Mathematics: Post your doubts here!

[notnek01]

0580/43/o/n/11
Q5-(b)

HELP!!

Imagine that the cost of a biscuit is 9 cents.

Then the number of biscuits Roshni could buy for 72 cents is 72/9 = 8.

But instead of 9 we have x so:

The number of biscuits Roshni can buy for 72 cents is 72/x

Also, the number of cakes she can buy for 72 cents is 72/(x+3).

So the equation is:

72/x = 72/(x+3) + 2

Can you continue from here?

 

[notnek01]

ppl... http://papers.xtremepapers.com/CIE/Cambridge IGCSE/Mathematics (0580)/0580_w12_qp_41.pdf
q 9 part c? y dont we divide since its in second so to make it into 365 Days we do... 300000/3600/24/365 but the ms shows 300000x3600x24x365??

Distance = Speed x Time

We need to work out the distance (light year) using the speed of light (300,000 km/s) and time (365 days).

But first we need to change 365 days into seconds:

365 days = 365 x 24 hours = 365 x 24 x 60 minutes = 365 x 24 x 60 x 60 seconds


Distance = Speed x Time:

1 Light year = (300,000) x (365 x 24 x 60 x 60)

 

[tweetie]

ok i didn't get what u said.. the answer is 8.2 as given so u should find the slant height of the whole cone then of the small come using Pythagoras theorem and just subtract to get the slant height S

well..i meant we can do this slant height of big cone - slant height of small cone
nd we find slant height l(2)=height(2)+ radius (2)
so here we will take the radius 4.5 as its the radius of the big cone nd should be the radius of the cup but in the marking scheme they subtract 2.7 (small cone's radius) from 4.5 nd if i do this way i get right answer
but actually why do they subtract?>

 

[tweetie]

ok and q6 is a really good question.. but read it very carefully.. it says OBTUSE angle and when we get 48.5 thats acute so minus it from 180 to get the obtuse part.. usually in math when a word is written in bold.. u should stress on that part.. many make this mistake

oooo its tricky! thnx a lot

 

[Bobby Jing]

Can someone please explain to me question 22. c) in this past paper?
http://papers.xtremepapers.com/CIE/Cambridge%20IGCSE/Mathematics%20(0580)/0580_w09_qp_22.pdf

Is not the distance travelled by the car whilst travelling faster than the truck the smaller triangle?

 

[notnek01]

Can someone please explain to me question 22. c) in this past paper?
http://papers.xtremepapers.com/CIE/Cambridge%20IGCSE/Mathematics%20(0580)/0580_w09_qp_22.pdf

Is not the distance travelled by the car whilst travelling faster than the truck the smaller triangle?

I think you mean Q21b.

The distance travelled is always the area under the graph, from the line to the x-axis.

The car is travelling faster than the truck between 15 and 55 seconds. So the distance travelled in this time by the car is the area from the car line to the x-axis between t=15 and t=55. Ignore the horizontal truck line.

 

[Master Minds]

if you want any private lessons in any topic in IGCSE Math, just send me a email me at [email protected] or start a conversation. I will let you know when I will be available on skype for online lessons. I can give you some useful tips also for the upcoming exams. You may trust me fully as I have already given this subject and AlhamdullilAllah achieved an A* in november 2012 session.
My skype id is talhanaveed53

 

[tweetie]

http://papers.xtremepapers.com/CIE/Cambridge IGCSE/Mathematics (0580)/0580_s11_qp_22.pdf
16-b
18-a
19-b and c
PLEASE TELL ME IN DETAIL HOW TO DO THESE..PLEASE HELP ASAP

 

[Master Minds]

http://papers.xtremepapers.com/CIE/Cambridge IGCSE/Mathematics (0580)/0580_s11_qp_22.pdf
16-b
18-a
19-b and c
PLEASE TELL ME IN DETAIL HOW TO DO THESE..PLEASE HELP ASAP

