Mathematics: Post your doubts here!

[CaptainDanger]

please help me solve this question. and a little bit on how these questions are normally solved.
___________________________________________________________________________
the area of a rectangle is 52cm^2 , the breath is 1.5cm less than its length. find the length and the breadth.
___________________________________________________________________________
i'd really appreciate it if someone would answer this. thank you.

L=L
B=L-1.5
L x B = Area
L ( L-1.5) = 52
L^2-1.5L-52=0
Solve quadratically and you get the answer...

 

[GlitteringLights]

omg so this does not require you to solve it simultaneously ? that's where my mistake was :O thank youu!

 

[ktc]

can someone please explain how question 9(d) is done please? i've tried it ever so many times but my result is always wrong :/

(d) A card is chosen at random, without replacement, from Set B until the letter shown is either
I or U.
Find the probability that this does not happen until the 4th card is chosen.
_
Thank you

You should remember probability should be either I or U.

The probability that the 1st card is neither I nor U = 3/5 [out of 5 cards, 3 of them are neither I nor U]

The probability that the 2nd card is neither I nor U = 2/4 [out of the remaining 4 cards now, 2 of them are neither I nor U]

The probability that the 3rd card is neither I nor U = 1/3 [now out of the remaining 3 cards, only 1 card is neither I nor U]

The probability that the 4th card is either I or U = 1 [out of the remaining 2 cards, both of the cards are either I or U]

Hence, [3/5] x [2/4] x [1/3] x 1 = 6/60

 

[shaheer salim]

heyy guys need help here . 0580/22/MJ/08 q20 its lyk i tried a million times and the answer to part a i get is 12933 which in the answer paper is 12900 theres no where in the question written to round it up or sth ..... pleese help me... i know its a simple stupid dout but reeli need help.. pleeaase

 

[XPFMember]

heyy guys need help here . 0580/22/MJ/08 q20 its lyk i tried a million times and the answer to part a i get is 12933 which in the answer paper is 12900 theres no where in the question written to round it up or sth ..... pleese help me... i know its a simple stupid dout but reeli need help.. pleeaase

assalamoalaikum wr wb!!

all non exact answers shud be rounded to 3 sf...
check first page (instructions) of the question papers

 

[Osama$]

http://www.xtremepapers.com/CIE/Cambridge IGCSE/0580 - Mathematics/0580_w05_qp_3.pdf
can anyone solve Q4d,e

 

[Osama$]

http://www.xtremepapers.com/CIE/Cambridge IGCSE/0580 - Mathematics/0580_s11_qp_33.pdf
and can anyone explain Q3f

 

[ktc]

http://www.xtremepapers.com/CIE/Cambridge IGCSE/0580 - Mathematics/0580_w05_qp_3.pdf
can anyone solve Q4d,e

Arrange the data given in ascending order:

0, 0, 2, 5, 6, 6, 6, 10, 10, 12, 13, 15, 16, 16.

d) frequency for the number of calls between 0 – 4 = 3

[because from the data, between 0 – 4, there are 0 number of calls for 2 days and 2 calls for 1 day.]

frequency for the number of calls between 5 – 9 = 4.

[because from the data, between 5 – 9, there are 5 calls for 1 day and 6 calls for 3 days.]

frequency for the number of calls between 10 – 14 = 4

[because from the data, between 10 - 14, there are 10 calls for 2 days, 12 calls for 1 day and 13 calls for 1 day.]

frequency for the number of calls between 15 – 19 = 3

[because from the data, between 15 – 19, there are 15 calls for 1 day and 16 calls for 2 days.]

e) i) 7/14 [out of the total frequency, 14 days, Jane receives 10 or more calls for 7 days]

ii) 3/14 [out of the total frequency, 14 days, Jane receives less than 5 calls for 3 days.]

http://www.xtremepapers.com/CIE/Cambridge IGCSE/0580 - Mathematics/0580_s11_qp_33.pdf
and can anyone explain Q3f

From the frequency found in the table, we know, out of the 288 students, 80 students scored 3 points.

So, if there were 1440 students, how many of them will score 3 points?

288 -> 80
1440 -> x
X = [1440 x 80]/288

x = 400 students.

