Mathematics: Post your doubts here!

[iKhaled]

find equation in form y=x^2+bx+c of a parabola which passes thru (-3,0) and (1,-16).....plz help!

excuse me, is it x^2 + bx + c or ax^2 + bx + c ?

 

[whitecorp]

find equation in form y=x^2+bx+c of a parabola which passes thru (-3,0) and (1,-16).....plz help!

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]

excuse me, is it x^2 + bx + c or ax^2 + bx + c ?

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]

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.

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]

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]

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.

my question is y 2+3? thats the part i cant get :/

 

[whitecorp]

my question is y 2+3? thats the part i cant get :/

The 2 comes from the 2sin^2 x part , and the 3 at the back gives you a total of 2+3=5.

 

[minhal]

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.

but the answer is x^2-2x-15

 

[ZohaibAsad]

What is going on ?

 

[iKhaled]

What is going on ?

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

can u explain this question pls? that's whats going on.. :/ i cant get iiittt

 

[Amy Bloom]

Help!

 

[iKhaled]

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/...S Level/Mathematics (9709)/9709_s10_qp_41.pdf

 

[parthrocks]

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

I havent taken mechanics!!so sorry!!!anyone who has taken p3 and can solve my doubt!!!help

 

[Amy Bloom]

Needs a helping hand!

 

[whitecorp]

but the answer is x^2-2x-15

My bad, I misread your second set of coordinates as (-1,-16) when it should be (1, -16).
Simply make the adjustments and you will get the correct answer.

Hope this helps. Peace.

 

[whitecorp]

Help!
View attachment 15411

Here are the full detailed workings:





Hope this helps. Peace.

 

[whitecorp]

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

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]

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.

y the string has slackened ?

 

[bamteck]

Please someone help me for no. 11 (iii) whitecorp

http://www.xtremepapers.com/papers/...AS Level/Mathematics (9709)/9709_s09_qp_1.pdf

 

[whitecorp]

y the string has slackened ?

The particle B is resting on the ground, hence as A moves forward, the string is no longer taut. You can try to visualize this to convince yourself.
Hope it helps. Peace.