Post by: astarmathsandphysics on October 06, 2008, 10:59:23 am
Post by: milton on November 18, 2008, 02:35:55 pm
Post by: astarmathsandphysics on November 19, 2008, 09:47:59 am
The equation of the line from the point to the plane is r=(2,3,1)+t(9,8,7)
so for this line x=2+9t
y=3+8t
z=1+7t
Substitute these into equation of plane
9(2+9t)+8(3+8t)+7(1+7t)=4
49+194t=4 so t=-45/194
and r intersects the plane at the point (2,3,1)-45/194(9,8,7)=(-17/194,222/194,-121/194)
Now find the distance between (2,3,1) and the point just found using the formula sqrt((x1-x2)^2+(y1-y2)^2+(z1-z2)^2)
Post by: milton on November 20, 2008, 09:38:24 am
Serious car accidents on a road follow a Poisson distribution and have a mean of 2 per week.
a) What is the mean in a 4-week period
b) Given that 1 year has 13 four-week periods, what is the probability of nine of those 13 four-week periods having atleast 1 serious car accidents
c) If the probability of a car crash in 'n' weeks is 0.99, what is n
Post by: astarmathsandphysics on November 20, 2008, 09:52:19 am
My memory is sketchy, but I found a tough question, worth 15 marks.
Serious car accidents on a road follow a Poisson distribution and have a mean of 2 per week.
a) What is the mean in a 4-week period
b) Given that 1 year has 13 four-week periods, what is the probability of nine of those 13 four-week periods having atleast 1 serious car accidents
c) If the probability of a car crash in 'n' weeks is 0.99, what is n
a)8
b)The poisson is scalable so the probability distribution now is Po(eight) so P(X=k)=e^-8*8/k!. the probability that x>=1 accident in 1 month is 1-e^-8*8^0/0!=1-e^-8
Now we have a binomial B(13,8e^-1) so P(x=9)=13!/(4!9!)*(e^-8)^4(1-(e^-8))^9
C) The probability of a no car crashese is 1-e^-2
So solve 0.01=(1-e^-2)^n
n=ln(0.01)/ln(1-e^-2)
Post by: omarsubei on November 22, 2008, 05:05:59 pm
Post by: astarmathsandphysics on November 22, 2008, 05:10:18 pm
Hi, I'd be grateful if you can tell me how to convert metres per minute squared (acceleration) to feet per second squared (also acceleration) and vice versa. I just can't do it!!! Thanks
Divide by 3600 to change meters per minute squared into meters per second squared then Times by 3.3 since there are 3.3 feet in every meter
Post by: omarsubei on November 22, 2008, 05:14:23 pm
Post by: lolo on February 01, 2009, 09:30:00 am
i dont get it ???
so if i have a question like this how do i solve it?
pointA and B have coordinates (4, 1) and (2, -5), respectively. Find a vector equation for the line which passes through the point A, and which is perpendicular to the line AB.
really appreaciate ur help
thanks :)
Post by: astarmathsandphysics on February 01, 2009, 01:25:22 pm
heyy am still stuck in vectors
i dont get it ???
so if i have a question like this how do i solve it?
pointA and B have coordinates (4, 1) and (2, -5), respectively. Find a vector equation for the line which passes through the point A, and which is perpendicular to the line AB.
really appreaciate ur help
thanks :)
gradient of line from A to B =change in y/change in x=(-5-1)/(2-4)=3 so gradient of perpendicular =-1/3
y=mx+c
y=-1/3x+c Put point A in to this equation.
1=(_1/3)4+c
so c=7/3 and y=-1/3x+7/3
Post by: lolo on February 01, 2009, 02:01:16 pm
u r smart lol
thankooo soooo much :D
really appreciate it ..
