A9=A1+(n-1)
-2.75+(8*0.25)
-2.75+2=-0.75
Answer:
B 3.78 seconds
Step-by-step explanation:
On the graph, the line ends when it hits (0,3.78) This means that the object was in the air for 3.78 seconds before hitting the ground (0 height)
Answer:
g(g(-2))
g(-8)
-32
This is the correct solution
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:
(x +4)^2 + (y-2)^2 = 16
Step-by-step explanation:
We can write the equation for a circle with the formula
(x-h)^2 + (y-k)^2 = r^2
Where (h,k) is the center and r is the radius
We know the center is at (-4,2) and the radius is 4
(x- -4)^2 + (y-2)^2 = 4^2
(x +4)^2 + (y-2)^2 = 16
