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HELP ME ! MATHX PAPER 2

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21A ) you can either find the gradient of the line to find out the acceleration or use the acceleration formula : final speed - initial speed / time taken

if you use the gradient method you get two co-ordinates of that line so in this case :

(0,0) and (18,30) then substitute these values into the formula

y2-y1 / x2 - x1

30-0 / 18- 0

30/18 = 0.6m/s squared is the answer.

The other method is by using the acceleration formula so it would be

18-0/30

which = 0.6 m/s squared
 
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In these graphs the distance traveled is found by the area under the lines , so in this case there are two small triangles on top and a large trapezium

triangle 1 - 0.5 x 20 x 6 = 60
triangle 2 - 0.5 x 10 x 6 =30
trapezium - 0.5 x (100+80) x 12 = 1080

1080+60+30 =1170 m
 
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23 ) diameter of semi-circle=12
radius of semi-circle=6
diameter of circle=6
radius of circle=3

area of circle ( the one inside the semi circle )
pi x 3squared =28.27433388

area of semi circle
pi x 6squared =113.0973355

shaded area =

area of semi circle - circle / 2

= 84.82300165 / 2
= 42,41150082
=42.4 cm2
 
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23b)

perimeter =

half circumference of circle + half circumference of semi circle + 6
=
34.3 cm
23 ) diameter of semi-circle=12
radius of semi-circle=6
diameter of circle=6
radius of circle=3

area of circle ( the one inside the semi circle )
pi x 3squared =28.27433388

area of semi circle
pi x 6squared =113.0973355

shaded area =

area of semi circle - circle / 2

= 84.82300165 / 2
= 42,41150082
=42.4 cm2
In these graphs the distance traveled is found by the area under the lines , so in this case there are two small triangles on top and a large trapezium

triangle 1 - 0.5 x 20 x 6 = 60
triangle 2 - 0.5 x 10 x 6 =30
trapezium - 0.5 x (100+80) x 12 = 1080

1080+60+30 =1170 m
YAAAAAAAAAAAAAR U R AWESOME !=D
 
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Area of circle inside a semicircle= pi x (r²/4), where r is the radius of the semicircle.
area of the circle = 28.274

pi x rsquared = 28.27
r = underoot 28.27 / pi
r= 3
 
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