 

[Ahmed Khider]

I have 4 questions to be solved:
1-w10_qp_23>>Qno 18
2-w10_qp_23>>Qno 20
3-w10_qp_23>>Qno 21
4-w10_qp_23>>Qno 23

 

[ktc]

I have 4 questions to be solved:
1-w10_qp_23>>Qno 18
2-w10_qp_23>>Qno 20
3-w10_qp_23>>Qno 21
4-w10_qp_23>>Qno 23

Q. 18]
The radius of the circle = 8cm.

The angle of the sector = x°

They’ve given the whole perimeter of the sector = 16 + 14π

But to calculate angle x, you need only the arc length of the sector.

Therefore, toget the arc length, you need to subtract the radii from the given perimeter.
[16 + 14 π] – 16.
Arc length of the sector = 14 π

360 -> 2 π (8)
x ->14 π
x = 360 [14 π] / 2 π (8)
x = 315°
_________
Q.20]
a) To calculate angle KML , use the Tan rule.Tan θ = opposite/adjacent
Hence, Tan θ = [5-1] / [3-1]
Tan θ = 4/2
θ = Tan- 2
θ = 63.4°

b) The 2*2 matrix for shear, when the shear factor is 3 and invariant line is X axis:
(1 3)
(0 1)
By multiplying this 2*2 matrix with all three co ordinates of the triangle in the diagram (1, 1) (1, 3) (5, 1), you get your corresponding image co ordinates. Then plot them, accordingly.
(1, 1) -> (4, 1)
(1, 3) -> (10, 3)
(5, 1) -> (8, 1)
___________
Q.21]
a) deceleration = [v – u]/t
= [20 – 8]/ 5
= 2.4m/s2

b) Divide the graph into two trapeziums and one rectangle.
[1/2] x 12 x [15 + 10] = 150m.
[1/2] x 12 x [25 + 10] = 210m.
40 x 8 = 320m.
150 + 210 + 320 = 680m.
_____________
Q.23]
a) angles subtended by the same arc are equal.

b) i) ADC = 100°[57 + 43]
ii) BDC = 43° [angle subtended from the same are as the central angle, touching the circumference of the circle, is half of the central angle]
iii) OBD = 3° [DBC = 50, OBC = 47. Hence, 50 – 47]

 

[AymanB]

Assalamoalaikum!!

Stuck somewhere in Maths?? Post your queries here! If you have any doubt in the pastpper questions, then kindly post the link to the paper!

P.S. I'm busy these days, so I can't promise to be there for help.

May Allah give us all success in this world as well as the HereAfter...Aameen!!

SEQUENCES-Points to remember:
Many people find it hard, but to be honest it's just more of logic..that's all!

Mathematics: Post your doubts here!

Hello! I am having some difficulty with one math question in the attached file. The year is October/November 2008, Paper 4.
Thank you.

 

[Ahmed Khider]

Q. 18]
The radius of the circle = 8cm.

The angle of the sector = x°

They’ve given the whole perimeter of the sector = 16 + 14π

But to calculate angle x, you need only the arc length of the sector.

Therefore, toget the arc length, you need to subtract the radii from the given perimeter.
[16 + 14 π] – 16.
Arc length of the sector = 14 π

360 -> 2 π (8)
x ->14 π
x = 360 [14 π] / 2 π (8)
x = 315°
_________
Q.20]
a) To calculate angle KML , use the Tan rule.Tan θ = opposite/adjacent
Hence, Tan θ = [5-1] / [3-1]
Tan θ = 4/2
θ = Tan- 2
θ = 63.4°

b) The 2*2 matrix for shear, when the shear factor is 3 and invariant line is X axis:
(1 3)
(0 1)
By multiplying this 2*2 matrix with all three co ordinates of the triangle in the diagram (1, 1) (1, 3) (5, 1), you get your corresponding image co ordinates. Then plot them, accordingly.
(1, 1) -> (4, 1)
(1, 3) -> (10, 3)
(5, 1) -> (8, 1)
___________
Q.21]
a) deceleration = [v – u]/t
= [20 – 8]/ 5
= 2.4m/s2

b) Divide the graph into two trapeziums and one rectangle.
[1/2] x 12 x [15 + 10] = 150m.
[1/2] x 12 x [25 + 10] = 210m.
40 x 8 = 320m.
150 + 210 + 320 = 680m.
_____________
Q.23]
a) angles subtended by the same arc are equal.

b) i) ADC = 100°[57 + 43]
ii) BDC = 43° [angle subtended from the same are as the central angle, touching the circumference of the circle, is half of the central angle]
iii) OBD = 3° [DBC = 50, OBC = 47. Hence, 50 – 47]

Thank you for taking the time to answer my questions. I disnt really understand question no 18. As for question no 20(b) i did wat u did in an exam but my teacher said it was wrong!!

