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Mathematics: Post your doubts here!

Thought blocker

Thought blocker

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Well, sun is down. Wish you all the very best, do well in exams, don't be in hurry, if you think paper is hard then u r mistaking check all questions again and try to do your best, Be neat and precise ;)

Again, best of luck. And if you think I help you a lot, do remember me in your prayers, IF YOU WANT ! :)


-Phew- Heartbeats start rising. :(
 
daredevil

daredevil

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A star did u get the answer to that question?? if not then send me the ms aik daffa just so i can confirm the answer and then see if i can do it...

all i could come up with was subtract the leaking rate from the filling rate.. but ajeeb sa ho gaya hai queston kuch O_O` :p
 
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saifeddin.moh

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can someone please help with the range and domain questions with pure math 1
i hv a problem with finding which is the domain and which is the range in a completed square expression

any help is appreciated :D
 
A

A star

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daredevil said:
A star did u get the answer to that question?? if not then send me the ms aik daffa just so i can confirm the answer and then see if i can do it...

all i could come up with was subtract the leaking rate from the filling rate.. but ajeeb sa ho gaya hai queston kuch O_O` :p
Click to expand...
asaan sawal hae i did it before made a silly mistake. integral of (80-kV)-1 dV ko mae -ln(80-kV) lae raha tha is lea bhand horaha tha :p mayjune 2013 p31 daikhlo wahaen sae hae
 
daredevil

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A star said:
asaan sawal hae i did it before made a silly mistake. integral of (80-kV)-1 dV ko mae -ln(80-kV) lae raha tha is lea bhand horaha tha :p mayjune 2013 p31 daikhlo wahaen sae hae
Click to expand...
OOOoohh righhtt!!
I used the k but multiplied it for some twisted reason rather than dividing it x_X
 
omaaaar

omaaaar

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Namehere said:
Please someone!!

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_w12_qp_11.pdf

if you integrate the curve, are u integrating both the positive and negative part of the curve or just the positive one?

I just randomly forgot to integrate, so if anyone could help me out on this one!!

Thanks!
Click to expand...
Q2
dy/dx=-8x^-3 -1
Integrating
-8(1/-3+1 x^-3+1)-1(1/0+1 x^0+1)
y=4x^-2 -x+k
Now put the x and y values given in the question to find k and then write the equation in x and y with the value of k
Hope u understand it is so hard to show working here
 
Namehere

Namehere

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omaaaar said:
Q2
dy/dx=-8x^-3 -1
Integrating
-8(1/-3+1 x^-3+1)-1(1/0+1 x^0+1)
y=4x^-2 -x+k
Now put the x and y values given in the question to find k and then write the equation in x and y with the value of k
Hope u understand it is so hard to show working here
Click to expand...

Sorry, I forgot to put the question, its Q8(ii), and a=5.

I dont need any calculations just an explanation of what i wrote:

"if you integrate the curve, are u integrating both the positive and negative part of the curve or just the positive one?

I just randomly forgot to integrate, so if anyone could help me out on this one!!"

Thanks!
 
L

Luisis1337

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Namehere said:
Sorry, I forgot to put the question, its Q8(ii), and a=5.

I dont need any calculations just an explanation of what i wrote:

"if you integrate the curve, are u integrating both the positive and negative part of the curve or just the positive one?

I just randomly forgot to integrate, so if anyone could help me out on this one!!"

Thanks!
Click to expand...
Well there are two ways to do this, in terms of dy or dx, in this problem it would be easier to do in terms of dy because of the y^2

To integrate here you need to find the limits, since you know a=5 but you are integrating in terms of y, you plug in 5 and 1/2 in either formula you get the new bounds, which is 9 and 0.

To integrate in terms of y you take the rightmost curve subtracted by the leftmost curve

so isolate the x and you get :
the integral from 9 to 0 ((3y+10)/2) - ((y^2+1)/2)

Then you integrate and plug in 9 in the new equation and subtract from that what you get by plugging in 0 to the new equation and you get the answer
 
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