X²+6x+8=0
x²+2*3x+9-1=0
(x+3)²-1=0
(x+3)²-1²=0
(x+3-1)(x+3+1)=0
(x+2)(x+4)=0
x₁=-2
x₂=-4
Answer: x₁=-2, x₂=-4
Answer: 4,000
Step-by-step explanation:
The only logical choice there because of the rate of others
A(3,4) B(-2.7) C(-3,2) D(3,0)
Add 1 to the 1st # of each
Add 2 to the 2nd #of each
We begin with an unknown initial investment value, which we will call P. This value is what we are solving for.
The amount in the account on January 1st, 2015 before Carol withdraws $1000 is found by the compound interest formula A = P(1+r/n)^(nt) ; where A is the amount in the account after interest, r is the interest rate, t is time (in years), and n is the number of compounding periods per year.
In this problem, the interest compounds annually, so we can simplify the formula to A = P(1+r)^t. We can plug in our values for r and t. r is equal to .025, because that is equal to 2.5%. t is equal to one, so we can just write A = P(1.025).
We then must withdraw 1000 from this amount, and allow it to gain interest for one more year.
The principle in the account at the beginning of 2015 after the withdrawal is equal to 1.025P - 1000. We can plug this into the compound interest formula again, as well as the amount in the account at the beginning of 2016.
23,517.6 = (1.025P - 1000)(1 + .025)^1
23,517.6 = (1.025P - 1000)(1.025)
Divide both sides by 1.025
22,944 = (1.025P - 1000)
Add 1000 to both sides
23,944 = 1.025P
Divide both by 1.025 for the answer
$22,384.39 = P. We now have the value of the initial investment.
