IGCSE/GCSE/O & A Level/IB/University Student Forum
Qualification => Subject Doubts => GCE AS & A2 Level => Math => Topic started by: halosh92 on April 07, 2010, 08:02:40 pm
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A circle has the equation
x2 + y2 + 8x ? 4y + k = 0,
where k is a constant.
(a) Find the coordinates of the centre of the circle. (2)
Given that the x-axis is a tangent to the circle,
(b) find the value of k.
for part b...how to get the radius exactly?
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Because the exapression completes the square as (y+2)^2 it is two from the x acxis to the centre so the radius of the circle is 2.
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can u explain how they got the angle here on wat basis?
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Sketch on the same diagram the graphs of y = sin 2x and y = tan
2
x for x
in the interval 0 ? x ? 360°.
(b) Hence state how many solutions exist to the equation
sin 2x = tan
2
x ,
for x in the interval 0 ? x ? 360° and give a reason for your answer.
I have the answers i just need an explanation for part a
thx
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When I get up
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could someone plz answer this question?? :-\ ??? ??? ???
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4 only include one or roots at 360, 0
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4 only include one or roots at 360, 0
i dont get why u chose such places to places the tan curve :S
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They asked for answers in the range 0-360
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heres a way you can try
sin2x = tan2x
sin2x = sin2x / cos2x
sin2x cos2x = sin2x
sin2x cos2x - sin2x = 0
sin2x ( cos 2x - 1) = 0
sin 2x = 0 and cos 2x = 1
you come out with
x = 0, 90, 180, 270, 360
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a circle with centre O and radius 6 cm.
the chord PQ divides the circle into a minor segemnt R1 of area A1
and a major segment R2 of area A2. the chord PQ subtends an angle thita radians at O/
a) show that A1= 18 (thita- sin thita)
given that A2= 3A1 and f(thita) = 2 thita - 2sin thita - pie
b) prove that f(thita)=0
c) evaluate f(2.3) and f(2.32) and deduce that 2.3<thita>2.32
how to do part b and c
??
thxxx
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can sme one solve this ques
the coefficient of x3 in the epansion of (2+x)(3-ax)4 is 30
find the values of the constant a.
please show step by step working
Thanks
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can sme one solve this ques
the coefficient of x3 in the epansion of (2+x)(3-ax)4 is 30
find the values of the constant a.
please show step by step working
Thanks
a=1
a= sqrt 30
is this the answer? if not post the correct answer please :)
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A=1, (5+or-sqrt 105)/8
please show workings i need to know how to get da ans
Thanks
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give me 2 mins
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A=1, (5+or-sqrt 105)/8
please show workings i need to know how to get da ans
Thanks
which paper is this?
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can sme one solve this ques
the coefficient of x3 in the epansion of (2+x)(3-ax)4 is 30
find the values of the constant a.
please show step by step working
Thanks
here u go..:)
@halosh, i answered ur angle question, check the post
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here u go..:)
@halosh, i answered ur angle question, check the post
thx AF but i did the whole thing its correct except when i got stuck at solving the cubic function..how do i do that?
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For C2, you will always get easy cubics and thus you will always be able to find one factor!!
In this case, you can see that if a=1 the equation is equal so you realise that (a-1) is a factor.
now divide by (a-1) and you will get a quadratic which you can easily factorise again...
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For C2, you will always get easy cubics and thus you will always be able to find one factor!!
In this case, you can see that if a=1 the equation is equal so you realise that (a-1) is a factor.
now divide by (a-1) and you will get a quadratic which you can easily factorise again...
thx aloot ;D ;D
but isnt this a C2 question?
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You tell me :P You asked it haha
It looks like a C2 but it's been more than 2 years since I did C2 so I can't be sure!!
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You tell me :P You asked it haha
It looks like a C2 but it's been more than 2 years since I did C2 so I can't be sure!!
haha lool no i didnt ask the question someone else did
XD thx alot anyways :) ;D ;D
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lool yes past papers is right..
firt you have to find the factors of the constant, in this case 30:
1,30,15,2,5,6,and the negatives of each..
and we know that (a+y) is a factor when f(-y) =0..therefore, a=1 obviously gives us zero,
and so (a-1) is a factor...
do long division, get the quadratic function, and find the solutions..
sorry 4 the late reply, i had to go lol ;D