Can any kind soul help me

a) 4.50m/s²

b) v= 21

c) 11.25m/s

Step-by-step explanation:

a) rate of change of speed in first 2 seconds

= gradient of graph from t=0s to t=2s

The two points are (0,0) and (2,9).

Rate of change of speed

Total distance= area under graph

Distance travelled for first 6s

= area of traingle +area of rectangle

= ½(2)(9) +(6-2)(9)

= 9 + 4(9)

= 9 +36

= 45m

Distance travelled from t=6s to t=12s

= 135 -45

= 90m

Area under graph from t=6s to t=12s =90

Area of trapezium= 90

½(A +B)(h)= 90

½(9+ v)(6)= 90

3(9 +v)= 90 (simplify)

9 +v= 30 (÷3 throughout)

v= 30 -9

v= 21

c) Average speed

= total distance ÷total time

= 135 ÷12

= 11.25 m/s

i believe it is c.y = 9x

rise over run. 9 units up over 1 unit right or 9/1

the equation cos(35) =a/25 can be used to find the length of bc what is the length of bc? round to the nearest tenth

step-by-step explanation: