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Discussion in 'International A And AS Level' started by XPFMember, Sep 8, 2011.
so cuz the triangle is isoscles both angle are equal k thx alot
Can anyone help me? I'm in A2 and it's just 4 months left my subjects are Phy, Maths and C.S. Any hope ?
Post the complete question
It being a one mark question is a hint at a clever solution:
From main diagram:
AreaTotal = 2*AreaB + AreaCircle (or 2*AreaSemiCircle)
AreaA = (i) - AreaB
Was your answer to the first part y = 70e^[e^(-3t) - 1] ?
As t goes to infinity,
e^-3t tends to zerothus,
y tends to 70e^[0 - 1] = 70e^-1
so we have:
yFinal = 70e^-1
yStart = 70 (from main question)
p = (yFinal/yStart)*100
p = 100e^-1
help required !!!!!!! in part b
How are the values of α and β calculated in the last part? Anyone plz help.....
For b (i), AC is the diameter of the circle so APC will be the angle opposite to the diameter, i.e. angle in a semicircle=90°
Or you can think of it by the other rule...ABCP is a quadrilateral with its vertices touching the circumference so opposite angles will be equal. Hence, angle APC=angle ABC
In b (ii), angle APB is subtended by the chord AB, whereas angle BPC by the chord BC.Angles subtended by chords of equal length are equal and since AB=BC, the stated angles are equal.
In b (iii) (a) APC=90 and APB and BPC are equal, so APB=90/2=45
In b(iii) (b), angle CPD is also subtended by an equal chord so it will also equal angle APB i.e. 45. Hence, APD= 45+45+45=135
For b(iv) see the diagram
So angle PDC=90+8=98
What are the answers? Is alpha = 1.23?
This makes sense?
Depends on how well are you doing right now. You can inbox me for discussing about this topic.
need help in part b and c
Having two sides be the same makes it an isosceles triangle
360 = 66 + 2x
ADE and EFC are equal (alternate angles)
ECF you already have
FEC and AED are equal
Better to post in O level thread
dont be angry sis
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