If g is the gallons of gas, and the truck holds up to 20 gallons, then g can be anything from 0 to 20.
If there is zero gallons of gas (g = 0) then the truck will travel zero miles because M(0) = 17(0) = 0.
The max miles depends on the max gallons of gas, so the miles possible with 20 gallons is M(20) = 17(20) = 340.
So, g ranges from 0 to 20 and M(g) ranges from 0 to 340.
Therefore your answer is the last option.
Answer:
Organisms that break down dead or decaying organisms
Step-by-step explanation:
Hope this helps :b
Answer:
18 hours for Dwight, 9 hours for Jake
Step-by-step explanation:
If J is the time it takes for Jake working alone, and D is the time it takes for Dwight working alone, then:
J = D − 9
Working together, Jake's speed plus Dwight's speed equals 1 cord per 6 hours.
1 / J + 1 / D = 1 / 6
Solve the system of equations by substituting for J:
1 / (D − 9) + 1 / D = 1 / 6
Multiply both sides by D − 9.
1 + (D − 9) / D = (D − 9) / 6
Multiply both sides by D:
D + D − 9 = D (D − 9) / 6
Simplify and solve:
2D − 9 = (D² − 9D) / 6
12D − 54 = D² − 9D
0 = D² − 21D + 54
0 = (D − 3) (D − 18)
D = 3 or 18
Since J = D − 9, D must be greater than 9. So D = 18 and J = 9.
That means it takes Dwight 18 hours working alone, and it takes Jake 9 hours working alone.
Wow !
OK. The line-up on the bench has two "zones" ...
-- One zone, consisting of exactly two people, the teacher and the difficult student.
Their identities don't change, and their arrangement doesn't change.
-- The other zone, consisting of the other 9 students.
They can line up in any possible way.
How many ways can you line up 9 students ?
The first one can be any one of 9. For each of these . . .
The second one can be any one of the remaining 8. For each of these . . .
The third one can be any one of the remaining 7. For each of these . . .
The fourth one can be any one of the remaining 6. For each of these . . .
The fifth one can be any one of the remaining 5. For each of these . . .
The sixth one can be any one of the remaining 4. For each of these . . .
The seventh one can be any one of the remaining 3. For each of these . . .
The eighth one can be either of the remaining 2. For each of these . . .
The ninth one must be the only one remaining student.
The total number of possible line-ups is
(9 x 8 x 7 x 6 x 5 x 4 x 3 x 2 x 1) = 9! = 362,880 .
But wait ! We're not done yet !
For each possible line-up, the teacher and the difficult student can sit
-- On the left end,
-- Between the 1st and 2nd students in the lineup,
-- Between the 2nd and 3rd students in the lineup,
-- Between the 3rd and 4th students in the lineup,
-- Between the 4th and 5th students in the lineup,
-- Between the 5th and 6th students in the lineup,
-- Between the 6th and 7th students in the lineup,
-- Between the 7th and 8th students in the lineup,
-- Between the 8th and 9th students in the lineup,
-- On the right end.
That's 10 different places to put the teacher and the difficult student,
in EACH possible line-up of the other 9 .
So the total total number of ways to do this is
(362,880) x (10) = 3,628,800 ways.
If they sit a different way at every game, the class can see a bunch of games
without duplicating their seating arrangement !
B. less: because he wants to make money, if he produces less wheat ha can supply more corn which costs more.
