• We need your support!

    We are currently struggling to cover the operational costs of Xtremepapers, as a result we might have to shut this website down. Please donate if we have helped you and help make a difference in other students' lives!
    Click here to Donate Now (View Announcement)

Mathematics: Post your doubts here!

Gémeaux

Gémeaux

Messages
1,558
Reaction score
2,428
Points
273
ahmed abdulla said:
yep bro its correct!!
how did you get it? or was it a guess? :p
Click to expand...
LOL no, not a guess :p
Basically, for question like this put a negative sign for the opposite direction. Doing this the geometric progression becomes: 0.6, -0.36, 0.216 ... where ratio r is (-.36/.6) = -3/5. In this case, r is less than 1 so the formula used for the sum would be
[a(1/r^n)]/1-r
Take n to be infinite, which means r^∞ approaches to 0.
[0.6(1-0)]/ 1-(-3/5) = 0.6/1.6 = 0.375. As the answer is +ve, we take it to be towards the East.
 
ahmed abdulla

ahmed abdulla

Messages
415
Reaction score
252
Points
53
Gémeaux said:
LOL no, not a guess :p
Basically, for question like this put a negative sign for the opposite direction. Doing this the geometric progression becomes: 0.6, -0.36, 0.216 ... where ratio r is (-.36/.6) = -3/5. In this case, r is less than 1 so the formula used for the sum would be
[a(1/r^n)]/1-r
Take n to be infinite, which means r^∞ approaches to 0.
[0.6(1-0)]/ 1-(-3/5) = 0.6/1.6 = 0.375. As the answer is +ve, we take it to be towards the East.
Click to expand...
i hope same question comes tomorrow .. just pray hh
thanks
 
Dug

Dug

Messages
227
Reaction score
515
Points
103
Gémeaux said:
I need help with this question. If anyone could please explain, I'd be grateful.
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s12_qp_33.pdf
Q6 (i). tan3x = ktanx where k is a constant and tanx ≠0.
By first expanding tan(2x + x), show that (3k-1)tan^2(x) = k-3
It's simple trigonometry, I know, but I can not come up with the expression they've given :/
Click to expand...
gqbKt89.jpg
 
Gémeaux

Gémeaux

Messages
1,558
Reaction score
2,428
Points
273
Dug Thankyou so much! I see I wasn't doing it wrong, but made it so messy that stuck half way through it.
 
  • Like
Reactions: Dug
Maz

Maz

Messages
302
Reaction score
665
Points
103
I've got another doubt in the same paper...
Q10 (iii)...
I got the critical values= 0 and 64/9
but then what would be the inequality... i reckoned that since "a" in negative in -kx^2 + 3kx -16 i.e. -k, the graph would have an upward parabola, but it turned out to be wrong.
 
Maz

Maz

Messages
302
Reaction score
665
Points
103
syed1995 said:
Wouldn't the graph be something like which starts at 3 and the curve cutting x axis at 2Pi instead of pi?
Click to expand...
Actually the curve cannot cut the x- axis at pi, tan is undefined for pi. Nor 2pi since they haven't asked for it.
 
Dug

Dug

Messages
227
Reaction score
515
Points
103
gary221 said:
Dug..if u rnt busy, could u help me out with ths pls??
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s08_qp_3.pdf
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s08_ms_3.pdf

ques 3 n 4 ii)

pls??
Click to expand...
Q3)
i) Perimeter of sector = ½(Perimeter of rectangle)
2r + rx = 4a
Sinx = a/r
a = rSinx
2r + rx = 4(rSinx)
2 + x = 4Sinx
Sinx = ¼(2 + x)

ii) x1 = 0.8
x2 = 0.7754
x3 = 0.7668
x4 = 0.7638
x5 = 0.7628
x6 = 0.7625

x = 0.76 (2 dp)

Q4)
i)
q4eGgvM.png


ii) tan²x + (6√3)tanx - 5 = 0
Let y = tanx
y² + (6√3)y - 5 = 0

Now you have to use quadratic formula. I am sure you can do that!
 
Nurul Huda sabid

Nurul Huda sabid

Messages
3
Reaction score
0
Points
11
i have problem with question 7(ii).....papers.xtremepapers.com/CIE/Cambridge%20International%20A%20and%20AS%20Level/Mathematics%20(9709)/9709_w12_qp_12.pdf
 
You must log in or register to reply here.
Top