:)
Post by: astarmathsandphysics on February 01, 2009, 02:11:18 pm
Post by: saifalan on February 01, 2009, 03:32:16 pm
Post by: astarmathsandphysics on February 01, 2009, 09:07:00 pm
Post by: saifalan on February 02, 2009, 01:05:05 pm
Post by: lolo on February 05, 2009, 12:11:37 pm
there were 1240 tickets sold for a concert. Of these, 780 were sold at $24.50 each, while the remainder were sold at a reduction of 20%. Find the total money taken in ticket sales.
thanks :)
Post by: astarmathsandphysics on February 05, 2009, 02:33:12 pm
i have another question plz:(780x24.5)+ 0.8x(1240-780)x24.5=28136
there were 1240 tickets sold for a concert. Of these, 780 were sold at $24.50 each, while the remainder were sold at a reduction of 20%. Find the total money taken in ticket sales.
thanks :)
Post by: lolo on February 08, 2009, 05:50:09 am
thanks :)
Post by: astarmathsandphysics on February 08, 2009, 07:35:13 am
The remainder have a discount of 20% iethey cost 24.50x0.8=19.60. There are 1240-780=460 of these so revenue =19.60x460=9016
Total revenue =19110+9016=28126
Post by: omarsubei on February 11, 2009, 11:45:41 pm
A force of magnitude F1 accelerates a body of mass m from rest to a speed v. A force of
magnitude F2 accelerates a body of mass 2m from rest to a speed 2v.
The ratio: (work done by F2)/(work done by F1) is:
A. 2.
B. 4.
C. 8.
D. 16.
Thanks
Post by: astarmathsandphysics on February 12, 2009, 08:38:02 am
Please help me with this question and explain to me why the answer is what it is:
A force of magnitude F1 accelerates a body of mass m from rest to a speed v. A force of
magnitude F2 accelerates a body of mass 2m from rest to a speed 2v.
The ratio: (work done by F2)/(work done by F1) is:
A. 2.
B. 4.
C. 8.
D. 16.
Thanks
The accelerations are eqaul sinceF2/2m=F1/m therefore from v^2=u^"+2as, s=v^2/2a sobody 2 travels 4 times as far as body 1.
(work done by F2)/(work done by F1)=F2*4s/F1*s=4
Post by: astarmathsandphysics on February 12, 2009, 08:45:15 am
One question about StatisticsI had this question a couple of days ago from one of my sudents.
The height of certain plants are normally distributed. The plants are classified into three categories.
The shortest 12,92% are in cat a
The tallest 10,38% are in cat C
All other plants are in between r and t cm and are cat B
Given that the mean height is 6.84 cm and the standard deviation is 0.25 find t and r
PS> With explanations if possible, i m not an expert on statistics
For 12.92 % z=-1.13 so -1.13=(r-6.84)/025 so r=0.25*-1.13+6.84=6.5575
For10.38 we must do 1-.1038=0.8962 and z=1.26 s0 1.26=(t-6.84)/0.25 so t=1.26*0.25+6.84=7.155
Post by: omarsubei on February 12, 2009, 10:14:17 am
Please help me with this question and explain to me why the answer is what it is:
A force of magnitude F1 accelerates a body of mass m from rest to a speed v. A force of
magnitude F2 accelerates a body of mass 2m from rest to a speed 2v.
The ratio: (work done by F2)/(work done by F1) is:
A. 2.
B. 4.
C. 8.
D. 16.
Thanks
The accelerations are eqaul sinceF2/2m=F1/m therefore from v^2=u^"+2as, s=v^2/2a sobody 2 travels 4 times as far as body 1.
(work done by F2)/(work done by F1)=F2*4s/F1*s=4
That's the answer I came up with, but strangely the mark scheme says the answer is C (8). I was thinking maybe because it's asking about the "Work" done by F, and that there is distance involved (W=Fd), that the distance in F2 is double that of F1 because it's twice as fast?
Post by: astarmathsandphysics on February 12, 2009, 10:49:39 am
Post by: omarsubei on February 12, 2009, 11:09:47 am
Post by: astarmathsandphysics on February 12, 2009, 11:19:22 am
0.5(2M)V^2/0.mvv^"=8 THEquestion is badlty wordedPlease help me with this question and explain to me why the answer is what it is:
A force of magnitude F1 accelerates a body of mass m from rest to a speed v. A force of
magnitude F2 accelerates a body of mass 2m from rest to a speed 2v.
