A.
this is assuming it grows by 20% of the houses it has at that time (compound interest)
A(x)=30(1.2)ˣ⁻¹ or A(x)=25(1.2)ˣ
B(x)=3(x-1)+45 or B(x)=3x+42
B.
A(5)=62.28 or about 62 homes
B(5)=60
C.
this is tricky so I think it should have been only 20% of 30
start over
A.
20% of 30=6
A(x)=6(x-1)+30 or A(x)=6x+24
B(x)=3(x-1)+45 or B(x)=3x+42
B.
A(5)=54 homes
B(5)=60
C.
6x+24=3x+42
minus 3x
3x+24=42
minus 24 both sides
3x=18
divide by 3
x=6
after 6 years
both will have 63 homes
Answer:
Step-by-step explanation:
Because the cost (c) depends on the pounds (p), we have to multiple the weight by 6, because it is $6 per pound.
Answer:
30 m
Step-by-step explanation:
51-21=30
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Answer: The company should produce 7 skateboards and 16 rollerskates in order to maximize profit.
Step-by-step explanation: Let the skateboards be represented by s and the rollerskates be represented by r. The available amount of labour is 30 units, and to produce a skateboard requires 2 units of labor while to produce a rollerskate requires 1 unit. This can be expressed as follows;
2s + r = 30 ------(1)
Also there are 40 units of materials available, and to produce a skateboard requires 1 unit while a rollerskate requires 2 units. This too can be expressed as follows;
s + 2r = 40 ------(2)
With the pair of simultaneous equations we can now solve for both variables by using the substitution method as follows;
In equation (1), let r = 30 - 2s
Substitute for r into equation (2)
s + 2(30 - 2s) = 40
s + 60 - 4s = 40
Collect like terms,
s - 4s = 40 - 60
-3s = -20
Divide both sides of the equation by -3
s = 6.67
(Rounded up to the nearest whole number, s = 7)
Substitute for the value of s into equation (1)
2s + r = 30
2(7) + r = 30
14 + r = 30
Subtract 14 from both sides of the equation
r = 16
Therefore in order to maximize profit, the company should produce 7 skateboards and 16 rollerskates.
9 2/3 - 2 8/10 = 16 13/15
