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Online Classes Malayalam Mother Tongue (Native)
English Proficient
Hindi Basic
VIT University 2013
Master of Engineering - Master of Technology (M.E./M.Tech.)
Labour Colony Road Thammanam, Kochi, India - 682032
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Class Location
Online class via Zoom
Student's Home
Tutor's Home
Years of Experience in Class 10 Tuition
3
Board
IGCSE
Subjects taught
Geography, Physics
Taught in School or College
Yes
Answered on 08/07/2020
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5x2 + 5y2 + 5z2 = 4xy + 4yz + 4zx
==> (4x2 + x2) + (4y2 + y2) + (4z2 +y2) = 4xy + 4yz + 4zx
==> [(2x)2 - 2(2x)(y) + y2] + [(2y)2 - 2(2y)(z) +z2] + [(2z)2 - 2(2z)(x) +x2] = 0
==> (2x-y)2 + (2y-z)2 + (2z-x)2 = 0
This is possible only when each component is zero.
==> (2x-y)2 = 0 ; (2y-z)2 = 0 ; (2z-x)2 = 0
==> 2x =y ; 2y = z; & 2z = x
==> x/y = 1/2 ; y/z = 1/2 ; z/x = 1/2
==> x = y/2 = z/4; y = z/2
x:y:z = 1:2:4
Answered on 08/07/2020
Ask a Question
5x2 + 5y2 + 5z2 = 4xy + 4yz + 4zx
==> (4x2 + x2) + (4y2 + y2) + (4z2 +y2) = 4xy + 4yz + 4zx
==> [(2x)2 - 2(2x)(y) + y2] + [(2y)2 - 2(2y)(z) +z2] + [(2z)2 - 2(2z)(x) +x2] = 0
==> (2x-y)2 + (2y-z)2 + (2z-x)2 = 0
This is possible only when each component is zero.
==> (2x-y)2 = 0 ; (2y-z)2 = 0 ; (2z-x)2 = 0
==> 2x =y ; 2y = z; & 2z = x
==> x/y = 1/2 ; y/z = 1/2 ; z/x = 1/2
==> x = y/2 = z/4; y = z/2
x:y:z = 1:2:4
Ask a Question
Also have a look at
Class Location
Online class via Zoom
Student's Home
Tutor's Home
Years of Experience in Class 10 Tuition
3
Board
IGCSE
Subjects taught
Geography, Physics
Taught in School or College
Yes
Answered on 08/07/2020
Ask a Question
5x2 + 5y2 + 5z2 = 4xy + 4yz + 4zx
==> (4x2 + x2) + (4y2 + y2) + (4z2 +y2) = 4xy + 4yz + 4zx
==> [(2x)2 - 2(2x)(y) + y2] + [(2y)2 - 2(2y)(z) +z2] + [(2z)2 - 2(2z)(x) +x2] = 0
==> (2x-y)2 + (2y-z)2 + (2z-x)2 = 0
This is possible only when each component is zero.
==> (2x-y)2 = 0 ; (2y-z)2 = 0 ; (2z-x)2 = 0
==> 2x =y ; 2y = z; & 2z = x
==> x/y = 1/2 ; y/z = 1/2 ; z/x = 1/2
==> x = y/2 = z/4; y = z/2
x:y:z = 1:2:4
Answered on 08/07/2020
Ask a Question
5x2 + 5y2 + 5z2 = 4xy + 4yz + 4zx
==> (4x2 + x2) + (4y2 + y2) + (4z2 +y2) = 4xy + 4yz + 4zx
==> [(2x)2 - 2(2x)(y) + y2] + [(2y)2 - 2(2y)(z) +z2] + [(2z)2 - 2(2z)(x) +x2] = 0
==> (2x-y)2 + (2y-z)2 + (2z-x)2 = 0
This is possible only when each component is zero.
==> (2x-y)2 = 0 ; (2y-z)2 = 0 ; (2z-x)2 = 0
==> 2x =y ; 2y = z; & 2z = x
==> x/y = 1/2 ; y/z = 1/2 ; z/x = 1/2
==> x = y/2 = z/4; y = z/2
x:y:z = 1:2:4
Ask a Question
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