x -y = 1
5x + 3y = 45
Solve the first equation for x: x = y + 1
Substitute x in the second with y + 1
5x + 3y = 45
5(y + 1) + 3y = 45
5y + 5 + 3y = 45
8y + 5 = 45
8y = 40
y = 5
Substitute 5 for y in x = y + 1
x = y + 1
x = 5 + 1
x = 6
Answer: (6, 5)
Answer:
(a) 74553
(b) 172120
(c) 234802
Step-by-step explanation:
Given

Solving (a): 1998
Year 1998 means that:


So, we have:




--- approximated
Solving (b): 2003
Year 2003 means that:


So, we have:




--- approximated
Solving (c): 2006
Year 2006 means that:


So, we have:




--- approximated
Answer:
You spend $55.78 more buying luxury items then store brand items.
Step-by-step explanation:
You have to add up all the items costs for the store branded items. And then you have to add up all the items costs for the luxury branded items. Then you subtract the luxury item costs, from the store brand items and you get, $55.78.
I HOPE THIS HELPED YOU!!!!
I am pretty sure the value of ‘n’ is 2.5!
9514 1404 393
Answer:
(12480 -3600)/48
Step-by-step explanation:
Assuming a 0% interest rate, the monthly payment will be 1/48 of the amount remaining after the down payment:
(12480 -3600)/48
_____
= 185