Mathematics: Post your doubts here!

[whitecorp]

3^(x+2)=3^x+3^2

anyone?

Consider the substitution y= 3^x. Hope this helps. Peace.

 

[kronix6]

Your identity is incorrect, the RHS should be 8 (cosx)^4 -3.
I have provided the solution for this:



For (ii), what equation exactly are you referring to? Note that (i) is an identity, there is nothing to solve for.

Peace.

THANK U SO MUCH> I APPRECIATE IT ALOT> ILIKED UR FB PAGE

 

[whitecorp]

THANK U SO MUCH> I APPRECIATE IT ALOT> ILIKED UR FB PAGE

Appreciate the gesture, peace.

 

[bloody_mary]

Did you mean to say one fourth?

and you really didn't know how to answer the questions because of a typo? -_-

thanks anyways

 

[whitecorp]

and you really didn't know how to answer the questions because of a typo? -_-

thanks anyways

Oh no, I can't, my math is awful, terrible. Peace.

 

[VelaneDeBeaute]

Okay, 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

 

[XPFMember]

Okay, 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 Wb

Alright, let me check..

 

[XPFMember]

Okay, 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 Wb


tan (A + B) = (tan A + tan B)/(1 - (tan A)(tan B))

from the info given, rearrange, and you get: x = tan A ; y = tan B

just substitute, and you get x + y / 1 - xy

 

[bamteck]

http://www.xtremepapers.com/papers/...S Level/Mathematics (9709)/9709_w09_qp_42.pdf

whitecorp please help me for no. 6 (ii) & (iii)
Thanks

 

[whitecorp]

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.

 

[bamteck]

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.

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/...S Level/Mathematics (9709)/9709_w09_ms_42.pdf

 

[leadingguy]

the way i attemptd the qstn is totally different can anyone explain me how this to be done???? the highlighted part. down is the ms of the qstn as well thanks

 

[iKhaled]

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

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

 

[bamteck]

(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

Yeah thank you very much !
I'll pm you ? Ok ?

 

[iKhaled]

Yeah thank you very much !
I'll pm you ? Ok ?

no problem and yeah ok, feel free to !

 

[GorgeousEyes]

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?

 

[iKhaled]

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?

from where did u get this question ?

 

[GorgeousEyes]

from where did u get this question ?

my teacher's note .

 

[iKhaled]

my teacher's note .

do u have the answer to the question though ?

 

[GorgeousEyes]

do u have the answer to the question though ?

the particle collide 4.5 sec after the projection of p at a height 78.75 m .