Answer:
the first one
Step-by-step explanation:
Suppose the larger pump alone can empty the tank in L hours, and the smaller pump can finish the job in S hours, then each hour the large pump empties 1/L portion of the tank, and the small pump empties 1/S per hour
Working together for three hours, they empty the whole tank, which is 100% of it, so 3/L+3/S=100%=1
Larger pump can empty the tank in 4 hours less than the smaller one, so L=S-4
replace L: 3/(S-4)+3/S=1
Make the denominator the same to solve for:
3S/[S(S-4)] +3(S-4)/[S(S-4)]=1
(3S+3S-12)/[S(S-4)]=1
(3S+3S-12)=[S(S-4)]
S^2-10s+12=0
use the quadratic formula to solve for S
S is about 8.6
The answer is not whole hour.
Step-by-step explanation:
SOHCAHTOA
sin=opp/hyp
cos=adj/hyp
tan=opp/adj
option B is correct
Answer: dude i wanna help you soooo bad but my stupid computar wont let me see the image can you say the question so i can try?
Step-by-step explanation:
Answer:
Remember that:
Speed = distance/time.
Then we can calculate the average speed in any segment,
Let's make a model where the average speed at t = t0 can be calculated as:
AS(t0) = (y(b) - y(a))/(b - a)
Where b is the next value of t0, and a is the previous value of t0. This is because t0 is the middle point in this segment.
Then:
if t0 = 100s
AS(100s) = (400ft - 0ft)/(200s - 0s) = 2ft/s
if t0 = 200s
AS(200s) = (1360ft - 50ft)/(300s - 100s) = 6.55 ft/s
if t0 = 300s
AS(300s) = (3200ft - 400ft)/(400s - 200s) = 14ft/s
if t0 = 400s
AS(400s) = (6250s - 1360s)/(500s - 300s) = 24.45 ft/s
So for the given options, t = 400s is the one where the velocity seems to be the biggest.
And this has a lot of sense, because while the distance between the values of time is constant (is always 100 seconds) we can see that the difference between consecutive values of y(t) is increasing.
Then we can conclude that the rocket is accelerating upwards, then as larger is the value of t, bigger will be the average velocity at that point.
