math specimen paper 4, help please !

[emi]

please help some one ,, i just want the first page

[pedrojosetrujillo]

Whatever u do, dont draw your graphs in pen. I knw its obvious, but once your in da exam u seem to forget as your so focused on solving da questions.

[X Abdulrahman X]

can any one solve Question 1 from SP 4 ? and tell me how you did it ?  please 

ok, 20,000,000 cm/ 100 (to get cm-m)    ==== 200,000m
     200,000 m/ 1000 (to get m-km)        ===== 200km
 
    200km x 12    =   2400km

OR  20,000,000 x 12 = 240,000,000
      240,000,000 / 100 = 2,400,000
      2,400,000 / 1000 = 2400km

Done 
 

lol sorry i forgot to finish the rest of it, anyway.

1 ii)    13x(20,000,000)^2 / 100,000^2  = 520,000km squared.

b i) 1580/31.6 = 1/n  = 50,     50 x 100 x 1000 = 5,000,000 ( because answer needs to be in cm )

b ii) 1155 to 1403 = 128 minutes
      S=D/T    1580/128 = 12.343...
      12.343... x 60 = 741 km/h

NOW ITS DONE 








           

[MB]

Hey, could someone please help me with number 8?  : )

And how do you rotate about a centre, and how do you reflect about a line? :S

Thanks!

[X Abdulrahman X]

Hey, could someone please help me with number 8?  : )

And how do you rotate about a centre, and how do you reflect about a line? :S

Thanks!

8. a) 1.2 x 0.3 x 3 x 60 = 64.8

    b) 1.2 x 0.8 x 15 x 60 = 864
        864-64.8=799.2
        799.2/ 64.8   =  1233%

   c) (( (pie) x r^2 x h ))  - (( (pie) x r^2 x (h-0.0013) = 2 ))     {0.0013 = metres}
       cancel out to get,    (pie) r^2 x 0.0013 = 2
       make R the subject,  2/0.0013 x (pie) = something and square root it to get 22.13m

   d) 50.40 / 0.12 = 420m^2 - surface area,
       420 = 0.8x +1.2x + 0.8x     =     420= 2.8x          x =  150m               
     

[X Abdulrahman X]

Hey, could someone please help me with number 8?  : )

And how do you rotate about a centre, and how do you reflect about a line? :S

Thanks!

umm hard to explain the rotation and reflection part, but imagine u have a point at 2,1 and u rotate it 90 degrees clockwise, the point will be at 1,2 ( i think tracing paper is allowed in the exam and will help alot)

and how dont u know reflections ? :S
imagine u have a line thats at 2,4  to  4,4
and its 4 lines above the X axis/line to reflect that u go down 4 lines and u get the reflection at 2,-4  to  4,-4

[MB]

Thanks. I get the direct reflections/rotations. You wouldn't happen to hav November 2001?

If not the question goes like this:

The coordinates of P, Q and R are (4,2), (6,2) and (8,4) respectively,
The points Q and R lie on the line l.
a- find the new coordinates for
i- P, after reflection in the line l,
iii- R, after the rotation of 90 degrees anticlockwise about centre P.

i don't get how to rotate about a centre (in this case P) and how to reflect about a line (apart from the x-axis/ y-axis).

I'm sorry, they're probably really easy but I can't seem to find the formulae.   

[X Abdulrahman X]

Thanks. I get the direct reflections/rotations. You wouldn't happen to hav November 2001?

If not the question goes like this:

The coordinates of P, Q and R are (4,2), (6,2) and (8,4) respectively,
The points Q and R lie on the line l.
a- find the new coordinates for
i- P, after reflection in the line l,
iii- R, after the rotation of 90 degrees anticlockwise about centre P.

i don't get how to rotate about a centre (in this case P) and how to reflect about a line (apart from the x-axis/ y-axis).

I'm sorry, they're probably really easy but I can't seem to find the formulae.   

thres no formulas needed, unfortuanetly i need to see the graph to help cuz i dont know whats meant be line l

[yaweng]

howd you guys calculated the 128 minutes?im not good at time..