16 b:
PQ: .5a - .5c
position vector of M = OM
= c + .5a + .5(.5a-.5c)
= c + .5a + .25a - .25c
= 0.75c + .75a
18 a:
A (7,0) B (5,3)
midpoint of AB: (1/2(7+5), 1/2(0+3)
= (6,1.5)

19:
b h(2x-3)
hh(x) = 2(2x-3)-3
= 4x-6-3
= 4x-9

c. fg(x+1) = f(2^x+(1))
= (2^x+1)^2
= 2^2(x+1)

 

[tiki-taka]

How do i calculate SHEAR FACTOR
????

 

[notnek01]

How do i calculate SHEAR FACTOR
????

Do you know how to find the invariant line?

If you know the invariant line then use the formula: Shear factor = (Distance moved by a point) / (Distance of point from invariant line).

Example 1:

Here I know the invariant line is the x-axis (ask me if you're not sure why). So look at the point C:

Distance of C from invariant line = +6
Distance moved from C to C' = +12

Shear factor = 12/6 = 2


Example 2:

Here I know the invariant line is the y-axis. So look at e.g. the point (-1,1):

Distance of (-1,1) from invariant line = -1
Distance moved from (-1,1) to (-1,2) = +1

Shear factor = +1 / -1 = -1

 

[Bobby Jing]

I think you mean Q21b.

The distance travelled is always the area under the graph, from the line to the x-axis.

The car is travelling faster than the truck between 15 and 55 seconds. So the distance travelled in this time by the car is the area from the car line to the x-axis between t=15 and t=55. Ignore the horizontal truck line.

Y

I think you mean Q21b.

The distance travelled is always the area under the graph, from the line to the x-axis.

The car is travelling faster than the truck between 15 and 55 seconds. So the distance travelled in this time by the car is the area from the car line to the x-axis between t=15 and t=55. Ignore the horizontal truck line.

Oh sweet, I got it now and yes q21)b). thanks yo

 

[tiki-taka]

Do you know how to find the invariant line?

If you know the invariant line then use the formula: Shear factor = (Distance moved by a point) / (Distance of point from invariant line).

Example 1:

Here I know the invariant line is the x-axis (ask me if you're not sure why). So look at the point C:

Distance of C from invariant line = +6
Distance moved from C to C' = +12

Shear factor = 12/6 = 2


Example 2:

Here I know the invariant line is the y-axis. So look at e.g. the point (-1,1):

Distance of (-1,1) from invariant line = -1
Distance moved from (-1,1) to (-1,2) = +1

Shear factor = +1 / -1 = -1

Thanx alot...

 

[tiki-taka]

Thanx alot...

 

[tiki-taka]

0580/42/MJ/11..Q-9c

 

[ZaqZainab]

guys!!! for sets how in the world are we suppose to know when we have to shade everything or only the circles? for ex A' U B do we shade the box or only B??? HELP PLX

only B but not the part where b is intersecting A

 

[notnek01]

0580/42/MJ/11..Q-9c

You need to substitute 2 value of n into the nth term formula and then solve the simultaneous equations.

E.g. When n=1 the number of lines in the first 1 diagram is 3 so:

a(1^3)+b(1^2)+1 = 3 --> a+b=2

When n=2, the number of lines in the first 2 diagrams is 9+3=12.

Can you carry on from here?

 

[notnek01]

only B but not the part where b is intersecting A

That's intersection not union.

 

[Aishy Vimal]

hi could someone help me with 2012 v21 MJ q14. if the line y<1/2x+4 where do i shade ?

 

[notnek01]

hi could someone help me with 2012 v21 MJ q14. if the line y<1/2x+4 where do i shade ?

For y < ax+b or y <= ax+b, the satisfied region is below the boundary line so the unwanted region is above the boundary line.

Similarly, for y > ax+b or y >= ax+b, the unwanted region is below the boundary line.