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if p+q+r=16, b+c=4 and x,y,z are the solution of linear equations


(x/p)+(y/p-b)+(z/p-c)=1


(x/q)+(y/q-b)+(z/q-c)=1


(x/r)+(y/r-b)+(z/r-c)=1 then x+y+z is

4 years ago

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plz can u give me solution of q48 and q49
 
 
q49. Sn will be equal to coff. of x in : (1+1/x) n .(1+x) n = ∑ n C i * n C i+1 (that is power x i+1 from second bracket multiplied by power x i of first, and i varying from 0 to n-1....
 
Akshay 9 months ago
 
will be the series in question. n . cofficient of x 2n-2i (-1) j (1+x) i C n = ∑ n -1) 2 q48. Assume ((x+1) Expand (x+1) 2 in LHS and solve. ATB.
 
Akshay 9 months ago
 
q48. Assume ((x+1) 2 -1) n = ∑ n C i (1+x) 2n-2i . cofficient of x n will be the series in question. Expand (x+1) 2 in LHS and solve. ATB.
 
Akshay 9 months ago
2 PLUS 2
 
 
answer is 2+2 = 4
 
A M S ARUN KRISHNA 9 months ago
 
2+2
 
KARTHIK 7 months ago
 
4
 
pa1 7 months ago
x 2 +2x+1=0,x=?
 
 
for this question is the expansion is x^2+2x+1=0 for u have put x= -1 in that equqtion you will get 0 so x= -1 value is satisfies the equation so that the answer is x= -1 thank u boss
 
AMMANANNA 10 days ago
 
roots for the equation ( x 2 +2x+1) is x=-1 this equation also written as (x+1)^2=0 that implies x+1=0 therefore x=-1
 
RAMCHANDRARAO 10 days ago
 
-1=x.
 
OUTPASS 7 months ago
A straight line x=y+2 touches the circle (x^2 + y^2) = r^2. The value of r is
 
 
Equation of circle :4(x 2 +y 2 )=r 2 centre of the circle = (0, 0) equation of tangent :x=y+2 [y=x-2].....(i) slope of the tangent = 1 slope of radius = -1 using one-point form :(y-y 1 ) =...
 
BALAJI SANKARAN 5 months ago
 
We have the equation of the circle as 4( x^2+y^2)=r^2 X^2+y^2=(r/2)^2 Now x=y+2 is a tangent to the circle.... Centre of the circle is (0,0) Tangent is always perpendicular to the radius... ...
 
Trina 6 months ago
hi I’m in tenth and i cant understand trigonometry . can you please help by telling me a way to learn this chapter?
 
 
trigonometry....yeah its a tough but an easy topic....am not an expert advice you....but still...the only way of learning things is be very clear with your basics and concepts plus formulas ...
 
darshini 8 months ago
 
tahnks alot :D i will try that
 
tenzin norzin 8 months ago
FIND UNIT VECTORS PERPENDICULAR TO THE PLANE DETERMINED BY THE POINTS P(1,-1,2) , Q(2,0,-1) AND R(0,2,1).
 
 
The two vectors in the given plane is PQ & PR PQ = (2 – 1) i + (0 + 1) j + (-1 – 2)k = i + j – 3k PR = (0 – 1)i + (2 + 1) j + (1 – 2k = –i + 3j – k The vector perpendicular to PQ & PR is PQ ...
 
Y RAJYALAKSHMI one year ago
 
The two vectors in the given plane is PQ & PR PQ = (2 – 1) i + (0 + 1) j + (-1 – 2)k = i + j – 3k PR = (0 – 1)i + (2 + 1) j + (1 – 2k = –i + 3j – k The vector perpendicular to PQ & PR is PQ ...
 
SHAIK HAFEEZUL KAREEM one month ago
 
THANKS!!!!
 
Bharat Makkar one year ago
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