Answer:
(y + 14)(y - 14)
Step-by-step explanation:
A difference of squares has the general form
a³ - b² and factors as
(a + b)(a - b)
Given
y² - 196
y² = (y)² ⇒ a = y and 196 = 14² ⇒ b = 14
y² - 196
= y² - 14² = (y + 14)(y - 14) ← difference of squares
Answer:
-12,-17
Step-by-step explanation:
I just added five to each of the numbers
Answer:
The nearest hundredth would be in the second decimal place. It would be 967.99 as you would round down to get there.
Step-by-step explanation
You need to use basic algebra for this.
For this I’ll use o as the items and p for the payment. First you need to find out how long it took for all the items to scan, so if it took each item 2 seconds to be scanned you need to times the total number of items (o) by two e.g. o x 2 = 62 items times two seconds which is equivalent to 62 seconds (1.02 minutes) after this step you need to minus the total time it took to scan the items for the transaction time (2 minutes) e.g. 2.00 - 1.02 = 2.58 minutes.
Hope this helped :)
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>