The ratio: (work done by F2)/(work done by F1) is:
A. 2.
B. 4.
C. 8.
D. 16.
Thanks
The accelerations are eqaul sinceF2/2m=F1/m therefore from v^2=u^"+2as, s=v^2/2a sobody 2 travels 4 times as far as body 1.
(work done by F2)/(work done by F1)=F2*4s/F1*s=4
That's the answer I came up with, but strangely the mark scheme says the answer is C (8). I was thinking maybe because it's asking about the "Work" done by F, and that there is distance involved (W=Fd), that the distance in F2 is double that of F1 because it's twice as fast?
Post by: omarsubei on February 12, 2009, 11:31:28 am
Post by: astarmathsandphysics on February 12, 2009, 11:32:49 am
Post by: astarmathsandphysics on February 12, 2009, 11:34:03 am
Post by: omarsubei on February 12, 2009, 11:38:24 am
The ratio of F2/F1 is 4. How does the "work" make that 8?
Post by: astarmathsandphysics on February 12, 2009, 11:41:08 am
Post by: astarmathsandphysics on February 12, 2009, 11:42:42 am
Post by: omarsubei on February 12, 2009, 11:43:29 am
Post by: omarsubei on February 13, 2009, 11:25:19 am
A container is filled with a mixture of nitrogen and oxygen. What is the ratio of the rms (root mean squared) speed of oxygen molecules to that of nitrogen molecules? (Molar mass of oxygen=32 g mol-1 ; molar mass of nitrogen=28 g mol-1
I'll tell you the answer this time :P : 0.935
Post by: astarmathsandphysics on February 13, 2009, 11:37:26 am
I just can't get this question:All the molecules have the same kinetic energy
A container is filled with a mixture of nitrogen and oxygen. What is the ratio of the rms (root mean squared) speed of oxygen molecules to that of nitrogen molecules? (Molar mass of oxygen=32 g mol-1 ; molar mass of nitrogen=28 g mol-1
I'll tell you the answer this time :P : 0.935
therefore0.5m(ox)v(ox)^2=0.5M(nit)v(nit)^2
therefore rms(ox)/rms(nit)=sqrt(m(nit)/(ox))=sqrt(14/16)=0.93
Post by: omarsubei on February 13, 2009, 11:59:49 am
Post by: omarsubei on February 19, 2009, 12:15:09 pm
I HATE Permutations and Combinations. Here it is:
"Mr Blue, Mr Black, Mr Green, Mrs White, Mrs Yellow and Mrs Red sit around a circular table for a meeting. Mr Black and Mrs White must not sit together.
Calculate the number of different ways these six people can sit at the table without Mr Black and Mrs White sitting together."
Post by: crucio on February 19, 2009, 12:46:13 pm
Post by: astarmathsandphysics on February 19, 2009, 12:59:31 pm
Post by: omarsubei on February 19, 2009, 01:36:29 pm
Soppose mr black site down fist then mrs white can site in one of 3 possible seats - draw it!. The other 4 guests can sit in 4! different ways with respect to oder. 3x24 =72
Wait wait wait. I DID DRAW IT, a few million times too. Ya but what if Mr. Black changes his seat. He can also sit in six different places can't he. It's a permutation isn't it? Please explain more. There is actually a solution, but it's even more complex than yours:
"The number of different ways six people can sit around a circular table
is 5! = 120. (M1)
The number of different ways these six people can sit around a circular
table with Mr Black and Mrs White together is 4! × 2 = 48. (A1)
Therefore, the number of different ways these six people can sit around
a circular table with Mr Black and Mrs White NOT together is
120 – 48 = 72 "
Why can they sit in 5! ways, not 6! ???????
Post by: omarsubei on February 19, 2009, 01:39:59 pm
Post by: astarmathsandphysics on February 19, 2009, 01:59:26 pm
Post by: astarmathsandphysics on February 19, 2009, 02:00:42 pm
Post by: astarmathsandphysics on February 19, 2009, 02:02:06 pm
Oooooooh, wait I think I get it. Is it because it doesn't matter where Mr. Black sits first, because the question is asking about the different "ways" they can sit i.e. their positions with respect to each other?
