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Mathematics: Post your doubts here!

whitecorp

whitecorp

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minhal said:
find equation in form y=x^2+bx+c of a parabola which passes thru (-3,0) and (1,-16).....plz help!
Click to expand...
Substitute in the two set of coordinates:

0=(-3)^2 - 3b+c ======> c=3b-9-------(1)
-16= (-1)^2 -b + c ======> c =b-17 ---------(2)
Equating (1) and (2) gives 3b-9 = b-17 or b=-4, c= -21

Therefore, the equation of the required parabola is y=x^2 -4x -21 . (shown)

Hope this helps. Peace.
 
whitecorp

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iKhaled said:
excuse me, is it x^2 + bx + c or ax^2 + bx + c ?
Click to expand...

It should be x^2 +bx+c, not ax^2+bx+c.
The question only cited two sets of coordinates, which mean you can only have two unknowns, not three. Hope this helps. Peace.
 
iKhaled

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whitecorp said:
It should be x^2 +bx+c, not ax^2+bx+c.
The question only cited two sets of coordinates, which mean you can only have two unknowns, not three. Hope this helps. Peace.
Click to expand...
yeah thats y i asked, i thought it had 3 unknowns which made me feel confused anyway, can u pls answer my question if u can ?

need explanation to the last part part of question 6 ( oct/nov 04). how do we find the maximum value in the range? y the mark scheme says " 3 + 2 " ?
http://www.xtremepapers.com/papers/...AS Level/Mathematics (9709)/9709_w04_qp_1.pdf
 
whitecorp

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5sin^x +3 cos ^x = 2 sin^2 x + ( 3sin^x +3 cos ^x) = 2sin^2 x + 3
Clearly, the maximum value of sin^2 x is 1 when x=pi/2, so the overall maximum value is simply 2+3 =5. (shown)

Hope this helps. Peace.
 
iKhaled

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whitecorp said:
5sin^x +3 cos ^x = 2 sin^2 x + ( 3sin^x +3 cos ^x) = 2sin^2 x + 3
Clearly, the maximum value of sin^2 x is 1 when x=pi/2, so the overall maximum value is simply 2+3 =5. (shown)

Hope this helps. Peace.
Click to expand...
my question is y 2+3? :(:(:( thats the part i cant get :/
 
M

minhal

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whitecorp said:
Substitute in the two set of coordinates:

0=(-3)^2 - 3b+c ======> c=3b-9-------(1)
-16= (-1)^2 -b + c ======> c =b-17 ---------(2)
Equating (1) and (2) gives 3b-9 = b-17 or b=-4, c= -21

Therefore, the equation of the required parabola is y=x^2 -4x -21 . (shown)

Hope this helps. Peace.
Click to expand...
but the answer is x^2-2x-15
 
parthrocks

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whitecorp

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iKhaled said:
hey guys, i need help in mechanics m/j 10

question 6 the 2nd part, why the acceleration is -3 and different than the first part which was 6? why the tension wasn't included in the calculation of the second part?

http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s10_qp_41.pdf
Click to expand...

For the second part, the string has slackened and hence there will be no more tension in it. Clearly, since A is part of the original connected system,
when T=0, its acceleration will also change accordingly.

Hope this helps. Peace.
 
iKhaled

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whitecorp said:
For the second part, the string has slackened and hence there will be no more tension in it. Clearly, since A is part of the original connected system,
when T=0, its acceleration will also change accordingly.

Hope this helps. Peace.
Click to expand...
y the string has slackened ?
 
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