if 4x^2 +y^2=40,xy= 6 find the value of (2 x +y)
Answers
Answered by
7
Here is the solution to your questions..
4x^2 + y^2 = 40
( 2x )^2 + y^2 = 40
( 2x + y )^2 - 2*2x*y = 40
( 2x + y )^2 - 2*2*6 = 40
(2x + y )^2 - 24 = 40
( 2x + y )^2 = 40 + 24
( 2x + y )^2. = 64
2x + y = √ 64 = 8 Ans
4x^2 + y^2 = 40
( 2x )^2 + y^2 = 40
( 2x + y )^2 - 2*2x*y = 40
( 2x + y )^2 - 2*2*6 = 40
(2x + y )^2 - 24 = 40
( 2x + y )^2 = 40 + 24
( 2x + y )^2. = 64
2x + y = √ 64 = 8 Ans
Anonymous:
??
nothing
But the answer is 8 -_-
yes
i get that
see my amswer now
yes
now yours is correct
prove ( a- b) ^2 + (b-c ) ^2 + (c- a) ^2 = 2a^2 +2b^2 +2c^2- 2ab -2bc -2ca
can't answer here, pls make a separate question
Answered by
5
answer is 8
hope I helped :)
hope I helped :)
Attachments:
Similar questions