Exactly
Post by: avrila on February 24, 2009, 10:55:03 am
1. Find the fourth vertex of the parallelogram:
A(2,1,0) B(3,2,-1) C(0,-1,1)
2. Find the shortest distance between point A (3,5,6) and line (x-7)/2=y-5=(z-12)/6
Thanks in advance...
Post by: astarmathsandphysics on February 24, 2009, 12:19:05 pm
Can you please help me with this:
1. Find the fourth vertex of the parallelogram:
A(2,1,0) B(3,2,-1) C(0,-1,1)
2. Find the shortest distance between point A (3,5,6) and line (x-7)/2=y-5=(z-12)/6
Thanks in advance...
AD=BC=(0,-1,1)-(3,2,-1)=(-3,-3,2) so D=A+(-3,-3,2)=(-1,-2,2)
2.eequation of line=(7,5,12)+t(2,1,6). Suppose they intersect at (a,b,c) then (a-3,b-5,c-6) is at right angles to (2,1,6) but (a,b,c) is on line so (a,b,c)= (7+2t,5+t,12+6t)
and (4+2t,t,6+6t).(2,1,6)=0 ie 44+41t=o so t=-44/41 sthe sub into (7+2t,5+t,12+6t) to find a and use pythagoras theorem in 3D to find distance between A and (a,b,c)
Post by: avrila on February 27, 2009, 10:19:21 pm
How would i solve this system of equations by hand of course. Is it possible?
625A -125B +25C -5D +E =4.5
112A -34.3B +10.6C -3.25D+E=2.25
0.00391A -0.0156B +0.0625C -0.25D +E=6.25
92.4A + 29.8B +9.61C + 3.10 D + E= 1.40
625A +125B +25C +5D +E =3.85
Help would be very much appreciated. Thanks.
Post by: astarmathsandphysics on February 27, 2009, 10:24:27 pm
Post by: avrila on February 27, 2009, 11:11:12 pm
Well, I'll keep on trying. Do tell me if you manage to solve this.. Thanks.
Post by: avrila on February 28, 2009, 06:17:12 am
A question ask, Find the position vector of an arbitrary point R which lies on the line.
I just want to know what is arbitrary point at the moment. ..
Post by: astarmathsandphysics on February 28, 2009, 08:25:26 am
Post by: astarmathsandphysics on February 28, 2009, 08:35:02 am
Yes use augmented matrices. Row reduce and the inverse is on the right hand side. The multiple the vector on right hand side of the original system of equations by the inverse you have found and the answer is the vector you have foundsuppose the equation of a line is r(t) =2i+4j-k+t(5i-2j+7k) the arbitrary point vector is (2+5t,4-2t,-1+7t)
Post by: avrila on March 01, 2009, 01:05:33 am
Find C so that QR and PR are orthogonal.
I know that I have to use dot product. QR.PR=0
But I got a weird value of C. Is my method correct? How do you do this.
Post by: astarmathsandphysics on March 01, 2009, 07:54:44 am
P=(4,1,-1) Q=(3,3,5) R=(1,0,2c) S=(1,1,2)QR=(-2,-3,2c-5) and PR=(-3,-1,2c+1)
Find C so that QR and PR are orthogonal.
I know that I have to use dot product. QR.PR=0
But I got a weird value of C. Is my method correct? How do you do this.
QR.PR=6+3+(2c-5)(2c+1)=9+4c^2-8c-5=4c^2-8c+4=0 solution is c=1
Post by: avrila on March 01, 2009, 02:08:23 pm
Post by: avrila on March 02, 2009, 02:43:57 am
Point B and C has position vectors (i +3 +2k) and (3i +2j + k) respectively.
>Find the position vector of the point where the line OB meets pi and find the perpendicular distance from B to pi.
Post by: astarmathsandphysics on March 02, 2009, 11:33:31 pm
Plane pi has equation r.(2i +j + 2k)= 36
Point B and C has position vectors (i +3 +2k) and (3i +2j + k) respectively.
>Find the position vector of the point where the line OB meets pi and find the perpendicular distance from B to pi.
We have
Post by: avrila on March 03, 2009, 04:30:40 am
But to find the distance from B to pi, why we just use the same coordinate? B has it owns coordinate right?...
Can you please explain.
