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3^(x+2)=3^x+3^2
anyone?
Consider the substitution y= 3^x. Hope this helps. Peace.
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3^(x+2)=3^x+3^2
anyone?
THANK U SO MUCH> I APPRECIATE IT ALOT> ILIKED UR FB PAGE
Did you mean to say one fourth?
and you really didn't know how to answer the questions because of a typo? -_-
thanks anyways
AsSalamoAlaikum Wr WbOkay, so there was a question of Trigonometry.
If arctan A = x and arctan B = y, express tan(A+B) in terms of x and y.
The answer is x + y / 1 - xy. I'm missing a tiny concept but I dunno where ofcourse.
XPFMember smzimran OakMoon! whitecorp BadRobot14
AsSalamoAlaikum Wr WbOkay, so there was a question of Trigonometry.
If arctan A = x and arctan B = y, express tan(A+B) in terms of x and y.
The answer is x + y / 1 - xy. I'm missing a tiny concept but I dunno where ofcourse.
XPFMember smzimran OakMoon! whitecorp BadRobot14
http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_w09_qp_42.pdf
whitecorp please help me for no. 6 (ii) & (iii)
Thanks
For (i) I obtained an acceleration value of 4 m/s^2.
(ii) Maximum height reached by A =distance B falls through before reaching the ground
Let this be s.
Then using v^2 =u^2 + 2as, substituting in the values, we have
1.6^2 = 0^2 +2 (4)(s) =====> s= 0.32 m (shown)
(iii) Using s=ut + 0.5*a*t^2, substituting in the values, we have
0.32 = 0.5*(4)*t^2 =2 t^2
Hence, t^2 = 0.16 =====> t=0.4 s (shown)
Hope this helps. Peace.
http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_w09_qp_42.pdf
whitecorp please help me for no. 6 (ii) & (iii)
Thanks
(ii)the distance moved by B is,Yeah I agree with you !
But the answer isn't right I guess
Have a look in the marking scheme !
http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_w09_ms_42.pdf
(ii)the distance moved by B is,
1.6^2 = 0+2(4)s
s= 0.32 m
this is NOT the maximum height reached by A, this is only the the distance A moved above the ground while b was moving downard. from this instant B will be touchin the ground so there will be no tension in the string anymore because the string is slack which means from now on A is moving upward under gravity (g ms^-2 ). so the maximum height A has moved is the distance it moved while b was moving downward PLUS the distance it moved till it reaches 0 velocity ( max height )
V^2 = u^2 + 2as
0 = 1.6^2 + 2(-10)(s)
s = 0.128
total distance moved by A = 0.32 + 0.128 = 0.448m
(iii) now for the total time taken, in the first stage when it was moving with an acceleration of 4ms^-2 it was:
v = u + at
1.6 = o + 4t
t= 0.4 s
now for the second stage when it was moving with a DECELERATION of 10 ms^-2
v = u + at
0 = 1.6 + -10t
t= 0.16 s
total time taken = 0.4 + 0.16
total time taken = 0.56 s
hope this makes sense to u, anymore questions feel free to ask me
no problem and yeah ok, feel free to !Yeah thank you very much !
I'll pm you ? Ok ?
from where did u get this question ?a particle p is projected vertically upwards from o with velocity 40 m/s one second later another particle is projected from O with the same vertical velocity .after what time and at what heoght will the two particles collide?
my teacher's note .from where did u get this question ?
do u have the answer to the question though ?my teacher's note .
the particle collide 4.5 sec after the projection of p at a height 78.75 m .do u have the answer to the question though ?
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