Mathematics: Post your doubts here!

[Iridescent]

No, it does not.
Lets check...
The equation of AB is:
3y = 5x - 1
Lets substitute (1.5,1.5) in this
3(1.5) = 5(1.5) -1
4.5 = 7.5 - 1
4.5 = 6.5
See, the equation is not satisfied.

Okay its (0.5,0.5). I got it. Thanx

 

[smzimran]

Okay its (0.5,0.5). I got it. Thanx

 

[hredoymohammad]

http://www.xtremepapers.com/CIE/ind...E/0580 - Mathematics/&file=0580_w08_qp_02.pdf
can someone do question 5 about locus pls and draw it in somewhere..i m not getting the question at all and thx ktc and captaindanger

 

[Rockzstar]

plz THE bearing...........

 

[XPFMember]

Assalamoalaikum wr wb!

Angle ABC= 180-110 = 70
Angle ACB = (180-70) / 2 [since AB = BC]
=55


N point north of C

angle NCB =180 - 110 = 70

reflex angle NCB=360 - 70 = 290

290 - 55 = 235

 

[Rockzstar]

thanxx

 

[Rockzstar]

how did u do this ????????

ᴊυѕт ᴀ ғᴇω мᴀɢɪсᴀʟ ωᴏяᴅѕ​

 

[Rockzstar]

this is another doubt

 

[сᴏᴏʟ сяуѕтᴀʟѕ]

this is another doubt

Assalam-o-alaikum wr wb
Well…in part (a) we will use the Pythagoras theorem..i.e
BC2=192+142
BC2=557
Therefore, BC=23.6 cm

in part (b) Surface area=2*base area + (perimeter of base) * height
So 2*(0.5*14*19) + (14+19+23.6) * 36
=2303.6, rounding it off to 3 significant figures, 2300 cm2

hope it helps

 

[Osama$]

http://www.xtremepapers.com/CIE/Cambridge IGCSE/0580 - Mathematics/0580_w09_qp_22.pdf
can anyone solve Q6

 

[ktc]

http://www.xtremepapers.com/CIE/Cambridge IGCSE/0580 - Mathematics/0580_w09_qp_22.pdf
can anyone solve Q6

Determinant of A = [0 x -12] – [-4 x 1]

= 0 + 4

= 4

Determinant of B = [3 x -4] – [0 x 4]

= -12 – 0

= - 12

Hence, [5 x 4] + [-12]

= 8

 

[Osama$]

thanks for solving it

 

[Osama$]

http://www.xtremepapers.com/CIE/Cambridge IGCSE/0580 - Mathematics/0580_s10_qp_42.pdf
can you plz solve Q2a plz

 

[ktc]

http://www.xtremepapers.com/CIE/Cambridge IGCSE/0580 - Mathematics/0580_s10_qp_42.pdf
can you plz solve Q2a plz

Q.2]

a)
i)3 + [2 x 6]

2 + [2 x 3]

= [15, 8]

ii) [ (15^2) + (8^2)]^ 1/2

= [ (225 + 64)]^ 1/2

= 289^ 1/2

= 17

 

[Osama$]

did not understood part 2

 

[ktc]

did not understood part 2

To calculate the of value of I p + 2q I ->

The square root of I [p^2] + [2q^2] I is the answer:

= [15^2 + 8^2]

= 225 + 64

= 289^½

= 17

( ^ ½ -> square root, btw)

 

[hredoymohammad]

http://www.xtremepapers.com/CIE/ind...E/0580 - Mathematics/&file=0580_w08_qp_02.pdf
can someone do question 5 about locus pls and draw it in somewhere..i m not getting the question at all and thx ktc and captaindanger can someone pls do dis?​

 

[GlitteringLights]

http://www.xtremepapers.com/CIE/Cambridge IGCSE/0580 - Mathematics/0580_w10_qp_22.pdf

i need help on question number 7 please

vector AB= a + tb and vector CD = a + (3t – 5)b where t is a number.
Find the value of t when AB=CD

Thanks a bunch : )

 

[CaptainDanger]

http://www.xtremepapers.com/CIE/Cambridge IGCSE/0580 - Mathematics/0580_w10_qp_22.pdf

i need help on question number 7 please

View attachment 5773


Thanks a bunch : )

AB=CD
a+tb = a +(3t-5)b
a-a +tb = (3t-5)b
(tb/b)=3t-5
t=3t-5
5=3t-t
5=2t
t=5/2

 

[GlitteringLights]

AB=CD
a+tb = a +(3t-5)b
a-a +tb = (3t-5)b
(tb/b)=3t-5
t=3t-5
5=3t-t
5=2t
t=5/2

OH Thanks SO much