Post by: astarmathsandphysics on March 03, 2009, 07:50:48 am
Post by: avrila on March 03, 2009, 10:02:40 am
For the distance, instead of 3,6,9 I got 6,3,6. After substitute this in the distance formula, I got 9 as the answer.
Hope I am in the right track.
Post by: radam on March 26, 2009, 06:29:29 pm
"A digital camera is used to take a photograph of a plant. The CCD in the camera has 1.6*10^-7 square pixels. Each pixel has an area of 2.3*10^-10m^2. A particular leaf of the plant has an area of 3.5*10^-2 m^2. The image formed on the CCD is 1*10^-3 m^2. Two indentations on the lead are seperated by 0.50mm. Deduce that it is unlikely that the images of the two indentations will be resolved"
The marking scheme states that you should calculate the magnification as (1*10^-3/2.5*10^-2)=0.04, and so my question is, isnt this formula the area magnication, in which case the linear magnification would be sqrt(0.04)= 0.2? The only problem is that if you use the square-root then you get the distance of seperation as greater than the minimum distance to be just-resolved... is their a problem in the question or is my method incorrect?
Thanks a lot in advance :)
Post by: astarmathsandphysics on March 26, 2009, 06:55:59 pm
Post by: astarmathsandphysics on March 27, 2009, 08:30:26 am
This is an extract from the Physics specimen 09 HL paper 2:The distance between the indentations is 0.2*0.5mm=0.1mm=
"A digital camera is used to take a photograph of a plant. The CCD in the camera has 1.6*10^-7 square pixels. Each pixel has an area of 2.3*10^-10m^2. A particular leaf of the plant has an area of 3.5*10^-2 m^2. The image formed on the CCD is 1*10^-3 m^2. Two indentations on the lead are seperated by 0.50mm. Deduce that it is unlikely that the images of the two indentations will be resolved"
The marking scheme states that you should calculate the magnification as (1*10^-3/2.5*10^-2)=0.04, and so my question is, isnt this formula the area magnication, in which case the linear magnification would be sqrt(0.04)= 0.2? The only problem is that if you use the square-root then you get the distance of seperation as greater than the minimum distance to be just-resolved... is their a problem in the question or is my method incorrect?
Thanks a lot in advance :)
Post by: radam on March 27, 2009, 01:08:36 pm
Post by: omarsubei on April 10, 2009, 09:28:39 pm
As usual I'm getting a different answer to the markscheme,
"A circle has equation x^2 + (y-3)^2 = 4 . Knowing that y= +/- (4 - x^2)^0.5 , find the volume of revolution of the shaded region about the x-axis.
The mark scheme says 144, and I got 237
PLEASE HELP!
Post by: astarmathsandphysics on April 10, 2009, 09:30:27 pm
Hi it's been a while.
As usual I'm getting a different answer to the markscheme,
"A circle has equation x^2 + (y-3)^2 = 4 . Knowing that y= +/- (4 - x^2)^0.5 , find the volume of revolution of the shaded region about the x-axis.
The mark scheme says 144, and I got 237
PLEASE HELP!
Surely y= 3+/- (4 - x^2)^0.5
Post by: omarsubei on April 10, 2009, 09:33:24 pm
Post by: astarmathsandphysics on April 10, 2009, 09:37:53 pm
Post by: omarsubei on April 10, 2009, 09:56:05 pm
Post by: omarsubei on April 10, 2009, 10:04:18 pm
Post by: omarsubei on April 10, 2009, 10:08:56 pm
Post by: astarmathsandphysics on April 10, 2009, 10:22:17 pm
Use the substition x=2sint then dx=2costdt
The limits are changed from
We have V=
Post by: omarsubei on April 10, 2009, 10:27:29 pm
Post by: astarmathsandphysics on April 10, 2009, 10:32:51 pm
I looked up the volume of a tours here. It is consistent with my answer.
http://whistleralley.com/torus/torus.htm
Post by: omarsubei on April 10, 2009, 10:38:24 pm
Post by: astarmathsandphysics on April 10, 2009, 10:42:30 pm
Post by: omarsubei on April 10, 2009, 10:47:09 pm
Post by: omarsubei on April 10, 2009, 10:50:19 pm
Post by: astarmathsandphysics on April 10, 2009, 10:51:49 pm
Post by: astarmathsandphysics on April 10, 2009, 10:55:03 pm
I don' get how you went from step one to two, how'd you put the integrals together?
