so e^i=cos(1)+isin(1)
=0.5403+0.8415i
so, i*e^i=i*(0.5403+0.8415i)
=0.5403i+0.8415*i^2
=0.5403i-0.8415

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U.k. 01/05/2016 in  Class XI-XII Tuition (PUC), Mathematics(Class XI-XII Tuition (PUC))

find the directrix of the parabola x^2 - 4x - 8y + 12 = 0

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Eti replied | 06/05/2016

y=0

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Kamal replied | 03/06/2016

y=-1

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U.k. 01/05/2016 in  Class XI-XII Tuition (PUC), Mathematics(Class XI-XII Tuition (PUC))

find the number of words that can be formed out of the letters of the word "ARTICLE" so that the vowels occupy even places ?

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Kalidasan replied | 06/05/2016

among 7 places, number of even places = 3, number of odd places = 4
given number of vowels =3 (a, e i) . number of consonants = 4 ( r,t, c l)
number of ways filling 3 even places with 3 vowels = 3! = 6 and number of ways filling 4 odd places with 4 consonants = 4! = 24 therefore total number of ways =( 6).( 24 ) = 144.

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Sahil replied | 07/05/2016

4!*3!=144

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Lakshya 12/04/2016 in  Class XI-XII Tuition (PUC), Mathematics(Class XI-XII Tuition (PUC))

I am in 11th and please tell should i take coching classes or tution classes . please reply to as fast as you can.

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Neel replied | 15/04/2016

Take one to one tuition class because in coaching class you'll not get enough attention as there will be multiple students....& also most coaching classes ash hefty amount without any quality education. I take both one to one and coaching classes in academy and so I am well aware of both...

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Kamal replied | 03/06/2016

try on your convenience

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Yeswanth 08/04/2016 in  Class XI-XII Tuition (PUC), Mathematics(Class XI-XII Tuition (PUC))

How many triangles can be constructed so that lengths of the sides are three consecutive odd integers and the perimeter is less than 1000?

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N

Neel replied | 15/04/2016

Let the sides be x,x+2,x+4 ,x being an odd integer then , x+(x+2)+(x+4)<1000 , i.e, 3*x<994 ,x being an odd integer. Max value of x can 331
Now x can be any odd no from 1 to 329...like 1,3,5,7,..............,331 and for every such x, you will get one triangle..
From arithmetic progression......331=1+(n-1)*2, i.e.
(n-1)*2 = 330
(n-1) = 165
n=166
which...  more»
Let the sides be x,x+2,x+4 ,x being an odd integer then , x+(x+2)+(x+4)<1000 , i.e, 3*x<994 ,x being an odd integer. Max value of x can 331
Now x can be any odd no from 1 to 329...like 1,3,5,7,..............,331 and for every such x, you will get one triangle..
From arithmetic progression......331=1+(n-1)*2, i.e.
(n-1)*2 = 330
(n-1) = 165
n=166
which gives n=166
So 166 possible triangles «less

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Kamal replied | 03/06/2016

166

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