Mathematics: Post your doubts here!

[Ayesha Asif333]

O580/21/M/j/11 QUESTION13,question 16 part (a),18PART(B)

q13) a) 1 : 20000

0.000027: x


b) 1 :20000square

x : 644000000



q16) a) the shaded area is A

and A =’s the area of the square minus area of the circle

area of the square is k square

the area of the circle is π(k/2)whole square ….where k/2 is the radius of the circle

now you can make the equation


q18) b) MX = MR + RX

r/2 + 3/4 q – r

Simplify it further and you'll get the ans

 

[Ayesha Asif333]

0580/21/0/n/11 QUESTION 18
0580/21/m/J/12 question 19 full and question 20 and21
o580/21/mj/2013 QUESTION 20(B) AND 26
0580/21/O/n/2013 QUESTION 18 AND 21

i'm tired now
maybe i'll do them some other day

 

[Ayesha Asif333]

wait i have something for you which can make it easier for you
download the PDF file
it contains all the formulas

God bless whoever made that formula sheet
was really helpful

check these too
Mathematics (Extended) Flashcards.pdf

gcse-mathematics-instant-revision book.pdf

 

[jamesosama]

i'm tired now
maybe i'll do them some other day

so can u please help

 

[Ayesha Asif333]

so can u please help

Will do those_In sha Allah

 

[Ayesha Asif333]

0580/21/0/n/11 QUESTION 18

Q18) a) by T 23 they mean that the nth term is 23 and you need to work of the value of 23rth term in the sequence, so just replace the n’s in the equation they have given with 23 and calculate

b) 1) calculate the T values then substitute it in the equation they have given….ans 4 and 9

2) (n+1)^2….notice that the value of U1 is 4 and and the term is 1 so if you put it in the equation (1+1)^2….u’ll get 2

C) compare the values of the two equations

T sequence: 1 , 5, 14, 30

V sequence: 4, 20, 56, 120

So now if you observe if you multiply the T values with 4 you get the value’s of V eg 1* 4= 4

Then 5*4 = 20, 14*4=56 , 30*4=120

And they have made our work easy with given us the formula of sequence T! so we simply multiply 4 with the fraction 1/6 to get 2/3 …rest whole equation remains the same

 

[Ayesha Asif333]

0580/21/m/J/12 question 19 full

a) PT= PO + OT
-p +t

b) PR= PS + SR
2t + p

c) when they ask the position vector they want you to find OR
so you would have to take this route....OP(p) + PS ( 2t) + SR(p)
this simplifies to 2(p+t)

there should little confusion in this ques for it just requires you to have the basic concepts in vectors
r u sure u practiced well from the book or something

 

[Ayesha Asif333]

and question 20

first step...read the ques thrice ...trace the rope in the diag so it becomes clear for u what you need to find

now the ques becomes more easy and clear
so find the length of the major arc = 17.8
find length PT, use trig ratio...tan78* 5 = 23.52
PT = PR so 23.52 + 23.52 = 47.04

last steo...add the two straight lengths with the length of the major arc

hope u get this

 

[kity way]

Math paper 21, june 2012, q19

i got part a, t-p but part b..
the answer is p+2t but i dont get it, shouldnt it be 2t? we r only going up not left or right so why is p included? bcz of this i donr get part c too, can someone plzz help

 

[Ayesha Asif333]

Math paper 21, june 2012, q19

i got part a, t-p but part b..
the answer is p+2t but i dont get it, shouldnt it be 2t? we r only going up not left or right so why is p included? bcz of this i donr get part c too, can someone plzz help

Assume the center to be X
if OT is t than PX should be t ...coz the lines r parallel
if PX is t then XS should be t, too
now we have a diagonal PS = 2t ...which is by adding the two t's
then you move to the right so PS + SR = 2t+ p

now assume the midpoint of the line XQ to be Y
to move vertically straight up u would be needing vector PY , which we don't have so your method isn't correct

 

[whyareallnamestaken]

Can anyone solve Q9 of this paper?
http://papers.gceguide.com/IGCSE/Mathematics (0580)/0580_m16_qp_42.pdf

 

[Ayesha Asif333]

Can anyone solve Q9 of this paper?
http://papers.gceguide.com/IGCSE/Mathematics (0580)/0580_m16_qp_42.pdf

U want an explaination for the whole ques?
I'll see if i get time

 

[whyareallnamestaken]

U want an explaination for the whole ques?
I'll see if i get time

Yes, only if you can manage it. Thanks in advance!

 

[Ayesha Asif333]

 

[whyareallnamestaken]

View attachment 62149 View attachment 62150 View attachment 62151

Thanks so much! This was really helpful

 

[Afza M]

http://papers.gceguide.com/IGCSE/Mathematics (0580)/0580_m17_qp_22.pdf

I have a doubt in Question 4 of this paper. Please help. Thank you.

 

[Ayesha Asif333]

Use the formula for the exponential growth and decay, which is similar to the compound inter formula
For growth u use. y= a( 1+ r%) ^n
For decay use y= a (1- r%) ^n
Where a is the initial value...r is rate....and n time

here we have to find the time so
7.31=7(1+1.1/100)^n

 

[Afza M]

Use the formula for the exponential growth and decay, which is similar to the compound inter formula
For growth u use. y= a( 1+ r%) ^n
For decay use y= a (1- r%) ^n
Where a is the initial value...r is rate....and n time

here we have to find the time so
7.31=7(1+1.1/100)^n

That's what I am not getting! How do I find the time? The markscheme says it is 4 years..how do I calculate it?

 

[Ayesha Asif333]

Method 1


Simplify......7.31=7(1.011)^n

Divide both sides by 7, u will get 731/700= 1.011^n

Take log on both the sides....log (731/700) = log (1.011)^n

Simplify....log(731/700)= n log(1.011)

Find n by division.....log(731/700) / log(1.011)

The final ans in your calculator will be 3.96... which u have to round off

Method 2

If you do not know how to work out with the log method, you can take a much simpler approach that is try the time with different values to see if u get 7.31...start with 1,2, 3 and so on

I hope its clear now!

 

[Afza M]

Thank you so much!!!!