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secondary 4 | A Maths
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Qn 5 and 6, thanks for the help
Date Posted: 6 years ago
Views: 269
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Date Posted:
6 years ago
You mean starting the dy/dt = 0 (rej) onwards?
You may bring the 1 to RHS then cross multiply...
A) From dy /dt = 7/(x+3)² * 2 dy/dt, you get
dy /dt = 14/(x+3)² * dy /dt (times the 7 with the 2), then bring the RHS to LHS you get,
B) dy/dt - 14/(x+3)² * dy /dt = 0, factorise the dy /dt becomes,
C) dy/dt [ 1 - 14/(x+3)² ] = 0, this is a product of two terms each term = 0, so
dy/dt = 0 (definitely reject),
D) the other term = 0 get 1 - 14/(x+3)² = 0...
dy /dt = 14/(x+3)² * dy /dt (times the 7 with the 2), then bring the RHS to LHS you get,
B) dy/dt - 14/(x+3)² * dy /dt = 0, factorise the dy /dt becomes,
C) dy/dt [ 1 - 14/(x+3)² ] = 0, this is a product of two terms each term = 0, so
dy/dt = 0 (definitely reject),
D) the other term = 0 get 1 - 14/(x+3)² = 0...
There is a rate of change in x, the rate of change in y cannot be 0, so reject dy/dt = 0.
You solve the other term, see part D)...
Qn says the rate of change in x is twice the rate of change in y, then dx/dt also = 0... Cannot find x...
You cannot find x using or solving dy/dt = 0....
I can't see any Qn7 from you.. Maybe someone has answered... Or you repost...
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