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can anyone do questions 5 part ii) and 6 ii) from the Oct/Nov 2014 paper 32
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Once again, sorry for the late reply, was really busy with revision. Uhmm, the first two part of the question is a bit blurry. Could u just type it again...? Please?As level ?
View attachment 57252
(i) OB = 6 cmOnce again, sorry for the late reply, was really busy with revision. Uhmm, the first two part of the question is a bit blurry. Could u just type it again...? Please?
Ya, AS level
Thanks a lot!(i) OB = 6 cm
AO = 6 cm
AB = sqrt(36 + 36)
AB = 6sqrt(2)
(ii) tan^-1 (6/6) = pi/4
Ali pushes a block by applying a force of F = -10i + 10j in units of newtons. What is the direction of this force (the angle of the vector with the +x direction)?
10cos(37) = 7.98 m/sA ball is kicked with an initial speed v0=10 m/s at an angle of θ= 37◦above the horizontal. Its speed in m/s at maximum height is:
I am guessing its 45Ali pushes a block by applying a force of F = -10i + 10j in units of newtons. What is the direction of this force (the angle of the vector with the +x direction)?
tan^2(2x) = 1 + sec^2(2x)
tan^2(2x) = 1 + sec^2(2x)
Substitute this to get
integrate : 5 + sec^2(2x)
this gives 5x + tan2x
simplify tan2x
Ohh yes it would be -1IN step 1 .. it should be -1 according to the formula ????
can u tell me the paper year ?View attachment 57295
Help! How to do part i and part b??
S15 p21can u tell me the paper year ?
part (I) .. first simplify 2cosec2x into 2/sin2x . simply tan x into sinx/cosx ... simply it u will the answerView attachment 57295
Help! How to do part i and part b??
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