Exapnad the brackets and simplify. You can just add the integrands. The squared terms cancel and for one term I get--=+
Post by: radam on April 23, 2009, 01:11:11 pm
I'd appreciate some help with this question;
(from the previous subsection b is an imaginary root of the equation z^5-1=0 and b^4+b^3+b^2+b+1=0)
________________________________________
Q 13, PART B, MATH HL NOV 08
c) If u=b+b^4 and v= b^2+b^3, show that
i) u+v=uv=-1
ii) u-v= sqrt(5) , given that u-v>0
_______________________________________
for i) Proving that u+v=-1 is not difficult... but i have no idea how to prove that uv=-1, or for ii) how u-v is sqrt(5).
Help would be greatly appreciated :)
Post by: astarmathsandphysics on April 23, 2009, 01:53:03 pm
Hey;
I'd appreciate some help with this question;
(from the previous subsection b is an imaginary root of the equation z^5-1=0 and b^4+b^3+b^2+b+1=0)
________________________________________
Q 13, PART B, MATH HL NOV 08
c) If u=b+b^4 and v= b^2+b^3, show that
i) u+v=uv=-1
ii) u-v= sqrt(5) , given that u-v>0
_______________________________________
for i) Proving that u+v=-1 is not difficult... but i have no idea how to prove that uv=-1, or for ii) how u-v is sqrt(5).
Help would be greatly appreciated :)
i)uv=(b+b^4)(b^2+b^3)=b^3+b^4+b^6+b^7 now use b^5=1 ie B^6=b and B^7=b^2 so we have b^3+b^4+b +b^2=-1 too.
I will answer the second part as soon as I come back, but I have to go RIGHT NOW and teach some maths.
ii)(u-v)^2=u^2-2uv+v^2=(b+b^4)^2-2(b+b^4)(b^2+b^3)+(b^2+b^3)^2=b^2+2b^5+b^8-2b^3-2b^4-2b^6-2b^7_+b^4+2b^5+b^6
b^5=1 so we have b^2+2+b^3-2b^3-2b^4-2b-2b^2+b^4+b=(b^2+b^3-2b^3-2b^4-2b-2b^2+b^4+b)+4
=-(b^3+b^4+b +b^2)+4=5
Post by: radam on April 23, 2009, 02:04:40 pm
I forgot that b^5=1;
I was thinking about the question with that bit of insight; and i was wondering if my method below is correct:
(u-v) is the same as sqroot (u-v)^2 (since u-v>0)
(U-v)=u^2-2uv+v^2= b^8+2b^5+b^2+2+b^4+2b^5+b^6
since uv=-1, and b^5=1
This gives b^4+b^3+b^2+b-6= 5
Hence (u-v)^2= 5
u-v= sqrt(5)
The answer seems correct, but I'm wary of squaring as it normally adds complications to the matter... is there an easier and less time consuming way to do it?
Again thanks a lot for your help! With my math exam 2 weeks away, its a life-saver!
Post by: Clerk on April 28, 2009, 11:00:43 pm
Post by: astarmathsandphysics on April 28, 2009, 11:06:34 pm
Ah thanks!
I forgot that b^5=1;
I was thinking about the question with that bit of insight; and i was wondering if my method below is correct:
(u-v) is the same as sqroot (u-v)^2 (since u-v>0)
(U-v)=u^2-2uv+v^2= b^8+2b^5+b^2+2+b^4+2b^5+b^6
since uv=-1, and b^5=1
This gives b^4+b^3+b^2+b-6= 5
Hence (u-v)^2= 5
u-v= sqrt(5)
The answer seems correct, but I'm wary of squaring as it normally adds complications to the matter... is there an easier and less time consuming way to do it?
Again thanks a lot for your help! With my math exam 2 weeks away, its a life-saver!
I can't think of a simpler way to do it.
