Answer:
hrs/house
Step-by-step explanation:
Rate is the amount of work done per unit of time;
Let Cara's rate = C
Let Jena's rate = J
C + J = 4 ------ i
Now;
Working alone:
Jena takes twice as long as Cara
J = 2C ---- i
How long does it take Cara to clean a house alone
C + J = 4
J = 2C
C + 2C = 4
3C = 4
C =
hrs/house
Answer:
x = -1
Step-by-step explanation:
Set is a sum of two intervals :(1,4) u [2,6]
In the first interval it is open on both sides so 1 and 4 don't belong to this interval, the second is closed so both 2 and 6 belong to interval and numbers between 2 and 6
We can write it as 1<x

summing both sets we have one set: (1,6]
so
a) represent our set
b) also represent because all numbers between 4-6 and 6 are in our set
c)also represent, all numbers are in main set
d)also represent
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.
So this is how we are going to solve for the given problem above.
Given that x = number of large boxes
and 120-x = number of small boxes.
So here is the solution:
50x + (120-x)20 = 4050
50x + 2400 - 20x = 4050
30x + 2400 = 4050
30x = 4050 - 2400
30x = 1650 <<divide both sides by 30
x = 55.
Therefore, there are 55 large boxes
120 - x = small boxes
120 - 55 = 65 small boxes.
Hope this is the answer that you are looking for.
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