Answer:
S varies partly directly as M and Q.
S=C.
S=KMQ+C.
For the first one...
speed=80,m=220,Q=30.
80=K20×30+C.
80=600K+C......(I).equation one.
For the second one....
speed=60,m=300,Q=40.
60=K300×40+C.
60=12000K+C.....(ii). equation two.
Minus eqtn(I) from eqtn(ii).
80=600K+C.
- 60=12000K+C.
K=0.01754~0.018.
Substitute K=0.018 into eqtn(I).
80=600K+C
80=600×0.018+C.
80=10.8+C.
C=80-10.8=69.2.
The relation is S=0.018MQ+69.2
when speed is 100 and mass is 250 find the volume.
100=0.018×250×Q+69.2.
100=4.5Q+69.2.
4.5Q=100-69.2
4.5Q=30.8.
Q=30.8/4.5.
Q=6.8~7litres.
Answer:
-4x-34
Step-by-step explanation:=−4x+−28+−6
=(−4x)+(−28+−6)
=−4x+−34
Answer:
See attachment
Step-by-step explanation:
You can obtain any two points on the graph of 
and use it to draw its graph.
When x=0, 
So you plot (0,2)
When x=1, 
You again plot (1,-71).
With a straight edge you can now draw a straight line through the two points.
See attachment
