Answer:
B. $715
Step-by-step explanation:
By multiplying your starting value, in this case $550, and your (simple) interest, 6%, or by 1.06 as 0.06 being your interest value and the 1.00 accommodating your starting value you will multiply, $550 × 1.06 = $583. Repeat these steps for however many years you are account for, in this case, five years. Giving you a total of $715
Answer:
$2355.06
Step-by-step explanation:
Use the compound interest formula, filling in the numbers you know. Then solve for the number you don't know.
A = P(1 +r/n)^(nt)
where A is the account balance, P is the amount invested, r is the annual rate, n is the number of times per year interest is compounded, and t is the number of years.
Filling in the given values, we have ...
4000 = P(1 +.053/52)^(52·10) = P(1.6984738)
P = 4000/1.6984738 ≈ 2355.06
You would need to deposit $2355.06 in order to have $4000 in 10 years.
Answer:
6/x^4
Step-by-step explanation:
Answer:
Terms (Variables) = x , d . Their corresponding coefficients = n^2 , 1/2 . Constant = 6
Step-by-step explanation:
6 + n x n + 1/2d
6 + n^2 x + 1/2d
Terms (Variables) = x , d . Their corresponding coefficients = n^2 , 1/2 . Constant = 6
Answer:
(-2,-3) and (3,2)
Step-by-step explanation:
sub in x-1 into y
x^2 + (x-1)^2 = 13
x^2 + (x-1)(x-1)=13
x^2 + x^2 -2x +1 = 13
2x^2 -2x-12=0
solve for x by factoring (quadratic formula, product sum etc..)
x= -2 and 3
plug in those values into y=x-1 and solve for y