Post by: astarmathsandphysics on April 28, 2009, 11:07:12 pm
Post by: IGSTUDENT on June 08, 2009, 10:11:13 am
Post by: omega007 on June 08, 2009, 11:43:55 am
http://en.wikibooks.org/wiki/IB_Textbook_Reviews
Post by: IGSTUDENT on June 08, 2009, 12:58:37 pm
All the same some of you who are already doing it..any suggestions?
Post by: IGSTUDENT on June 08, 2009, 12:59:18 pm
All the same some of you who are already doing it..any suggestions?
Post by: sippah on October 23, 2009, 01:24:10 pm
Post by: astarmathsandphysics on October 23, 2009, 01:52:24 pm
Post by: sippah on October 23, 2009, 02:40:47 pm
thanx in advance
Post by: astarmathsandphysics on October 23, 2009, 03:30:17 pm
Post by: sweet777 on May 12, 2010, 03:13:06 am
Post by: Monica on May 12, 2010, 03:15:52 am
Post by: astarmathsandphysics on May 12, 2010, 08:47:48 am
They make you pay or do a survey
Post by: astarmathsandphysics on May 12, 2010, 08:49:03 am
Post by: sweet777 on May 13, 2010, 04:02:31 am
Post by: astarmathsandphysics on May 13, 2010, 09:35:31 am
Post by: azhaan on May 15, 2010, 06:48:51 pm
im an igcse grade 10 student
and i have to choose my IB subjects now,
so im taking
HL-math, physics, economics
SL- english, french abinition, history
any suggestion?
are these subjects to hard or not helpful
PS: i want to do business or finance in university
Post by: Burnside on May 18, 2010, 01:20:55 pm
HL maths and physics are considered the two hardest courses in IB. I'm not saying they will be too hard for you, however if you're taking those two subjects I'd suggest taking up chemistry instead of the double humanities (chemistry and physics have quite a bit of overlap).
Post by: azhaan on May 20, 2010, 03:09:58 pm
@azhaanthanks but is it easier to score in history HL than physics HL???
HL maths and physics are considered the two hardest courses in IB. I'm not saying they will be too hard for you, however if you're taking those two subjects I'd suggest taking up chemistry instead of the double humanities (chemistry and physics have quite a bit of overlap).
Post by: Burnside on May 23, 2010, 01:02:25 pm
Subjects that have a large gap between SL and HL content:
Physics
Mathematics
Chemistry
Biology
All Languages
Small gap between SL and HL content:
Economics
Geography
English
Subjects that are hard conceptually:
HL maths
HL Physics
Subjects that are hard due to the sheer content you need to remember:
Economics
Biology
History
HL Chemistry (SL chem is easy)
If you do HL Physics, SL Chemistry will be a breeze due to alot of the concepts already being covered in physics.
Post by: sasukeuchiha on May 25, 2010, 09:54:04 pm
http://www.freeexampapers.com/FreeExamPapers.com_.php?__lo=TyBMZXZlbC9QaHlzaWNzL0NJRS8yMDA3IEp1bi81MDU0X3MwN19xcF8xLnBkZg== (http://www.freeexampapers.com/FreeExamPapers.com_.php?__lo=TyBMZXZlbC9QaHlzaWNzL0NJRS8yMDA3IEp1bi81MDU0X3MwN19xcF8xLnBkZg==)
Post by: astarmathsandphysics on May 25, 2010, 10:45:00 pm
31. Total power =0.5 so cost =0.5*8*2=8 cents
33 Use right jhand grip rule and remeber like currents attract
Post by: azhaan on May 26, 2010, 05:53:50 pm
I'm not doing history so it's hard to judge. However from my experience I can make a few comments about certain subjects.thanks a lot you are really helpful!!
Subjects that have a large gap between SL and HL content:
Physics
Mathematics
Chemistry
Biology
All Languages
Small gap between SL and HL content:
Economics
Geography
English
Subjects that are hard conceptually:
HL maths
HL Physics
Subjects that are hard due to the sheer content you need to remember:
Economics
Biology
History
HL Chemistry (SL chem is easy)
If you do HL Physics, SL Chemistry will be a breeze due to alot of the concepts already being covered in physics.
taking history HL and physics SL finally :)
hope it will help