PART A:
Find the rate of change between 1980 and 1989
d for P₁ = 80 - 60
d for P₁ = 20
d for P₂ = 76 - 82
d for P₂ = -6
The rate of change in P₁ is 20 hundred per year. The rate of change in P₂ is -6 hundred per year.
PART B:
Find the rate of change between 1989 and 1996
d for P₁ = 100 - 80
d for P₁ = 20
d for P₂ = 70 - 76
d for P₂ = -6
The rate of change in P₁ is 20 hundred per year. The rate of change in P₂ is -6 hundred per year.
PART C:
Find the rate of change between 1980 and 1996
d for P₁ = 100 - 60
d for P₁ = 40
d for P₂ = 70 - 82
d for P₂ = -12
The rate of change in P₁ is 40 hundred per year. The rate of change in P₂ is -12 hundred per year.
The sum of all edges is 108 cm.
There are two bases whose edges are all 8 cm; this gives us 8(4)+8(4) = 32+32=64.
There are 4 edges between bases that are 11 cm each; 11(4) = 44
Together, this gives us 64+44=108 cm.
Let a, b, and c be the times each pump will fill the tank when working alone.
Therefore, in 1 hour;
1/a +1/b = 1/(6/5) = 5/6 ---- (1)
1/a+1/c = 1/(3/2) = 2/3 ---- (2)
1/b+1/c = 1/(2) = 1/2 ---- (3)
From equation (1)
1/a = 5/6-1/b
Substituting for 1/a in eqn (2)
5/6-1/b+1/c = 2/3
-1/b +1/c = -1/6 => 1/c = 1/b - 1/6 --- (4)
Using eqn (4) in eqn (3)
1/b+1/b-1/6 = 1/2
2/b-1/6 = 1/2
2/b =1/2+1/6 = 2/3
1/b = 1/3
Then,
1/c = 1/3 - 1/6 = 1/6
1/a = 5/6 - 1/3 = 1/2
This means, in 1 hour and with all the pumps working together, the tank will be filled to;
1/a+1/b+1/c = 1/2+1/3+1/6 = 1 (filled fully).
Therefore, it will take 1 hour to fill the tank when all pumps are working together.
Answer: $31.25
altogether she spends... 12.5×$4.50= $56.25
in total, she makes 5 beds ( 12.5÷2.5=5)
so she makes 5×$17.50= $87.50 from selling the beds.
however, $87.50-$56.25= $31.25
