Answer:
-y^(2)+x+y
Step-by-step explanation:
It's the first one lmk if you want an explanation
Answer:
5 gallons of water must be added.
10 gallons to start with
adding 'x' gallons of water final volume = 10 + x
.9 (10) = amount of vinegar
.1(10) = amount of water
amount of vinegar before and after water addition is constant
.9(10) = .6(10+x)
9 = 6 + .6x
3 = .6x
x = 5 gallons of water to be added
(i truly hope i helped you
Answer:
1st problem: b) 
2nd problem: c) 
Step-by-step explanation:
1st problem:
The formula/equation you want to use is:

where
t=number of years
A=amount he will owe in t years
P=principal (initial amount)
r=rate
n=number of times the interest is compounded per year t.
We are given:
P=2500
r=12%=.12
n=12 (since there are 12 months in a year and the interest is being compounded per month)

Time to clean up the inside of the ( ).


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2nd Problem:
Compounded continuously problems use base as e.

P is still the principal
r is still the rate
t is still the number of years
A is still the amount.
You are given:
P=2500
r=12%=.12
Let's plug that information in:
.