Fit Fast: a set feet per class => y = Ax
Stepping Up: a monthly fee plus an additioal fee per class => h = Bx + C
You can discard the second and the fourth systems because they do not have the form established from the statement.
The first system produce an obvious result given that is represents an option that is always better than the other 5.5x will be lower than 7.5x + 10 for any positive value of x, and so there is no need to make any comparission.
The third system is
y = 7.5x and y = 5.5x + 10 which need to be solved to determine when one rate is more convenient than the other.
Answer: y = 7.5x and y = 5..5x + 10
Answer:
what is the quastion
Step-by-step explanation:
plz
Answer:
32
Step-by-step explanation:
We can write an algebraic equation to solve this situation: 
, where x = first integer (small number) and x + 1 = the following integer.
Step 1: Combine like terms.
Step 2: Subtract 1 from both sides.
Step 3: Divide both sides by 2.
Therefore, the smaller number is 32 while the larger number is 33.
Have a lovely rest of your day/night, and good luck with your assignments! ♡
Answer:
b(b/a)^2
Step-by-step explanation:
Given that the value of the car depreciates such that its value at the end of each year is p % less than its value at the end of the previous year and that car was worth a dollars on December 31, 2010 and was worth b dollars on December 31, 2011, then
b = a - (p% × a) = a(1-p%)
b/a = 1 - p%
p% = 1 - b/a = (a-b)/a
Let the worth of the car on December 31, 2012 be c
then
c = b - (b × p%) = b(1-p%)
Let the worth of the car on December 31, 2013 be d
then
d = c - (c × p%)
d = c(1-p%)
d = b(1-p%)(1-p%)
d = b(1-p%)^2
d = b(1- (a-b)/a)^2
d = b((a-a+b)/a)^2
d = b(b/a)^2 = b^3/a^2
The car's worth on December 31, 2013 = b(b/a)^2 = b^3/a^2
