Part A: For 100 miles, the cost will be same for both companies.
Part B: Alex should use Drive and Move company as it will cost him less money.
Step-by-step explanation:
Given,
Starting amount of YMI= $40
Per mile charges = $1.99
Let,
x be the number of miles.
y be the total charges.
y = 1.99x+40 Eqn 1
Starting amount of DM = $60
Per mile charges = $1.79
y=1.79x+60 Eqn 2
Part A:
For same charges;
Eqn 1 = Eqn 2

Dividing both sides by 0.20

For 100 miles, the cost will be same for both companies.
Part B:
Total miles = x = 150
Putting x=150 in Eqn 1
y=1.99(150)+40
y=298.5+40
y=$338.50
Putting x=150 in Eqn 2
y=1.79(150)+60
y=268.5+60
y=$328.50
Alex should use Drive and Move company as it will cost him less money.
r = continuous growth rate
revenue t years after 2013 = 54.5×10⁹×e^(rt)
74.6×10⁹ = 54.5×10⁹×e^(2r)
e^(2r) = 74.6/54.5
2r = ln(74.6/54.5)
r = ½ln(74.6/54.5) ≅ 0.15697 = 15.697% per year
Hope this helps!
Assume that the number of apples is x and the number of oranges is y.
For the first given, we know that each apple costs $0.24 and each orange costs $0.8, therefore:
amount paid for apples = 0.24x and amount paid for oranges = 0.8y
we also know that the total amount spent is $12, therefore the first equation is as follows:
0.24x + 0.8y = 12
For the second given, we know that the total number of fruit bought is 20, therefore, the second equation is:
x + y = 20
You can easily graph these two functions and find a possible combination from the graph (the correct combination would be the intersection between the two lines).
Markup is the amount added to the cost price of goods to cover overhead and profit.
Sue’s Corner Market has a markup of 60% on bottled water.
Let us say original price was $x.
Now price after markup is $2.
So we can make an equation like:
original price + markup price = price after markup
x + 60% of x =2


dividing both sides by 1.6
x= 1.25
So original price was 1.25 dollars.
Answer:
The domain of the function (cd)(x) will be all real values of x except x = 2.
Step-by-step explanation:
The two functions are
and d(x) = x + 3
So, (cd)(x) = 
Then, for x = 2 the function (cd)(x) will be undefined as zero in the denominator will make the function (cd)(x) undefined.
Therefore, the domain of the function (cd)(x) will be all real values of x except x = 2. (Answer)