Mathematics: Post your doubts here!

[anooshraja]

http://www.mediafire.com/?440dtidvkdvi282 question 9, totally confused >_< how to solveE ?

 

[always-smile :)]

http://www.mediafire.com/?440dtidvkdvi282 question 9, totally confused >_< how to solveE ?

use the formula a+(n-1)d ((( but only with arthematic seq. { adding or subtracting } )))​

a= first term​

d= common difference​

s0 it will be --> 2n-1 then put the root​

then for b) (2 x 29)-1 = 57 = root 57 ​

hope i helped ...check it if it is ok ​

 

[Mohammad Ahmad]

what tym is ur exams guys ( i mean paper 2 maths ) and when !?

 

[bluefa1con]

part b can u explain plzzzzzz

-2/3 means that -2 is the power and 1/3 is cuberoot
any negative power will change the number to a fraction of type 1/x where x is the number u squared

 

[bluefa1con]

the image is not opening

it is now
try opening in another tab

 

[bluefa1con]

http://www.xtremepapers.com/papers/CIE/Cambridge IGCSE/Mathematics (0580)/0580_s10_qp_41.pdf​

​

8 part D III) please :! please guys im in a hurry if you dont mind ! ​

inverse of a function make a mirrored graph so plot a rough reflection and check the value

 

[bluefa1con]

There are only 2 linees EC and CD :/

here i posted here so every one can see it

 

[GlitteringLights]

f in verse is log(base10)X
take x as 1 answer will be zero
because anything with the power zero answer is 1

for one, on my calculator i get the answer as 1 0.0
and two, do you know how to do it manually? please explain if you do
thankss

 

[natsume hyuga]

helpppppppp

ok question 12 first of the paper u gave.
12 a) corresponding interior angles in parallel lines are supplementary. In English, it means that if u have two angles between two parallel lines( DC and EF ), their sum will be 180. So FED = 109, so EDC= 180-109=71
Because u got EDC u can also find BCD using the same concept: 180-71= 109

For the next question u gotto know this formula (n-2)*180 = sum of all interior angles. (n is the number of sides the polygon has, in this case 6 cuz the question says its a hexagon)
So u got all angles except FAB and ABC. The question also says FAB and ABC are equal.
(6-2)*180=540
540=95+109+71+109+2x
x=78
FAB= 78 ............ hope the answer is correct according to the mark scheme


Btw before i answer the area thingy which past paper is it in?

 

[bluefa1con]

for one, on my calculator i get the answer as 1 0.0
and two, do you know how to do it manually? please explain if you do
thankss

i did my add maths in oct/nov11
you guys dont have logs in normal maths

think of it as what power of 10 will give you the answer 1
anything with power 0 is 1

take a function like 2power x
the inverse will be log base 2 of given x
if they give x as 4 for this function
the question means what power of 2 will give the answer 4 so the answer is 2

 

[natsume hyuga]

A rectangular photograph measures 23.3 cm by 19.7 cm, each correct to 1 decimal place.​

Calculate the lower bound for​

the perimeter and the area.​

http://www.xtremepapers.com/papers/...cs (with coursework) (0581)/0581_w02_qp_2.pdf question 12​

lower bound 0.3----> 0.25
0.7------>0.65
perimeter: (2*23.25)+(2*19.65)=85.8

area: 23.25*19.65= 456.86

 

[GlitteringLights]

i did my add maths in oct/nov11
you guys dont have logs in normal maths

think of it as what power of 10 will give you the answer 1
anything with power 0 is 1

take a function like 2power x
the inverse will be log base 2 of given x
if they give x as 4 for this function
the question means what power of 2 will give the answer 4 so the answer is 2

ahh it's just too confusing >.< i get the 1st part but not the rest ..
i wish you'd explain it IG-wise,anyways thanks for tryin to help

 

[bluefa1con]

ahh it's just too confusing >.< i get the 1st part but not the rest ..
i wish you'd explain it IG-wise,anyways thanks for tryin to help

the thing is its not part of the IG syllabus
thats why its for only 1 mark
such questions are included just to separate normal candidates from those expecting distinctions
there is always 1 or 2 such questions in the paper that are a bit tricky
if there is no such question than the threshold is very high for those years

you can google to see how to use logs
these questions just need a basic idea

 

[GlitteringLights]

the thing is its not part of the IG syllabus
thats why its for only 1 mark
such questions are included just to separate normal candidates from those expecting distinctions
there is always 1 or 2 such questions in the paper that are a bit tricky
if there is no such question than the threshold is very high for those years

you can google to see how to use logs
these questions just need a basic idea

oh,, alriiiight :O smartt
thanks !!

 

[FATIMA1996]

Hello there okay so i was doing o/n 22 number 12 and i got 200 by doing this( 0.5*10*40) but when i looked at the mark scheme it says its 80 can somebody explain it to me howw

 

[IGCSE STUDENT]

--------------------------------
i think i got it
u can get RS which is (( -r+3q )) then M in the midpoint so it needs only half the distance so divide it all by 2 not only -r
plz tell me if u got it hope i helped

thanks for your answer of course, but wait, so according to you, MS is half of RS? I don't think that's right...

 

[kboss]

Hello there okay so i was doing o/n 22 number 12 and i got 200 by doing this( 0.5*10*40) but when i looked at the mark scheme it says its 80 can somebody explain it to me howw

ur supposed to find the distance travelled by the small car then the distance travelled by the large car and subtract.
1) distance of small car = area of trapezoid
therefore: distance = 0.5*(4+10)*40
= 280
2) distance of large car = area of triangle
therefore: distance = 0.5*40*10
= 200
3) subtract both
therefore: =280-200
= 80
hope u got it!!

 

[cigspiring]

http://www.xtremepapers.com/CIE/Cambridge IGCSE/0581 - Mathematics Extended/0581_w05_qp_2.pdf

Question 18a & b, simplifying. How do you go about doing this?

 

[MERIHAN]

Hello there okay so i was doing o/n 22 number 12 and i got 200 by doing this( 0.5*10*40) but when i looked at the mark scheme it says its 80 can somebody explain it to me howw

you should first find the distance of the small car :
distance = area under the graph
area of triangle + area of rectangle
1/2 *6*40 + 4*40 =280

now find the distance of large car

area of triangle = 1/2 *10*40 = 200

now subtract both the car distance = 280 -200 =80

 

[bluefa1con]

http://www.xtremepapers.com/CIE/Cambridge IGCSE/0581 - Mathematics Extended/0581_w05_qp_2.pdf

Question 18a & b, simplifying. How do you go about doing this?

part a

(xpower 9)cube/3 cube
the 3 from 2/3 will be cancelled
than square them

part b
this is similar
take square out of them the final answer will be 2x