In an installment loan, a lender loans a borrower a principal amount P, on which the borrower will pay a yearly interest rate of i (as a fraction, e.g. a rate of 6% would correspond to i=0.06) for n years. The borrower pays a fixed amount M to the lender q times per year. At the end of the n years, the last payment by the borrower pays off the loan.
After k payments, the amount A still owed is
<span>A = P(1+[i/q])k - Mq([1+(i/q)]k-1)/i,
= (P-Mq/i)(1+[i/q])k + Mq/i.
</span>The amount of the fixed payment is determined by<span>M = Pi/[q(1-[1+(i/q)]-nq)].
</span>The amount of principal that can be paid off in n years is<span>P = M(1-[1+(i/q)]-nq)q/i.
</span>The number of years needed to pay off the loan isn = -log(1-[Pi/(Mq)])/(q log[1+(i/q)]).
The total amount paid by the borrower is Mnq, and the total amount of interest paid is<span>I = Mnq - P.</span>
Answer: £225
Step-by-step explanation:
Let Sarah's amount be represented by x.
Since Caroline gets three times as much as Sarah, Caroline will get: 3x
Bridget gets twice as much as Caroline, therefore Bridget will get: 2 × 3x = 6x
Sarah = x
Caroline = 3x
Bridget = 6x
Total = x + 3x + 6x = 10x
Caroline's faction is 3/10. We then multiply the fraction by £750. This will be:
= 3/10 × £750
= 0.3 × £750
= £225
Caroline will get £225
Answer:
5π
Step-by-step explanation:
just need to find half of the circumference
9514 1404 393
Answer:
F. 30·tan(52°)
Step-by-step explanation:
The mnemonic SOH CAH TOA reminds you of the relationship between the opposite and adjacent sides in a right triangle.
Tan = Opposite/Adjacent
tan(52°) = BD/DL
BD = DL·tan(52°)
The boat-to-dock distance is ...
30·tan(52°) . . . . matches F