 

[ktc]

Thank you for taking the time to answer my questions. I disnt really understand question no 18. As for question no 20(b) i did wat u did in an exam but my teacher said it was wrong!!

Q.18] Perimeter of the sector and arc length of the sector are two different things.

Angle x can only be found by using the arc length of the sector; only the length of the curve is required to find angle x.

But the question gave the whole perimeter of the sector, which means, the arc length + the two radii.

Now, we only need the arc length to find angle x. So, we subtract the two radii from the perimeter of the sector to get the arc length of the sector.

Perimeter - two radii = arc length.

(16 + 14π)–(8 x 2)

= 16 + 14π – 16

Hence, arc length = 14π

We can now use 14π, the arc length, to find angle x:

[2π (8)x] / 360 = 14 π

[2π (8)x] = 360 [14 π]

[2π (8)x] = 5040 π

[2π (8)x] = [5040 π]

x = [5040 π] / [2 π (8)]

x = 315

Q.20] b) Just checked the marking scheme and the Examiner report.

(check out the uploaded files)

My/your answer matches with the answer given in both the MS and the ER:

Vertices at (4, 1), (8, 1) and (10, 3).

I think we could safely say that is the right answer.

Check with your teacher again. Perhaps it was a misunderstanding.

 

[Osama$]

http://www.xtremepapers.com/papers/CIE/Cambridge IGCSE/Mathematics (0580)/0580_w08_qp_04.pdf
Q6b

 

[suhaib05]

http://www.xtremepapers.com/papers/CIE/Cambridge IGCSE/Mathematics (0580)/0580_s11_qp_21.pdf
Q6 - I thought this question was REALLY straightforward, but then I was surprised to see that the answer is 108g. Why do we have to cube the scale factor for the mass of the large egg?

 

[Ahmed Khider]

T

Q.18] Perimeter of the sector and arc length of the sector are two different things.

Angle x can only be found by using the arc length of the sector; only the length of the curve is required to find angle x.

But the question gave the whole perimeter of the sector, which means, the arc length + the two radii.

Now, we only need the arc length to find angle x. So, we subtract the two radii from the perimeter of the sector to get the arc length of the sector.

Perimeter - two radii = arc length.

(16 + 14π)–(8 x 2)

= 16 + 14π – 16

Hence, arc length = 14π

We can now use 14π, the arc length, to find angle x:

[2π (8)x] / 360 = 14 π

[2π (8)x] = 360 [14 π]

[2π (8)x] = 5040 π

[2π (8)x] = [5040 π]

x = [5040 π] / [2 π (8)]

x = 315

Q.20] b) Just checked the marking scheme and the Examiner report.

(check out the uploaded files)

My/your answer matches with the answer given in both the MS and the ER:

Vertices at (4, 1), (8, 1) and (10, 3).

I think we could safely say that is the right answer.

Check with your teacher again. Perhaps it was a misunderstanding.

Thank you verrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrry much, I really appreciate it

 

[Razo513]

can u give me an easy way to do this ?

 

[Yousif Mukkhtar]

can u give me an easy way to do this ?

For question 11 (a), use your calculator. If you are having a CASIO calculator of the fx brand use this method:
1. Use the In function
2. Enter the following into your calculator: In 0.125/In 2

For (b):
1. Change 512 to 2^9 so that all the numbers have the number 2 with a power.
2. Then do an equation:
4n+2n=9
3. Find the value of n
6n=9
n=9/6
n=1.5

Hope this was clear to you.

 

[Razo513]

yes

For question 11 (a), use your calculator. If you are having a CASIO calculator of the fx brand use this method:
1. Use the In function
2. Enter the following into your calculator: In 0.125/In 2

For (b):
1. Change 512 to 2^9 so that all the numbers have the number 2 with a power.
2. Then do an equation:
4n+2n=9
3. Find the value of n
6n=9
n=9/6
n=1.5

Hope this was clear to you.

yes thank u soo much

 

[amir]

how can i get all the maths formulas,,.