Mathematics: Post your doubts here!

[weirdoatlarge77]

i am so sorry i wuz wrong u hav 2 make d angle bisector not perp. bisector..once again i am truly sorry

Don't worry dude, everyone makes mistakes!

 

[notnek01]

Wait I actually don't get the scales
Could anyone just solve this question :
The scale of a map is 1:500000.
(a) The actual distance between two towns is 172km.
Calculate the distance, in centimetres, between the towns on the map.

Thank you!

The map distance is 500000 times smaller than the real distance. So you just need to divide 172km by 500,000.

The answer needs to be in cm so you should convert from km to cm first:

172km = 17,200,000 cm

Now divide by 500,000:

Map distance = 17,200,000 / 500,000 = 34.4cm

 

[weirdoatlarge77]

Thanks Notnek101, if you have any bio/chem questions let me know!

 

[weirdoatlarge77]

Thanks to sj0007 too!

 

[Aishy Vimal]

Hi can someone help me with summer 2010 v 23 q17 b. I got 3.91 but i have no idea how to get the answer! HELP PLEASE :c

 

[sj0007]

Don't worry dude, everyone makes mistakes!

thanxx...

 

[sj0007]

Hi can someone help me with summer 2010 v 23 q17 b. I got 3.91 but i have no idea how to get the answer! HELP PLEASE :c

it is OK+KL+OL=5.6+3.91+5.6=15.1 cm

 

[sj0007]

Thanks to sj0007 too!

welcome...welcom...

 

[notnek01]

Thanks Notnek101, if you have any bio/chem questions let me know!

I actually teach IGCSE maths and it's been a long time since I did any bio/chem

 

[Aishy Vimal]

it is OK+KL+OL=5.6+3.91+5.6=15.1 cm

Thank you soo much <3

 

[Aishy Vimal]

btw guys could yall help s 2010 v23 q16. i kinda really suck at switching the formula around

 

[sj0007]

I actually teach IGCSE maths and it's been a long time since I did any bio/chem

lol...........i also thought u were a student....

 

[sj0007]

btw guys could yall help s 2010 v23 q16. i kinda really suck at switching the formula around

5A=r(y+2)
y+2=5A/r
y=5A/r - 2

 

[Princess Raven]

GUYS!!1 MAY JUNE 2011 V 43 Q 11 PART A V??????? y cant we just subsitute 800 in the foumula and divide it by 2???? y do we have to do 2 x 400?/ plz explain?

 

[notnek01]

GUYS!!1 MAY JUNE 2011 V 43 Q 11 PART A V??????? y cant we just subsitute 800 in the foumula and divide it by 2???? y do we have to do 2 x 400?/ plz explain?

If we substitute 800 it will give us the answer to

1 + 2 + 3 + ... + 798 + 799 + 800

Then if we divide this sum by 2 we get:

1/2 (1 + 2 + 3 + ... + 798 + 799 + 800) = 0.5 + 1 + 1.5 + ... + 399.5 + 400

But this is not what the question is asking for.


Instead, if we substitute 400 into the formula and multiply by 2 we have:

2(1 + 2 + 3 + ... + 398 + 399 + 400) = 2 + 4 + 6 + ... + 796 + 798 + 800


Does this make sense?

 

[weirdoatlarge77]

If we substitute 800 it will give us the answer to

1 + 2 + 3 + ... + 798 + 799 + 800

Then if we divide this sum by 2 we get:

1/2 (1 + 2 + 3 + ... + 798 + 799 + 800) = 0.5 + 1 + 1.5 + ... + 399.5 + 400

But this is not what the question is asking for.


Instead, if we substitute 400 into the formula and multiply by 2 we have:

2(1 + 2 + 3 + ... + 398 + 399 + 400) = 2 + 4 + 6 + ... + 796 + 798 + 800


Does this make sense?

Could you please explain part (iv) of that same question?

 

[notnek01]

Could you please explain part (iv) of that same question?

121+122+...+199+200 = (1+2+...+199+200) - (1+2+...+119+120)

Does this make sense? So you can substitute in n=200 and n=120 into the formula and then subtract the answers.

 

[notnek01]

Question 19 b)
http://papers.xtremepapers.com/CIE/Cambridge IGCSE/Mathematics (0580)/0580_s12_qp_21.pdf

Why is the answer p+2t? How? :O

Call M the centre of the hexagon.

Then P->R = P->M + M->Q + Q->R = t + p + t = p + 2t

 

[weirdoatlarge77]

shouldn't the answer be 2/12 for Question 21a in this paper?

http://papers.xtremepapers.com/CIE/Cambridge IGCSE/Mathematics (0580)/0580_s12_qp_21.pdf

 

[weirdoatlarge77]

Call M the centre of the hexagon.

Then P->R = P->M + M->Q + Q->R = t + p + t = p + 2t

Thanks!